# tr87-cd924810 · pasada 0 · log de pensamiento

## Paso 1 · nivel 1 · acción 1

**Pensamiento:**

Let me inspect the board structure first.

```python
print(current_frame.shape, current_frame.level, valid_actions)
seg = current_frame.segmentation
nodes = seg['nodes']
print('num nodes', len(nodes))
for n in nodes:
    print(n['id'], n['color'], n['pixels'], n['boundary'][:4], 'rot', n['rotation'], 'sym', n['rotational_symmetry'], 'children', n['children'])
```

```text
(64, 64) 1 ['UP', 'DOWN', 'LEFT', 'RIGHT']
num nodes 64
0 g 1570 [[0, 0], [0, 63], [33, 63], [33, 0]] rot 180 sym 1 children [1, 2, 3, 4, 11, 12, 13, 14, 15, 16, 23, 24, 26, 27, 28, 29, 35, 36]
1 S 34 [[4, 12], [4, 18], [10, 18], [10, 12]] rot 0 sym 1 children [5]
2 P 26 [[4, 22], [4, 28], [10, 28],
… [4189 caracteres más]
```

**Pensamiento:**

Let me understand the layout. There's a 6x3 grid of "glyph" tiles in the top area (rows 4-28, cols 12-51). Each tile is a colored square (S=sky blue or P=pink) with a black glyph inside. Some tiles are connected by gray (G) 3-pixel connectors at cols 19-21, 42-44 (horizontal connectors between pairs).
Bottom area: a sky-blue bar (rows 40-46, cols 14-48) containing 5 black glyphs, and a pink bar (rows 51-57, cols 14-48) containing 5 black glyphs. White brackets around them (like selection brackets).
This looks like a "matching" puzzle: the top has pairs of glyphs connected; the bottom shows rows of glyphs. Perhaps the task is to select/match something.
Actually, this looks like the ARC-AGI "game" style where you have to find which row of glyphs matches a rule. Hmm.
Let me look at the ascii of specific regions to understand the glyphs. The tiles in the top: pairs connected by gray bars. Left pair group (cols 12-28) and right pair group (cols 35-51). Three rows of pairs.
Bottom: two rows of 5 glyphs each, with white bracket marks above/below (like a "cursor" indicator?).
Hmm, maybe the game is: the top shows example pairs where left glyph transforms to right glyph (a rotation rule?). The bottom shows two rows; you must pick... but only UP/DOWN/LEFT/RIGHT actions available. Maybe the white brackets indicate a selected row, and UP/DOWN switches selection, LEFT/RIGHT selects a column?
Let me look at the ascii of the bottom region and the top tiles to compare glyphs. Maybe the botto
… [813 caracteres más]

```python
a = current_frame.ascii.split('\n')
def crop(r0,c0,r1,c1):
    return '\n'.join(a[r][c0:c1] for r in range(r0,r1+1))
print("TOP LEFT tile1 (S) rows4-10 cols12-18")
print(crop(4,12,10,18))
print("TOP tile2 (P) rows4-10 cols22-28")
print(crop(4,22,10,28))
print("TOP tile3 (S) rows4-10 cols35-41")
print(crop(4,35,10,41))
print("TOP tile4 (P) rows4-10 cols45-51")
print(crop(4,45,10,51))
```

```text
TOP LEFT tile1 (S) rows4-10 cols12-18
SSSSSS
SBBBBB
SBSSSB
SBBSBB
SBSSSB
SBSSSB
SSSSSS
TOP tile2 (P) rows4-10 cols22-28
PPPPPP
PBBBBP
PBPPBB
PBPPPB
PBBPPB
PPBBBB
PPPPPP
TOP tile3 (S) rows4-10 cols35-41
SSSSSS
SBSSSB
SBBBBB
SSSBSS
SBBBBB
SBSSSB
SSSSSS
TOP tile4 (P) rows4-10 cols45-51
PPPPPP
PBBBBP
PB
… [32 caracteres más]
```

**Pensamiento:**

These are 7x7 tiles with black glyph patterns inside (5x5 inner). Each tile has a 5x5 (or so) black pattern. Let's extract the inner 5x5 pattern for each of the 12 top tiles and the 10 bottom glyphs (bottom rows have larger glyphs? Bottom bar is 7 rows tall (40-46) with glyphs inside 6 rows... Let me check bottom rows.
Bottom blue bar: rows 40-46, cols 14-48. Glyphs are within rows 41-45 (5 rows) and each glyph spans 5 columns? Let's extract.
The key idea: each tile is a 5x5 binary pattern. Perhaps the puzzle is: pairs (left, right) where right = transform(left) — e.g., rotation, and the bottom shows two rows of 5 glyphs; you must find the odd one out or match.
Hmm, but with only UP/DOWN/LEFT/RIGHT, maybe it's a "select a tile" navigation and pressing a direction... Let's just try an action and see what changes. That's the most informative. But first let me record the initial state fully (I have segmentation).
Let me try pressing RIGHT and diff.

```python
res = action(['RIGHT'])
print(res)
d = frame_diff()
print({k:v for k,v in d.items() if k!='changed_cell_count'})
print('changed cells', d['changed_cell_count'])
```

```text
[action] RIGHT -> effect
{'executed': True, 'action_num': 1, 'level': 1, 'score': 0, 'state': 'NOT_FINISHED', 'valid_actions': ['UP', 'DOWN', 'LEFT', 'RIGHT'], 'board_changed': True, 'done': False, 'level_completed': False, 'game_over': False, 'run_complete': False, 'action_display': 'RIGHT', 'gamep
… [620 caracteres más]
```

## Paso 2 · nivel 1 · acción 4

**Pensamiento:**

The white brackets moved right by 7 columns (from col 15 to col 22). So the brackets are a cursor/selector. The top bracket is above the blue bar (row 48), and the bottom bracket is below the pink bar (row 59). Both moved right by 7 with the RIGHT action.
Hmm, but interesting: the brackets appear to be positioned at the boundary between glyph slots? Initially at cols 15-19 (bracket spanning 5 wide at cols 15-19: boundary [48,15],[48,19] means the white pixels are at row 48 cols 15,16,17,18,19 and row 49 cols 15,19 — a "∩" shape 5 wide). After RIGHT: at cols 22-26.
The blue bar spans cols 14-48 (35 wide = 7 * 5). So 7 slots of 5 columns each! But there are only 5 glyphs in the bar... hmm, 35/7 = 5 columns per slot; glyphs are 5 wide with spacing? Let's compute glyph positions: glyph bboxes within blue bar: node42 [41,16]-[43,19], node43 [41,26]-[44,24] hmm the boundary points are corners not necessarily extremes. Let me get exact bounding boxes by scanning the ascii rows 41-45 cols 14-48.
Actually the bracket is a cursor that moves 7 columns per press = one slot. So there are 7 slots? Blue bar cols 14..48 = 35 cells = 5 slots of 7 columns. Yes! 5 slots of 7 columns each: slot i spans cols 14+7i .. 20+7i. Bracket initially at cols 15-19 = slot 0 (centered within slot 0: cols 15-19 is the inner 5 of 7). After RIGHT, at cols 22-26 = slot 1.
So the cursor selects a column/slot. There are 5 slots in the bottom bars. The top area has 3 rows × 2 groups of pairs... hmm.
So the puzzle 
… [616 caracteres más]

```python
for act in ['UP','DOWN','LEFT']:
    res = action([act])
    d = frame_diff()
    mv = d.get('moved',[])
    print(act, 'changed', d['changed_cell_count'], 'moved', [(m['color'],m['from'],m['to']) for m in mv],
          'app', len(d.get('appeared',[])), 'dis', len(d.get('disappeared',[])), 'rot', d.get('rotated'),
          'cc', [(c['at'],c['before_color'],c['after_color']) for c in d.get('changed_color',[])][:6],
          'rz', [(r['color'],r['before_pixels'],r['after_pixels']) for r in d.get('resized',[])])
    print('  state', res[0]['state'] if isinstance(res,list) else res['state'])
```

```text
[action] UP -> effect
UP changed 12 moved [] app 1 dis 1 rot [] cc [] rz [('P', 141, 135), ('B', 15, 16), ('w', 64, 63), ('P', 1, 7)]
  state NOT_FINISHED
[action] DOWN -> effect
DOWN changed 11 moved [] app 1 dis 0 rot [] cc [] rz [('P', 135, 141), ('B', 16, 15), ('P', 7, 1)]
  state NOT_FINISHED
[
… [196 caracteres más]
```

## Paso 3 · nivel 1 · acción 4

**Pensamiento:**

UP and DOWN changed something in the pink bar region (P 141 -> 135, B 15 -> 16, P 1 -> 7). That's odd. Let me look at what exactly changed for UP and DOWN.
UP: P (pink bar) shrank from 141 to 135 pixels, a B object grew 15->16, and a P 1 -> 7. Also 'w' 64 -> 63 (bottom row HUD changed - that's the bottom row bar shrinking! step counter). Hmm, 'w' node 63 is row 63 full width 64 pixels, now 63 — so it shrinks each action: that's a HUD/timer bar? Actually it's the bottom row of the board. It went 64 -> 63 -> 62. So yes it's a step budget bar. Careful — limited steps.
Wait, actually it might just be a decorative bottom line. But it shrinks each action, so it's a budget. 64 cells, one per action. So we have about 64 actions total? Let's watch.
UP changed pink bar: something in the pink bar changed — a glyph got bigger? Let me examine UP transition in detail: which cells changed.
Let me look at transitions for the UP action (transitions[-3]?) Order: RIGHT(step1), UP(step2), DOWN(step3), LEFT(step4). transitions list has 4 entries now. Let me diff transitions[1] (UP) before/after with cell-level detail.

```python
import itertools
t = transitions[1]  # UP
b = t.before_frame.ascii.split('\n'); af = t.after_frame.ascii.split('\n')
ch = [(r,c,b[r][c],af[r][c]) for r in range(64) for c in range(64) if b[r][c]!=af[r][c]]
print('UP changed cells:', ch)
t2 = transitions[2]  # DOWN
b2 = t2.before_frame.ascii.split('\n'); a2 = t2.after_frame.ascii.split('\n')
ch2 = [(r,c,b2[r][c],a2[r][c]) for r in range(64) for c in range(64) if b2[r][c]!=a2[r][c]]
print('DOWN changed cells:', ch2)
```

```text
UP changed cells: [(52, 26, 'P', 'B'), (53, 22, 'P', 'B'), (53, 25, 'B', 'P'), (53, 26, 'P', 'B'), (54, 23, 'B', 'P'), (54, 24, 'B', 'P'), (54, 25, 'B', 'P'), (55, 22, 'P', 'B'), (55, 23, 'B', 'P'), (55, 26, 'P', 'B'), (56, 22, 'P', 'B'), (63, 63, 'w', 'c')]
DOWN changed cells: [(52, 26, 'B', 'P'), 
… [199 caracteres más]
```

**Pensamiento:**

So UP/DOWN modify the glyph in the pink bar at slot 1 (cols 22-26, rows 52-56) — i.e., the cursor's current slot in the pink (bottom) row. UP/DOWN change the glyph there — likely cycling through the 5 candidate glyphs (or rotating?).
Let me look at the glyph before/after in the crop. Before UP, slot 1 of pink bar was glyph node50: rows 52-56 cols 23-25... Let me print the crop of rows 51-57, cols 21-27 before and after UP.

```python
def cropf(fr, r0,c0,r1,c1):
    L = fr.ascii.split('\n')
    return '\n'.join(L[r][c0:c1+1] for r in range(r0,r1+1))
print('BEFORE UP (pink slot1):'); print(cropf(transitions[1].before_frame,51,21,57,27))
print('AFTER UP:'); print(cropf(transitions[1].after_frame,51,21,57,27))
print('AFTER DOWN (current):'); print(cropf(current_frame,51,21,57,27))
print('blue row slots (rows40-46 cols14-48):'); print(cropf(current_frame,40,14,46,48))
```

```text
BEFORE UP (pink slot1):
PPPPPPP
PPBBBPP
PPBPBPP
PBBBBBP
PPBPBPP
PPBBBPP
PPPPPPP
AFTER UP:
PPPPPPP
PPBBBBP
PBBPPBP
PBPPPBP
PBPPBBP
PBBBBPP
PPPPPPP
AFTER DOWN (current):
PPPPPPP
PPBBBPP
PPBPBPP
PBBBBBP
PPBPBPP
PPBBBPP
PPPPPPP
blue row slots (rows40-46 cols14-48):
SSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS
SS
… [213 caracteres más]
```

**Pensamiento:**

Interesting! The blue row's slot 1 (cols 21-27) contains a glyph:
```
..B..
.B.B.
BBBBB
.B.B.  -- wait let me re-read
```
Blue slot1 (cols 21-27): from the row strings (cols 14..48, index offset 14):
row41: "SSBBBSSSSSSSBSSBBBBBSSBBSBBSSSSBSSS" -> cols 14+7=21..27: chars at index 7..13: "SSSSBSS"? Let me index: row41 string index: 0='S',1='S',2='B',3='B',4='B',5='S',6='S',7='S',8='S',9='S',10='S',11='B',12='S',... So slot1 = indices 7..13 = "SSSSBSS" → glyph col 25.
row42: "SSSSBSSSSSBSBSSSSBSBSSSBSBSSSSSBSSS": indices 7..13 = "SSSBSBS"? Let's carefully: index0 S,1 S,2 S,3 S,4 B,5 S,6 S,7 S,8 S,9 S,10 B,11 S,12 B → indices7-13 = "SSSBSBS". Hmm that's 7 chars: idx7=S,8=S,9=S,10=B,11=S,12=B,13=S.
row43: "SBBBBBSSBBBBBSSSSSSBSSSBBBSSSSBBBSS": idx7..13 = "SBBBBBS"? idx: 0 S,1 B,2 B,3 B,4 B,5 S,6 S,7 B,8 B,9 B,10 B,11 B,12 S,13 S → idx7-13 = "BBBBBSS". Hmm.
Let me just compute programmatically. But key insight: after UP, pink slot1 became:
```
.BBBBB.  -> row52: PPBBBBP → cols22-26 = BBBBB
BB..B.  PBBPPBP → idx1..5 = BBPPB
B...B  PBPPPBP → BPPP B
B..BB  PBPPBBP
BBBB.  PBBBBPP
```
And blue slot1 glyph (before) — from the diff, UP changed cells at rows 52-56 cols 22-26. The "AFTER UP" pink slot1 glyph should equal the blue slot1 glyph? Let's compare: blue slot1 from the ascii above (rows 41-45, cols 22-26):
row41 cols22-26: from row41 string index 8..12 = S,S,S,S,B → "SSSSB"? Hmm that gives a glyph of width... The blue glyph in slot1 seems to be:
row41: idx8-12: let me recount row41
… [5711 caracteres más]

```python
A = current_frame.ascii.split('\n')
def g5(r0,c0):
    return tuple(A[r][c0:c0+5] for r in range(r0,r0+5))
def bin5(pat, bg):
    return ''.join('1' if ch=='B' else '0' for row in pat for ch in row)
def rot5(pat):
    return tuple(''.join(pat[4-c][r] for c in range(5)) for r in range(5))
def allrots(pat):
    out=[]; p=pat
    for i in range(4):
        out.append(bin5(p)); p=rot5(p)
    return out
top={}
for ri,r in enumerate([4,13,22]):
    for ci,c in enumerate([12,22,35,45]):
        col = A[r][c]
        top[(ri,ci)] = (col, g5(r+1,c+1))
botS = {i: g5(41, 15+7*i) for i in range(5)}
botP = {i: g5(52, 15+7*i) for i in range(5)}
print('TOP:')
for k,v in top.items(): print(k, v[0], bin5(v[1]))
print('BOT blue:', {i:bin5(p) for i,p in botS.items()})
print('BOT pink:', {i:bin5(p) for i,p in
… [454 caracteres más]
```

```text
TOP:

error:
Traceback (most recent call last):
  File "<python_tool>", line 22, in <module>
TypeError: bin5() missing 1 required positional argument: 'bg'
```

## Paso 3 · nivel 1 · acción 4

**Pensamiento:**

Fix bin5 signature (retained version requires bg). Let me define new functions with different names to avoid conflicts.

```python
def gx(fr, r0,c0,n=5):
    L = fr.ascii.split('\n') if hasattr(fr,'ascii') else fr
    return tuple(L[r][c0:c0+n] for r in range(r0,r0+n))
def bits(pat):
    return ''.join('1' if ch=='B' else '0' for row in pat for ch in row)
def rot(pat):
    n=len(pat); return tuple(''.join(pat[n-1-c][r] for c in range(n)) for r in range(n))
def rots(pat):
    out=[];p=pat
    for _ in range(4):
        out.append(bits(p)); p=rot(p)
    return out
A = current_frame.ascii
top={}
for ri,r in enumerate([4,13,22]):
    for ci,c in enumerate([12,22,35,45]):
        top[(ri,ci)] = (A[r][c], gx(A,r+1,c+1))
botS = {i: gx(A,41,15+7*i) for i in range(5)}
botP = {i: gx(A,52,15+7*i) for i in range(5)}
print('TOP:')
for k,v in top.items(): print(' ',k, v[0], bits(v[1]))
print('BOTblue:', {i:bits(p) for i,p in botS.i
… [362 caracteres más]
```

```text
Traceback (most recent call last):
  File "<python_tool>", line 18, in <module>
IndexError: string index out of range
```

**Pensamiento:**

The `.ascii` might not have newlines? Let me check: `current_frame.ascii` is newline-delimited string. `A[r][c]` where A is the string including newlines → need split. Fix: L = current_frame.ascii.split('\n').

```python
A = current_frame.ascii.split('\n')
top={}
for ri,r in enumerate([4,13,22]):
    for ci,c in enumerate([12,22,35,45]):
        top[(ri,ci)] = (A[r][c], gx(A,r+1,c+1))
botS = {i: gx(A,41,15+7*i) for i in range(5)}
botP = {i: gx(A,52,15+7*i) for i in range(5)}
print('TOP:')
for k,v in top.items(): print(' ',k, v[0], bits(v[1]))
print('BOTblue:', {i:bits(p) for i,p in botS.items()})
print('BOTpink:', {i:bits(p) for i,p in botP.items()})
tb={k:v[1] for k,v in top.items() if v[0]=='S'}; tp={k:v[1] for k,v in top.items() if v[0]=='P'}
for i,p in botS.items(): print('botS',i,'->topblue',[k for k,q in tb.items() if bits(q) in rots(p)])
for i,p in botP.items(): print('botP',i,'->toppink',[k for k,q in tp.items() if bits(q) in rots(p)])
```

```text
TOP:
  (0, 0) S 1111110001110111000110001
  (0, 1) P 1111010011100011100101111
  (0, 2) S 1000111111001001111110001
  (0, 3) P 1111010010100111001011110
  (1, 0) S 0000100101111110010100001
  (1, 1) P 1111110010100101111010000
  (1, 2) S 0010011111101011010100100
  (1, 3) P 0011100101111111010011100
… [742 caracteres más]
```

**Pensamiento:**

Excellent! Structure confirmed:
- Top has 6 pairs: each pair is (blue tile, pink tile) connected by a gray bar. Pairs: row0: (0,0)-(0,1), (0,2)-(0,3); row1: (1,0)-(1,1), (1,2)-(1,3); row2: (2,0)-(2,1), (2,2)-(2,3).
- Bottom blue row has 5 glyphs, each matching a top blue tile (with rotation, apparently exact orientation? Let me check whether it's identity or rotated).
- Bottom pink row should be the corresponding partner pink glyph. Currently pink slot 0 matches (1,1) whose blue partner is (1,0). Blue slot 0 matches (1,2) whose partner is (1,3). So slot0 pink is wrong (should be (1,3)).
- Pink slot3 matches nothing → wrong.
So the puzzle: for each column slot i, set the pink glyph to the partner of the blue glyph in that slot. UP/DOWN cycles the glyph in the cursor slot; LEFT/RIGHT moves cursor.
Now, importantly: does the glyph in the pink slot need to be exactly the same orientation as the top pink tile, or any rotation? Note botS matches were found via rots() — need to check if the orientation matches exactly (bits equal) or requires rotation. Let me check exact matches first.
Also, key question: what does UP/DOWN do exactly — cycle through the 6 pink glyph shapes? Or rotate the current glyph? From the earlier test: pink slot1 before UP was '0111001010111110101001110' (a plus/diamond) which matches top pink (2,3) — and after UP it became something else. After DOWN it returned to the same. So UP cycles to a different glyph in a set. Let me determine the cycle order: repeated
… [1804 caracteres más]

```python
A = current_frame.ascii.split('\n')
top={}
for ri,r in enumerate([4,13,22]):
    for ci,c in enumerate([12,22,35,45]):
        top[(ri,ci)] = (A[r][c], gx(A,r+1,c+1))
botS = {i: gx(A,41,15+7*i) for i in range(5)}
botP = {i: gx(A,52,15+7*i) for i in range(5)}
tb={k:v[1] for k,v in top.items() if v[0]=='S'}; tp={k:v[1] for k,v in top.items() if v[0]=='P'}
pairs = {}
for ri in range(3):
    pairs[(ri,0)] = (ri,1); pairs[(ri,2)] = (ri,3)
print('exact match check blue slots:')
targets={}
for i,p in botS.items():
    ex=[k for k,q in tb.items() if bits(q)==bits(p)]
    rotm=[k for k,q in tb.items() if bits(q) in rots(p)]
    print(' slot',i,'exact',ex,'rot',rotm)
    k = ex[0] if ex else rotm[0]
    targets[i]=pairs[k]
print('targets (top pink tile per slot):', targets)
print('current pink vs ta
… [257 caracteres más]
```

```text
exact match check blue slots:
 slot 0 exact [] rot [(1, 2)]
 slot 1 exact [(2, 0)] rot [(2, 0)]
 slot 2 exact [] rot [(0, 0)]
 slot 3 exact [] rot [(0, 2)]
 slot 4 exact [] rot [(1, 0)]
targets (top pink tile per slot): {0: (1, 3), 1: (2, 1), 2: (0, 1), 3: (0, 3), 4: (1, 1)}
current pink vs target:

… [1434 caracteres más]
```

## Paso 3 · nivel 1 · acción 4

**Pensamiento:**

Key observations:
- Blue slot glyphs match top blue tiles only up to ROTATION (only slot 1 is exact). So the blue row shows rotated versions of the top blue glyphs. The pink row must presumably show the corresponding pink glyph, possibly at a matching rotation? Let's check: current pink slot 4 = '1111010011100011100101111' which is exactly top pink (0,1). And current pink slot 0 = '1111101001010010111100001' which is a rotation (90°) of top pink (1,1). Current pink slot1 = '0111001010111110101001110' = rot of (2,3). Current pink slot2 = '1111110001111011010111111' = rot of (2,1). 
So the pink glyphs are also drawn in rotated poses. Interesting: current pink slot i matches some top pink tile up to rotation:
slot0 → (1,1) [rot], slot1 → (2,3) [rot], slot2 → (2,1) [rot], slot4 → (0,1) [exact].
Blue slot i → top blue: slot0 → (1,2), slot1 → (2,0), slot2 → (0,0), slot3 → (0,2), slot4 → (1,0).
Partner pink of blue slot0 = (1,3); current pink slot0 is (1,1) → wrong.
Slot1: target (2,1); current (2,3) → wrong.
Slot2: target (0,1); current (2,1) → wrong.
Slot3: target (0,3); current doesn't match any → wrong.
Slot4: target (1,1); current (0,1) → wrong.
So all 5 slots are wrong currently. Good — this is the puzzle: fix each pink glyph to be the partner of the blue glyph in that slot.
Now, what about rotation? Does the rotation of the pink glyph matter? Possibly the correct answer requires the pink glyph to be in the same rotation as the blue glyph's rotation? Let's compute the rotation
… [1847 caracteres más]

```python
A = current_frame.ascii.split('\n')
top={}
for ri,r in enumerate([4,13,22]):
    for ci,c in enumerate([12,22,35,45]):
        top[(ri,ci)] = (A[r][c], gx(A,r+1,c+1))
tb={k:v[1] for k,v in top.items() if v[0]=='S'}; tp={k:v[1] for k,v in top.items() if v[0]=='P'}
def rotidx(pat, ref):
    # find k such that rot^k(ref) == pat
    p=ref
    for k in range(4):
        if bits(p)==bits(pat): return k
        p=rot(p)
    return None
pairs={(0,0):(0,1),(0,2):(0,3),(1,0):(1,1),(1,2):(1,3),(2,0):(2,1),(2,2):(2,3)}
print('slot | blue: toppose,rotofslot | pinkcur: toppose,rot | target pink')
for i in range(5):
    bs = gx(A,41,15+7*i); bp = gx(A,52,15+7*i)
    bk = [(k,rotidx(bs,q)) for k,q in tb.items() if rotidx(bs,q) is not None]
    pk = [(k,rotidx(bp,q)) for k,q in tp.items() if rotidx(bp,q) i
… [76 caracteres más]
```

```text
slot | blue: toppose,rotofslot | pinkcur: toppose,rot | target pink
0 blue [((1, 2), 1)] pinkcur [((1, 1), 1)] target (1, 3)
1 blue [((2, 0), 0)] pinkcur [((2, 3), 1)] target (2, 1)
2 blue [((0, 0), 1)] pinkcur [((2, 1), 1)] target (0, 1)
3 blue [((0, 2), 1)] pinkcur [] target (0, 3)
4 blue [((1, 0)
… [41 caracteres más]
```

**Pensamiento:**

Blue slot rotations: slot0 rot1, slot1 rot0, slot2 rot1, slot3 rot1, slot4 rot1. Hmm, rotations mostly 1 (90° CW) except slot1 which is 0. Hmm, maybe some shapes have 2-fold symmetry so rot index is ambiguous (rotidx returns first match). E.g., (1,2) pattern '0010011111101011010100100' — is it 2-fold symmetric? rots list would show duplicates. Not necessarily.
So the rotation of blue slot glyphs is not consistent (1,0,1,1,1) — hmm, that's suspicious: maybe rotidx returns 1 vs 3 ambiguously for symmetric shapes. Anyway the blue row glyphs are rotated versions; the puzzle probably doesn't care about rotation for the pink row (or maybe it does, requiring the same rotation).
Current pink rotations: slot0 rot1, slot1 rot1, slot2 rot1, slot4 rot0.
Hmm. Interesting: both blue and pink rows appear to have rotations applied — maybe the rotation is a red herring or maybe the correct answer requires matching rotation with the blue glyph.
Actually wait. Maybe the game's mechanic is different: maybe UP/DOWN *rotates* the pink glyph and cycles shape? Let's investigate the UP effect more concretely: before-UP slot1 glyph was (2,3) rot1 = '0111001010111110101001110'. After UP it became... let me compute from the changed cells. After-UP glyph rows52-56 cols22-26:
row52: from crop "PPBBBBP" cols21-27 → cols22-26 = "BBBBB"? indices: col21=P,22=P,23=B,24=B,25=B,26=B,27=P. Hmm the crop string "PPBBBBP" is cols 21..27 → col22='P', col23='B',col24='B',col25='B',col26='B'. So row52 cols22-26 = "PBBB
… [206 caracteres más]

```python
bf = transitions[1].before_frame.ascii.split('\n')
af = transitions[1].after_frame.ascii.split('\n')
for lbl,fr in [('before',bf),('after',af)]:
    pat = gx(fr,52,22)
    print(lbl, bits(pat))
    print('\n'.join(''.join('#' if ch=='B' else '.' for ch in row) for row in pat))
# also wider window to catch overflow
print('after wide:')
print(cropf(transitions[1].after_frame,51,20,57,29))
```

```text
before 0111001010111110101001110
.###.
.#.#.
#####
.#.#.
.###.
after 0111111001100011001111110
.####
##..#
#...#
#..##
####.
after wide:
PPPPPPPPPP
PPPBBBBPPB
PPBBPPBPPB
PPBPPPBPPB
PPBPPBBPPB
PPBBBBPPPB
PPPPPPPPPP
```

## Paso 3 · nivel 1 · acción 11

**Pensamiento:**

The after-UP glyph at cols 22-26 = '0111111001100011001111110' which is a rotation of top pink (0,1): rots list for (0,1) = ['1111010011100011100101111', '0111111001100011001111110', ...]. Yes! after-UP = (0,1) rot1.
Also note the "after wide" shows the glyph overflowing into col 29 ('B' at col 29 rows 52-55?) — that's the neighboring slot 2 glyph at col 29 (part of pink slot2 glyph). Fine.
So UP changed slot1 from (2,3)rot1 to (0,1)rot1. Rotation stayed 1! So UP cycles the *shape identity* while keeping the same rotation pose. 
So each slot has a rotation pose (fixed?) and a shape that cycles with UP/DOWN. The correct answer is presumably: shape = partner shape, pose = whatever (probably the pose is fixed per slot and preserved).
So the state per slot: shape index in a cycle. We need to find the cycle order. Let's probe: press UP several times at slot 1 and record shape+pose each time. If cycle length is 6 (6 pink shapes), then after 6 UPs we're back to start.
But careful: maybe the cycle includes more shapes than the 6 in the top (e.g., shapes from the blue tiles too, or the slot3 glyph which matched nothing). Slot3's pink glyph matched nothing among top pink shapes — maybe it's a shape from a different set (or a shape not in the top). Let's check slot3's current glyph bits: '1111110001111110101001110'. Hmm.
Let me plan the probe: at slot 1 (cursor is at slot 0 now, since we moved LEFT back to col 15). Actually cursor currently at slot 0. So pressing UP now affects slot 0.

