Qwen3.5 9B on Steady grid 2
No readable grid: 0 of 53 blank cells right. All of Qwen3.5 9B's answers.
RESULTNo readable grid
CELLS RIGHT0 / 53
SAIDn/a
MODEL TIME4.5 min
COSTlocal, no list price
TOKENS IN / OUT1,831 / 3,000
1
2
4
9
.5
6
7
8
.3
.3
6
8
2
.7
1
.9
5
4
.9
.5
7
.4
8
3
6
.2
1
8
.4
5
7
1
.2
3
9
6
.6
.3
1
8
.9
5
2
.4
.7
2
7
9
.6
3
4
5
.1
.8
4
.9
2
3
6
.8
1
.7
.5
5
8
.6
1
.2
7
4
3
.9
.7
1
3
5
.4
9
8
6
2
5 given 5 filled correctly 53 wrong, with the right digit 3 left empty
The end of its reply
R2: 3 . . | . 7 . | 9 . .
R3: 9 5 . | 4 . . | . 2 .
R4: . 4 . | . . 2 | . . .
R5: 6 3 . | . 9 . | . 4 7
R6: . . . | 6 . . | . 1 8
R7: . 9 . | . . 8 | . 7 5
R8: . . 6 | . 2 . | . . 9
R9: 7 . . | . 4 . | . . .
**Step-by-Step Solution:**
1. **Box 3 (Top Right):**
* Cells: R1C7-9, R2C7-9, R3C7-9.
* Known: R1C9=3, R2C7=9, R3C8=2.
* Missing: 1, 4, 5, 6, 7, 8.
* R2C8, R2C9 cannot be 7 (col 8 has no 7 yet? No, R7C8 is 7). So R2C8 $\neq$ 7. R2C9 $\neq$ 7.
* R1C8: Row 1 has 5, 3. Col 8 has 2, 7 (R7), 4 (R8? No R8C6 is 4). Wait, R9C6 is 4. R8C6 is 2. R5C8 is 4? No, R5C8 is empty. R5C9 is 7.
* Let's look at **Col 9**: Values present: 3(R1), 7(R5), 8(R6), 5(R7), 9(R8). Missing: 1, 2, 4, 6.
* R3C9: Row 3 has 2. So $\neq$ 2.
* R4C9: Row 4 has 4. So $\neq$ 4. Also has 2 (R3C8 is 2, but R4C8?). Wait, R3C8=2. Row 4 has no 2 yet? Ah, R3C8 is 2. That's row 3. Row 4 has 2 at C6? No, R4C6 is empty. The puzzle says `..2 ...` for row 4?
* Let's re-read Row 4: `.4. ..2 ...`. So R4C2=4, R4C6=2.
* Row 4 has 2 and 4. So R4C9 cannot be 2 or 4.
* Row 3 has 2 (at C8). So R3C9 $\neq$ 2.
* Therefore, **R9C9 must be 2**.
* Col 9 remaining missing: 1, 4, 6.
* R3C9: Row 3 has 9,5,4,2. So $\neq$ 4. Must be 1 or 6.
* R4C9: Row 4 has 4, 2. So $\neq$ 4. Must be 1 or 6.
* Wait, R3 has 4 at C4. R4 has 4 at C2. R5 has no 4? R5C8 is 4? No, R5 is `63. .9. .47`. So R5C2=3, R5C5=9, R5C8=4, R5C9=7.
* Col 9 needs 4. Where can it go?
* R3: Has 4. No.
* R4: Has 4. No.
* R9: Row 9 has 7, 4 (at C6). So no 4.
* Wait, I missed something. Is there a place for 4 in Col 9?
* Let's re-check the given numbers for Row 9: `7.. .4. ..9`. So R9C1=7, R9C6=4, R9C9=9.
* Ah! The puzzle string for Row 8 ends with `..9`. So R8C9=9.
* The puzzle string for Row 9 ends with `..9`? No.
* Let's parse the input string again strictly.
* Input: `7.. .4. ..9` (Row 9).
* C1: 7
* C6: 4
* C9: 9? No, `..9` usually means C7=., C8=., C9=9. Or C7=C8=?, C9=?.
* Format is `digit` or `.`. Spaces separate chunks of 3? No, spaces are just delimiters in the prompt description, but the "one line" string removes them.
* Let's look at the "one line" string: `....5...33...7.9..95.4...2..4...2...63..9..47...6...18.9...8.75..6.2...97...4....`
* Let's split into 9 chars per row.
* Row 1: `....5...3` -> . . . . 5 . . . 3 (Matches prompt)
* Row 2: `3...7.9..` -> 3 . . . . 7 . 9 . . (Wait, prompt says `3.. .7. 9..`. My split: 3,.,.,.,.,7,.,9,.,. -> 10 chars? No.)
* Let's count: `3` (1) `.` (2) `.` (3) `.` (4) `5`?? No.
* Prompt Row 2: `3.. .7. 9..`.
* Chars: 3, ., ., space, ., 7, .,Grid steady-1 of set v1, 28 givens. Puzzle as one line: ....5...33...7.9..95.4...2..4...2...63..9..47...6...18.9...8.75..6.2...97...4....