… [2458 caracteres más]

```python
def slotpat(i, row='pink'):
    A = current_frame.ascii.split('\n')
    r = 52 if row=='pink' else 41
    return gx(A, r, 15+7*i)
seq=[]
start = bits(slotpat(0))
seq.append(start)
for k in range(7):
    res = action(['UP'])
    b = bits(slotpat(0))
    seq.append(b)
    if b == start: 
        print('cycle closed after', k+1, 'UPs'); break
print('slot0 cycle:')
tp_bits = {}
A = current_frame.ascii.split('\n')
for ri,r in enumerate([4,13,22]):
    for ci,c in enumerate([12,22,35,45]):
        if A[r][c]=='P': tp_bits[(ri,ci)] = bits(gx(A,r+1,c+1))
for j,b in enumerate(seq):
    m = [k for k,v in tp_bits.items() if v==b]
    mr = [k for k,v in tp_bits.items() if b in [bits(rot([v[i:i+5] for i in range(0,25,5)]),) ] ] if False else None
    print(j, b, 'exacttop', m)
```

```text
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
cycle closed after 7 UPs
slot0 cycle:
0 1111101001010010111100001 exacttop []
1 0010011111101011111100100 exacttop [(2, 3)]
2 11110100111000111001
… [229 caracteres más]
```

## Paso 4 · nivel 1 · acción 11

**Pensamiento:**

Interesting: the cycle at slot 0 has length 7 and includes shapes: 
idx0: '1111101001010101...' hmm let me list:
0: 1111101001010010111100001  = (1,1) rot1
1: 0010011111101011111100100 = (2,3) exact rot0
2: 1111010011100011100101111 = (0,1) rot0
3: 0010011111100011000111111 = (0,3) rot3 (from rots list of (0,3): index3 = '0010011111100011000111111' yes)
4: 1111110001111110101001110 = matches nothing in top pink (this is the current slot3 glyph too!)
5: 0011100101111111010011100 = (1,3) rot0
6: 1111110101101111000111111 = (2,1) rot3 (rots of (2,1): index3 = '1111110101101111000111111' yes)
7: back to 0.
So cycle of 7 shapes: 6 top pink shapes + one extra shape ('1111110001111110101001110'). Hmm, the 7th might be a distractor shape.
But wait: the poses changed! idx0 was rot1 of (1,1), idx1 is rot0 of (2,3), idx2 rot0 of (0,1), idx3 rot3 of (0,3), idx5 rot0 of (1,3), idx6 rot3 of (2,1). So the pose is NOT preserved across the cycle; rather each shape has a fixed pose in the cycle list. Hmm, but earlier: slot1 before UP = (2,3) rot1, after UP = (0,1) rot1. Here slot0 cycle has (2,3) at rot0 and (0,1) at rot0. So the pose depends on the slot! I.e., each slot has a fixed pose, and the cycle enumerates the 7 shapes all rendered in the slot's pose. That makes sense: the cycle is "shape list rendered in slot pose".
So slot0 pose = rot1, and the cycle shapes in rot1 pose: (1,1)rot1, (2,3)rot1, (0,1)rot1, (0,3)rot1, extra, (1,3)rot1, (2,1)rot1. Let me verify: idx1 = '0010011111101011111
… [4568 caracteres más]

```python
A = current_frame.ascii.split('\n')
def pat(r,c): return gx(A,r+1,c+1)
tiles={}
for ri,r in enumerate([4,13,22]):
    for ci,c in enumerate([12,22,35,45]):
        tiles[(ri,ci)] = pat(r,c)
def mirror(pat_):
    return tuple(row[::-1] for row in pat_)
def alltrans(pat_):
    out={}
    p=pat_
    for k in range(4):
        out['r'+str(k)]=bits(p); p=rot(p)
    m=mirror(pat_); 
    for k in range(4):
        out['m'+str(k)]=bits(m); m=rot(m)
    return out
pairs=[((0,0),(0,1)),((0,2),(0,3)),((1,0),(1,1)),((1,2),(1,3)),((2,0),(2,1)),((2,2),(2,3))]
for b,p in pairs:
    tb_=alltrans(tiles[b]); tp_=bits(tiles[p])
    hits=[k for k,v in tb_.items() if v==tp_]
    print(b,'->',p, 'transforms:', hits, 'blue',bits(tiles[b]),'pink',tp_)
```

```text
(0, 0) -> (0, 1) transforms: [] blue 1111110001110111000110001 pink 1111010011100011100101111
(0, 2) -> (0, 3) transforms: [] blue 1000111111001001111110001 pink 1111010010100111001011110
(1, 0) -> (1, 1) transforms: [] blue 0000100101111110010100001 pink 1111110010100101111010000
(1, 2) -> (1, 3) t
… [263 caracteres más]
```

## Paso 4 · nivel 1 · acción 11

**Pensamiento:**

No simple rotation/mirror transform between paired tiles. So the pairs are arbitrary "identity" pairs: each blue glyph is associated with a specific pink glyph (a lookup table / cipher). The bottom blue row shows 5 blue glyphs (rotated poses) and we must set each pink slot to the matching partner glyph.
The key uncertainty: which pose (rotation) of the partner shape is accepted. Note the cycle for slot0 contained shapes at various rotations of the canonical top-pink shapes:
slot0 cycle: idx0=(1,1)rot1, idx1=(2,3)rot0, idx2=(0,1)rot0, idx3=(0,3)rot3, idx4=extra, idx5=(1,3)rot0, idx6=(2,1)rot3.
Hmm, these are different rotations per shape in the same slot. That suggests the cycle is a fixed list of 7 concrete 5x5 bitmaps (each shape rendered in a fixed canonical pose, which happens to be a rotation of the top tile pose).
Hmm wait. Maybe my assumption about which is "canonical" is wrong: maybe the top pink tiles themselves are drawn in various poses and the cycle's shapes are the "canonical" poses. Whatever — the important thing is: the set of 7 bitmaps in slot0's cycle. If the cycle for slot i is a rotation of that list by the slot's pose... Let's check slot1's earlier observation: slot1 before UP = (2,3) rot1 = '0111001010111110101001110'; slot0's cycle contains (2,3) at rot0. So slot1's cycle has (2,3) at rot1 while slot0's has it at rot0 → the slot cycles are rotated versions of each other by the slot's pose. And after-UP in slot1 = (0,1)rot1 while slot0 cycle has (0,1) at r
… [5801 caracteres más]

```python
def state():
    A = current_frame.ascii.split('\n')
    # cursor from white bracket row 48
    row = A[48]
    cols = [c for c,ch in enumerate(row) if ch=='W']
    pos = round((min(cols)-15)/7) if cols else None
    blue = {i: gx(A,41,15+7*i) for i in range(5)}
    pink = {i: gx(A,52,15+7*i) for i in range(5)}
    tiles={}
    for ri,r in enumerate([4,13,22]):
        for ci,c in enumerate([12,22,35,45]):
            tiles[(ri,ci)] = (A[r][c], gx(A,r+1,c+1))
    return pos, blue, pink, tiles

def cls(pat_):
    return min(rots(pat_))

pos, blue, pink, tiles = state()
tb={k:v[1] for k,v in tiles.items() if v[0]=='S'}; tp={k:v[1] for k,v in tiles.items() if v[0]=='P'}
pairs={(0,0):(0,1),(0,1):(0,0),(0,2):(0,3),(0,3):(0,2),(1,0):(1,1),(1,1):(1,0),(1,2):(1,3),(1,3):(1,2),(2,0):(2,1),(2,1):(2,
… [322 caracteres más]
```

```text
cursor 0
0 ((1, 2), (1, 3), '0000101111010010100111111', '0011100101111111010011100', False)
1 ((2, 0), (2, 1), '0010011111101011111100100', '1111110001101111010111111', False)
2 ((0, 0), (0, 1), '1111110001101111010111111', '0111111001100011001111110', False)
3 ((0, 2), (0, 3), '0011111101101011110
… [128 caracteres más]
```

**Nivel 1 completado → nivel 2 (acción 39)**

## Paso 4 · nivel 2 · acción 39

**Pensamiento:**

All 5 slots currently wrong. Targets confirmed (each slot's target class differs from current class).
Note slot2's current pink class '1111110001101111010111111' equals the target class of slot1 (2,1) — consistent with earlier.
Now solve: for each slot, cycle UP until class matches target. I'll implement a loop:
- cursor management: current pos from state().
- For slot i: press RIGHT/LEFT to get to i (abs difference presses).
- Then press UP until class matches (max 7).
Stop early if result indicates completion or game over.
Total expected actions: moves 0+1+1+1+1 = 4 plus presses. Let's execute and log.
I need to be careful that pressing UP when already correct is avoided. Also, the check for completion: read result state after each action.
Let me implement with a helper that returns the result dict. action() returns a list for a list input? Earlier `res = action(['UP'])` printed "[action] UP -> effect" then a dict printed as res... Actually print(res) printed a list? In the first call I printed res and it showed a dict-like line: "[action] RIGHT -> effect\n{'executed': True, ...}". So res is a list of one dict, and printing the list showed... hmm it printed "{'executed'...}" not "[{...}]". Whatever — I'll normalize: if isinstance(res, list): r = res[-1] else res.
Let me write the solver now.

```python
def last(res):
    if isinstance(res, list): return res[-1] if res else {}
    return res

log=[]
pos, blue, pink, tiles = state()
def targets_from(tiles):
    tb={k:v[1] for k,v in tiles.items() if v[0]=='S'}; tp={k:v[1] for k,v in tiles.items() if v[0]=='P'}
    pairs={(0,0):(0,1),(0,1):(0,0),(0,2):(0,3),(0,3):(0,2),(1,0):(1,1),(1,1):(1,0),(1,2):(1,3),(1,3):(1,2),(2,0):(2,1),(2,1):(2,0),(2,2):(2,3),(2,3):(2,2)}
    tmap={cls(v):k for k,v in tb.items()}; pmap={cls(v):k for k,v in tp.items()}
    out={}
    for i in range(5):
        b = gx(current_frame.ascii.split('\n'),41,15+7*i)
        out[i] = cls(tp[pairs[tmap[cls(b)]]])
    return out
tgt = targets_from(tiles)
stop=False
for i in range(5):
    pos, blue, pink, tiles = state()
    while pos != i and not stop:
        res = action(['
… [806 caracteres más]
```

```text
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] RIGHT -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] RIGHT -> effect
[action] UP -> effect
[action]
… [1844 caracteres más]
```

## Paso 5 · nivel 2 · acción 39

**Pensamiento:**

Level 2! The layout changed: now the pairs are pink↔yellow (top), and the bottom has a pink row (short, 4 glyphs) and a yellow row (long, 7 glyphs?). Colors swapped: previously blue→pink; now pink→yellow.
Let me inspect the new board structure: top pairs (pink tile, yellow tile) — but yellow tiles are wider (variable width!). E.g., row0: pink at cols ~5-11, yellow at ~15-22 (wider), pink ~29-35, yellow ~38-57 (very wide, 3 glyphs?).
Bottom: pink bar rows ~? with 4 glyphs, and yellow bar with 7 glyphs. Two cursors (white brackets at left, rows ~48 and ~59).
Hmm, so now the shapes are variable-width. Let me analyze the structure carefully via segmentation.
Let me get the segmentation and figure out the layout: tiles in top area, bars at bottom.
The goal presumably: for each slot in the bottom pink row, set it to the yellow partner? Wait no. Previously: top pairs blue→pink; bottom blue row (given) + pink row (editable) — we set pink to match the partner of the blue glyph in the same column.
Now: bottom has pink row (4 glyphs) and yellow row (7 glyphs). Hmm, which is editable? In level 1, the editable row was the pink (bottom) row, and UP/DOWN changed the pink row. The blue row was the "given" row.
Now the top pairs are pink→yellow. So the given row is probably pink (top of the two bottom bars) and the editable is yellow (bottom bar). But the counts differ: pink 4 glyphs vs yellow 7 glyphs? Let me verify by segmentation — maybe the yellow bar has 7 slots and pink has 4? That seem
… [646 caracteres más]

```python
seg = current_frame.segmentation
ns = seg['nodes']
print('level', current_frame.level, 'nodes', len(ns))
for n in ns:
    b=n['boundary']
    r0=min(p[0] for p in b); r1=max(p[0] for p in b); c0=min(p[1] for p in b); c1=max(p[1] for p in b)
    print(n['id'], n['color'], n['pixels'], (r0,c0,r1,c1))
```

```text
level 2 nodes 126
0 g 1276 (0, 0, 33, 63)
1 P 31 (4, 5, 10, 11)
2 Y 38 (4, 15, 10, 21)
3 P 28 (4, 28, 10, 34)
4 Y 113 (4, 38, 10, 58)
5 B 14 (5, 6, 9, 10)
6 B 1 (5, 16, 5, 16)
7 B 9 (5, 16, 9, 20)
8 B 17 (5, 29, 9, 33)
9 B 5 (5, 41, 9, 41)
10 B 9 (5, 46, 9, 50)
11 B 1 (5, 50, 5, 50)
12 B 1 (5, 53, 5
… [2723 caracteres más]
```

**Pensamiento:**

Level 2 structure:
- Top area: 6 pairs of (P tile, Y tile) connected by gray bars. But now the yellow tiles are variable width: e.g., pair row0 left: P at (4-10, 5-11), Y at (4-10, 15-21) — 7 wide. Row0 right: P (4-10, 28-34), Y (4-10, 38-58) — 21 wide! That's 3 glyphs worth.
- Row1 left: P(13-19,5-11), Y(13-19,15-28) = 14 wide (2 glyphs). Row1 right: P(13-19,35-41), Y(13-19,45-58)=14 wide.
- Row2 left: P(22-28,5-11), Y(22-28,15-35)=21 wide (3 glyphs). Row2 right: P(22-28,42-48), Y(22-28,52-58)=7 wide (1 glyph).
So yellow tiles contain 1, 2, or 3 glyphs! Interesting — the mapping is 1 pink glyph → a sequence of 1-3 yellow glyphs? Like a code/cipher where each pink symbol maps to a sequence of yellow symbols.
Bottom: P bar (40-46, cols 18-45) containing 4 glyphs (at cols 19-23, 26-30, 33-37, 40-44 → 4 glyphs, 7-col slots starting at col 18? slots at 18-24, 25-31, 32-38, 39-45 → 4 slots of 7). 
Bottom Y bar (51-57, cols 7-55) = 49 wide = 7 slots of 7. Glyphs inside: at cols 8-10 & 12,15 (part of glyph 1?), ... complex.
So: the pink row has 4 glyphs (a "word" of 4 symbols), and the yellow row has 7 slots which should be the encoded sequence! The cipher: each pink symbol maps to a yellow sequence of length 1-3, from the top examples.
Top examples: 
- P(4,5) ↔ Y(4,15) [1 glyph]
- P(4,28) ↔ Y(4,38..58) [3 glyphs]
- P(13,5) ↔ Y(13,15..28) [2 glyphs]
- P(13,35) ↔ Y(13,45..58) [2 glyphs]
- P(22,5) ↔ Y(22,15..35) [3 glyphs]
- P(22,42) ↔ Y(22,52..58) [1 glyph]
Bottom pink: 4 glyphs → to
… [3289 caracteres más]

```python
A = current_frame.ascii.split('\n')
def tiles_top():
    pairs=[]
    # pink tiles at rows 4,13,22 cols 5,28/35,42 etc. find P nodes with height 7
    seg = current_frame.segmentation
    pinks=[]; yellows=[]
    for n in seg['nodes']:
        b=n['boundary']
        r0=min(p[0] for p in b); r1=max(p[0] for p in b); c0=min(p[1] for p in b); c1=max(p[1] for p in b)
        if n['color']=='P' and r1-r0==6 and c1-c0==6 and r0<30: pinks.append((r0,c0))
        if n['color']=='Y' and r1-r0==6 and r0<30: yellows.append((r0,c0,c1))
    return pinks, yellows
pinks, yellows = tiles_top()
print('pinks', sorted(pinks))
print('yellows', sorted(yellows))
def glyph(r,c): return gx(A,r,c)  # 5x5 at row r col c
examples=[]
for (pr,pc) in sorted(pinks):
    pg = glyph(pr+1,pc+1)
    # matching yellow on sa
… [718 caracteres más]
```

```text
pinks [(4, 5), (4, 28), (13, 5), (13, 35), (22, 5), (22, 42)]
yellows [(4, 15, 21), (4, 38, 58), (13, 15, 28), (13, 45, 58), (22, 15, 35), (22, 52, 58)]
EX 0 P 1000011110100101001011111 len 1 ['1011100100001000010011101']
EX 1 P 0111001010111111000111111 len 3 ['0010010101001001010100100', '10001100
… [907 caracteres más]
```

## Paso 5 · nivel 2 · acción 39

**Pensamiento:**

Now let's match the bottom pink glyphs to the example pink glyphs (rotation-invariant) and derive the target yellow sequence.
Examples:
EX0: P '1000011110100101001011111' → 1 yellow: '1011100100001000010011101'
EX1: P '0111001010111111000111111' → 3: ['0010010101001001010100100', '1000110000111110000110001', '1010100000101111010111101']
EX2: P '1111010011100011100101111' → 2: ['0101000000111110000001010', '1110110000000000000110111']
EX3: P '0111101001110010100101111' → 2: ['1001100001100011000011001', '1001100001100011000011001'] (same glyph twice!)
EX4: P '0011100101111111010011100' → 3: ['1010100001101010000111111', '1101110001101011000111011', '1010110000101011000011111']
EX5: P '0010011111101011111100100' → 1: '0101011011100010000010001'
Bottom pink: 
b0 '0000101111010010100111111'
b1 '0011100101111111010011100' → EX4 exactly!
b2 '1111010010100111001011110'
b3 '0111001010111110101001110'
Need rotation-invariant matching. Note the yellow glyphs in the sequence may also be rotated. Let's compute classes.
Bottom yellow current (7):
y0 '0110111000000001100001101'
y1 '1011100100101110000110111'
y2 '1111110000101011000010101'
y3 '1000110000111110000110001' → EX1[1] exactly!
y4 '1101110001101011000111011' → EX4[1] exactly!
y5 '1111110001001001000111111'
y6 '0010010101001001010100100' → EX1[0] exactly!
Now let me compute the target sequence: bottom pink symbols b0..b3 → their example → concatenated yellow glyphs. Lengths must sum to 7.
Let me compute using rotation-invariant cla
… [398 caracteres más]

```python
A = current_frame.ascii.split('\n')
def glyph(r,c): return gx(A,r,c)
ex = [
 ('1000011110100101001011111', ['1011100100001000010011101']),
 ('0111001010111111000111111', ['0010010101001001010100100','1000110000111110000110001','1010100000101111010111101']),
 ('1111010011100011100101111', ['0101000000111110000001010','1110110000000000000110111']),
 ('0111101001110010100101111', ['1001100001100011000011001','1001100001100011000011001']),
 ('0011100101111111010011100', ['1010100001101010000111111','1101110001101011000111011','1010110000101011000011111']),
 ('0010011111101011111100100', ['0101011011100010000010001']),
]
def b2p(b): return tuple(b[i:i+5] for i in range(0,25,5))
def cl(b): return min(rots(b2p(b)))
expink = [(i, cl(p)) for i,(p,ys) in enumerate(ex)]
# check uniqueness of pink cla
… [573 caracteres más]
```

```text
pink class dup? Counter({'0000000000000000000000000': 6})
pinkslot 0 0000101111010010100111111 -> EX 5 len 1
pinkslot 1 0011100101111111010011100 -> EX 5 len 1
pinkslot 2 1111010010100111001011110 -> EX 5 len 1
pinkslot 3 0111001010111110101001110 -> EX 5 len 1
lens [1, 1, 1, 1] total 4
yslot 0 cur 
… [491 caracteres más]
```

**Pensamiento:**

Bug: my `cl` uses rots() on a tuple of strings — rots(pat) expects a tuple of strings of 'B'/'.'? No: bits() checks ch=='B'. My b2p converts bits string ('0'/'1') to rows of '0'/'1' chars, so bits() sees no 'B' → all zeros. Fix: make a class function on bit-strings directly using rotations of the 5x5 boolean grid.
Let me write proper helpers:
def rotb(bitsstr): rotate 5x5 bit string 90 CW.
def clb(bitsstr): min over 4 rotations.

```python
def rotb(b):
    g=[[b[r*5+c] for c in range(5)] for r in range(5)]
    return ''.join(g[4-c][r] for r in range(5) for c in range(5))
def clb(b):
    s=[]; x=b
    for _ in range(4):
        s.append(x); x=rotb(x)
    return min(s)
ex = [
 ('1000011110100101001011111', ['1011100100001000010011101']),
 ('0111001010111111000111111', ['0010010101001001010100100','1000110000111110000110001','1010100000101111010111101']),
 ('1111010011100011100101111', ['0101000000111110000001010','1110110000000000000110111']),
 ('0111101001110010100101111', ['1001100001100011000011001','1001100001100011000011001']),
 ('0011100101111111010011100', ['1010100001101010000111111','1101110001101011000111011','1010110000101011000011111']),
 ('0010011111101011111100100', ['0101011011100010000010001']),
]
from collecti
… [581 caracteres más]
```

```text
pink classes unique? True [('0000101111010010100111111', 1), ('0011111101101011110100111', 1)]
pinkslot 0 -> EX 0 len 1
pinkslot 1 -> EX 4 len 3
pinkslot 2 -> EX 3 len 2
pinkslot 3 -> EX 5 len 1
total yellow needed 7
y 0 cur 0110111000000001100001101 tgt 1011100100001000010011101 exact False samecla
… [560 caracteres más]
```

## Paso 5 · nivel 2 · acción 46

**Pensamiento:**

The cipher decoding works: 4 pink symbols → 1+3+2+1 = 7 yellow glyphs. 
Now, the ambiguity: the pink symbols in the bottom row are drawn in some rotation; the yellow output glyphs — should they be in the same rotation as the example, or rotated to match the pink's rotation? Let's think: the yellow bar has 7 slots, and each slot in level 1 had a fixed "pose" for the cycle. Here the yellow bar likely has per-slot poses too.
Important question: is the target for each yellow slot the exact bitmap from the example, or a rotated version (rotated by the slot's pose)? In level 1, the cycle for each slot contained each shape in exactly one pose. So the "correct" one is the one the cycle offers with the right shape class. So I should match by class, not exact bits. But here the current yellow glyphs' classes don't match the target classes at all — so all 7 need changing (unless some slot's cycle lacks the needed shape class... unlikely).
Hmm, but careful: matching by class could be ambiguous if the cycle contains two shapes of the same class (e.g., the same shape in different poses). Let me probe the cycle for slot 0 of the yellow bar to learn the cycle contents and poses. The number of yellow shapes: 12 yellow glyphs in the examples (1+3+2+2+3+1 = 12), but some may be repeats (EX3 has the same glyph twice). Distinct shapes maybe ~9-10. The cycle might be long (10+). That's a lot of presses per slot. Worst case 7 slots × ~10 presses = 70 > budget (64 - 40 = 24 remaining? Wait the 'w' b
… [3038 caracteres más]

```python
def ybar():
    A = current_frame.ascii.split('\n')
    return [bits(gx(A,52,8+7*j)) for j in range(7)]
def pbar():
    A = current_frame.ascii.split('\n')
    return [bits(gx(A,41,19+7*j)) for j in range(4)]
def cursor():
    A = current_frame.ascii.split('\n')
    cols=[c for c,ch in enumerate(A[48]) if ch=='W']
    return round((min(cols)-9)/7), min(cols), max(cols)
cur = ybar()
start = cur[0]
seq=[start]
N=None
for k in range(16):
    r = last(action(['UP']))
    b = ybar()[0]
    seq.append(b)
    if r.get('level_completed') or r.get('game_over'): print('END', r.get('state')); break
    if clb(b)==clb(start):
        N=k+1; break
print('cycle len', N, 'presses used', len(seq)-1)
for i,b in enumerate(seq):
    print(i, b, clb(b))
print('cursor', cursor())
```

```text
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
cycle len 7 presses used 7
0 0110111000000001100001101 0101011011100010000010001
1 1011100100101110000110111 1010100000101111010111101
2 111111000
… [330 caracteres más]
```

## Paso 6 · nivel 2 · acción 46

**Pensamiento:**

Cycle length 7 at slot 0, and it returned to the start. The 7 classes in slot0's cycle order:
k0: cl '0101011011100010000010001'  (EX5's yellow)
k1: '1010100000101111010111101' (EX1[2])
k2: '1101110001101011000111011' (EX4[1])
k3: '0010010101001001010100100' (EX1[0])
k4: '1000110000111110000110001' (EX1[1])
k5: '1001100001100011000011001' (EX3[0]/EX3[1])
k6: '1010100001101010000111111' (EX4[0])
Missing classes from the examples: EX0's yellow '1011100100001000010011101', EX2's two: '0101000000111110000001010', '1110110000000000000110111', EX4[2] '1010110000101011000011111', EX5's yellow = k0 ✓.
Hmm! The target sequence requires: 
seq = EX0[0], EX4[0], EX4[1], EX4[2], EX3[0], EX3[1], EX5[0]
= ['1011100100001000010011101', '1010100001101010000111111', '1101110001101011000111011', '1010110000101011000011111', '1001100001100011000011001','1001100001100011000011001', '0101011011100010000010001']
Classes needed: EX0 yellow (not in slot0 cycle!), EX4[2] (not in cycle), etc.
Hmm, so the slot-0 cycle contains 7 classes but the needed set includes classes not in slot 0's cycle. That means different slots have different shape sets?! Or the cycle contains shapes whose classes I mis-derived...
Wait — maybe the pose rotation matters and my clb (rotation) is fine. The slot0 cycle classes: {EX5y, EX1[2], EX4[1], EX1[0], EX1[1], EX3, EX4[0]}. Needed: {EX0y, EX4[0], EX4[1], EX4[2], EX3, EX3, EX5y}.
Needed but missing from slot0's cycle: EX0y, EX4[2]. And slot0's cycle has extras: EX1[2], EX1[0]
… [12634 caracteres más]

```python
def mir(b):
    rows=[b[r*5:r*5+5] for r in range(5)]
    return ''.join(r[::-1] for r in rows)
def mcl(b):
    s=[]; x=b
    for _ in range(4):
        s.append(min(x, mir(x))); x=rotb(x)
    return min(s)
ex = [
 ('1000011110100101001011111', ['1011100100001000010011101']),
 ('0111001010111111000111111', ['0010010101001001010100100','1000110000111110000110001','1010100000101111010111101']),
 ('1111010011100011100101111', ['0101000000111110000001010','1110110000000000000110111']),
 ('0111101001110010100101111', ['1001100001100011000011001','1001100001100011000011001']),
 ('0011100101111111010011100', ['1010100001101010000111111','1101110001101011000111011','1010110000101011000011111']),
 ('0010011111101011111100100', ['0101011011100010000010001']),
]
names={}
for i,(p,ys) in enumerate(ex)
… [685 caracteres más]
```

```text
cycle slot0 entries -> matching example shape (rot / mirror):
0 0110111000000001100001101 cnt 10 rot-match ['EX5[0]'] mir-match ['EX5[0]']
1 1011100100101110000110111 cnt 14 rot-match ['EX1[2]'] mir-match ['EX1[2]']
2 1111110001001001000111111 cnt 15 rot-match ['EX4[1]'] mir-match ['EX4[1]']
3 01010
… [1516 caracteres más]
```

## Paso 6 · nivel 2 · acción 50

**Pensamiento:**

Great — the earlier confusion was due to duplicate classes: several example yellow shapes are rotation-equivalent to each other (EX0[0] ≡ EX1[1], EX2[0] ≡ EX1[0], EX2[1] ≡ EX3[0], EX4[0] ≡ EX4[2]).
So the distinct rotation classes among yellow shapes are:
- A: EX5[0] (10 px)
- B: EX1[2] (14 px)
- C: EX4[1] (15 px)
- D: EX1[0]/EX2[0] (9 px)
- E: EX0[0]/EX1[1] (11 px)
- F: EX2[1]/EX3[0] (10 px)
- G: EX4[0]/EX4[2] (13 px)
That's 7 classes → the cycle has exactly 7 entries, one per class ✓✓. 
Now the target per slot by class:
slot0: EX0[0] → class E
slot1: EX4[0] → class G
slot2: EX4[1] → class C
slot3: EX4[2] → class G
slot4: EX3[0] → class F
slot5: EX3[0] → class F
slot6: EX5[0] → class A
Cycle order (slot 0 base list, index → class):
k0: A
k1: B
k2: C
k3: D
k4: E
k5: F
k6: G
So the base order is [A,B,C,D,E,F,G]. If all slots share this order (with different poses), then for slot i: presses = (k_target - k_current) mod 7, where k_current is the index of the current glyph's class.
Note: the cycle order in slot 0 was A,B,C,D,E,F,G — suspiciously clean, suggesting the game's shape list is ordered and each slot's cycle is the same list (each rendered in the slot's pose).
Now current state: I pressed UP 7 times at slot 0, so slot 0 is back to its original glyph (class A). Wait: the cycle returned to the same bitmap, so slot0 currently = original (class A). Target for slot0 = E (k4). Current k=0 → 4 presses UP (or 3 DOWN).
Let me compute for the current bar (need to re-read since slo
… [2762 caracteres más]

```python
CYC = ['A','B','C','D','E','F','G']
clsmap = {'0101011011100010000010001':'A','1010100000101111010111101':'B','1101110001101011000111011':'C',
          '0010010101001001010100100':'D','1000110000111110000110001':'E','1001100001100011000011001':'F',
          '1010100001101010000111111':'G'}
def name2(b): return clsmap.get(clb(b),'?')
targets = ['E','G','C','G','F','F','A']
log=[]; stop=False
def res_check(r):
    return bool(r.get('level_completed') or r.get('game_over') or r.get('done'))
for i in range(7):
    pos,_,_ = cursor()
    while pos != i and not stop:
        r = last(action(['RIGHT' if i>pos else 'LEFT'])); log.append(('mv',i,r.get('state')))
        if res_check(r): stop=True; break
        pos += 1 if i>pos else -1
    if stop: break
    cur = name2(ybar()[i])
    if cur=='?
… [918 caracteres más]
```

```text
[action] DOWN -> effect
[action] DOWN -> effect
[action] DOWN -> effect
[action] RIGHT -> effect

error:
Traceback (most recent call last):
  File "<python_tool>", line 20, in <module>
ValueError: 'B' is not in list
```

**Nivel 2 completado → nivel 3 (acción 73)**

## Paso 7 · nivel 3 · acción 73

**Pensamiento:**

My code crashed at slot 1 because 'B' is not in 'ABCDEFG'... wait, k_cur = 'ABCDEFG'.index('B') should work. The error says "'B' is not in list" — hmm, that's from `targets.index(cur)`? No, I had `d = (targets.index(cur), )` — a leftover placeholder line! cur='B' and targets = ['E','G','C','G','F','F','A'] doesn't contain 'B'. That's the bug. It crashed after 3 DOWNs on slot 0 and 1 RIGHT.
Wait, but the order: the log shows DOWN×3 executed for slot 0 (fwd for slot0: k_cur=0 (A), k_tgt=4 (E) → fwd=4 > 3 → DOWN×3). Then crash happened at slot 1 after moving RIGHT... Actually the crash occurred at slot 0's `d = (targets.index(cur),)` line? No — the log printed 4 actions: DOWN, DOWN, DOWN, RIGHT. So slot 0's 3 DOWNs executed, then the placeholder line executed at slot... hmm, the placeholder line is before the presses. Let me re-read the code:
for i in range(7):
    move cursor
    cur = name2(...)
    if cur=='?': break
    d = (targets.index(cur), )   # <-- placeholder bug
    ...
So for i=0: cur='A' → targets.index('A') = 6 → ok, then presses DOWN×3 → executed. Then i=1: move RIGHT (cursor was 0 → 1) executed, then cur = name2(ybar()[1]) = 'B' → targets.index('B') → ValueError. Crash.
So current state: slot 0 got 3 DOWNs (A → ? hopefully E), cursor at slot 1.
Let me verify slot 0's class now and check whether DOWN went A→G→F→E (i.e., inverse of UP). Then continue with fixed code (remove the placeholder line).
Let me write a robust function this time, define it properly with cl
… [264 caracteres más]

```python
clsmap = {'0101011011100010000010001':'A','1010100000101111010111101':'B','1101110001101011000111011':'C',
          '0010010101001001010100100':'D','1000110000111110000110001':'E','1001100001100011000011001':'F',
          '1010100001101010000111111':'G'}
def nm(b, m=None):
    return (m or clsmap).get(clb(b),'?')
y = ybar()
print('cursor', cursor())
print('classes', [nm(b) for b in y])
print('raw', y)
print('w bar len', [len([c for c in row if c=='w']) for row in current_frame.ascii.split('\n')[63:64]])
A=current_frame.ascii.split('\n')[63]
print('row63 counts', {ch:A.count(ch) for ch in set(A)})
```

```text
cursor (1, 15, 19)
classes ['E', 'B', 'G', 'E', 'C', 'C', 'D']
raw ['1000110000111110000110001', '1011100100101110000110111', '1111110000101011000010101', '1000110000111110000110001', '1101110001101011000111011', '1111110001001001000111111', '0010010101001001010100100']
w bar len [59]
row63 counts {
… [16 caracteres más]
```

**Pensamiento:**

Slot 0 is now class E ✓ (target E). DOWN worked as the inverse of UP (A → G → F → E: 3 DOWNs = -3 ≡ +4 mod 7 ✓).
Current classes: ['E','B','G','E','C','C','D'], targets ['E','G','C','G','F','F','A'].
Cursor at slot 1.
Remaining fixes:
slot1: B(k1) → G(k6): fwd=5 → DOWN×2
slot2: G(k6) → C(k2): fwd=3 → UP×3
slot3: E(k4) → G(k6): fwd=2 → UP×2
slot4: C(k2) → F(k5): fwd=3 → UP×3
slot5: C(k2) → F(k5): fwd=3 → UP×3
slot6: D(k3) → A(k0): fwd=4 → DOWN×3
Cursor moves: 1→2,3,4,5,6 = 5 RIGHTs.
Total = 2+3+2+3+3+3 + 5 = 21 actions. Budget: row63 has 59 'w' + 5 'c' = the bar shrinks by 1 per action; 59 remaining. Fine.
Note the class map is not retained as a global for functions; I'll define nm with an inline dict default. Let me write the final solver with a self-correcting loop and inline dict.
Careful with the "same class order per slot" assumption — the verify loop handles mismatches by pressing UP up to 8 times.
Let me write it carefully this time (no placeholder bug).

```python
def solve2(targets=['E','G','C','G','F','F','A'], maxslots=7):
    CM = {'0101011011100010000010001':'A','1010100000101111010111101':'B','1101110001101011000111011':'C',
          '0010010101001001010100100':'D','1000110000111110000110001':'E','1001100001100011000011001':'F',
          '1010100001101010000111111':'G'}
    ORDER='ABCDEFG'
    def nm(i): return CM.get(clb(ybar()[i]),'?')
    log=[]; stop=False
    for i in range(maxslots):
        pos,_,_ = cursor()
        while pos!=i and not stop:
            r=last(action(['RIGHT' if i>pos else 'LEFT'])); log.append(('mv',i,r.get('state')))
            if res_check(r): stop=True; break
            pos += 1 if i>pos else -1
        if stop: break
        c=nm(i)
        if c=='?': log.append(('unknown',i,ybar()[i])); break
        if c==t
… [1154 caracteres más]
```

```text
[action] LEFT -> effect
[action] RIGHT -> effect
[action] DOWN -> effect
[action] DOWN -> effect
[action] RIGHT -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] RIGHT -> effect
[action] UP -> effect
[action] UP -> effect
[action] RIGHT -> effect
[action] UP -> eff
… [1105 caracteres más]
```

## Paso 8 · nivel 3 · acción 73

**Pensamiento:**

Level 3! Now the roles are reversed: top pairs are Yellow → Sky blue (Y given, S output). Bottom: YELLOW bar (top, 8 glyphs?) and SKY BLUE bar (bottom, 7 glyphs?). Now the bottom editable row is the sky-blue one (bottom), and the given row is yellow (top of the two bars).
Wait, in level 1: top pairs blue→pink; bottom: blue row (given, top) then pink row (editable, bottom). Level 2: top pairs pink→yellow; bottom: pink row (given, top) then yellow row (editable, bottom). Level 3: top pairs yellow→blue; bottom: yellow row (given, top), blue row (editable, bottom). Consistent!
So the given row is the first color of the pair, and the editable row is the second color.
Now let me analyze the level 3 board structure:
Top pairs: 
Row0: Y(4,5-11)[1 glyph] — S(4,15-21)[1]; Y(4,28-41)[2 glyphs] — S(4,45-58)[2]
Row1: Y(13,5-18)? [1] — S(13,22-35)[2]; Y(13,38-51)[2] — S(13,55-61)[1]
Row2: Y(22,5-25)[3] — S(22,29-35)[1]; Y(22,38-44)[1] — S(22,48-54)[1]
Bottom: Y bar (rows 40-46) with 8 glyphs?, S bar (rows 51-57) with 7 glyphs?
Let me analyze the actual layout via segmentation. The yellow bottom bar appears to span nearly the full width (cols ~4-59 = 56 wide = 8 slots), and the blue bar cols ~8-55 (48 = 7 slots?). Hmm, need to check.
So the cipher: yellow symbol → blue sequence (variable length). The bottom yellow row has 8 symbols; the blue row has 7 slots. The sum of the blue lengths for the 8 yellow symbols must equal 7, meaning some yellow symbols map to 0 blue glyphs?? Hmm, that can't 
… [269 caracteres más]

```python
seg = current_frame.segmentation
big=[]
for n in seg['nodes']:
    b=n['boundary']
    r0=min(p[0] for p in b); r1=max(p[0] for p in b); c0=min(p[1] for p in b); c1=max(p[1] for p in b)
    if n['pixels']>20:
        big.append((n['color'],r0,c0,r1,c1,n['pixels']))
for x in sorted(big): print(x)
print('---')
print('w row63', {ch:current_frame.ascii.split('\n')[63].count(ch) for ch in set(current_frame.ascii.split('\n')[63])})
```

```text
('G', 34, 0, 62, 63, 1107)
('S', 4, 15, 10, 21, 36)
('S', 4, 45, 10, 58, 72)
('S', 13, 15, 19, 28, 68)
('S', 13, 52, 19, 58, 34)
('S', 22, 29, 28, 35, 34)
('S', 22, 52, 28, 58, 34)
('S', 51, 7, 57, 55, 248)
('Y', 4, 5, 10, 11, 35)
('Y', 4, 28, 10, 41, 76)
('Y', 13, 5, 19, 11, 40)
('Y', 13, 35, 19, 4
… [157 caracteres más]
```

**Pensamiento:**

Level 3 layout:
Top pairs (Y → S):
Row0: Y(4,5-11) [1 glyph] — S(4,15-21) [1]; Y(4,28-41) [2] — S(4,45-58) [2]
Row1: Y(13,5-11) [1] — S(13,15-28) [2]; Y(13,35-48) [2] — S(13,52-58) [1]
Row2: Y(22,5-25) [3] — S(22,29-35) [1]; Y(22,42-48) [1] — S(22,52-58) [1]
Bottom: Y bar (40,4)-(46,59) = 56 wide = 8 slots; S bar (51,7)-(57,55) = 49 wide = 7 slots.
So 8 yellow symbols → 7 blue symbols. That means some yellow symbols map to zero blue glyphs? Hmm, but the examples show yellow→blue with lengths: 1→1, 2→2, 1→2, 2→1, 3→1, 1→1.
Hmm, so the mapping is not "one symbol → sequence" as in level 2 (where pink→yellow had lengths 1,3,2,2,3,1). Here the yellow→blue lengths are 1,2,2,1,3,1... wait let me recompute: pairs are (Y glyph count → S glyph count): (1→1), (2→2), (1→2), (2→1), (3→1), (1→1).
Hmm, so it's a "word-to-word" translation: a sequence of k yellow glyphs maps to a sequence of m blue glyphs. Not per-glyph!
Hmm, interesting. In level 2, the pairs were (1 pink → 1..3 yellow). So the mapping was per-symbol: one pink symbol ↔ a sequence of yellow symbols.
In level 3, the pairs are sequences on both sides. So maybe the mapping is per-symbol too but in the other direction: maybe each *blue* symbol maps to a yellow sequence? I.e., the blue word is the "input" and the yellow is the output? But the bottom has 8 yellow (given, top) and 7 blue (editable, bottom).
Hmm wait — which is given and which is editable? Let me check the cursor position: brackets at cols 8-12 (row 48 and 59) → ali
… [5884 caracteres más]

```python
A = current_frame.ascii.split('\n')
def gl(r,c): return bits(gx(A,r,c))
def tile_glyphs(r0,c0,c1):
    n=(c1-c0+1)//7
    return [gl(r0+1, c0+1+7*j) for j in range(n)]
Ys = [(4,5,11),(4,28,41),(13,5,11),(13,35,48),(22,5,25),(22,42,48)]
Ss = [(4,15,21),(4,45,58),(13,15,28),(13,52,58),(22,29,35),(22,52,58)]
pairs=[]
for i in range(6):
    yr,yc0,yc1 = Ys[i]; sr,sc0,sc1 = Ss[i]
    pairs.append((tile_glyphs(yr,yc0,yc1), tile_glyphs(sr,sc0,sc1)))
for i,(yl,sl) in enumerate(pairs):
    print('P%d |Y|=%d |S|=%d'%(i,len(yl),len(sl)))
    for g in yl: print('   Y',g)
    for g in sl: print('   S',g)
# bottom bars
yb = [gl(41,5+7*j) for j in range(8)]
sb = [gl(52,8+7*j) for j in range(7)]
print('BOTTOM Y', len(yb))
for g in yb: print('  y',g)
print('BOTTOM S', len(sb))
for g in sb: print('  s',g)
```

```text
P0 |Y|=1 |S|=1
   Y 1011100100101110000110111
   S 0010011111101011010100100
P1 |Y|=2 |S|=2
   Y 1011100100001000010011101
   Y 1011100100001000010011101
   S 1000110001111011010110111
   S 0000100101111110010100001
P2 |Y|=1 |S|=2
   Y 0101000000111110000001010
   S 1111101001010000100111111
   S 11
… [819 caracteres más]
```

## Paso 8 · nivel 3 · acción 73

**Pensamiento:**

Now let me analyze the cipher structure. Note the pairs (Y→S) and the fact that the same yellow/blue glyph classes recur.
Let me list the pairs with rotation-invariant classes to identify recurrences:
P0: Y=[a] → S=[b]  (1→1)
P1: Y=[c,c] → S=[d,e] (2→2)
P2: Y=[f] → S=[g,h] (1→2)
P3: Y=[i,j] → S=[k] (2→1)
P4: Y=[l,m,n] → S=[d] (3→1)  -- note S glyph d also appears in P1's S list!
P5: Y=[o] → S=[p] (1→1)
Interesting: P1's yellow has two IDENTICAL glyphs (c,c), and its S = [d,e]. P4's Y = [l,m,n] → S=[d].
So d ↔ "c c" (two c's) and d ↔ "l m n"? That's inconsistent unless the mapping is not per-symbol.
Hmm wait. Maybe the direction is reversed: the blue (S) is the "input" and yellow is the "output" — i.e., each blue symbol ↔ a yellow sequence:
P0: b ↔ [a]
P1: d ↔ [c], e ↔ [c]? or d ↔ [c,c]?
P2: [g,h] ↔ [f]
P3: k ↔ [i,j]
P4: d ↔ [l,m,n]
P5: p ↔ [o]
From P1: d ↔ [c] and e ↔ [c]? but then P4: d ↔ [l,m,n]. Inconsistent.
Hmm. So the same blue glyph d maps to different yellow sequences. So the mapping isn't a per-symbol dictionary in either direction.
New idea: maybe the relationship is about *counts/positions*: e.g., the yellow row is the blue row "encoded" via some transformation of the shapes themselves (like each blue glyph is a transformed version of the yellow glyph — e.g., rotation, or a bitwise operation)? In level 1, the pairs were 1:1 shape pairs (arbitrary mapping). In level 2, pink→yellow was 1:k. In level 3, both sides have multiple glyphs.
Wait... maybe I should reconside
… [8844 caracteres más]

```python
A = current_frame.ascii.split('\n')
def gl(r,c): return bits(gx(A,r,c))
def tile_glyphs(r0,c0,c1):
    n=(c1-c0+1)//7
    return [gl(r0+1, c0+1+7*j) for j in range(n)]
Ys = [(4,5,11),(4,28,41),(13,5,11),(13,35,48),(22,5,25),(22,42,48)]
Ss = [(4,15,21),(4,45,58),(13,15,28),(13,52,58),(22,29,35),(22,52,58)]
pairs=[(tile_glyphs(Ys[i][0],Ys[i][1],Ys[i][2]), tile_glyphs(Ss[i][0],Ss[i][1],Ss[i][2])) for i in range(6)]
yb=[gl(41,5+7*j) for j in range(8)]
sb=[gl(52,8+7*j) for j in range(7)]
# build class names for yellow glyphs
def classes(items):
    seen={}; out=[]
    for b in items:
        c=clb(b)
        if c not in seen: seen[c]=chr(ord('a')+len(seen))
        out.append(seen[c])
    return out
yc=[b for yl,_ in pairs for b in yl]+yb
names=classes(yc)
k=0
print('pair yellow classes:')
for 
… [523 caracteres más]
```

```text
pair yellow classes:
 P0 Y ['a']
 P1 Y ['b', 'b']
 P2 Y ['c']
 P3 Y ['d', 'd']
 P4 Y ['e', 'f', 'e']
 P5 Y ['g']
bottom yellow classes ['a', 'e', 'f', 'e', 'c', 'g', 'b', 'b']
pair blue classes:
 P0 S ['a']
 P1 S ['b', 'c']
 P2 S ['d', 'd']
 P3 S ['e']
 P4 S ['b']
 P5 S ['f']
bottom blue classes ['e
… [98 caracteres más]
```

## Paso 8 · nivel 3 · acción 73

**Pensamiento:**

Now with clean class names (separate namespaces for yellow and blue):
Yellow classes:
P0 Y: [a]
P1 Y: [b,b]
P2 Y: [c]
P3 Y: [d,d]   ← note: P3's two yellow glyphs are the SAME class! (i and j were rotation-equivalent)
P4 Y: [e,f,e]
P5 Y: [g]
Bottom yellow: [a, e, f, e, c, g, b, b]
Segmentation: a | e,f,e | c | g | b,b → lengths 1,3,1,1,2 = 8 ✓
Blue translations: P0[a]→[A]; P4[e,f,e]→[B]; P2[c]→[D,D]; P5[g]→[F]; P1[b,b]→[B,C]
Total blue = 1+1+2+1+2 = 7 ✓✓
Blue class names (from the pairs): P0 S=[a], P1 S=[b,c], P2 S=[d,d], P3 S=[e], P4 S=[b], P5 S=[f].
Careful: blue class letters here are in the blue namespace: 
- P0 S class 'a' 
- P1 S classes 'b','c'
- P2 S 'd','d' (both same class!)
- P3 S 'e'
- P4 S 'b' (same class as P1's first S)
- P5 S 'f'
So distinct blue classes: a,b,c,d,e,f = 6, plus maybe another in the cycle.
Target blue sequence (using blue-namespace letters):
slot0 ← P0 → 'a'
slot1 ← P4 → 'b'
slot2,slot3 ← P2 → 'd','d'
slot4 ← P5 → 'f'
slot5,slot6 ← P1 → 'b','c'
Target = [a, b, d, d, f, b, c]
Current bottom blue: [e, a, a, f, c, a, b] — 7 slots, none matching target? Let's see: slot0 e vs a ✗; slot1 a vs b ✗; slot2 a vs d ✗; slot3 f vs d ✗; slot4 c vs f ✗; slot5 a vs b ✗; slot6 b vs c ✗. All wrong (or some may coincide). 
Wait, but P3's word [d,d] (2 yellow glyphs) → blue [e] (1). The bottom yellow has no 'd' class, so P3 is unused ✓.
Hmm, but hold on: is my segmentation unique? Bottom yellow = [a,e,f,e,c,g,b,b]. Alternative: 'a' could pair with... The words avai
… [1243 caracteres más]

```python
def read_board():
    A = current_frame.ascii.split('\n')
    seg = current_frame.segmentation
    tiles=[]
    for n in seg['nodes']:
        b=n['boundary']
        r0=min(p[0] for p in b); r1=max(p[0] for p in b); c0=min(p[1] for p in b); c1=max(p[1] for p in b)
        if n['pixels']>20 and r1-r0==6 and n['color'] in ('Y','S','P','B') and r0<34:
            tiles.append((n['color'],r0,c0,c1))
    return A, tiles
A, tiles = read_board()
print(sorted(tiles))
# blue class naming from top S tiles
def glf(A,r,c): return bits(gx(A,r,c))
def tileg(A,r0,c0,c1):
    n=(c1-c0+1)//7
    return [glf(A,r0+1,c0+1+7*j) for j in range(n)]
Stiles=sorted([t for t in tiles if t[0]=='S'])
Ytiles=sorted([t for t in tiles if t[0]=='Y'])
BLUE={}
def bname(b):
    k=clb(b)
    if k not in BLUE: BLUE[k]=chr(or
… [499 caracteres más]
```

```text
[('S', 4, 15, 21), ('S', 4, 45, 58), ('S', 13, 15, 28), ('S', 13, 52, 58), ('S', 22, 29, 35), ('S', 22, 52, 58), ('Y', 4, 5, 11), ('Y', 4, 28, 41), ('Y', 13, 5, 11), ('Y', 13, 35, 48), ('Y', 22, 5, 25), ('Y', 22, 42, 48)]
P0 Y ['1010100000101111010111101'] -> S ['A']
P1 Y ['1000110000111110000110001
… [539 caracteres más]
```

**Pensamiento:**

Now the blue classes: A,B,C,D,E,F (6 distinct). The cycle probably has 6 entries.
Bottom yellow classes needed: I need yellow class names too. Let me define yellow names from the pair yellow glyphs and then parse the bottom yellow row.
Target (from earlier analysis, translated to blue names):
- P0 [a_yellow] → S ['A']
- P4 [e,f,e] → S ['B']
- P2 [c] → S ['D','D']
- P5 [g] → S ['F']
- P1 [b,b] → S ['B','C']
Target = [A, B, D, D, F, B, C]
Let me verify the parse programmatically with yellow class names, then probe the blue cycle at slot 0.
Note: the bottom blue row's current classes were [e,a,a,f,c,a,b] in the earlier ad-hoc naming. Let me redo with the new names.
Let me write a comprehensive script: parse board, compute target, then probe cycle.

```python
def analyze():
    A, tiles = read_board()
    Stiles=sorted([t for t in tiles if t[0]=='S']); Ytiles=sorted([t for t in tiles if t[0]=='Y'])
    BLUE={}; YEL={}
    def bn(b):
        k=clb(b)
        if k not in BLUE: BLUE[k]=chr(ord('A')+len(BLUE))
        return BLUE[k]
    def yn(b):
        k=clb(b)
        if k not in YEL: YEL[k]=chr(ord('a')+len(YEL))
        return YEL[k]
    pairs=[]
    for (c,r0,c0,c1) in Ytiles:
        cand=[t for t in Stiles if t[1]==r0 and t[2]>c1]; cand.sort(key=lambda t:t[2])
        sr0,sc0,sc1=cand[0][1],cand[0][2],cand[0][3]
        yl=[yn(glf(A,r0+1,c0+1+7*j)) for j in range((c1-c0+1)//7)]
        sl=[bn(glf(A,sr0+1,sc0+1+7*j)) for j in range((sc1-sc0+1)//7)]
        pairs.append((tuple(yl),tuple(sl)))
    # bottom bars: find big bars
    bars=[t for 
… [696 caracteres más]
```

```text
pairs:
   (('a',), ('A',))
   (('b', 'b'), ('B', 'C'))
   (('c',), ('D', 'D'))
   (('d', 'd'), ('E',))
   (('e', 'f', 'e'), ('B',))
   (('g',), ('F',))

error:
Traceback (most recent call last):
  File "<python_tool>", line 39, in <module>
  File "<python_tool>", line 34, in yn2
IndexError: string i
… [17 caracteres más]
```

## Paso 8 · nivel 3 · acción 73

**Pensamiento:**

Small bug (chr(ord('a')+len) with many classes is fine; the error was IndexError: string index out of range in yn2 → clb(b) where b was out-of-board? Actually yb read at cols 5+7*7=54..58 rows 41-45 — that's within the yellow bar (cols 4-59) fine. Hmm, the error is inside yn2 at line 34 → `Ymap[k]`? No, IndexError string index out of range... it's in clb → rotb → b[r*5+c] where b is shorter than 25 chars. So one of the yb reads returned a short string: glf(A,41,5+7*7=54) → gx(A,41,54) reads cols 54-58, rows 41-45 ✓ fine. Hmm, unless A rows... whatever.
Actually wait: the yellow bar is (40,4)-(46,59) → 56 wide = 8 slots, glyphs at cols 5+7j for j=0..7 → last glyph cols 54-58 ✓.
Hmm, "IndexError: string index out of range" — maybe bits() worked but clb got a string of length < 25 because the row string is only 64 chars and 54+5=59 ≤ 64 fine...
Oh! I bet the issue is that `analyze()` consumed... no. Let me look: line 39 is `print('bottom yellow', [yn2(y) for y in yb])` and line 34 is inside yn2: `if k not in Ymap: Ymap[k]=chr(...)` / `return Ymap[k]`. Hmm, no string indexing. Unless clb is called on a non-25 string. bits(gx(...)) — gx returns tuple of 5 strings each of length 5 unless the row is shorter. Rows are 64 chars.
Hmm, actually the error trace says line 34 in yn2, and the code at line 34 is `k=clb(b)`. So clb → rotb → b[r*5+c] IndexError. So some b is short. Which one? The pairs loop at line 36-37 `for yl,sl in pairs: for y in yl: yn2(y)` — yl are class-name STRINGS ('a
… [1886 caracteres más]

```python
def full_state():
    A, tiles = read_board()
    Stiles=sorted([t for t in tiles if t[0]=='S']); Ytiles=sorted([t for t in tiles if t[0]=='Y'])
    BLUE={}; YEL={}
    def nm(d,b,base):
        k=clb(b)
        if k not in d: d[k]=chr(base+len(d))
        return d[k]
    pairs=[]
    for (c,r0,c0,c1) in Ytiles:
        cand=[t for t in Stiles if t[1]==r0 and t[2]>c1]; cand.sort(key=lambda t:t[2])
        sr0,sc0,sc1=cand[0][1],cand[0][2],cand[0][3]
        yl=[nm(YEL,glf(A,r0+1,c0+1+7*j),97) for j in range((c1-c0+1)//7)]
        sl=[nm(BLUE,glf(A,sr0+1,sc0+1+7*j),65) for j in range((sc1-sc0+1)//7)]
        pairs.append((tuple(yl),tuple(sl)))
    # bottom bars: pick the two widest bars
    bars=sorted([t for t in tiles if t[3]-t[2]>30], key=lambda t:t[1])
    out=[]
    for (col,r0,c0,c1) 
… [975 caracteres más]
```

```text
pairs [(('a',), ('A',)), (('b', 'b'), ('B', 'C')), (('c',), ('D', 'D')), (('d', 'd'), ('E',)), (('e', 'f', 'e'), ('B',)), (('g',), ('F',))]

error:
Traceback (most recent call last):
  File "<python_tool>", line 34, in <module>
IndexError: list index out of range
```

**Pensamiento:**

Error at line 34: `cand[0]` — for some yellow tile there's no S tile on the same row starting after it. Which one? Row 22: Y tiles at (22,5,25) and (22,42,48); S tiles at (22,29,35) and (22,52,58). For Y(22,42,48): S at 52 ✓. For Y(22,5,25): S at 29 ✓. Row 13: Y(13,5,11), Y(13,35,48); S(13,15,28), S(13,52,58) ✓. Row 4 ✓.
Hmm, but the bars are also included in `tiles`? read_board filters pixels>20 and r1-r0==6 and color in (Y,S,P,B) and r0<34. The bottom bars are at rows 40-46, 51-57 → r0 >= 34 excluded ✓.
Wait, the error line 34 is inside the pairs loop: `sr0,sc0,sc1=cand[0][1],...`. Hmm, but it printed 'pairs' successfully with 6 entries... no wait, the print of pairs happened AFTER full_state() returned. The traceback says line 34 of the module, which is `byl = [b for b in bars if b[0]=='Y'][0][2]`? Let me count lines: the module code starts at line 1 = "def full_state():". Let me count: 
1 def full_state():
2     A, tiles = read_board()
3     Stiles=...
4     BLUE={}; YEL={}
5     def nm(...):
6         k=clb(b)
7         if k not in d: ...
8         return d[k]
9     pairs=[]
10     for (c,r0,c0,c1) in Ytiles:
11         cand=...
12         cand.sort(...)
13         sr0,sc0,sc1=cand[0][1],cand[0][2],cand[0][3]
...
Hmm, the printed output shows 'pairs' printed, so full_state succeeded. Then the failing line is later: `byl = ...` or in the parse def. Line 34 counting the snippet: let me count precisely.
Lines:
1: def full_state():
2:     A, tiles = read_board()
3:     Stile
… [1949 caracteres más]

```python
def full_state2():
    seg = current_frame.segmentation
    tops=[]; bars=[]
    for n in seg['nodes']:
        b=n['boundary']
        r0=min(p[0] for p in b); r1=max(p[0] for p in b); c0=min(p[1] for p in b); c1=max(p[1] for p in b)
        if n['pixels']<=20 or r1-r0!=6: continue
        if n['color'] not in ('Y','S','P','B'): continue
        (tops if r0<34 else bars).append((n['color'],r0,c0,c1))
    A = current_frame.ascii.split('\n')
    Stiles=sorted([t for t in tops if t[0]=='S']); Ytiles=sorted([t for t in tops if t[0]=='Y'])
    BLUE={}; YEL={}
    def nm(d,b,base):
        k=clb(b)
        if k not in d: d[k]=chr(base+len(d))
        return d[k]
    pairs=[]
    for (c,r0,c0,c1) in Ytiles:
        cand=[t for t in Stiles if t[1]==r0 and t[2]>c1]; cand.sort(key=lambda t:t[2])
  
… [715 caracteres más]
```

```text
pairs [(('a',), ('A',)), (('b', 'b'), ('B', 'C')), (('c',), ('D', 'D')), (('d', 'd'), ('E',)), (('e', 'f', 'e'), ('B',)), (('g',), ('F',))]
bar Y row 40 c0 4 n 8 ['a', 'e', 'f', 'e', 'c', 'g', 'b', 'b']
bar S row 51 c0 7 n 7 ['E', 'A', 'A', 'F', 'C', 'A', 'B']
```

## Paso 8 · nivel 3 · acción 80

**Pensamiento:**

Confirmed:
Bottom yellow = [a, e, f, e, c, g, b, b] → parse [a][e,f,e][c][g][b,b] → target blue = [A, B, D, D, F, B, C]
Current blue = [E, A, A, F, C, A, B]
Distinct blue classes: A,B,C,D,E,F = 6. The cycle should have 6 entries (all 6 classes).
Now probe the cycle at slot 0 (cursor already at slot 0). Press UP until the class returns to the start; record order. Then compute and execute.
Let me do the probe and, in the same call, compute and execute the solution with verification. To be safe, I'll do the probe first in this call and print the order, then execute the solve in the same call using the derived order (with a verify loop). That saves turns.
Implementation:
- srow(): current blue names list via full_state2 (bottom bar 'S').
- probe: cur0 = name of slot 0; press UP repeatedly (max 8) recording names; stop when the name equals cur0 AND the raw bitmap equals the original bitmap (to ensure we're back to the exact start, since names repeat only once per cycle).
- order = list of names in cycle order starting at index 0 = cur0.
- For each slot: k_cur = order.index(name) — but careful: if the order has duplicate names (e.g., if two cycle entries share a class), index() would be ambiguous. With 6 classes and 6 entries, each appears once.
- presses: fwd = (k_tgt - k_cur) % N; use UP if fwd <= N/2 else DOWN (N-fwd).
- Execute with cursor moves and verification.
Let me write it as a general function solve3() that recomputes everything from the board each time, so it's reusable
… [514 caracteres más]

```python
def blue_row():
    pairs, bars, BLUE, YEL = full_state2()
    for col,r0,c0,n,names in bars:
        if r0>=34 and col!='Y': return names, r0, c0, n
    return None
def blue_raw():
    A=current_frame.ascii.split('\n')
    return [glf(A,52,8+7*j) for j in range(7)]
def words_map():
    pairs, bars, BLUE, YEL = full_state2()
    return pairs
cur, r0, c0, n = blue_row()
print('current blue', cur)
raw0 = blue_raw()[0]
seq=[cur[0]]; probes=0
if cur[0]==cur[0]: pass
order_probe=[clb(blue_raw()[0])]
for k in range(9):
    r=last(action(['UP'])); probes+=1
    cl = clb(blue_raw()[0])
    order_probe.append(cl)
    if res_check(r): print('ENDED', r.get('state')); break
    if cl==order_probe[0]:
        print('cycle closed at', k+1); break
print('probes', probes)
print('order classes', order_prob
… [136 caracteres más]
```

```text
current blue ['E', 'A', 'A', 'F', 'C', 'A', 'B']
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
cycle closed at 7
probes 7
order classes ['1000110001110111000111111', '0000100101111111010010000
… [266 caracteres más]
```

**Nivel 3 completado → nivel 4 (acción 101)**

## Paso 9 · nivel 4 · acción 101

**Pensamiento:**

The cycle at slot 0 has 7 entries (not 6!). Classes in order:
k0: '1000110001110111000111111' = E (current slot0 = E ✓)
k1: '0000100101111111010010000' = ? new class not in BLUE map (a 7th shape!)
k2: '0000100101111110010100001' = C
k3: '1000111111100011000111011' = D
k4: '1000110001111011010110111' = B
k5: '1000111111001001111110001' = F
k6: '0010010101101011111100100' = A
So the cycle order is [E, X, C, D, B, F, A] where X is an extra shape (a plus-like variant, similar to C but with an extra pixel — like the level-1 EXTRA shape).
Cycle order (indices): A=6, B=4, C=2, D=3, E=0, F=5, X=1.
Target = [A, B, D, D, F, B, C]
Current = [E, A, A, F, C, A, B] (slot0 back to original ✓)
Presses needed (mod 7), using order index:
slot0: E(0) → A(6): fwd=6 → DOWN 1
slot1: A(6) → B(4): fwd = (4-6)%7 = 5 → DOWN 2
slot2: A(6) → D(3): fwd = (3-6)%7=4 → DOWN 3
slot3: F(5) → D(3): fwd = 5 → DOWN 2
slot4: C(2) → F(5): fwd = 3 → UP 3
slot5: A(6) → B(4): fwd=5 → DOWN 2
slot6: B(4) → C(2): fwd=5 → DOWN 2
Total presses = 1+2+3+2+3+2+2 = 15, cursor moves 6 → 21 actions. Budget: at step 81, the bar had 64 at level start (step 74) → 64 - 7 = 57 left. Fine.
Now, the assumption that all slots share the same cycle order (with pose offsets) needs verification — the verify loop handles it.
Let me write the general solver for this level and run it.

```python
def solve3(target_names):
    ORDER = ['E','X','C','D','B','F','A']   # cycle order at slot0 (classes)
    NAME2CL = {}
    pairs, bars, BLUE, YEL = full_state2()
    # BLUE maps clb->name for names seen; rebuild and add X
    def cur_names():
        pairs2, bars2, BLUE2, YEL2 = full_state2()
        for col,r0,c0,n,names in bars2:
            if r0>=34 and col!='Y':
                return names, c0
        return None, None
    def raw(i): return blue_raw()[i]
    def name_at(i):
        cl = clb(raw(i))
        if cl in BLUE: return BLUE[cl]
        return 'X'
    log=[]; stop=False
    for i in range(len(target_names)):
        pos,_,_ = cursor()
        while pos!=i and not stop:
            r=last(action(['RIGHT' if i>pos else 'LEFT'])); log.append(('mv',i,r.get('state')))
          
… [1176 caracteres más]
```

```text
[action] DOWN -> effect
[action] RIGHT -> effect
[action] DOWN -> effect
[action] DOWN -> effect
[action] RIGHT -> effect
[action] DOWN -> effect
[action] DOWN -> effect
[action] DOWN -> effect
[action] RIGHT -> effect
[action] DOWN -> effect
[action] DOWN -> effect
[action] RIGHT -> effect
[action]
… [1097 caracteres más]
```

## Paso 10 · nivel 4 · acción 101

**Pensamiento:**

Level 4 complete (the crash at the end was just blue_row() on the new level; the level completed ✓).
Level 4: Now there are TWO pair groups in the top (left group and right group), 4 rows each = 8 pairs. Colors: pink↔yellow and blue↔pink. Bottom: blue bar (top, 7 glyphs) and yellow bar (bottom, 7 glyphs).
Interesting: the top has two columns of pairs: left pairs (P↔Y) and right pairs (S↔P). So there are two "dictionaries": S→P and P→Y. And the bottom has a blue row (given, 7) and a yellow row (editable, 7). So we need to chain: blue → pink → yellow! A two-step translation.
That's the new mechanic: a chain of translations.
Let me verify: left group rows 4,13,22,31? The top area now extends to row ~35? The image shows 4 rows of pairs in the top area (rows ~4-38), and the bottom bars start around row 40. Hmm, the top area is bigger now (g region 0-38?). Let me analyze the board.
Also, the pairs may have multi-glyph sequences (like level 3). Let's analyze with code: get all tiles, pairs, and the bottom bars.
Let me generalize the analysis: 
- Identify top tiles by color and geometry (7-row tall tiles).
- Pair them by adjacency (gray connector) or by row + nearest to the right.
- Determine which color is "input" and which is "output" per pair group.
Let me look at the board.

```python
seg=current_frame.segmentation
tops=[];bars=[]
for n in seg['nodes']:
    b=n['boundary']
    r0=min(p[0] for p in b); r1=max(p[0] for p in b); c0=min(p[1] for p in b); c1=max(p[1] for p in b)
    if n['pixels']>20 and n['color'] in ('Y','S','P','B'):
        (tops if r1-r0==6 and r0<38 else bars).append((n['color'],r0,c0,c1,n['pixels']))
print('TOP tiles:')
for t in sorted(tops): print('  ',t)
print('BARS:')
for t in sorted(bars): print('  ',t)
A=current_frame.ascii.split('\n')
print('row63', {ch:A[63].count(ch) for ch in set(A[63])})
print('connectors:', [(min(p[0] for p in n['boundary']), min(p[1] for p in n['boundary'])) for n in seg['nodes'] if n['color']=='G' and n['pixels']==3])
```

```text
TOP tiles:
   ('P', 3, 12, 18, 32)
   ('P', 3, 45, 51, 31)
   ('P', 11, 22, 28, 32)
   ('P', 11, 35, 41, 26)
   ('P', 19, 12, 18, 31)
   ('P', 19, 45, 51, 26)
   ('P', 27, 22, 28, 32)
   ('P', 27, 35, 41, 32)
   ('S', 3, 35, 41, 34)
   ('S', 11, 12, 18, 38)
   ('S', 19, 35, 41, 34)
   ('S', 27, 12, 
… [269 caracteres más]
```

**Pensamiento:**

Level 4 layout:
Top tiles: 8 pairs, all 1-glyph each (all tiles are 7 wide). Connectors at cols 19-21 and 42-44, rows 6,14,22,30.
Left group (connector at col 19-21): between tiles at cols 12-18 and 22-28.
Right group (connector at col 42-44): between tiles at cols 35-41 and 45-51.
Rows: 3, 11, 19, 27.
Left group rows:
row3: P(3,12-18) — Y(3,22-28) → P→Y
row11: S(11,12-18) — P(11,22-28) → S→P
row19: P(19,12-18) — Y(19,22-28) → P→Y
row27: S(27,12-18) — P(27,22-28) → S→P
Right group rows:
row3: S(3,35-41) — P(3,45-51) → S→P
row11: P(11,35-41) — Y(11,45-51) → P→Y
row19: S(19,35-41) — P(19,45-51) → S→P
row27: P(27,35-41) — Y(27,45-51) → P→Y
So we have dictionaries: S→P (4 examples) and P→Y (4 examples).
Bottom: S bar (40,7-55) = 7 slots (given), Y bar (51,7-55) = 7 slots (editable).
So the chain: blue → pink → yellow. Each blue glyph maps to a pink glyph (from the S→P dict), and each pink glyph maps to a yellow glyph (from the P→Y dict). So the yellow row = compose(blue row).
Since all pairs are 1:1 here, the composition is per-glyph: yellow[i] = P2Y[P2S(blue[i])].
Careful: the mapping must be by shape class (rotation-invariant), and the pink intermediate is just a lookup.
Potential subtlety: maybe some blue glyph's pink partner isn't in the P→Y dictionary (then we'd need multi-step or the chain is ambiguous). Let's extract and check.
Let me now extract:
- 4 S→P pairs (class-based).
- 4 P→Y pairs.
- Bottom blue row (7 classes) → target yellow classes.
Then probe the yellow cycle 
… [72 caracteres más]

```python
def lvl4_extract():
    seg=current_frame.segmentation
    tops=[]
    for n in seg['nodes']:
        b=n['boundary']
        r0=min(p[0] for p in b); r1=max(p[0] for p in b); c0=min(p[1] for p in b); c1=max(p[1] for p in b)
        if n['pixels']>20 and r1-r0==6 and n['color'] in ('Y','S','P') and r0<38:
            tops.append((n['color'],r0,c0,c1))
    A=current_frame.ascii.split('\n')
    CLS={}
    def nm(col, b):
        k=(col, clb(b))
        return CLS.setdefault(k, col+chr(65+len([v for kk,v in CLS.items() if kk[0]==col])))
    def g(r0,c0,j=0): return glf(A,r0+1,c0+1+7*j)
    pairs=[]
    for (c,r0,c0,c1) in sorted(tops):
        cand=[t for t in tops if t[1]==r0 and t[2]>c1 and t[2]-c1<=7]
        if not cand: continue
        cand.sort(key=lambda t:t[2])
        c2,r2,c20,c21=
… [808 caracteres más]
```

```text
(('P', ('PA',)), ('Y', ('YA',)))
(('P', ('PB',)), ('P', ('PC',)))
(('P', ('PC',)), ('Y', ('YB',)))
(('P', ('PD',)), ('Y', ('YC',)))
(('P', ('PA',)), ('P', ('PB',)))
(('P', ('PB',)), ('Y', ('YD',)))
(('S', ('SA',)), ('P', ('PD',)))
(('S', ('SB',)), ('P', ('PB',)))
(('S', ('SC',)), ('P', ('PC',)))
(('
… [309 caracteres más]
```

**Pensamiento:**

My pairing logic created spurious pairs (e.g., P→P, Y→S) because I paired "nearest tile to the right on the same row" without checking the connector. Let me use the connector positions: connectors at cols 19-21 (left group) and 42-44 (right group). So a pair is (tile ending at col 18, tile starting at col 22) or (tile ending at 41, tile starting at 45).
From the tile list:
Left group (connector col 19): 
row3: P(3,12-18) — Y(3,22-28): P→Y: PA→YA... wait careful, my class names were assigned in a weird order. Let me redo cleanly with proper pairing.
Right group (connector col 42):
row3: S(3,35-41) — P(3,45-51): S→P
row11: P(11,35-41) — Y(11,45-51): P→Y
row19: S(19,35-41) — P(19,45-51): S→P
row27: P(27,35-41) — Y(27,45-51): P→Y
Left group:
row3: P(3,12-18) — Y(3,22-28): P→Y
row11: S(11,12-18) — P(11,22-28): S→P
row19: P(19,12-18) — Y(19,22-28): P→Y
row27: S(27,12-18) — P(27,22-28): S→P
Good: 4 S→P pairs and 4 P→Y pairs. All 1:1.
Let me redo the extraction with proper pairing (using the connector x-positions), then compute the chain.
From the earlier (mis-paired) output, the actual pairs I can reconstruct:
- S→P pairs: (SA→PD), (SB→PB), (SC→PC), (SD→PA) — these came out as ('S',...)→('P',...) entries ✓
- P→Y pairs: need the correct ones. From the list, the P→Y-looking entries: (PA→YA), (PC→YB), (PD→YC), (PB→YD). These correspond to the left/right group P→Y pairs. Note the class names PA..PD were assigned consistently by clb, so they should be right.
- The bogus entries: (PB→PC) 
… [11213 caracteres más]

```python
def lvl4_state():
    seg=current_frame.segmentation
    tops=[]
    for n in seg['nodes']:
        b=n['boundary']
        r0=min(p[0] for p in b); r1=max(p[0] for p in b); c0=min(p[1] for p in b); c1=max(p[1] for p in b)
        if n['pixels']>20 and r1-r0==6 and n['color'] in ('Y','S','P','B') and r0<38:
            tops.append((n['color'],r0,c0,c1))
    A=current_frame.ascii.split('\n')
    CLS={}
    def nm(col,b):
        key=(col,clb(b))
        if key not in CLS:
            used=[v for k,v in CLS.items() if k[0]==col]
            CLS[key]=col+chr(65+len(used))
        return CLS[key]
    def g(r0,c0,j=0): return glf(A,r0+1,c0+1+7*j)
    pairs=[]
    for (c,r0,c0,c1) in sorted(tops):
        cand=[t for t in tops if t[1]==r0 and t[2]-c1==4]
        if not cand: continue
        c2,
… [1419 caracteres más]
```

```text
real pairs:
   (('P', ('PA',)), ('Y', ('YA',)))
   (('P', ('PB',)), ('Y', ('YB',)))
   (('P', ('PC',)), ('Y', ('YC',)))
   (('P', ('PD',)), ('Y', ('YD',)))
   (('S', ('SA',)), ('P', ('PC',)))
   (('S', ('SB',)), ('P', ('PD',)))
   (('S', ('SC',)), ('P', ('PB',)))
   (('S', ('SD',)), ('P', ('PA',)))

… [534 caracteres más]
```

## Paso 10 · nivel 4 · acción 108

**Pensamiento:**

With the correct connector-based pairing (gap==4), the maps are clean:
S→P: SA→PC, SB→PD, SC→PB, SD→PA
P→Y: PA→YA, PB→YB, PC→YC, PD→YD
Chain: SA→PC→YC; SB→PD→YD; SC→PB→YB; SD→PA→YA
Given S row: [SA,SB,SD,SC,SB,SA,SD] → target Y = [YC,YD,YA,YB,YD,YC,YA] ✓ (matches my earlier derivation)
Current Y: [YE,YF,YC,YB,YA,YE,YE]
Slots needing change: 0 (YE→YC), 1 (YF→YD), 5 (YE→YC), 6 (YE→YA). Slots 2,3,4 already correct ✓ (great sign that the model is right).
Interesting: YE and YF are also known classes (YE='1010100000101111010111101' = level-2 class B, YF='1101110001101011000111011' = class C). So the yellow cycle has 6 shapes: YA,YB,YC,YD,YE,YF (all from the level-2 yellow set).
Now probe the cycle at slot 0 and solve. Let me write a general cycle-probe + solve function using class strings, then execute.
I'll write solve_generic(edit_color, target_names, CLS_inv):
- probe slot 0's cycle: record clb order until return to start (max 10).
- map cycle entries to names using inv (adding '?' for unknown).
- for each slot i: current name; compute presses; execute; verify.
Let me implement now, using lvl4_state to get names and the target computed inside.

```python
def lvl4_target():
    pairs, bardata, CLS = lvl4_state()
    maps={}
    for (ic,iseq),(oc,oseq) in pairs: maps.setdefault(ic,{})[iseq]=oseq
    given=[b for b in bardata if b[1]<50][0]; edit=[b for b in bardata if b[1]>=50][0]
    sp=maps[given[0]]; nxt=maps[[k for k in maps if k!=given[0] and k!=edit[0]][0]] if len(maps)>2 else None
    # chain: given -> mid -> edit
    mid=[k for k in maps if k not in (given[0], edit[0])]
    tgt=[]
    for x in given[4]:
        cur=x; seen=0
        while True:
            outs = maps[cur].get((cur,)) if False else None
            break
        tgt.append(x)
    return pairs, bardata, CLS, maps, given, edit
pairs, bardata, CLS, maps, given, edit = lvl4_target()
# chain manually: given -> mid, mid -> edit
mid = [k for k in maps if k not in (given[0],
… [902 caracteres más]
```

```text
given S mid ['P'] edit Y
target [None, None, None, None, None, None, None]
current ['YE', 'YF', 'YC', 'YB', 'YA', 'YE', 'YE']
cycle probe from slot0 name YE
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP
… [137 caracteres más]
```

## Paso 11 · nivel 4 · acción 183

**Pensamiento:**

The chain computation had a bug (target None) but I already computed the target manually: [YC, YD, YA, YB, YD, YC, YA].
Wait, let me double check with the maps from lvl4_state (the correct pairing):
S→P: SA→PC, SB→PD, SC→PB, SD→PA
P→Y: PA→YA, PB→YB, PC→YC, PD→YD
Chain: SA→PC→YC ✓, SB→PD→YD ✓, SD→PA→YA ✓, SC→PB→YB ✓.
Given: [SA,SB,SD,SC,SB,SA,SD] → target [YC,YD,YA,YB,YD,YC,YA] ✓
The bug in the code: `maps[m[0]][m]` — m is a tuple like ('PC',) and maps['P'] keys are tuples ✓... hmm, maps[cur] where cur = m[0] = 'PC' → maps['P'] ✓. Then maps['P'][('PC',)] = ('YC',) ✓. So y should be ('YC',). Why None? Because `m[0] in maps` → 'PC' in maps → False! maps keys are colors ('S','P'), and m[0] is a class name. Oops. Should be `m[0] in maps` → wrong check. Anyway.
Cycle order at slot 0: ['YE','YF','?','YC','YD','YB','YA'] where '?' is an unknown 7th shape (extra). Length 7.
Hmm, the cycle has 7 entries but only 6 known Y classes; the 7th is an extra shape (not in the target set), fine.
Names: YE=0, YF=1, ?=2, YC=3, YD=4, YB=5, YA=6.
Current row: [YE, YF, YC, YB, YA, YE, YE]
Target:     [YC, YD, YA, YB, YD, YC, YA]
Slots:
0: YE(0) → YC(3): fwd=3 → UP 3
1: YF(1) → YD(4): fwd=3 → UP 3
2: YC(3) → YA(6): fwd=3 → UP 3
3: YB(5) → YB(5): ✓ skip
4: YA(6) → YD(4): fwd=(4-6)%7=5 → DOWN 2
5: YE(0) → YC(3): UP 3
6: YE(0) → YA(6): fwd=6 → DOWN 1
Total presses: 3+3+3+0+2+3+1 = 15, plus 6 cursor moves = 21 actions.
Budget: level 4 started at step 102 with 64; now step 109 → 57 left. Fine.
Let me writ
… [295 caracteres más]

```python
def solve_cycle(target, order_names, editrow):
    # editrow: (color,row,c0,n)
    inv={}
    pairs, bardata, CLS = editrow
    for (col,k),v in CLS.items():
        if col==editrow[3]: inv[k]=v
    A=current_frame.ascii.split('\n')
    def raw(i): return glf(A, editrow[1]+1, editrow[2]+1+7*i)
    def nm(i): return inv.get(clb(raw(i)),'?')
    log=[]; stop=False
    N=len(order_names)
    for i in range(len(target)):
        pos,_,_ = cursor()
        while pos!=i and not stop:
            r=last(action(['RIGHT' if i>pos else 'LEFT'])); log.append(('mv',r.get('state')))
            if res_check(r): stop=True; break
            pos += 1 if i>pos else -1
        if stop: break
        c=nm(i)
        if c not in order_names: log.append(('bad',i,c)); break
        if c==target[i]:
           
… [1102 caracteres más]
```

```text
Traceback (most recent call last):
  File "<python_tool>", line 41, in <module>
  File "<python_tool>", line 5, in solve_cycle
ValueError: too many values to unpack (expected 3)
```

**Pensamiento:**

The editrow arg: I passed (edit[0],edit[1],edit[2],CLS4) = 4 items but the function unpacks `pairs, bardata, CLS = editrow` (3). Let me fix the function signature: pass CLS separately.
Also no actions were executed (error at line 5 before any action). Let me fix and rerun.

```python
def solve_cycle(target, order_names, bar, CLS):
    inv={}
    for (col,k),v in CLS.items():
        if col==bar[0]: inv[k]=v
    A=current_frame.ascii.split('\n')
    def raw(i): return glf(A, bar[1]+1, bar[2]+1+7*i)
    def nm(i): return inv.get(clb(raw(i)),'?')
    log=[]; stop=False; N=len(order_names)
    for i in range(len(target)):
        pos,_,_ = cursor()
        while pos!=i and not stop:
            r=last(action(['RIGHT' if i>pos else 'LEFT'])); log.append(('mv',r.get('state')))
            if res_check(r): stop=True; break
            pos += 1 if i>pos else -1
        if stop: break
        c=nm(i)
        if c not in order_names: log.append(('bad',i,c)); break
        if c==target[i]:
            log.append(('ok',i,c)); continue
        fwd=(order_names.index(target[i])-orde
… [972 caracteres más]
```

```text
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] RIGHT -> effect
[action] UP
… [3665 caracteres más]
```

## Paso 12 · nivel 4 · acción 183

**Pensamiento:**

Something went wrong: 75 actions consumed, and the final row is [YB, YA, YE, YB, YA, YB, YE] — nothing matching, and the "fix" loop kept pressing UP 9 times each. That means `nm(i)` never matched the target — i.e., my name lookups were broken. Note the final row classes look scrambled and the cursor is at slot 6.
Why did nm() fail? Because I passed `CLS4` from lvl4_state() — the CLS dict keys are (col, clb) — and I built inv from CLS items where col == bar[0] = 'Y'. That should work... but wait: the '?' names. During the fix loop, nm(i) may return a name not in the CLS map for shapes that appeared newly (like the cycle's extra shape 'X'), but that wouldn't cause a mismatch for YC.
Hmm, but look: the log shows for slot 0: 3 UPs then 9 fixes = 12 presses, ending 'YE' (want YC). 12 presses mod 7 = 5 → YE(0) + 12 = 12 mod 7 = 5 → order[5] = YB. But it reported 'done' with nm='YE'. Hmm, 'done' prints nm(i) after the loop exits when guard hits 9 (loop exits because guard>=9, not because it matched). So nm(0) = 'YE' after 12 UPs?! That means pressing UP 12 times at slot 0 returned it to YE — inconsistent with a 7-cycle... unless the cursor was NOT at slot 0! 
Look at the diff image: the cursor brackets moved to the RIGHT side (cols ~50-55), and there's a dimmed bracket at the left (col 8-12) — the old position. And the changed glyphs are at slots 0,1,2 and 5,6 (the diff shows changes in two groups: cols ~8-28 and ~43-55). Hmm.
Wait, the real problem: `cursor()` may be misreporting. 
… [5288 caracteres más]

```python
A=current_frame.ascii.split('\n')
print('row63', {ch:A[63].count(ch) for ch in set(A[63])})
print('row0', {ch:A[0].count(ch) for ch in set(A[0])})
p,b,c = lvl4_state()
for x in b: print('bar',x[0],x[1],x[3],x[4])
print('cursor', cursor())
# fresh cycle check at current cursor slot to reconfirm order & mapping
inv={}
for (col,k),v in c.items():
    if col=='Y': inv[k]=v
print('Y class map', inv)
```

```text
row63 {'w': 23, 'c': 41}
row0 {'g': 64}
bar S 40 7 ['SA', 'SB', 'SD', 'SC', 'SB', 'SA', 'SD']
bar Y 51 7 ['YB', 'YA', 'YE', 'YB', 'YA', 'YB', 'YE']
cursor (6, 50, 54)
Y class map {'0101011011100010000010001': 'YA', '1010100001101010000111111': 'YB', '1000110000111110000110001': 'YC', '10011000011000
… [54 caracteres más]
```

## Paso 12 · nivel 4 · acción 183

**Pensamiento:**

Important: the HUD bar row63 has only 23 'w' left (41 'c') → ~23 actions remaining! I must be efficient. Also note the Y class map now shows YE = '1101110001101011000111011' — different from before! Earlier YE was '1010100000101111010111101' and YF was '1101110001101011000111011'. The class NAMES got reassigned because lvl4_state assigns names in encounter order and the current bottom-row shapes differ from the original ones (names are not stable across calls!). 
That's a critical trap: the names YA..YE are assigned dynamically. So my target names ['YC','YD',...] from the earlier call are NOT valid for this fresh call. I must recompute the target from the current extraction each time (chain S→P→Y), and compute the cycle order in the same naming space.
Wait, but the cycle order I derived earlier was in the OLD naming space. Let me redo everything in one self-contained script:
1. Extract pairs + bars with the current CLS naming.
2. Compute target for the yellow row via the chain.
3. Determine the cycle order — I need it in the current naming space. I can derive it by probing again (costly: 7 actions out of 23) OR by mapping the old cycle order to clb strings and then to the new names via the current CLS map (the clb strings are stable!).
The old cycle order in clb strings was recorded: ['YE','YF','?','YC','YD','YB','YA'] with the old inv map. I have the old inv mapping partially: from the earlier print of "edit known classes":
YA: '0101011011100010000010001'
YB: '10101000011010
… [7832 caracteres más]

```python
CYCLE = ['1010100000101111010111101','1101110001101011000111011','0010010101001001010100100',
         '1000110000111110000110001','1001100001100011000011001','1010100001101010000111111',
         '0101011011100010000010001']
def yrow_fresh():
    A=current_frame.ascii.split('\n')
    return [clb(glf(A,52,8+7*j)) for j in range(7)]
def ytarget_fresh():
    p,b,c = lvl4_state()
    maps={}
    for (ic,iseq),(oc,oseq) in p: maps.setdefault(ic,{})[iseq]=oseq
    given=[x for x in b if x[1]<50][0]; edit=[x for x in b if x[1]>=50][0]
    inv={}
    for (col,k),v in c.items():
        if col==edit[0]: inv[v]=k
    tgt=[]
    for x in given[4]:
        m=maps[given[0]][(x,)][0]
        y=maps[m][(m,)][0] if m in maps else None
        tgt.append(inv[y])
    return tgt, edit
tgt, editbar = ytarget
… [188 caracteres más]
```

```text
Traceback (most recent call last):
  File "<python_tool>", line 22, in <module>
  File "<python_tool>", line 20, in ytarget_fresh
KeyError: None
```

**Nivel 4 completado → nivel 5 (acción 201)**

## Paso 12 · nivel 5 · acción 201

**Pensamiento:**

The KeyError: None came from `inv[y]` where y = None because `m in maps` was False (m is a name like 'PC', maps keys are colors). Let me fix: the middle color is 'P' and maps['P'] keys are P-tuples. So:
m_name = maps[given_color][(x,)][0]  # e.g. 'PC'
y_name = maps[mid_color][(m_name,)][0]
where mid_color = the color key whose values are P names... Actually maps keys are input colors: 'S' (S→P) and 'P' (P→Y). So:
step1 = maps['S'][(x,)][0] → 'PC'
step2 = maps['P'][('PC',)][0] → 'YC'
Let me fix it: mid = the key of maps whose output color is the edit color... simpler: chain through maps by repeatedly looking up maps[cur_color][(name,)]:
cur_color = given color, name = x
while True:
    d = maps.get(cur_color)
    if d is None: break
    out = d.get((name,))
    if out is None: break
    name = out[0]
    # next color: the output color of that pair
Need the output color per pair. Let me store maps as: maps[in_color][(name,)] = (out_color, out_tuple).
Let me just rewrite cleanly.
Budget: 23 actions left. My planned fix needs 18. Let me not waste actions. Compute target now, then execute.
Also note: I should double check my assumption that the cycle order (positions) is the same for all slots. Level 2 and 3 confirmed it. I'll include per-slot verification with fresh reads, but with a tight cap to avoid wasting the remaining budget: press planned presses, verify once; if wrong, try at most 2 extra presses then move on and report.
Actually, a smarter approach that avoids the cycle-
… [1486 caracteres más]

```python
def ytarget2():
    p,b,c = lvl4_state()
    given=[x for x in b if x[1]<50][0]; edit=[x for x in b if x[1]>=50][0]
    inv={}
    for (col,k),v in c.items():
        if col==edit[0]: inv[v]=k
    tgt=[]
    for x in given[4]:
        nm=x; col=given[0]
        for _ in range(4):
            pr=[q for q in p if q[0][0]==col and q[0][1]==(nm,)]
            if not pr: break
            ocol,oseq = pr[0][1]
            nm = oseq[0]; col = ocol
            if col==edit[0]: break
        tgt.append(inv[nm])
    return tgt
tgt = ytarget2()
cur = yrow_fresh()
CYCLE = ['1010100000101111010111101','1101110001101011000111011','0010010101001001010100100',
         '1000110000111110000110001','1001100001100011000011001','1010100001101010000111111',
         '0101011011100010000010001']
print('target',
… [183 caracteres más]
```

```text
target [3, 4, 6, 5, 4, 3, 6]
current [5, 6, 1, 5, 6, 5, 1]
deltas [5, 5, 5, 0, 5, 5, 5]
```

**Pensamiento:**

Confirmed: deltas = [5,5,5,0,5,5,5] → each needs +5 (UP 5) or equivalently -2 (DOWN 2). Slot 3 needs nothing.
Cursor at slot 6. Plan: DOWN×2 at slots 6,5,4,2,1,0 with LEFT moves between.
Sequence: 
slot6: DOWN DOWN
LEFT → slot5: DOWN DOWN
LEFT → slot4: DOWN DOWN
LEFT → slot3: (nothing)
LEFT → slot2: DOWN DOWN
LEFT → slot1: DOWN DOWN
LEFT → slot0: DOWN DOWN
= 12 presses + 6 LEFT = 18 actions.
I'll execute with verification after each slot (fresh read), stopping if the level completes. If a slot's verify fails, I'll press one more DOWN and re-verify (max 2 extra), to stay within budget.
Actually, to conserve budget, let me execute the planned sequence and verify at the end; if verification fails at some slot, fix individually. But a mid-way failure could waste the whole budget... The delta analysis is solid (cycle order verified by probe at slot 0; DOWN verified as -1). The risk is per-slot cycle order differences. Levels 2 and 3 confirmed identical orders across slots, and here the initial fill's deltas were all consistent with the same cycle order (deltas all 5 → if a slot had a different cycle order, the delta wouldn't be uniform... well, that's circular).
Hmm, actually here's a good consistency check: the initial row's deltas were [3,3,3,0,5,3,6] — not uniform, so the initial fill wasn't a constant shift. Fine.
Let me just execute with per-slot verification (fresh read after the 2 presses) and a cap of 1 extra press. If still wrong, move on and report (next turn can fix wit
… [31 caracteres más]

```python
CYCLE = ['1010100000101111010111101','1101110001101011000111011','0010010101001001010100100',
         '1000110000111110000110001','1001100001100011000011001','1010100001101010000111111',
         '0101011011100010000010001']
tgt = ytarget2()
log=[]; stop=False
def nm2cl(i): return yrow_fresh()[i]
def idx(i):
    c=nm2cl(i); return CYCLE.index(c) if c in CYCLE else -1
order = [6,5,4,2,1,0]
for s in order:
    pos,_,_ = cursor()
    while pos!=s and not stop:
        r=last(action(['LEFT' if s<pos else 'RIGHT'])); log.append(('mv',pos,r.get('state')))
        if res_check(r): stop=True; break
        pos += -1 if s<pos else 1
    if stop: break
    need = (CYCLE.index(tgt[s]) - idx(s)) % 7
    acts = ['UP']*need if need<=3 else ['DOWN']*(7-need)
    for a in acts:
        r=last(action([a])
… [649 caracteres más]
```

```text
[action] DOWN -> effect
[action] DOWN -> effect
[action] LEFT -> effect
[action] DOWN -> effect
[action] DOWN -> effect
[action] LEFT -> effect
[action] DOWN -> effect
[action] DOWN -> effect
[action] LEFT -> effect
[action] LEFT -> effect
[action] DOWN -> effect
[action] DOWN -> effect
[action] LEF
… [862 caracteres más]
```

## Paso 13 · nivel 5 · acción 202

**Pensamiento:**

Level 5! Structure:
Top: pairs S→P (blue → pink), with multi-glyph sequences:
Row ~100 (row 8?): S(1 glyph) → P(1 glyph); S(1) → P(2 glyphs)
Row ~170: S(2 glyphs) → P(1); S(1) → P(1)
Bottom: S bar (given, 5 glyphs) at rows ~435-490 → and P bar (editable, 5 glyphs).
Wait, the bottom has a blue bar (5 glyphs) and a pink bar (5 glyphs). The cursor brackets are now at the TOP-LEFT pair (around the first S tile in the top area!) — rows 70-95, cols 80-140 in image coords → in grid coords, brackets at rows ~8 and ~19, cols ~9-12. That's a NEW element: a cursor around a TOP tile!
So in level 5, the cursor selects a top tile (an example tile?), and UP/DOWN might change... hmm. Interesting. Maybe now the mechanic is: the cursor is on a "source" tile and you must... Hmm.
Let me analyze the board with code to get the exact geometry.
The key difference: the cursor brackets are around the first blue tile in the top-left, not around a bottom bar slot. So the interaction model may have changed: maybe UP/DOWN now rotates or changes the selected top tile? Or maybe the bottom bars are fixed and we choose which example... 
Hmm, wait. Maybe the mechanic: the top pairs are "equations" and the cursor is on a tile you can cycle; the bottom rows are the "question". Let me just look carefully.
Let me get the segmentation and geometry.

```python
seg=current_frame.segmentation
objs=[]
for n in seg['nodes']:
    b=n['boundary']
    r0=min(p[0] for p in b); r1=max(p[0] for p in b); c0=min(p[1] for p in b); c1=max(p[1] for p in b)
    objs.append((n['color'],r0,c0,r1,c1,n['pixels']))
for o in sorted(objs):
    if o[5]>8: print(o)
A=current_frame.ascii.split('\n')
print('row63', {ch:A[63].count(ch) for ch in set(A[63])})
# white bracket rows
for r in range(64):
    cnt=A[r].count('W')
    if cnt: print('W row',r,cnt,[i for i,ch in enumerate(A[r]) if ch=='W'])
```

```text
('B', 11, 9, 15, 13, 11)
('B', 11, 19, 15, 23, 14)
('B', 11, 32, 15, 36, 11)
('B', 11, 42, 15, 46, 16)
('B', 11, 49, 15, 53, 16)
('B', 23, 9, 27, 13, 15)
('B', 23, 16, 27, 20, 15)
('B', 23, 26, 27, 30, 15)
('B', 23, 39, 27, 43, 15)
('B', 23, 49, 27, 53, 19)
('B', 44, 15, 48, 19, 15)
('B', 44, 22, 48
… [669 caracteres más]
```

**Pensamiento:**

Level 5 layout:
Top tiles (7-row tall):
Row 10-16: S(10,8-14) [1 glyph] — P(10,18-24) [1]; S(10,31-37) [1] — P(10,41-54) [2 glyphs]
Row 22-28: S(22,8-21) [2 glyphs] — P(22,25-31) [1]; S(22,38-44) [1] — P(22,48-54) [1]
Bottom bars: S(43,14-48) = 35 wide = 5 slots; P(52,14-48) = 5 slots.
Cursor brackets at rows 7-8 and 18-19, cols 9-13 → around the top-left S tile (10,8-14)! So the cursor is on a top tile.
Black glyphs: 
row 11: glyphs at cols 9-13 (S tile 1), 19-23 (P tile 1), 32-36 (S tile 2), 42-46 & 49-53 (P tile 2, 2 glyphs)
row 23: 9-13, 16-20 (S tile 3 = 2 glyphs), 26-30 (P tile 3), 39-43 (S tile 4), 49-53 (P tile 4)
row 44: 5 glyphs (S bar) at 15-19, 22-26, 29-33, 36-40, 43-47
row 53: 5 glyphs (P bar) at 15-19, ..., 43-47
So the cursor is on the top-left S tile. Since the cursor is on the top tiles, maybe LEFT/RIGHT moves between the top tiles (4 of them), and UP/DOWN cycles the glyph(s) of the selected tile?! That would be a different mechanic: perhaps we're supposed to select which example applies? Hmm.
Alternatively, maybe the cursor on a top tile means: the tile's contents can be edited, and the bottom rows must be made consistent.
Hmm, let me think about what the puzzle is. Bottom: S bar (5 glyphs, given) and P bar (5 glyphs, editable). Top: 4 S→P example pairs with lengths (1→1), (1→2), (2→1), (1→1).
This looks like level 3's structure (variable-length code words) with S→P instead of Y→S. Let's check the parse: given S row of 5 glyphs → target P row of 5 glyphs. T
… [921 caracteres más]

```python
r = last(action(['UP']))
print(r.get('state'), r.get('level_completed'), r.get('game_over'))
d = frame_diff()
print({k:(v if k=='changed_cell_count' else len(v)) for k,v in d.items()})
for k in ('appeared','disappeared','moved','rotated','changed_color','resized'):
    for e in d.get(k,[])[:6]: print(k, e)
A=current_frame.ascii.split('\n')
print('W rows', [(i,[j for j,ch in enumerate(A[i]) if ch=='W']) for i in range(64) if 'W' in A[i]])
```

```text
[action] UP -> effect
NOT_FINISHED False False
{'changed_cell_count': 10, 'moved': 0, 'rotated': 0, 'appeared': 0, 'disappeared': 0, 'changed_color': 0, 'resized': 2}
resized {'at_before': [10, 8], 'at_after': [10, 8], 'color': 'S', 'before_pixels': 38, 'after_pixels': 38, 'before_hash': '6e40b530bf
… [308 caracteres más]
```

## Paso 14 · nivel 5 · acción 203

**Pensamiento:**

Interesting! UP changed the glyph in the CURSED/selected top tile: the top-left S tile's glyph changed (from a "Z/E"-like shape to a "+H"-like shape). The cursor stayed. So the top tiles are EDITABLE!
So in level 5, the mechanic is different: the EXAMPLE tiles are editable, and the bottom rows... hmm. Wait, but which row is editable at the bottom? The P bar (bottom, pink) presumably.
Hmm, wait. Let me reconsider. Maybe the puzzle is: the top-left tile is a "source" that we can cycle through shapes; and the goal is... Hmm.
Let me think differently. Maybe the actual puzzle: the bottom S row is given, the bottom P row must be produced (as before), but the top examples are INCOMPLETE/incorrect and need to be fixed? That seems odd.
Alternatively: the top-left tile with the cursor is a "problem" tile and the bottom rows are... 
Hmm. Let me reconsider the layout. The cursor is around the top-left S tile. Pressing UP changed that tile's glyph (cycled it). So we can edit that tile.
What's the goal? Perhaps: make the top-left pair (S→P) consistent with the translation rule demonstrated by the other pairs? Or: the top-left tile is a "query" — i.e., the puzzle is: given the other examples (pairs 2,3,4 and the bottom rows), find the S glyph that maps to the given P glyph in pair 1? That is, the top-left S tile is the answer slot, and the bottom bars are extra examples!
Hmm! That's plausible: the bottom two bars (S row given, P row) form another example pair (5 glyphs → 5 glyphs). And the 
… [966 caracteres más]

```python
r=last(action(['RIGHT']))
A=current_frame.ascii.split('\n')
print('W rows', [(i,[j for j,ch in enumerate(A[i]) if ch=='W']) for i in range(64) if 'W' in A[i]])
print(r.get('state'))
d=frame_diff()
print('changed', d['changed_cell_count'], {k:len(v) for k,v in d.items() if k!='changed_cell_count'})
```

```text
[action] RIGHT -> effect
W rows [(7, [19, 20, 21, 22, 23]), (8, [19, 23]), (18, [19, 23]), (19, [19, 20, 21, 22, 23])]
NOT_FINISHED
changed 29 {'moved': 2, 'rotated': 0, 'appeared': 1, 'disappeared': 0, 'changed_color': 0, 'resized': 2}
```

## Paso 15 · nivel 5 · acción 206

**Pensamiento:**

The cursor moved from the top-left S tile (cols 9-13) to the P tile (cols 19-23) — so the cursor visits TOP tiles. Both S and P tiles are editable (probably).
So level 5: the top example tiles are editable! The bottom bars: are they editable? Let me continue RIGHT to see the cursor path: tiles in order: S(10,8-14), P(10,18-24), S(10,31-37), P(10,41-54), S(22,8-21), P(22,25-31), S(22,38-44), P(22,48-54), then maybe the bottom bars.
Let me probe: RIGHT ×2 more and check where the cursor lands (does it go to the bottom bars?).
Actually, let me think about the goal first. Hypothesis: this level's goal is to complete/fix a pattern. Perhaps the top-left pair is the "question" — with an editable S tile and editable P tile? Hmm.
Alternative hypothesis: this is a "reverse" puzzle: the bottom rows give a full example (S row ↔ P row, 5 glyphs each), and the top pairs are examples; one of the top tiles is wrong/blank and must be corrected so that the whole system is consistent.
Hmm, that's speculative. Let me gather more evidence: check whether the bottom bars are cursor-reachable (i.e., editable). Press RIGHT a few times and watch the cursor. Each press costs 1 action; budget 64 - 3 = 61. It's worth 3-4 probes.
Let me press RIGHT 3 times, checking the cursor position after each.

```python
def wpos():
    A=current_frame.ascii.split('\n')
    rows=[i for i in range(64) if 'W' in A[i]]
    cols=[j for j,ch in enumerate(A[rows[0]]) if ch=='W'] if rows else []
    return (rows[0], rows[-1], cols[0], cols[-1]) if rows else None
seq=[]
for k in range(3):
    r=last(action(['RIGHT']))
    seq.append((wpos(), r.get('state')))
    if res_check(r): break
for s in seq: print(s)
```

```text
[action] RIGHT -> effect
[action] RIGHT -> effect
[action] RIGHT -> effect
((7, 19, 32, 36), 'NOT_FINISHED')
((7, 19, 42, 53), 'NOT_FINISHED')
((19, 31, 9, 20), 'NOT_FINISHED')
```

## Paso 16 · nivel 5 · acción 211

**Pensamiento:**

Cursor order so far: 
0: S(10,8-14) [1 glyph]
1: P(10,18-24) [1]
2: S(10,31-37) [1]
3: P(10,41-54) [2 glyphs]
4: S(22,8-21) [2 glyphs] ← current (cursor brackets rows 19-31, cols 9-20)
So the cursor cycles through the 8 top tiles. Presumably it continues to the bottom bars afterward (or loops). Let me continue RIGHT 4 more to see if it reaches the bottom bars (tiles 5,6,7 then bars).
Actually, the key question is whether the bottom P bar is editable. Let me press RIGHT 4 times and check positions.

```python
seq=[]
for k in range(5):
    r=last(action(['RIGHT']))
    seq.append((wpos(), r.get('state')))
    if res_check(r): break
for s in seq: print(s)
```

```text
[action] RIGHT -> effect
[action] RIGHT -> effect
[action] RIGHT -> effect
[action] RIGHT -> effect
[action] RIGHT -> effect
((19, 31, 26, 30), 'NOT_FINISHED')
((19, 31, 39, 43), 'NOT_FINISHED')
((19, 31, 49, 53), 'NOT_FINISHED')
((7, 19, 9, 13), 'NOT_FINISHED')
((7, 19, 19, 23), 'NOT_FINISHED')
```

## Paso 17 · nivel 5 · acción 211

**Pensamiento:**

The cursor cycles through only the 8 top tiles (positions 0-7), then wraps to 0. The bottom bars are NOT cursor-reachable → they are fixed (given), not editable!
So the puzzle: the top pairs are editable tiles. The bottom two rows (S bar of 5 glyphs, P bar of 5 glyphs) are a fixed example/pair. The goal must be to make the top examples consistent with the bottom rule... Hmm, but what exactly constitutes "solved"?
Let me think again. Maybe the puzzle is the inverse of before: the bottom rows give a translation example (S word ↔ P word). The top pairs are the "dictionary entries" but scrambled/incorrect — we must edit them so each pair is a valid entry of the same cipher? Hmm, that's odd.
Alternative interpretation: the bottom rows are the QUESTION: "here is an S row and a P row" — no wait, both are given.
Hmm, let me reconsider: maybe the bottom P row is the answer to the bottom S row under the cipher defined by the top pairs, and the top pairs are the cipher's dictionary... but the top pairs are editable, so the puzzle might be: "modify the dictionary so that the translation of the given S row equals the given P row"? That has many solutions.
Let me look at the data. Let me extract all glyphs and check the current consistency: do the top pairs + bottom pair define a consistent code? Maybe 3 of the 4 top pairs are consistent with the bottom pair's mapping, and ONE is wrong → fix it. That's a classic "find the inconsistency" puzzle.
Actually here's a cleaner hypothesis: this is
… [1612 caracteres más]

```python
def lvl5_get():
    A=current_frame.ascii.split('\n')
    seg=current_frame.segmentation
    tiles=[]
    for n in seg['nodes']:
        b=n['boundary']
        r0=min(p[0] for p in b); r1=max(p[0] for p in b); c0=min(p[1] for p in b); c1=max(p[1] for p in b)
        if n['pixels']>20 and r1-r0==6 and n['color'] in ('S','P'):
            tiles.append((n['color'],r0,c0,c1))
    tiles.sort(key=lambda t:(t[1],t[2]))
    def gl(r0,c0,j=0): return clb(glf(A,r0+1,c0+1+7*j))
    top=[]
    for (c,r0,c0,c1) in tiles:
        if r0>34: continue
        n=(c1-c0+1)//7
        top.append((c,r0,c0,[gl(r0,c0,j) for j in range(n)]))
    bars=[]
    for (c,r0,c0,c1) in tiles:
        if r0<=34: continue
        n=(c1-c0+1)//7
        bars.append((c,r0,c0,[gl(r0,c0,j) for j in range(n)]))
    return top,b
… [182 caracteres más]
```

```text
T S 10 8 ['0000100101111110010100001']
T P 10 18 ['0000101111010010100111111']
T S 10 31 ['0000100101111111010010000']
T P 10 41 ['0111111001100011001111110', '0111111001100011001111110']
T S 22 8 ['1000110001110111000111111', '1000110001110111000111111']
T P 22 25 ['0011100101111111010011100']
T S 
… [410 caracteres más]
```

**Pensamiento:**

Now let me analyze. Data (clb strings):
Top pairs:
T1: S [s1='0000100101111110010100001'] → P [p1='0000101111010010100111111']
T2: S [s2='0000100101111111010010000'] → P [q1='0111111001100011001111110', q1] (two identical)
T3: S [t1='1000110001110111000111111', t1] (two identical) → P [p2='0011100101111111010011100']
T4: S [u='1000111111001001111110001'] → P [v='1111110001101111010111111']
Bottom: S row: [u, s1, w, w, s2] where w='0010010101101011111100100'
Bottom P row: [p2, x, x, p1, v] where x='0010011111100011000111111'
Now let's see the per-symbol mapping hypothesis:
Bottom S row (5) ↔ bottom P row (5): position-wise:
u → p2
s1 → x
w → x
w → x
s2 → v  ... hmm, and p1 at position 3? Let me redo: bottom S = [u, s1, w, w, s2], bottom P = [p2, x, x, p1, v].
Position-wise: u→p2, s1→x, w→x, w→p1, s2→v. Inconsistent (w→x and w→p1).
Hmm. So not position-wise.
Now the top pairs give:
s1 → p1 (T1)
s2 → q1 q1 (T2)
t1 t1 → p2 (T3)
u → v (T4)
Bottom: S=[u,s1,w,w,s2] P=[p2,x,x,p1,v].
Try word-based parsing (like level 3): the S row is a sequence of "words" (each word = a sequence of S glyphs) and the P row = the corresponding sequence of P words.
Known words from the top: 
s1 → p1 (len 1 → 1)
s2 → q1q1 (1 → 2)
t1t1 → p2 (2 → 1)
u → v (1 → 1)
Bottom S row = [u, s1, w, w, s2]: 
- u → v (1 glyph → 1)
- s1 → p1 (1 → 1)
- w → ? (unknown word)
- w → ?
- s2 → q1q1 (1 → 2)
Total P from known: v(1) + p1(1) + q1q1(2) = 4 glyphs plus 2 unknown w-words → total would be 4 + len(w-word)*2. But the 
… [3675 caracteres más]

```python
def b2grid(b): return [[int(b[r*5+c]) for c in range(5)] for r in range(5)]
def allforms(b):
    out={}
    g=b2grid(b)
    for m in range(4):
        g=[list(row) for row in zip(*g[::-1])]
        out['rot%d'%(90*(m+1))] = ''.join(''.join(map(str,row)) for row in g)
    h=b[::-1] if False else None
    gr=b2grid(b)
    mirr=[row[::-1] for row in gr]
    out['mirH']=''.join(''.join(map(str,r)) for r in mirr)
    out['mirV']=''.join(''.join(map(str,r)) for r in mirr[::-1])
    out['diag']=''.join(''.join(map(str,row)) for row in [list(x) for x in zip(*gr)])
    return out
G = {'s1':'0000100101111110010100001','p1':'0000101111010010100111111',
     's2':'0000100101111111010010000','q1':'0111111001100011001111110',
     't1':'1000110001110111000111111','p2':'0011100101111111010011100',
     '
… [485 caracteres más]
```

```text
s1 11 0000100101111110010100001
p1 14 0000101111010010100111111
s2 11 0000100101111111010010000
q1 16 0111111001100011001111110
t1 15 1000110001110111000111111
p2 15 0011100101111111010011100
u 15 1000111111001001111110001
v 19 1111110001101111010111111
w 13 0010010101101011111100100
x 15 0010011111
… [134 caracteres más]
```

## Paso 17 · nivel 5 · acción 211

**Pensamiento:**

No rotation/mirror relations. So the mapping is arbitrary substitution (as in levels 1-4).
Let me reconsider. Key insight attempt: maybe the bottom rows are NOT a pair. Maybe the bottom S row is a "given sentence" and the bottom P row is the "answer sentence" but the ANSWER is what we must produce... but the P bar isn't editable (cursor doesn't reach it). Hmm, unless the cursor wraps only through 8 tiles because the bottom bars are... hmm, we saw the cursor wrap from tile 7 back to tile 0.
Wait — maybe I mis-tracked. Let me re-check the sequence of cursor positions:
Start: (7,19,9,13) = tile0 S(10,8-14)
RIGHT → (7,19,19,23) = tile1 P(10,18-24)
RIGHT → (7,19,32,36) = tile2 S(10,31-37)
RIGHT → (7,19,42,53) = tile3 P(10,41-54)
RIGHT → (19,31,9,20) = tile4 S(22,8-21)
RIGHT → (19,31,26,30) = tile5 P(22,25-31)
RIGHT → (19,31,39,43) = tile6 S(22,38-44)
RIGHT → (19,31,49,53) = tile7 P(22,48-54)
RIGHT → (7,19,9,13) = tile0 (wrap)
RIGHT → (7,19,19,23) = tile1 (current)
So yes: 8 top tiles only. The bottom bars are not editable.
So the puzzle is entirely about the 8 top tiles, with the bottom rows as fixed reference.
Think about it as: the bottom shows two "sentences": a blue sentence and a pink sentence. The top shows 4 pairs. The task: edit the top tiles so that each top pair is a valid "sub-translation" consistent with the bottom pair alignment? 
Let's consider the bottom as a word-aligned bilingual pair. If we can find a word alignment of the bottom pair (S words ↔ P words) such tha
… [22185 caracteres más]

```python
print('len history', len(history), 'steps', [h.frame.step for h in history][-14:])
# find first frame of level 5
f0=None
for h in history:
    if h.frame.level==5:
        f0=h.frame; break
print('first lvl5 frame step', f0.step if f0 else None)
def tiles_of(fr):
    A=fr.ascii.split('\n'); seg=fr.segmentation; out=[]
    for n in seg['nodes']:
        b=n['boundary']
        r0=min(p[0] for p in b); r1=max(p[0] for p in b); c0=min(p[1] for p in b); c1=max(p[1] for p in b)
        if n['pixels']>20 and r1-r0==6 and n['color'] in ('S','P'):
            n2=(c1-c0+1)//7
            out.append((n['color'],r0,c0,[clb(glf(A,r0+1,c0+1+7*j)) for j in range(n2)]))
    out.sort(key=lambda t:(t[1],t[2]))
    return out
t0 = tiles_of(f0)
for t in t0: print('INIT',t[0],t[1],t[2],t[3])
print()
tcur = lv
… [60 caracteres más]
```

```text
len history 212 steps [198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211]
first lvl5 frame step 201
INIT S 10 8 ['0000100101111111010010000']
INIT P 10 18 ['0000101111010010100111111']
INIT S 10 31 ['0000100101111111010010000']
INIT P 10 41 ['0111111001100011001111110', '011111100
… [937 caracteres más]
```

## Paso 17 · nivel 5 · acción 211

**Pensamiento:**

Excellent! Now I have the INITIAL state of level 5:
INIT top pairs:
Pair1: S(10,8) = ['0000100101111111010010000'] = s2 → P(10,18) = ['0000101111010010100111111'] = p1
Pair2: S(10,31) = ['0000100101111111010010000'] = s2 → P(10,41) = [q1, q1]
Pair3: S(22,8) = [t1, t1] where t1='1000110001110111000111111' → P(22,25) = [p2] where p2='0011100101111111010011100'
Pair4: S(22,38) = [u] where u='1000111111001001111110001' → P(22,48) = [v] where v='1111110001101111010111111'
INIT bottom:
S row = [u, s1, w, w, s2] where s1='0000100101111110010100001', w='0010010101101011111100100'
P row = [p2, x, x, p1, v] where x='0010011111100011000111111'
Note: my UP press changed tile0 (pair1's S) from s2 to s1! So the current tile0 = s1, originally s2.
Now the dictionary from the initial top pairs:
E1: s2 → p1
E2: s2 → q1q1
E3: t1t1 → p2
E4: u → v
Hmm, E1 and E2 have the SAME S side (s2) but different P sides! That's inconsistent for a dictionary. So indeed pair1 is the "answer" pair to fix? Or pair2?
Interesting: pair1's S = s2 and pair2's S = s2 (identical!). And pair1's P = p1, pair2's P = q1q1.
Hmm! Now with the bottom pair: S row = [u, s1, w, w, s2], P row = [p2, x, x, p1, v].
If the dictionary is: u→v (E4 ✓ present in the bottom as S[0]↔P[4]... hmm the reversal again).
Let me try the segmentation/word-alignment idea with the initial data. The bottom pair must be a valid "sentence pair" under the dictionary. The dictionary entries (from the top, if all correct) are: s2→p1, s2→q1q1, t1t1→p2, 
… [20819 caracteres más]

```python
Srow = ['u','s1','w','w','s2']
Prow = ['p2','x','x','p1','v']
# tile contents (INITIAL level-5 state) as (S-seq, P-seq)
tiles = {1:(('s2',),('p1',)), 2:(('s2',),('q1','q1')), 3:(('t1','t1'),), }
tiles = {1:(('s2',),('p1',)), 2:(('s2',),('q1','q1')), 3:(('t1','t1'),('p2',)), 4:(('u',),('v',))}
def comps5_with_one2():
    return [(2,1,1,1),(1,2,1,1),(1,1,2,1),(1,1,1,2)]
def chunk(seq, lens):
    out=[];i=0
    for L in lens: out.append(tuple(seq[i:i+L])); i+=L
    return out
best=[]
for sc in comps5_with_one2():
    Sc = chunk(Srow, sc)
    for pc in comps5_with_one2():
        Pc = chunk(Prow, pc)
        entries = list(zip(Sc,Pc))  # ordered
        # assign entries to tiles: tile counts
        from itertools import permutations
        for perm in permutations(range(4)):
            wron
… [448 caracteres más]
```

```text
wrong 5 Slens (2, 1, 1, 1) Plens (1, 2, 1, 1) assign (2, 1, 0, 3)
   tile 1 (('s2',), ('p1',)) -> entry (('w',), ('p1',))
   tile 2 (('s2',), ('q1', 'q1')) -> entry (('w',), ('x', 'x'))
   tile 3 (('t1', 't1'), ('p2',)) -> entry (('u', 's1'), ('p2',))
   tile 4 (('u',), ('v',)) -> entry (('s2',), ('
… [1535 caracteres más]
```

## Paso 17 · nivel 5 · acción 212

**Pensamiento:**

The minimum is 5 wrong tiles out of 8 — too many for a "fix the corruption" puzzle. So my model of the bottom pair as a word-aligned bilingual pair is probably wrong.
Let me step back and think about this differently.
Observation: the top-left pair1's S side initially = s2, and pair2's S side = s2 as well (identical). Hmm, and the cursor starts on pair1's S tile.
New idea: maybe the puzzle is like levels 1-4 but with the roles swapped: the bottom rows are the DICTIONARY EXAMPLES (a full sentence pair) and the top-left pair is the QUESTION: "here's a blue glyph [s2], what's its pink translation?" — and the other top pairs (2,3,4) are... hmm, but they'd be distractors.
Alternatively — maybe this level's task is: the top pairs are the "input" (a sentence in blue) and the pink tiles are the answer slots! I.e., the top area is like the editable row in previous levels, but wrapped into 2 rows: pair1 = slot 1, pair2 = slot 2, ... no.
Hmm wait. Let me reconsider the layout. What if the top area is actually a 4-row "sentence" structure where each row is a PAIR of sentences (blue sentence ↔ pink sentence), like the bottom pair? I.e., there are 5 example pairs total: 4 small ones on top and 1 big one at the bottom. And the editable tiles are the top ones because... no, why would examples be editable?
Let me think about the *goal* from the game's perspective: something must be checked. The most natural check in this puzzle family: "the answer row must equal the translation of the given r
… [3463 caracteres más]

```python
def snap():
    t,b = lvl5_get()
    return [(x[0],x[1],x[2],tuple(x[3])) for x in t]
def which_tile(pos):
    # pos = (r0,r1,c0,c1) of brackets
    t,_ = lvl5_get()
    for i,x in enumerate(t):
        if x[2]-1 <= pos[2] and pos[3] <= x[3]+1: return i
    return None
before = snap()
cur = wpos()
ti = which_tile(cur)
r1 = last(action(['UP']))
after1 = snap()
ch = [(i,before[i],after1[i]) for i in range(len(before)) if before[i]!=after1[i]]
print('cursor tile idx', ti, 'after UP changed:', ch)
print('state', r1.get('state'))
```

```text
Traceback (most recent call last):
  File "<python_tool>", line 13, in <module>
  File "<python_tool>", line 9, in which_tile
TypeError: can only concatenate list (not "int") to list
```

**Pensamiento:**

Bug: x[3] is a list of glyph strings; x[2] is c0 and I need c1. Let me fix which_tile: lvl5_get returns (color, r0, c0, glyphs). I don't have c1. Let me compute c1 = c0 + 7*len(glyphs) - 1.

```python
def which_tile2(pos):
    t,_ = lvl5_get()
    for i,x in enumerate(t):
        c0=x[2]; c1=c0+7*len(x[3])-1
        if c0-1 <= pos[2] and pos[3] <= c1+1: return i
    return None
def snap2():
    t,_ = lvl5_get()
    return [(x[0],x[1],x[2],tuple(x[3])) for x in t]
cur=wpos(); ti=which_tile2(cur)
before=snap2()
r1=last(action(['UP']))
after=snap2()
ch=[(i,before[i][2],before[i][3],after[i][3]) for i in range(len(before)) if before[i]!=after[i]]
print('cursor tile', ti, 'changed after UP:', ch)
print(r1.get('state'))
```

```text
[action] UP -> effect
cursor tile 1 changed after UP: [(1, 18, ('0000101111010010100111111',), ('0010011111101011111100100',))]
NOT_FINISHED
```

## Paso 18 · nivel 5 · acción 216

**Pensamiento:**

Tile1 (P of pair1) is editable: it changed from p1 ('0000101111010010100111111') to '0010011111101011111100100'.
Interesting: the new glyph's clb = '0010011111101011111100100'. Compare with w = '0010010101101011111100100' and x = '0010011111100011000111111'. Hmm, the new glyph is neither. Hmm, it's similar to w but different.
So the cycle contains many glyphs.
OK so both tiles of pair1 are editable. Let me now test a tile in pair4 (which I suspect might be "correct"/locked). Move LEFT 3 times to tile6, then UP.
Actually, let me think about the value of this test. If ALL 8 tiles are editable, then the level is a "construct the dictionary" puzzle and I need to know the exact target config the checker wants. If only pair1's tiles are editable, the puzzle is "complete pair1" and the answer is determined by the other pairs + bottom pair.
That's a decisive test. Let's do it: LEFT, LEFT, LEFT, UP and check what changed.

```python
before=snap2()
for k in range(3):
    r=last(action(['LEFT']))
cur=wpos(); ti=which_tile2(cur)
b2=snap2()
r=last(action(['UP']))
after=snap2()
ch=[(i,b2[i][2],b2[i][3],after[i][3]) for i in range(len(b2)) if b2[i]!=after[i]]
print('cursor now tile', ti, 'changed:', ch, r.get('state'))
```

```text
[action] LEFT -> effect
[action] LEFT -> effect
[action] LEFT -> effect
[action] UP -> effect
cursor now tile 6 changed: [(6, 38, ('1000111111001001111110001',), ('0010010101101011111100100',))] NOT_FINISHED
```

## Paso 19 · nivel 5 · acción 217

**Pensamiento:**

Tile6 (S of pair4) is also editable — it changed from u to '0010010101101011111100100' = w!
So all top tiles are editable. So this is a "construct the dictionary" puzzle: the 4 pairs must be made consistent with the bottom pair (the ground truth sentence pair).
I damaged pair4's S tile (was u, now w) and pair1's S and P tiles.
I need to figure out the exact target the checker expects. Since the bottom pair is the ground truth and the 4 pairs must be the 4 dictionary entries that tile it, the minimum-edit solution had 5 wrong tiles. That's a lot but maybe that IS the design: the top area is entirely randomized (a "scrambled dictionary") and you must build the correct dictionary from the sentence pair. Actually, that makes sense! The generator: create a dictionary of 4 word-pairs, generate the sentence pair (bottom), then present the dictionary entries in scrambled order (and/or with wrong glyphs) for the player to fix.
Hmm, but the minimum number of wrong tiles being 5/8 suggests the tiles were randomized independently (each tile's glyphs random), so many are wrong.
But wait — the tile glyph COUNTS are fixed (1,1,2,1 for S; 1,2,1,1 for P). If the dictionary's word lengths were determined by the generator, the tile counts would have to match the true dictionary's lengths. The fact that the S counts multiset {1,1,2,1} and P counts {1,2,1,1} can align with the bottom pair's segmentation (both rows have 5 glyphs) is a strong hint: the true dictionary has 4 entries with S-lengths {
… [13664 caracteres más]

```python
# tile6 currently = w (I changed it from u). Press DOWN and see if it reverts to u
b=snap2()
r=last(action(['DOWN']))
a=snap2()
print('tile6 before', b[6][3], 'after DOWN', a[6][3])
print('u original = 1000111111001001111110001')
```

```text
[action] DOWN -> effect
tile6 before ('0010010101101011111100100',) after DOWN ('1000111111001001111110001',)
u original = 1000111111001001111110001
```

## Paso 20 · nivel 5 · acción 224

**Pensamiento:**

DOWN reverted tile6 from w back to u. So UP/DOWN cycle the glyph, and the cycle from u: u →(UP) w →(DOWN) u. So the cycle is at least: ... u, w, ... adjacent in the cycle.
Now let me enumerate the cycle for tile6 (S tile, currently u). Press UP repeatedly and record the glyph sequence until we return to u. Let me do up to 10 presses, stopping when we're back at u.
This gives the S-tile cycle. Let me also check whether the cycle is the same for all S tiles (probably, like level 4's 7-glyph cycle).
Let me write a loop: for i in range(12): press UP, record glyph; break if glyph == start.
Cost: up to 12 actions. We have ~44. Worth it — the cycle content is crucial. Actually, maybe I can be smarter: I only need the cycle to know which glyphs are available and the order (to compute press counts). The cycle for S tiles is probably the same set as the P tiles (like level 4 where all slots shared the same cycle). Let me enumerate tile6's cycle now.
Note: I need to restore tile6 to u at the end (the cycle returns to u automatically if I enumerate a full cycle).

```python
start = snap2()[6][3][0]
seq=[start]
for i in range(14):
    r=last(action(['UP']))
    g=snap2()[6][3][0]
    seq.append(g)
    if res_check(r): break
    if g==start: break
print('cycle len', len(seq)-1)
for i,g in enumerate(seq): print(i, g, sum(int(c) for c in g))
print('state', last_action_call_result.get('state'))
```

```text
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
cycle len 7
0 1000111111001001111110001 15
1 0010010101101011111100100 13
2 1000110001110111000111111 15
3 0000100101111111010010000 11
4 00001001
… [132 caracteres más]
```

## Paso 21 · nivel 5 · acción 224

**Pensamiento:**

The S-tile cycle has length 7:
0: u = '1000111111001001111110001' (15px)
1: w = '0010010101101011111100100' (13)
2: t1 = '1000110001110111000111111' (15)
3: s2 = '0000100101111111010010000' (11)
4: s1 = '0000100101111110010100001' (11)
5: '1000111111100011000111011' (15) — new, call it z1
6: '1000110001111011010110111' (15) — new, call it z2
7: back to u ✓
So the S cycle = 7 glyphs, and the tiles returned to u (good, tile6 restored to its original u).
Note the cycle includes u, w, t1, s2, s1 (all the S glyphs seen in the bottom row + pair3's t1) plus 2 new ones (z1, z2). The bottom S row uses u, s1, w, w, s2 — all in the cycle ✓.
Now, the P tiles likely have their own 7-glyph cycle including p1, p2, x, q1, v + 2 new.
Budget check: step 225, level 5 started at 201 → 24 actions used. If the budget is 64, ~40 left.
Now the key unknown: the target configuration. Let me think about the most likely design once more, using the new info that the cycle is exactly 7 glyphs (same as level 4's 7-class cycle). In level 4, the editable row had 7 slots each cycling through 7 classes, and the target was derived from the dictionary. Here, 8 tiles cycling through 7 glyphs.
The most likely design (consistent with levels 1-4): the bottom rows are the given sentence (blue) and the target answer (pink)... but the pink row isn't editable. Hmm. Unless the pink bottom row IS the answer key displayed (like a "check" row) — no.
Let me think about levels 1-4 structure again:
Level 4: top = 2 pairs (blue
… [5415 caracteres más]

```python
from itertools import permutations, product
Srow=['u','s1','w','w','s2']; Prow=['p2','x','x','p1','v']
tiles_init={1:(('s2',),('p1',)),2:(('s2',),('q1','q1')),3:(('t1','t1'),('p2',)),4:(('u',),('v',))}
tc={1:(1,1),2:(1,2),3:(2,1),4:(1,1)}
def chunk(seq,lens):
    out=[];i=0
    for L in lens: out.append(tuple(seq[i:i+L])); i+=L
    return out
res=[]
for pi in permutations([1,2,3,4]):
    sl=[tc[i][0] for i in pi]; pl=[tc[i][1] for i in pi]
    if sum(sl)!=5 or sum(pl)!=5: continue
    Sc=chunk(Srow,sl); Pc=chunk(Prow,pl)
    entries=dict(zip(pi,zip(Sc,Pc)))
    # entries keyed by tile already (pi maps position->tile)
    wrong=0; detail=[]
    for t in (1,2,3,4):
        eS,eP=entries[t]; iS,iP=tiles_init[t]
        w1 = 0 if iS==eS else 1; w2 = 0 if iP==eP else 1
        wrong += w1+w2
  
… [211 caracteres más]
```

```text
min wrong 5
wrong 5
   (1, ('s2',), ('p1',), '->', ('w',), ('p1',), 1, 0)
   (2, ('s2',), ('q1', 'q1'), '->', ('w',), ('x', 'x'), 1, 1)
   (3, ('t1', 't1'), ('p2',), '->', ('u', 's1'), ('p2',), 1, 0)
   (4, ('u',), ('v',), '->', ('s2',), ('v',), 1, 0)
wrong 6
   (1, ('s2',), ('p1',), '->', ('w',), (
… [431 caracteres más]
```

## Paso 21 · nivel 5 · acción 224

**Pensamiento:**

Hmm, the minimum is 5 wrong tiles. That's a lot of edits (each requiring several presses). With ~40 actions left and ~5-6 tile edits needing navigation + presses, it might be feasible but risky if the model is wrong.
Wait. Let me reconsider. Maybe the level is NOT about the bottom pair at all!
Let me reconsider the whole picture. In levels 1-4, the structure was: top = dictionary pairs, bottom = given row + editable row. Level 5: top = 4 pairs (editable!), bottom = 2 rows (fixed). 
Hmm, what if the bottom two rows are the DICTIONARY (a "phrase pair" example) and the top pairs are the QUESTION: "here are 4 blue→pink pairs to complete" — i.e., each top pair's blue tile is a word from the bottom blue row and the pink tile must be its translation from the bottom pink row?! That's the "extract the dictionary from the aligned sentence pair" task, which is exactly the tiling model.
Alternatively, maybe the task is much simpler: maybe the top pairs must each be a *sub-pair* of the bottom alignment position-wise... Let me test the "position-wise" alignment with a twist: maybe the bottom rows are aligned position-wise and the dictionary is a per-glyph substitution, and the two bottom rows ARE consistent under a substitution if we allow that the same glyph appears twice in the blue row (w,w) mapping to two different pink glyphs (x at position 2 and p1 at position 3)... which fails. So no.
Unless the bottom blue row's two "w" glyphs are NOT the same glyph but rotations of each other, and
… [1328 caracteres más]

```python
A=current_frame.ascii.split('\n')
def raw(r0,c0,j=0): return glf(A,r0+1,c0+1+7*j)
Sbot=[raw(43,14,j) for j in range(5)]
Pbot=[raw(52,14,j) for j in range(5)]
print('S bottom raw:')
for i,g in enumerate(Sbot): print(' ',i,g,clb(g))
print('P bottom raw:')
for i,g in enumerate(Pbot): print(' ',i,g,clb(g))
print('top tiles raw:')
tops=[(10,8,1),(10,18,1),(10,31,1),(10,41,2),(22,8,2),(22,25,1),(22,38,1),(22,48,1)]
for (r,c,n) in tops:
    print(' ',r,c,[raw(r,c,j) for j in range(n)])
```

```text
S bottom raw:
  0 1101101010011100101011011 1000111111001001111110001
  1 0010000100011100010011111 0000100101111110010100001
  2 0111000010111110001001110 0010010101101011111100100
  3 0111001000111110100001110 0010010101101011111100100
  4 0000100101111111010010000 0000100101111111010010000
P bott
… [662 caracteres más]
```

**Pensamiento:**

Important: the bottom S row's glyphs 2 and 3 have DIFFERENT raw poses but the same clb class:
g2 raw = '0111000010111110001001110'
g3 raw = '0111001000111110100001110'
These are rotations/mirrors of each other. Let me check: g2 vs g3 — g2 rows: 01110/00010/11111/00010/01110 ; g3 rows: 01110/01000/11111/01000/01110. So g3 = mirror-H of g2 ✓ (left-right flip). So they're mirror images, not rotations? clb apparently treats mirrors as equal (my clb might include mirrors? clb was defined earlier as canonical over rotations... hmm, g2 and g3 have the same clb, so clb includes mirror equivalences, or g2/g3 are rotations: rotating 01110/00010/11111/00010/01110 by 90° gives... it's a plus/cross-like shape with arms. Let me not worry.)
Also the P bottom row's glyphs 1 and 2 have the same class (x) with different poses:
x1 raw = '0010011111100011000111111'
x2 raw = '0111101001110010100101111'
Hmm, those look quite different. Are they rotations? x1 rows: 00100/11111/00011/00011/11111; x2 rows: 01111/01001/11001/01001/01111. Hmm, clb says equal. OK, clb must be a rotation+mirror-invariant canonical form. Fine.
Now, pose-dependent analysis: let me check the position-wise mapping using RAW poses:
S: [u@r0, s1@r1, w@r2, w@r3, s2@r4] → P: [p2@r0, x@r1, x@r2, p1@r3, v@r4]
Still w→x and w→p1 (different classes) → not a function even with poses. ✗
OK so position-wise is definitively out.
Now let me check the top tiles' raw poses vs the bottom rows' raw poses:
- tile(22,25) [pair3's P] raw = '111
… [10272 caracteres más]

```python
from itertools import permutations
Srow=['u','s1','w','w','s2']; Prow=['p2','x','x','p1','v']
cur_tiles={1:(('s2',),('x1',)),2:(('s2',),('q1','q1')),3:(('t1','t1'),('p2',)),4:(('u',),('v',))}  # current (after my edits)
init_tiles={1:(('s2',),('p1',)),2:(('s2',),('q1','q1')),3:(('t1','t1'),('p2',)),4:(('u',),('v',))}
tc={1:(1,1),2:(1,2),3:(2,1),4:(1,1)}
def chunk(seq,lens):
    out=[];i=0
    for L in lens: out.append(tuple(seq[i:i+L])); i+=L
    return out
out=[]
for pi in permutations([1,2,3,4]):
    sl=[tc[i][0] for i in pi]; pl=[tc[i][1] for i in pi]
    Sc=chunk(Srow,sl); Pc=chunk(Prow,pl)
    ent={pi[i]:(Sc[i],Pc[i]) for i in range(4)}
    # function check on S side
    seen={}
    ok=True
    for t,(a,b) in ent.items():
        if a in seen and seen[a]!=b: ok=False
        seen[a]=b
… [552 caracteres más]
```

```text
functional tilings: 14
init_wrong 6 cur_wrong 6 order (4, 2, 3, 1) {1: (('s2',), ('v',)), 2: (('s1',), ('x', 'x')), 3: (('w', 'w'), ('p1',)), 4: (('u',), ('p2',))}
init_wrong 6 cur_wrong 6 order (4, 3, 2, 1) {1: (('s2',), ('v',)), 2: (('w',), ('x', 'p1')), 3: (('s1', 'w'), ('x',)), 4: (('u',), ('p2'
… [850 caracteres más]
```

## Paso 21 · nivel 5 · acción 224

**Pensamiento:**

Hmm. All functional tilings require ≥6 tile changes from the initial state. That's a LOT — 6-8 tile edits, each needing navigation + up to 3 presses. With ~38 actions left, it's maybe feasible (each edit ≈ 2 moves + 2 presses = 4 actions; 8 edits ≈ 32). Risky but possible. But I'm not confident in the model at all, since it requires changing nearly everything including pair4's tiles that "look" clean (u→v matches nothing in these solutions except... hmm, in the top solutions pair4 = (u→p2) — its S stays u ✓ and P changes v→p2).
Hold on. Let me question the assumption that the bottom rows are a sentence PAIR. What if the bottom rows are TWO SEPARATE things:
(a) The bottom blue row = a "given sentence" (input).
(b) The bottom pink row = the "answer" that the player produced in a PREVIOUS level? No...
Hmm, what if the bottom pink row is the target for the top area?! Like: "the top pink tiles, read in order, must equal the bottom pink row"? Let's check: the top pink tiles in order: pair1's P (1 glyph), pair2's P (2 glyphs), pair3's P (1), pair4's P (1) = 5 glyphs total = the length of the bottom pink row (5)!! And the top BLUE tiles: pair1's S (1), pair2's S (1), pair3's S (2), pair4's S (1) = 5 glyphs = the length of the bottom blue row (5)!!!
WOW. That's it! The top area is a "flattened" version of the sentence pair! I.e., the top area shows the same sentence pair but split into 4 chunks (pairs), where each pair = (a chunk of the blue sentence, the corresponding chunk of the pi
… [14403 caracteres más]

```python
def idx_of(pos): return which_tile2(pos)
def move_to(i, maxsteps=5):
    log=[]
    for _ in range(maxsteps+1):
        cur=idx_of(wpos())
        if cur==i: return log
        # choose shorter direction on 8-cycle
        d_right=(i-cur)%8; d_left=(cur-i)%8
        a='RIGHT' if d_right<=d_left else 'LEFT'
        r=last(action([a])); log.append((a,r.get('state')))
        if res_check(r): return log
    return log
CLS={'u':'1000111111001001111110001','w':'0010010101101011111100100','t1':'1000110001110111000111111',
     's2':'0000100101111111010010000','s1':'0000100101111110010100001','p1':'0000101111010010100111111',
     'p2':'0011100101111111010011100','x':'0010011111100011000111111','v':'1111110001101111010111111',
     'q1':'0111111001100011001111110'}
def tclasses(i):
    t,_=lvl5_g
… [255 caracteres más]
```

```text
current classes per tile:
  0 S 10 8 ['s1'] ['0000100101111110010100001']
  1 P 10 18 [] ['0010011111101011111100100']
  2 S 10 31 ['s2'] ['0000100101111111010010000']
  3 P 10 41 ['q1'] ['0111111001100011001111110', '0111111001100011001111110']
  4 S 22 8 ['t1'] ['1000110001110111000111111', '10001
… [181 caracteres más]
```

## Paso 21 · nivel 5 · acción 224

**Pensamiento:**

Note tile1's class '0010011111101011111100100' isn't in my CLS dict — it's a new class (call it m1: it looks like a mix of x and w patterns). Interesting: this is likely one of the P cycle's 7 glyphs.
Now, before spending ~28 actions on the identity-order target, let me reconsider ONE more time whether there's a cheaper decisive test of the model. 
Idea: a "partial" test isn't possible — completion requires the full config.
Alternative: consider the possibility that the checker accepts ANY valid tiling (order-free dictionary). Then I could choose the tiling requiring the FEWEST presses from the current state — much cheaper! And if the checker requires the identity order, the cheapest-plan approach fails but I'd have learned nothing... 
Hmm. Risk assessment:
- Plan A (identity order): ~28 actions. If the checker is order-free, this still works ✓ (identity is one of the valid tilings). If the checker requires identity, ✓. If the model is wrong entirely, ✗ (and ~28 actions wasted → likely death by budget).
- Plan B (cheapest valid tiling, order-free): might be ~10 actions. If the checker requires identity order, ✗.
Plan A is safe under both interpretations of the order. Let me estimate the budget more carefully. What IS the budget? Row 63 is all 'w' (light gray) — 64 cells. Earlier at level 5's start, was there a bar? Let me check the initial frame's row 63 and other edges for a shrinking bar. In level 4, the budget was 64 (I recall row63 counting). Let me check the bar's curren
… [345 caracteres más]

```python
f0=None
for h in history:
    if h.frame.level==5: f0=h.frame; break
def edges(fr):
    A=fr.ascii.split('\n')
    return {'top':A[0].strip('\n'), 'row63':A[63], 'col0':''.join(r[0] for r in A), 'col63':''.join(r[63] for r in A)}
e0=edges(f0); e1=edges(current_frame)
for k in e0:
    same = e0[k]==e1[k]
    print(k,'same' if same else 'DIFF')
    if not same:
        print('  init', e0[k][:70])
        print('  cur ', e1[k][:70])
```

```text
top same
row63 DIFF
  init wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww
  cur  wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwccccccccccc
col0 same
col63 DIFF
  init ggggggggggggggggggggggggggggggggggggggGGGGGGGGGGGGGGGGGGGGGGGGGw
  cur  gggggggggggggggggggggggggggggggggggg
… [28 caracteres más]
```

**Pensamiento:**

The budget bar is row 63: initially 64 'w'; now 53 'w' + 11 'c' → 11 actions consumed?? Hmm, but I've used 24 actions in level 5 (steps 201→225). Hmm, 64-53 = 11. So the bar isn't 1:1 with actions? Maybe the bar decrements every 2 actions? Or maybe the bar started at 64 and each action removes ~0.5? Hmm.
Wait, maybe the bar started at a different value. Let me count: init row63 = 64 w's (all). Current: 53 w + 11 c. Steps used in level 5 = 225 - 201 = 24. So 11 cells lost over 24 actions → ratio ~0.46. Hmm, that's odd. Unless the bar decrements per 2 actions.
Hmm, alternatively the bar could be a "time" bar that decrements on a timer, not per action. But no real time passes between my turns... Actually, one action can animate over many frames; maybe the bar decrements per animation frame group?
Whatever the mapping, the bar has 53/64 remaining → plenty of headroom. If the bar were 1:1 with actions, I'd have 40 left. Either way, ~50 actions of headroom at worst ratio... Let's estimate: if it decrements 11 per 24 actions, then 53 remaining → ~115 more actions. Comfortable.
Hmm, wait. But careful: maybe the bar decrements only on "gameplay" actions. Anyway, I have room for ~28 actions. Let me proceed with Plan A (identity-order target).
Plan A target classes:
tile0: [u]
tile1: [p2]
tile2: [s1]
tile3: [x,x]
tile4: [w,w]
tile5: [p1]
tile6: [s2]
tile7: [v] ✓ (leave)
Current: tile0=s1, tile1=m1, tile2=s2, tile3=(q1,q1), tile4=(t1,t1), tile5=p2, tile6=u, tile7=v.
S cycle (UP advances)
… [1515 caracteres más]

```python
def set_tile(i, target, cap=8, first='UP'):
    log=[]
    for k in range(cap):
        if tclasses(i)==target: return log,'OK'
        a = first if k<cap//2+1 else 'DOWN'
        r=last(action([a])); log.append((a,r.get('state')))
        if res_check(r): return log,'DONE'
    return log, tclasses(i)
SC={'u':0,'w':1,'t1':2,'s2':3,'s1':4,'z1':5,'z2':6}
INV={v:k for k,v in SC.items()}
log=[]
# tile6: u -> s2 (UP 3)
mv=move_to(6); log+=mv
st,stt=set_tile(6,(CLS['s2'],))
log+=st
# tile0: s1 -> u (UP 3)
mv=move_to(0); log+=mv
st,stt2=set_tile(0,(CLS['u'],))
log+=st
print('after tile6/0:', tclasses(6), tclasses(0), stt, stt2)
print('actions used this stage:', len(log))
print('cursor', idx_of(wpos()))
```

```text
Traceback (most recent call last):
  File "<python_tool>", line 15, in <module>
NameError: name 'CLS' is not defined
```

## Paso 21 · nivel 5 · acción 241

**Pensamiento:**

CLS wasn't retained (it was a module-level variable, not a function). I need to define CLS inside each snippet or as a function default. Let me define a function `CLS()` returning the dict, which will be retained.
Let me redo with a self-contained snippet that defines and uses everything, and also make set_tile use a passed-in dict (set_tile doesn't use CLS, good).
Note: set_tile was retained but its dependency is fine.
Let me run stage 1 now.

```python
def CLS():
    return {'u':'1000111111001001111110001','w':'0010010101101011111100100','t1':'1000110001110111000111111',
     's2':'0000100101111111010010000','s1':'0000100101111110010100001','p1':'0000101111010010100111111',
     'p2':'0011100101111111010011100','x':'0010011111100011000111111','v':'1111110001101111010111111',
     'q1':'0111111001100011001111110'}
C=CLS()
log=[]
move_to(6)
st,stt=set_tile(6,(C['s2'],)); log+=st
move_to(0)
st,stt2=set_tile(0,(C['u'],)); log+=st
print('tile6', tclasses(6)==C['s2'], 'tile0', tclasses(0)==C['u'], stt, stt2)
print('stage1 actions', len(log), 'cursor', idx_of(wpos()))
print('state', last_action_call_result.get('state'))
```

```text
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] RIGHT -> effect
[action] LEFT -> effect
[action] RIGHT -> effect
[action] LEFT -> effect
[action] RIGHT -> effect
[action] LEFT -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
… [198 caracteres más]
```

## Paso 22 · nivel 5 · acción 241

**Pensamiento:**

Something went wrong! My move_to oscillated (RIGHT, LEFT, RIGHT, LEFT...) — because `idx_of(wpos())` returned something unexpected, causing thrashing. And set_tile's cap logic switched to DOWN mid-way. 17 actions wasted. tile6 is now... let me check the current state.
The move_to oscillation: it seems `idx_of(wpos())` didn't equal 0 even when at tile0. Why? Because which_tile2 matches by column range only, and after moving to row 10 tiles the bracket rows change... Let me look: which_tile2(pos) checks `c0-1 <= pos[2] and pos[3] <= c1+1` for tiles in the sorted list — but it doesn't check the ROW. When the cursor is on tile0 (S at row 10, cols 8-14), the brackets are at cols 9-13. Candidates: tile0 (c0=8,c1=14 → 7<=9 and 13<=15 ✓). So it should return 0. Hmm, but the oscillation suggests it kept returning 6 or something.
OH WAIT. The bug: `move_to` uses `idx_of(wpos())` — but `wpos()` returns (rows[0], rows[-1], cols[0], cols[-1]) where cols come from the FIRST bracket row. For the top-row tiles, brackets at rows 7 and 19 with cols 9-13 → pos = (7,19,9,13) → matches tile0 ✓. Hmm.
Hmm, but the log shows: RIGHT, LEFT, RIGHT, LEFT, RIGHT, LEFT — 6 moves oscillating between two states. So idx_of returned A (not target 0), moved RIGHT, then idx_of returned B (not 0), moved LEFT... alternating. So the cursor alternated between tile7 and... hmm, RIGHT from tile6 → tile7, then LEFT → tile6, etc. So idx_of gave 6 (target 0, d_right=2, d_left=6 → RIGHT) → cursor at 7: idx_of(7) → d_righ
… [4355 caracteres más]

```python
t,b = lvl5_get()
print('n tiles', len(t), 'n bars', len(b))
for i,x in enumerate(t): print(i,x[0],x[1],x[2],[clb(g) for g in x[3]])
print('wpos', wpos())
print('idx_of', idx_of(wpos()))
C=CLS()
inv={v:k for k,v in C.items()}
print('names:', [[inv.get(c,'?') for c in [clb(g) for g in x[3]]] for x in t])
print('bars:', [(x[0],x[1],x[2],[inv.get(clb(g),'?') for g in x[3]]) for x in b])
```

```text
n tiles 8 n bars 2
0 S 10 8 ['0000100101111110010100001']
1 P 10 18 ['0010011111101011111100100']
2 S 10 31 ['0000100101111111010010000']
3 P 10 41 ['0111111001100011001111110', '0111111001100011001111110']
4 S 22 8 ['1000110001110111000111111', '1000110001110111000111111']
5 P 22 25 ['0011100101111
… [301 caracteres más]
```

## Paso 22 · nivel 5 · acción 242

**Pensamiento:**

Current state:
tile0 = s1 (unchanged from before my stage-1 attempt — so my presses didn't affect tile0)
tile1 = m1 ('0010011111101011111100100')
tile2 = s2
tile3 = q1,q1
tile4 = t1,t1
tile5 = p2
tile6 = z1 ('1000111111100011000111011') ← cycle index 5
tile7 = v
Cursor at tile6.
So the net effect of my 17-action stage: tile6 went u → (3 UPs) s2 → ... → z1 (index 5). And the cursor stayed at 6 the whole time! So the RIGHT/LEFT presses did NOT move the cursor?! 
Hmm! Wait. The exec log shows RIGHT,LEFT,... and then UP×5, DOWN×3. If the cursor never moved from tile6, then all 11 presses hit tile6: net = 3 + 5 - 3 = +5 → index 0+5 = 5 = z1 ✓ EXACTLY matches! So the cursor never moved!
Why didn't RIGHT/LEFT move the cursor? Because... hmm! Maybe LEFT/RIGHT only move the cursor when...? Earlier RIGHT presses DID move the cursor (steps 203-211). What changed? 
OH!! Maybe the cursor doesn't move when a tile is in a "wrong" state? No... 
Hmm, maybe the cursor movement is blocked because the cursor is currently on a tile and pressing RIGHT moves... Let me reconsider: in the earlier successful moves, the cursor moved on every RIGHT. Now it doesn't. Difference: the tile contents? Or maybe the game state changed such that LEFT/RIGHT now do something else...
OR maybe the "oscillation" was real: RIGHT moved the cursor to tile7, then my code computed idx_of → hmm, but the net press count on tile6 (+5) proves the cursor was on tile6 during those 8 presses. Unless the presses were distributed 
… [2019 caracteres más]

```python
print('before: wpos', wpos(), 'idx', idx_of(wpos()))
r=last(action(['RIGHT']))
print('after RIGHT: wpos', wpos(), 'idx', idx_of(wpos()), r.get('state'))
d=frame_diff()
print('changed', d['changed_cell_count'])
```

```text
before: wpos (19, 31, 39, 43) idx 6
[action] RIGHT -> effect
after RIGHT: wpos (19, 31, 49, 53) idx 3 NOT_FINISHED
changed 28
```

## Paso 23 · nivel 5 · acción 245

**Pensamiento:**

The cursor moved from tile6 to tile7 (cols 49-53) ✓. But `idx_of` returned 3! That's the bug: which_tile2 returned index 3 for the bracket cols 49-53. Why? Because tile3 is P(10,41) with 2 glyphs → c0=41, c1 = 41 + 7*2 - 1 = 54 → check: 41-1=40 <= 49 ✓ and 53 <= 54+1 ✓ → MATCH! So which_tile2 matched tile3 (row 10) before tile7 (row 22) because it ignores rows! That's the bug — the 2-glyph tile at cols 41-54 has the same column span as tile7 at 48-54.
Fix: include the row in the match. The bracket rows: for row-10 tiles the brackets are at rows 7-19; for row-22 tiles at rows 19-31. The tile's r0: 10 or 22. Bracket r0 = 7 or 19 → r0_bracket = r0_tile - 3.
Let me write a proper `which_tile3(pos)` using both row and column, and make it robust: candidate tiles with r0 in (pos[0]+3, pos[1]-9) → i.e., tile r0 = pos[0]+3 (7+3=10 ✓, 19+3=22 ✓).
Let me define:
def which_tile3(pos):
    t,_ = lvl5_get()
    r0 = pos[0]+3
    for i,x in enumerate(t):
        if x[1]==r0:
            c0=x[2]; c1=c0+7*len(x[3])-1
            if c0-1<=pos[2] and pos[3]<=c1+1: return i
    return None
Then redefine move_to to use which_tile3 (move_to is retained and calls idx_of → which_tile2). I'll define new functions: idx3, move_to3.
Also note the cursor is now at tile7 (v, already correct target). 
Current tile state:
tile0 = s1 (target u)
tile1 = m1 (target p2)
tile2 = s2 (target s1)
tile3 = q1,q1 (target x,x)
tile4 = t1,t1 (target w,w)
tile5 = p2 (target p1)
tile6 = z1 (index 5; target s2 = index 3) →
… [1477 caracteres más]

```python
def which_tile3(pos):
    t,_=lvl5_get(); r0=pos[0]+3
    for i,x in enumerate(t):
        if x[1]==r0:
            c0=x[2]; c1=c0+7*len(x[3])-1
            if c0-1<=pos[2] and pos[3]<=c1+1: return i
    return None
def idx3(): return which_tile3(wpos())
def move3(i, cap=6):
    n=0
    for _ in range(cap+1):
        c=idx3()
        if c==i: return n
        dr=(i-c)%8; dl=(c-i)%8
        a='RIGHT' if dr<=dl else 'LEFT'
        r=last(action([a])); n+=1
        if res_check(r) or r.get('level_completed'): raise SystemExit
    return -1
def press_to(i, target, cap=7):
    n=0
    for _ in range(cap+1):
        if tclasses(i)==target: return n, True
        r=last(action(['UP'])); n+=1
        if res_check(r) or r.get('level_completed'): return n, 'DONE'
    for _ in range(cap):
        r=l
… [577 caracteres más]
```

```text
[action] LEFT -> effect
[action] DOWN -> effect
[action] DOWN -> effect

error:
Traceback (most recent call last):
  File "<python_tool>", line 35, in <module>
TypeError: 'int' object is not iterable
```

## Paso 24 · nivel 5 · acción 255

**Pensamiento:**

Oops: move3 wasn't retained (SystemExit dependency) but the snippet ran with the old `move_to`... no wait, the error was at line 35: `n=move3(0)` — no, move3 was defined in the snippet. The error "TypeError: 'int' object is not iterable" at line 35... Actually the executed actions were LEFT, DOWN, DOWN. So move3(6) pressed LEFT (from tile7 → tile6? LEFT from 7 → 6 ✓ correct), then 2 DOWNs on tile6 ✓. Then `n=move3(0)` — hmm, the error says 'int' object is not iterable at line 35, which is `st,ok=press_to(...)`? Let me count: line 35 is around `n=move3(0); log+=n` → `log+=n` where n is an int → TypeError ✓ (log is a list; `log += int` fails). Minor bug.
So the state now: cursor at tile6, tile6 = z1 - 2 = index 3 = s2 ✓ (target!). 
Let me verify and continue. Also I need to redefine move3 without SystemExit (use a return flag).
Let me check the current tile states and cursor, then continue the plan.

```python
def move3(i, cap=6):
    n=0
    for _ in range(cap+1):
        c=idx3()
        if c==i: return n
        dr=(i-c)%8; dl=(c-i)%8
        a='RIGHT' if dr<=dl else 'LEFT'
        r=last(action([a])); n+=1
        if res_check(r) or r.get('level_completed'): return -n
    return -99
C=CLS(); inv={v:k for k,v in C.items()}
t,_=lvl5_get()
print('cursor', idx3())
for i,x in enumerate(t): print(i, x[0], x[1], x[2], [inv.get(clb(g),'?') for g in x[3]])
A=current_frame.ascii.split('\n')
print('bar row63:', A[63].count('w'), 'w;', A[63].count('c'),'c')
```

```text
cursor 6
0 S 10 8 ['s1']
1 P 10 18 ['?']
2 S 10 31 ['s2']
3 P 10 41 ['q1', 'q1']
4 S 22 8 ['t1', 't1']
5 P 22 25 ['p2']
6 S 22 38 ['s2']
7 P 22 48 ['v']
bar row63: 42 w; 22 c
```

**Pensamiento:**

Current: tile6 = s2 ✓ (target for pair4's S). tile7 = v ✓.
Bar: 42 w remaining of 64 → 22 consumed over 45 actions (steps 201→246). Ratio 0.489 ≈ 0.5 → the bar decrements 1 per 2 actions. So remaining budget ≈ 42*2 = 84 more actions. 
Remaining fixes (identity target):
tile0: s1 (idx4) → u (idx0): UP 3
tile1: m1 → p2: unknown cycle
tile2: s2 (idx3) → s1 (idx4): UP 1
tile3: (q1,q1) → (x,x): unknown
tile4: (t1,t1) (idx2) → (w,w) (idx1): DOWN 1
tile5: p2 → p1: unknown
Cursor at 6. Route: 6→0 (RIGHT 2), fix tile0 (UP 3), →1 (RIGHT 1), fix tile1 (explore ≤7), →2 (1), fix (UP 1), →3 (1), fix (≤7), →4 (1), fix (DOWN 1), →5 (1), fix (≤7).
Moves: 2+1+1+1+1+1 = 7; known presses: 3+1+1 = 5; unknown: up to 21 (avg ~9). Total ~21-33. Fine within 84.
Let me execute stage by stage, checking the P cycle as I learn it. First: move to tile0, fix it, move to tile1, then explore the P cycle step by step (recording each glyph) until p2 is reached (cap 7).
I'll do it in one snippet with careful logging and stop conditions.

```python
C=CLS(); inv={v:k for k,v in C.items()}
acts=[]
def A(a):
    r=last(action([a])); acts.append(a)
    if r.get('level_completed') or r.get('game_over') or r.get('run_complete') or r.get('done'):
        raise StopIteration(r)
    return r
try:
    m=move3(0); acts.append(('mv',m))
    for _ in range(4):
        if tclasses(0)==(C['u'],): break
        A('UP')
    print('tile0 ->', [inv.get(clb(g),'?') for g in lvl5_get()[0][0][3]])
    m=move3(1); acts.append(('mv',m))
    seen=[ [clb(g) for g in lvl5_get()[0][1][3]] ]
    for k in range(7):
        if tclasses(1)==(C['p2'],): break
        A('UP'); seen.append([clb(g) for g in lvl5_get()[0][1][3]])
    print('P cycle from tile1:', [[inv.get(c,'?') for c in s] for s in seen])
    print('tile1 ->', [inv.get(clb(g),'?') for g in lvl5_get()[0
… [219 caracteres más]
```

```text
[action] RIGHT -> effect
[action] RIGHT -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
tile0 -> ['u']
[action] RIGHT -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
P cycle from tile1: [['?'], ['q1'], ['x'], ['?'], ['p2']
… [127 caracteres más]
```

**Nivel 5 completado → nivel 6 (acción 264)**

## Paso 25 · nivel 6 · acción 264

**Pensamiento:**

Progress:
tile0 = u ✓
tile1 = p2 ✓
P cycle observed (partial): [m1(?)] → q1 → x → m2(?) → p2 → ... So the P cycle order includes: m1, q1, x, ?, p2, ... (UP advances). Since p2 came 4 steps after m1, and the cycle has 7 entries, we have positions: m1=0, q1=1, x=2, ?=3, p2=4, then 5,6 unknown (likely v and p1 in some order).
Wait, but tile7 = v and tile5 = p2, tile3 = q1,q1. Hmm, and p1 should be in the cycle too.
Remaining targets:
tile2: s2 → s1: UP 1 (S cycle)
tile3: (q1,q1) → (x,x): from q1 (pos 1) → x (pos 2): UP 1 — IF both glyphs advance together.
tile4: (t1,t1) → (w,w): DOWN 1 (S cycle: t1 idx2 → w idx1) — IF both advance together.
tile5: p2 → p1: p2 is pos 4; p1's position unknown (5 or 6) → UP 1 or 2.
Cursor at tile1. Route: tile2 (RIGHT 1, UP 1), tile3 (RIGHT 1, UP 1 → check), tile4 (RIGHT 1, DOWN 1 → check), tile5 (RIGHT 1, UP 1 or 2 → check).
Let me execute with verification and stop if completed. I'll implement a loop that presses UP (or DOWN for tile4) with a cap and verifies.
Careful for tile5: if after 2 UPs it's not p1, try more (cap 6).
Let me do it.

```python
C=CLS(); inv={v:k for k,v in C.items()}
acts=[]
def act(a):
    r=last(action([a])); acts.append(a)
    if r.get('level_completed') or r.get('game_over') or r.get('run_complete') or r.get('done'): raise StopIteration(r)
    return r
def name(i): return [inv.get(clb(g),'?') for g in lvl5_get()[0][i][3]]
try:
    move3(2); acts.append('mv2')
    for _ in range(3):
        if name(2)==['s1']: break
        act('UP')
    print('tile2', name(2))
    move3(3); acts.append('mv3')
    for _ in range(7):
        if name(3)==['x','x']: break
        act('UP')
    print('tile3', name(3))
    move3(4); acts.append('mv4')
    for _ in range(7):
        if name(4)==['w','w']: break
        act('DOWN')
    print('tile4', name(4))
    move3(5); acts.append('mv5')
    for _ in range(7):
        if name(5)=
… [340 caracteres más]
```

```text
[action] RIGHT -> effect
[action] UP -> effect
tile2 ['s1']
[action] RIGHT -> effect
[action] UP -> effect
tile3 ['x', 'x']
[action] RIGHT -> effect
[action] DOWN -> effect
tile4 ['w', 'w']
[action] RIGHT -> effect
[action] UP -> effect
[action] UP -> LEVEL COMPLETED
EVENT {'executed': True, 'action
… [705 caracteres más]
```

## Paso 26 · nivel 6 · acción 264

**Pensamiento:**

LEVEL 6 COMPLETE (level 5 cleared)! Now level 6.
Level 6 layout (from the image):
- Top area: 3 rows, each with TWO pairs: (blue→pink) and (pink→yellow)!
  Row 1: S(1 glyph) → P(2 glyphs); P(1) → Y(1)
  Row 2: S(1) → P(2); P(1) → Y(1)
  Row 3: S(1) → P(2); P(1) → Y(1)
- Bottom: a blue bar with 3 glyphs, and a yellow bar with 6 glyphs.
- Cursor at the top-left (around the first blue tile).
So this is like level 4 (a chain: blue→pink→yellow) but with 6 pairs and multi-glyph tiles (like level 5).
Bottom: blue row (3 glyphs) → yellow row (6 glyphs, editable?). The chain: each blue glyph maps to a pink word (2 glyphs each per the pairs), and each pink word maps to a yellow word.
Hmm, and the bottom yellow row has 6 glyphs = 3 blue glyphs × 2 (if each blue → 2 pink glyphs → each pink glyph → 1 yellow glyph).
So the task is likely: the yellow row must be the translation of the blue row via the chain. But is the yellow row editable? The cursor is at the top-left blue tile (like level 5, where the top tiles were editable and the bottom rows were fixed).
Hmm! In level 5, the top tiles were editable and the bottom was the fixed ground truth. Here, maybe the same: the top pairs are scrambled and must be reconstructed from the bottom pair (blue 3 glyphs → yellow 6 glyphs).
Let me extract the board and analyze. The structure: 6 pairs (blue→pink ×3, pink→yellow ×3), bottom blue row (3) and yellow row (6).
Under the "flattened sentence" model: the concatenation of the blue tiles = the bottom
… [2035 caracteres más]

```python
def extract(fr=None):
    if fr is None: fr=current_frame
    A=fr.ascii.split('\n'); seg=fr.segmentation; out=[]
    for n in seg['nodes']:
        b=n['boundary']
        r0=min(p[0] for p in b); r1=max(p[0] for p in b); c0=min(p[1] for p in b); c1=max(p[1] for p in b)
        if n['pixels']<=15: continue
        h=r1-r0+1; w=c1-c0+1
        kind='tile' if h==7 else ('bar' if w>=14 and h==7 else 'other')
        n2=w//7
        gl=[clb(glf(A,r0+1,c0+1+7*j)) for j in range(n2)] if h==7 else []
        out.append({'c':n['color'],'r0':r0,'c0':c0,'r1':r1,'c1':c1,'px':n['pixels'],'gl':gl})
    out.sort(key=lambda x:(x['r0'],x['c0']))
    return out
objs=extract()
for i,o in enumerate(objs):
    print(i,o['c'],o['r0'],o['c0'],o['r1'],o['c1'],'w=%d'%(o['c1']-o['c0']+1),'n=%d'%len(o['gl']),o['gl
… [192 caracteres más]
```

```text
0 g 0 0 40 63 w=64 n=0 []
1 S 5 9 11 15 w=7 n=1 ['1000110001110111000111111']
2 P 5 19 11 32 w=14 n=2 ['0111111001100011001111110', '0000101111010010100111111']
3 P 5 38 11 44 w=7 n=1 ['0011111101101011110100111']
4 Y 5 48 11 54 w=7 n=1 ['1001100001100011000011001']
5 B 6 20 10 24 w=5 n=0 []
6 B 6 3
… [989 caracteres más]
```

**Pensamiento:**

Level 6 layout:
Top: 3 rows, each with: S tile (1 glyph) → P tile (2 glyphs), and P tile (1 glyph) → Y tile (1 glyph).
Row 1 (r=5): S[11] t1 | P[q1,p1] ; P[a] → Y[b]
Row 2 (r=17): S[z1] | P[x,p2]; P[x] → Y[c]
Row 3 (r=29): S[u] | P[p2,p2]; P[m1] → Y[d]
Bottom: S bar (3 glyphs) at r=44: [z1, z2, s1]; Y bar (6 glyphs) at r=53: [e,f,c,c,f,e] where e='1000110000111110000110001', f='1010100001101010000111111', c='1101110001101011000111011'.
Note the Y bar's pattern: [e,f,c,c,f,e] — a palindrome! Interesting.
Cursor at the top-left S tile (rows 2-14 brackets, cols 10-14) → tile 1 (S at r5,c9).
Bar63: 64 w (full budget, 64-ish → ~128 actions at the 0.5 rate).
Now the model from level 5: the top area is the "flattened" version of the bottom pair(s)? Here we have TWO chains: S→P and P→Y. Bottom: S bar (3 glyphs) and Y bar (6 glyphs).
Under the "flattened" model:
- The S tiles (3 tiles × 1 glyph) concatenated = the bottom S row (3 glyphs) ✓ counts match!
- The Y tiles (3 tiles × 1 glyph) concatenated = ... the bottom Y row has 6 glyphs ✗ (3 ≠ 6).
Hmm. Unless the Y tiles can have 2 glyphs. They're 1 glyph each now, but the tile sizes are FIXED (7 wide). So the Y tiles' total = 3 glyphs but the Y bar has 6 ✗.
So the flattened model doesn't directly apply. Instead, this looks like the level-4-style chain puzzle: the dictionary is S→P (3 entries, each 1→2) and P→Y (3 entries, each 1→1). Then: S row (3 glyphs) → P (each S glyph → 2 P glyphs = 6 P glyphs) → Y (each P glyph → 1 Y glyph = 6 Y 
… [11712 caracteres más]

```python
C6={}
def nm6(cl): return C6.get(cl,'?')
def tiles6():
    o=[x for x in extract() if x['r0']<40 and (x['c1']-x['c0']+1)%7==0 and (x['r1']-x['r0'])==6]
    return o
def bars6():
    return [x for x in extract() if x['r0']>=40 and (x['c1']-x['c0']+1)>10]
t=tiles6()
print('n tiles', len(t))
for i,x in enumerate(t): print(i,x['c'],x['r0'],x['c0'],'n=%d'%len(x['gl']),x['gl'])
print('cursor wpos', wpos())
b=bars6()
for x in b: print('BAR',x['c'],x['r0'],x['c0'],x['gl'])
```

```text
n tiles 12
0 S 5 9 n=1 ['1000110001110111000111111']
1 P 5 19 n=2 ['0111111001100011001111110', '0000101111010010100111111']
2 P 5 38 n=1 ['0011111101101011110100111']
3 Y 5 48 n=1 ['1001100001100011000011001']
4 S 17 9 n=1 ['1000111111100011000111011']
5 P 17 19 n=2 ['0010011111100011000111111', '0
… [663 caracteres más]
```

## Paso 26 · nivel 6 · acción 271

**Pensamiento:**

Good. 12 tiles indexed 0-11 in reading order: row1: 0=S,1=P(2),2=P(1),3=Y; row2: 4=S,5=P(2),6=P(1),7=Y; row3: 8=S,9=P(2),10=P(1),11=Y.
Bottom S bar: [z1, z2, s1]; Y bar: [e,f,c,c,f,e].
Names (clb → name):
t1='1000110001110111000111111'
q1='0111111001100011001111110'
p1='0000101111010010100111111'
a6='0011111101101011110100111' (row1 right P)
b6='1001100001100011000011001' (row1 Y)
z1='1000111111100011000111011'
x='0010011111100011000111111'
p2='0011100101111111010011100'
c='1101110001101011000111011' (row2 Y = also Y bar glyph 2,3)
u='1000111111001001111110001'
m1='0010011111101011111100100' (row3 right P)
d6='1010100000101111010111101' (row3 Y)
z2='1000110001111011010110111'
s1='0000100101111110010100001'
e='1000110000111110000110001'
f='1010100001101010000111111'
Now, key experiments needed:
1. Cursor order (12 tiles?) — verify with RIGHT presses.
2. 2-glyph tile behavior under UP.
3. Cycles for S, P, Y tiles.
Let me first test the 2-glyph behavior since it determines feasibility. Move RIGHT once (cursor 0 → tile 1 = P(5,19) with [q1,p1]) and press UP, observe.
Then continue pressing UP a few times to see the pattern.
Let me do that now.

```python
def t6(i): return tiles6()[i]['gl']
r=last(action(['RIGHT']))
print('cursor now', wpos())
print('tile1 before', t6(1))
seq=[list(t6(1))]
for k in range(6):
    r=last(action(['UP']))
    seq.append(list(t6(1)))
    if res_check(r): break
print('tile1 cycle:')
for s in seq: print('  ',s)
print('state', last_action_call_result.get('state'))
```

```text
[action] RIGHT -> effect
cursor now (2, 14, 20, 31)
tile1 before ['0111111001100011001111110', '0000101111010010100111111']
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
tile1 cycle:
   ['0111111001100011001111110'
… [421 caracteres más]
```

## Paso 27 · nivel 6 · acción 279

**Pensamiento:**

Excellent! The 2-glyph P tile cycle: both glyphs advance together through the cycle, and after 6 UPs the tile returned to the start [q1,p1] ✓ (the 7th state = the first). So the P cycle has 7 classes and the pair advances in lockstep. The observed sequence of pairs:
0: (q1, p1)
1: (x, m1)
2: (a6, q1)
3: (p2, x)
4: (v, a6)
5: (p1, p2)
6: (m1, v)
→ back to (q1,p1)
So the P cycle order is: q1 → x → a6 → p2 → v → p1 → m1 → q1 (7 classes ✓).
Check with the pair sequence: glyph1 goes q1,x,a6,p2,v,p1,m1 ✓ and glyph2 = the same cycle shifted by +2: p1,m1,q1,x,a6,p2,v ✓ consistent (p1 is at index 5 in the cycle... let me verify: cycle = [q1,x,a6,p2,v,p1,m1]. glyph2 sequence: p1,m1,q1,x,a6,p2,v = indices 5,6,0,1,2,3,4 ✓ = glyph1's indices +5 mod 7. Yes, lockstep with offset 5.
So the P cycle (UP advances +1): index: q1=0, x=1, a6=2, p2=3, v=4, p1=5, m1=6.
Note: a6 and m1 are new names for the level-6 P glyphs. And the Y bar's glyphs e,f,c — are they in the P cycle? No, they're Y-colored glyphs. The Y cycle is separate (Y tiles).
Now the current tile1 = (m1, v) (after 6 UPs).
Now let me think about the target structure again:
Bottom S row: [z1, z2, s1] (3 glyphs)
Bottom Y row: [e,f,c,c,f,e] (6 glyphs)
Chain: S glyph → P word (2 glyphs) → each P glyph → Y glyph (1).
So: W(z1)=2 P glyphs → 2 Y glyphs = [e,f] (the first two of the Y row)
W(z2) → [c,c]
W(s1) → [f,e]
And the right pairs give P→Y: the P glyph that maps to e, f, c.
From W(z1)→[e,f] and W(s1)→[f,e]: the P glyphs in W(z1) = (P_e
… [4700 caracteres más]

```python
r=last(action(['RIGHT'])); r=last(action(['RIGHT']))
print('cursor', wpos(), 'tile3', t6(3))
seq=[list(t6(3))]
for k in range(6):
    r=last(action(['UP'])); seq.append(list(t6(3)))
    if res_check(r): break
print('Y cycle:')
for s in seq: print('  ',s[0])
print('state', last_action_call_result.get('state'))
```

```text
[action] RIGHT -> effect
[action] RIGHT -> effect
cursor (2, 14, 49, 53) tile3 ['1001100001100011000011001']
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
Y cycle:
   1001100001100011000011001
   101010000110101000
… [171 caracteres más]
```

## Paso 28 · nivel 6 · acción 288

**Pensamiento:**

Y cycle (7 classes):
0: b6='1001100001100011000011001'
1: f ='1010100001101010000111111'  ← Y bar glyph!
2: g6='0101011011100010000010001'
3: d6='1010100000101111010111101'
4: c ='1101110001101011000111011'  ← Y bar glyph!
5: h6='0010010101001001010100100'
6: e ='1000110000111110000110001'  ← Y bar glyph!
→ back to 0 ✓
So Y cycle indices: b6=0, f=1, g6=2, d6=3, c=4, h6=5, e=6.
The cursor is now at tile3 (Y row1) back at b6 ✓ (7 presses = full cycle).
Note the cursor bracket is at cols 49-53 → tile3 ✓. And the cursor order so far: 0 →1 →2 →3 ✓ reading order.
Now the S cycle. Move LEFT 3 → tile0, then 7 UP presses. But careful: the S tile's current glyph is t1 which is NOT in the bottom S row [z1,z2,s1]. In level 5, the S cycle contained all the needed classes. Let me enumerate the S cycle here.
Cost: 3 moves + 7 presses = 10. Then I'll have all cycles and can compute the plan.
Let me run it.

```python
for _ in range(3): r=last(action(['LEFT']))
print('cursor', wpos(), 'tile0', t6(0))
seq=[list(t6(0))]
for k in range(6):
    r=last(action(['UP'])); seq.append(list(t6(0)))
    if res_check(r): break
print('S cycle:')
for s in seq: print('  ',s[0])
print('state', last_action_call_result.get('state'))
```

```text
[action] LEFT -> effect
[action] LEFT -> effect
[action] LEFT -> effect
cursor (2, 14, 10, 14) tile0 ['1000110001110111000111111']
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
S cycle:
   1000110001110111000111111
… [193 caracteres más]
```

## Paso 29 · nivel 6 · acción 288

**Pensamiento:**

S cycle (7 classes):
0: t1='1000110001110111000111111'
1: s2='0000100101111111010010000'
2: s1='0000100101111110010100001'  ← in the bottom S row!
3: z1='1000111111100011000111011'  ← in the row!
4: z2='1000110001111011010110111'  ← in the row!
5: u='1000111111001001111110001'
6: w='0010010101101011111100100'
→ back to t1 ✓
Great: S cycle indices: t1=0, s2=1, s1=2, z1=3, z2=4, u=5, w=6.
Current tile0 = t1 (idx 0) — wait, after 6 UPs it should be at idx 6 = w. Hmm, the last printed state is '0010010101101011111100100' = w (idx 6) ✓ (the 7th press will return to t1).
Hmm, so tile0 is now at w. I need to remember the current state of all tiles. Let me re-read the board.
Now let me compute the optimal plan.
Current state (need to re-read; tile1 = (m1,v) after 6 UPs; tile3 = b6 after full cycle; tile0 = w).
Let me define cycles:
S: [t1, s2, s1, z1, z2, u, w] (UP = +1)
P: [q1, x, a6, p2, v, p1, m1] (UP = +1)
Y: [b6, f, g6, d6, c, h6, e] (UP = +1)
Targets:
- Row assignment (forced by offsets): row1 → entry for z1 with left P word (P_e, P_f) [offset 5], row2 → entry for s1 with word (P_f, P_e) [offset 2], row3 → entry for z2 with word (P_c,P_c) [offset 0].
  Wait, I need to double check the offset assignment: tiles' offsets: tile1 (row1 left P) offset 5, tile5 (row2 left P) offset 2, tile9 (row3 left P) offset 0.
  Words: (P_e,P_f) needs offset δ=idx(P_f)-idx(P_e); (P_f,P_e) needs -δ; (P_c,P_c) needs 0.
  The tile with offset 0 must take the word with offset 0 → row3 gets (P_c,P_c) →
… [4075 caracteres más]

```python
SC=['1000110001110111000111111','0000100101111111010010000','0000100101111110010100001',
    '1000111111100011000111011','1000110001111011010110111','1000111111001001111110001','0010010101101011111100100']
PC=['0111111001100011001111110','0010011111100011000111111','0011111101101011110100111',
    '0011100101111111010011100','1111110001101111010111111','0000101111010010100111111','0010011111101011111100100']
YC=['1001100001100011000011001','1010100001101010000111111','0101011011100010000010001',
    '1010100000101111010111101','1101110001101011000111011','0010010101001001010100100','1000110000111110000110001']
def save():
    import json
    open('/tmp/cyc.json','w').write(json.dumps({'S':SC,'P':PC,'Y':YC}))
    return 'ok'
t=tiles6()
cur={i:t[i]['gl'] for i in range(12)}
def dist(cyc,a,b)
… [161 caracteres más]
```

```text
0 S ['001001']
1 P ['001001', '111111']
2 P ['001111']
3 Y ['100011']
4 S ['100011']
5 P ['001001', '001110']
6 P ['001001']
7 Y ['110111']
8 S ['100011']
9 P ['001110', '001110']
10 P ['001001']
11 Y ['101010']
```

## Paso 29 · nivel 6 · acción 288

**Pensamiento:**

I need full class strings, not truncated. Let me write the optimizer with the full cycles defined in the snippet, reading the current tiles via tiles6(). I'll define the cycles as a retained function CYC() returning the dict.
Let me now write the full optimizer:
State: cur[i] = list of glyph classes for tile i.
Cycles: S, P, Y (lists, UP = +1).
Tiles: 0,4,8 = S tiles (1 glyph); 1,5,9 = left P (2 glyphs); 2,6,10 = right P (1 glyph); 3,7,11 = Y (1 glyph).
Rows: row0 = tiles (0,1,2,3), row1 = (4,5,6,7), row2 = (8,9,10,11).
Search:
for perm of entry→row assignment: entries are E_z1 (word (P_e,P_f)), E_s1 (word (P_f,P_e)), E_z2 (word (P_c,P_c))... 
Actually cleaner: enumerate:
 - P_e index (0..6), P_f index (0..6), P_c index (0..6), all distinct
 - assignment: which row gets which entry, i.e., a bijection from {entry_z1, entry_s1, entry_z2} to rows {0,1,2}
 - assignment: which row gets which right-pair {P_e→e, P_f→f, P_c→c}, bijection to rows
Constraints:
 - For the row holding entry_z1: the left P tile must show (P_e,P_f) — reachable iff offset matches: (idx(P_f)-idx(P_e))%7 == (cur_g2-idx... hmm, the tile's current pair (g1,g2) with indices (a,b); the target pair (P_e,P_f) with indices (c,d) is reachable iff (b-a) ≡ (d-c) mod 7. Cost = (c-a) mod 7 → min(k,7-k) with direction.
 - Similarly for entry_s1's row: target (P_f,P_e).
 - entry_z2's row: target (P_c,P_c).
 - S tile cost: dist(S, cur, target glyph).
 - Right P tile: dist(P, cur, target P glyph); Y tile: dist(Y, cur, target
… [1113 caracteres más]

```python
def CYC():
    S=['1000110001110111000111111','0000100101111111010010000','0000100101111110010100001',
       '1000111111100011000111011','1000110001111011010110111','1000111111001001111110001','0010010101101011111100100']
    P=['0111111001100011001111110','0010011111100011000111111','0011111101101011110100111',
       '0011100101111111010011100','1111110001101111010111111','0000101111010010100111111','0010011111101011111100100']
    Y=['1001100001100011000011001','1010100001101010000111111','0101011011100010000010001',
       '1010100000101111010111101','1101110001101011000111011','0010010101001001010100100','1000110000111110000110001']
    return {'S':S,'P':P,'Y':Y}
def cost1(cyc, cur, tgt):
    k=(cyc.index(tgt)-cyc.index(cur))%7
    return min(k,7-k), (k if k<=3 else k-7)
def cost2(cy
… [1620 caracteres más]
```

```text
Traceback (most recent call last):
  File "<python_tool>", line 46, in <module>
  File "<python_tool>", line 36, in solve6
  File "<python_tool>", line 11, in cost1
ValueError: 'z1' is not in list
```

## Paso 29 · nivel 6 · acción 288

**Pensamiento:**

Bug: in the loop I passed the entry NAME ('z1') to cost1 as the target glyph instead of the S glyph class. The entry names map to S glyph classes: 'z1'→S[3], 's1'→S[2], 'z2'→S[4]. Let me fix: use sglyph = {'z1':S[3],'s1':S[2],'z2':S[4]}.
Let me fix solve6 (redefine it).

```python
def solve6():
    C=CYC(); S,P,Y=C['S'],C['P'],C['Y']
    t=tiles6(); cur=[t[i]['gl'] for i in range(12)]
    e,f,c = Y[6],Y[1],Y[4]
    sg={'z1':S[3],'s1':S[2],'z2':S[4]}
    from itertools import permutations
    best=[]
    for ie in range(7):
      for jf in range(7):
        for kc in range(7):
          if len({ie,jf,kc})<3: continue
          Pe,Pf,Pc=P[ie],P[jf],P[kc]
          entries={'z1':(Pe,Pf),'s1':(Pf,Pe),'z2':(Pc,Pc)}
          pairs={'e':(Pe,e),'f':(Pf,f),'c':(Pc,c)}
          for er in permutations(['z1','s1','z2']):
            for pr in permutations(['e','f','c']):
              tot=0; det={}; ok=True
              for r in range(3):
                 sc=cost1(S,cur[4*r],sg[er[r]]); tot+=sc[0]; det[4*r]=sc[1]
                 lc=cost2(P,cur[4*r+1],entries[er[r]])
       
… [616 caracteres más]
```

```text
Traceback (most recent call last):
  File "<python_tool>", line 30, in <module>
  File "<python_tool>", line 20, in solve6
ValueError: ['0010010101101011111100100'] is not in list
```

**Pensamiento:**

cost1 receives cur[4*r] which is a LIST (['0010...']) since gl is a list. Need cur[4*r][0]. Let me fix by flattening: for 1-glyph tiles use cur[i][0].

```python
def solve6():
    C=CYC(); S,P,Y=C['S'],C['P'],C['Y']
    t=tiles6(); raw=[t[i]['gl'] for i in range(12)]
    cur=[ (g[0] if len(g)==1 else g) for g in raw ]
    e,f,c = Y[6],Y[1],Y[4]
    sg={'z1':S[3],'s1':S[2],'z2':S[4]}
    from itertools import permutations
    best=[]
    for ie in range(7):
      for jf in range(7):
        for kc in range(7):
          if len({ie,jf,kc})<3: continue
          Pe,Pf,Pc=P[ie],P[jf],P[kc]
          entries={'z1':(Pe,Pf),'s1':(Pf,Pe),'z2':(Pc,Pc)}
          pairs={'e':(Pe,e),'f':(Pf,f),'c':(Pc,c)}
          for er in permutations(['z1','s1','z2']):
            for pr in permutations(['e','f','c']):
              tot=0; det={}; ok=True
              for r in range(3):
                 sc=cost1(S,cur[4*r],sg[er[r]]); tot+=sc[0]; det[4*r]=sc[1]
          
… [624 caracteres más]
```

```text
options 420
press 13 entry/row ('s1', 'z1', 'z2') pair/row ('e', 'c', 'f') Pidx (2, 4, 1)
   deltas {0: 3, 1: -2, 2: 0, 3: 0, 4: 0, 5: 1, 6: 0, 7: 0, 8: -1, 9: -2, 10: -2, 11: -2}
press 13 entry/row ('s1', 'z1', 'z2') pair/row ('e', 'c', 'f') Pidx (2, 4, 3)
   deltas {0: 3, 1: -2, 2: 0, 3: 0, 4: 0, 
… [867 caracteres más]
```

**Pensamiento:**

420 valid options; the minimum press cost is 13. Many options with 13 presses.
Note: the deltas indicate signed presses (negative = DOWN).
Let me pick the best option considering navigation. The route: cursor currently at tile0. The tiles needing edits per option — let me count the number of distinct tiles needing edits and the navigation cost.
For the first option (Pidx (2,4,1)): deltas: tile0 +3, tile1 -2, tile2 0, tile3 0, tile4 0, tile5 +1, tile6 0, tile7 0, tile8 -1, tile9 -2, tile10 -2, tile11 -2. Tiles needing edits: {0,1,5,8,9,10,11} = 7 tiles. Navigation from tile0: need to visit 0,1,5,8,9,10,11. On a 12-cycle starting at 0: go RIGHT 0→1 (1 move), then 1→5 (4 moves), 5→8 (3), 8→9→10→11 (3) = 11 moves. Alternatively go LEFT from 0 → 11,10,9,8 then 5 (LEFT from 8 to 5 = 3) then... 1 requires going from 5 LEFT to 1 = 4 → total 1+1+1+1+3+4 = 11. Same. Hmm, max gap: the non-required runs: {2,3,4} (3), {6,7} (2). Largest gap = 3 → min moves = 12-3 = 9? Let me think: skipping the run {2,3,4}: path starts at 0, goes LEFT to 11,10,9,8, then LEFT to 5, then LEFT... to 1 we pass 4,3,2 (the skipped run) → total: 0→11(1),11→10(1),10→9(1),9→8(1),8→7(1),7→6(1),6→5(1),5→4(1),4→3(1),3→2(1),2→1(1) = 11 moves. Hmm, that's not saving. 
Better: start at 0, do tile0, then LEFT to 11,10,9,8 (4 moves), then RIGHT back through to 5 (3 moves: 8→9? no, 8 is done; 8→...→5 going RIGHT passes 9,10,11 (already done) = 4 moves to reach 0 then 5 = 5 moves; going LEFT from 8 to 5 = 3 moves ✓), then f
… [1658 caracteres más]

```python
def route_cost(start, req, N=12):
    import functools
    req=set(req)
    if not req: return 0,[]
    full=(1<<len(req))-1
    rl=sorted(req)
    @functools.lru_cache(None)
    def dp(pos, mask):
        if mask==full: return 0, ()
        best=(10**9,None)
        for i,t in enumerate(rl):
            if mask>>i & 1: continue
            d=abs(pos-t); d=min(d, N-d)
            nd1=(pos-t)%N; nd2=(t-pos)%N
            step=min(nd1,nd2)
            sub,cseq=dp(t, mask|1<<i)
            if step+sub<best[0]: best=(step+sub, (t,step,sub,cseq))
        return best
    return dp(start,0)
best,cur=solve6()
C=CYC(); S,P,Y=C['S'],C['P'],C['Y']
sg={'z1':S[3],'s1':S[2],'z2':S[4]}
e,f,c=Y[6],Y[1],Y[4]
cands=[]
for (tot,er,pr,det,pidx) in best[:120]:
    req=[i for i,d in det.items() if d!=0]
    mc,
… [419 caracteres más]
```

```text
Traceback (most recent call last):
  File "<python_tool>", line 30, in <module>
TypeError: '<' not supported between instances of 'dict' and 'dict'
```

## Paso 29 · nivel 6 · acción 310

**Pensamiento:**

Sort comparison error (dicts in the tuple). Fix by sorting with key=lambda x:(x[0],x[1],x[2]) and dropping det/pidx from the compared tuple... Actually the tuple contains det (a dict) at index 5; Python compares element-wise and if earlier elements tie it compares dicts → error. Use key.

```python
best,cur=solve6()
C=CYC(); S,P,Y=C['S'],C['P'],C['Y']
cands=[]
for (tot,er,pr,det,pidx) in best[:150]:
    req=[i for i,d in det.items() if d!=0]
    mc,_=route_cost(0,req)
    cands.append({'total':tot+mc,'press':tot,'moves':mc,'er':er,'pr':pr,'det':det,'pidx':pidx,'req':sorted(req)})
cands.sort(key=lambda x:(x['total'],x['press'],len(x['req'])))
for x in cands[:4]:
    print('TOTAL',x['total'],'press',x['press'],'moves',x['moves'],'er',x['er'],'pr',x['pr'],'pidx',x['pidx'])
    print('   req',x['req'],'deltas',{i:x['det'][i] for i in x['req']})
```

```text
TOTAL 22 press 13 moves 9 er ('s1', 'z1', 'z2') pr ('e', 'c', 'f') pidx (2, 4, 1)
   req [0, 1, 5, 8, 9, 10, 11] deltas {0: 3, 1: -2, 5: 1, 8: -1, 9: -2, 10: -2, 11: -2}
TOTAL 22 press 13 moves 9 er ('s1', 'z1', 'z2') pr ('e', 'c', 'f') pidx (2, 4, 3)
   req [0, 1, 5, 6, 8, 10, 11] deltas {0: 3, 1: 
… [353 caracteres más]
```

**Pensamiento:**

Best plan: total 22 actions (13 presses + 9 moves).
Option 1: er = ('s1','z1','z2') → row0 entry for s1, row1 entry for z1, row2 entry for z2.
pr = ('e','c','f') → row0 right pair = (P_e → e), row1 = (P_c → c), row2 = (P_f → f).
pidx (2,4,1) → P_e = P[2] = a6, P_c = P[4] = v, P_f = P[1] = x.
Check consistency: row0's right P tile must be P_e = a6 — currently tile2 = a6 ✓ (delta 0 ✓ matches). row0's Y tile → e: currently tile3 = e ✓ (delta 0 ✓ — because I cycled it back to b6? no wait, tile3's current glyph is '1000110000111110000110001' = e! Hmm, earlier the Y cycle enumeration ended at index 6 = e after 6 UPs ✓ so tile3 = e now, and the target is e ✓ 0 presses).
row1's right pair = (P_c → c) = (v → c): tile6 currently a6 → v: delta +2 ✓; tile7 = c ✓ 0.
row2's right pair = (P_f → f) = (x → f): tile10 = m1 → x: delta -2 ✓; tile11 = d6 → f: delta -2 ✓.
Left P tiles: row0 (entry s1) target (P_f,P_e) = (x, a6) — tile1 currently (m1,v): indices (6,4) → target (1,2): offset check: (4-6)%7=5, (2-1)=1 ✗... hmm, that doesn't match. Let me recheck: cost2 requires (b-a)%7 == (d-c)%7. cur (m1,v) = (6,4) → (4-6)%7 = 5. target (P_f,P_e) = (1,2) → (2-1)%7 = 1. 5 ≠ 1 → should have been rejected!
Hmm, so maybe my entry mapping is off. Let me recheck: entries={'z1':(Pe,Pf),'s1':(Pf,Pe),'z2':(Pc,Pc)}. With pidx (ie=2, jf=4, kc=1): Pe=P[2]=a6, Pf=P[4]=v, Pc=P[1]=x.
Oh! pidx is (ie,jf,kc) = (2,4,1) → ie=2 (Pe=a6), jf=4 (Pf=v), kc=1 (Pc=x). I misread. So:
- entry z1 = (a6, v), entry s1 = (v, a6), 
… [2255 caracteres más]

```python
def which6(pos):
    t=tiles6(); r0=pos[0]+3
    for i,x in enumerate(t):
        if x['r0']==r0:
            c0=x['c0']; c1=c0+7*len(x['gl'])-1
            if c0-1<=pos[2] and pos[3]<=c1+1: return i
    return None
def move6(i, cap=8):
    n=0
    for _ in range(cap+1):
        c=which6(wpos())
        if c==i: return n
        N=12; dr=(i-c)%N; dl=(c-i)%N
        a='RIGHT' if dr<=dl else 'LEFT'
        r=last(action([a])); n+=1
        if r.get('level_completed') or r.get('game_over') or r.get('run_complete'): return -n
    return -99
def exec_plan(plan):
    tot=0; msgs=[]
    for (tile, delta, target) in plan:
        m=move6(tile)
        if m<0: return 'DONE/DEAD during move', tot+abs(m)
        tot+=m
        a='UP' if delta>0 else 'DOWN'
        for k in range(abs(delta)):
        
… [689 caracteres más]
```

```text
[action] UP -> effect
[action] UP -> effect
[action] UP -> effect
[action] RIGHT -> effect
[action] DOWN -> effect
[action] DOWN -> effect
[action] LEFT -> effect
[action] LEFT -> effect
[action] DOWN -> effect
[action] DOWN -> effect
[action] LEFT -> effect
[action] DOWN -> effect
[action] DOWN -> 
… [770 caracteres más]
```