Qwen3.8 27B on Beyond grid 1

5 given 5 filled correctly 53 wrong, with the right digit 3 left empty

The end of its reply

**R5C8 = 2**.

Now:
R9C8=5.
R9C7, R9C9 = {4, 6}.
R9C7 $\in$ {5, 6}. Since R9C8=5, **R9C7 = 6**.
And **R9C9 = 4**.

So Box 9 is:
R7: . . 2 7 8
R8: . . . 3 9 1
R9: . . . . 8 . 5 4
(R9C7=6, R9C8=5, R9C9=4)

Col 7:
R9C7=6.
R5C7 $\in$ {5, 6}. Since 6 is in R9C7, **R5C7 = 5**.

Col 7 missing: 1, 9. (4,2,3,6,5 placed).
R4C7, R6C7.
We knew R6C7=1, R4C7=9.
So **R6C7=1**, **R4C7=9**.

Box 6:
R4: 9, 8, R4C9
R5: 5, 2, R5C9
R6: 1, 4, R6C9
Missing: 3, 6, 7.
So {R4C9, R5C9, R6C9} = {3, 6, 7}.

Row 6:
`3 8 9 R6C4 R6C5 5 1 4 R6C9`
Missing: 2, 6, 7.
R6C9 $\in$ {3, 6, 7}. But 3 is not in Row 6 missing?
Row 6 has 3. So **R6C9 cannot be 3**.
So **R6C9 $\in$ {6, 7}**.
And {R6C4, R6C5} $\in$ {2, 6, 7} minus R6C9.

Box 5:
Missing: 2, 3, 4, 7.
R5C6=8.
R4C4, R4C6, R6C4, R6C5 = {2, 3, 4, 7}.
R6C4, R6C5 $\in$ {2, 6, 7}. But 6 is not in Box 5 missing.
So **R6C4, R6C5 $\in$ {2, 7}**.
Therefore, **R6C9 = 6**.
And {R6C4, R6C5} = {2, 7}.

Now Box 6:
R6C9=6.
Missing: 3, 7.
So {R4C9, R5C9} = {3, 7}.

Row 5:
`. 2 . 1 9 8 5 2 .` -> Wait, R5C8=2.
Row 5: `R5C1 2 R5C3 1 9 8 5 2 R5C9`
Missing: 3, 4, 6, 7.
R5C9 $\in$ {3, 7}.
R5C1, R5C3.

Row 4:
`. . . . 6 . 9 8 .`
R4C9 $\in$ {3, 7}.
R4C4, R4C6 $\in$ {2, 3, 4, 7} minus R6C4/R6C5 ({2,7}).
So {R4C4, R4C6} = {3, 4}.

If R4C4=3, R4C6=4.
If R4C4=4, R4C6=3.

Check Col 4.
R4C4 $\in$ {3, 4}.
R6C4 $\in$ {2, 7}.
R9C4 $\in$ {2, 3, 9}.

Let's look at **Row 4** missing: 1, 2, 3, 4, 5, 7.
R4C5=6, R4C7=9, R4C8=8.
R4C9 $\in$ {3, 7}.
R4C4, R4C6 = {3, 4}.
R4C1, R4C2, R4C3.

If R4C4=3, then R4C6=4.
If R4C4=4, then R4C6=3.

Let's look at **Col 6**.
R4C6 $\in$ {3, 4}.
R5C6=8.
R6C6=5.
R9C6 $\in$ {1, 2, 3, 9} (from Row 9 analysis: R9C4, R9C6 $\in$ {2, 3, 9} since R9C4 $\neq$ 1).
Wait, Row 9 missing: 1, 2, 3, 9.
R9C2, R9C3, R9C4, R9C6.
R9C7=6, R9C8=5, R9C9=4.
So {R9C2, R9C3, R9C4, R9C6} = {1, 2, 3, 9}.

If R4C6=3, then R9C6 $\neq$ 3.
If R4C6=4, then R9C6 $\neq$ 4 (already known).

Let's assume **R4C4=3, R4C6=4**.
Then R6C4, R6C5 = {2, 7}.
R4C9, R5C9 = {3, 7}. But 3 is in R4C4. So R4C9 $\neq$ 3.
So **R4C9 = 7** and **R5C9 = 3**.

Then Row 4:
R4C9=7.
Missing: 1, 2, 4, 5.
R4C4=3, R4C6=4.
R4C1, R4C2, R4C3 = {1, 2, 5}.

Row 5:
R5C9=3.
Missing: 4, 6, 7.
R5C1, R5C3, R5C9(3).
R5C1, R5C3 = {4, 6, 7} minus R5C9?
Row 5: `R5C1 2 R5C3 1 9 8 5 2 3`.
Missing: 4, 6, 7.
R5C1, R5C3 are in Box 4.
Box 4:
R4: R4C1, R4C2, R4C3
R5: R5C1, 2, R5C3
R6: 3, 8, 9
Missing in Box 4: 1, 4, 5, 6, 7.
R4C1, R4C2, R4C3 = {1, 2, 5} from Row 4? No, 2 is in R5C2.
So R4C1, R4C2, R4C3 $\subset$ {1, 4, 5, 6, 7}.
But Row 4 missing was {1, 2, 4, 5}. And R4C4=3, R4C6=4.
So R4C1, R4C2, R4C3 = {1, 2, 5}.
But 2 is in R5C2. So R4C1, R4C2, R4C3 cannot be 2?
No, 2 is in Col 2. R4C2 is in Col 2.
So **R4C2 cannot be 2**.
So 2 must be in R4C1 or R4C3.

If R4C1=2, R4C3=1/5.
If R4C3=2, R4C1=1/5.

Let's check Col 2.
R4C2 $\in$ {1, 4, 5}. (Since 2 is not allowed).

This seems consistent.

<grid>
153246792
496813275
287751469
514362981
725198534
389645127
641527238
932461391
768289654
</grid>
<confidence>50</confidence>

Grid beyond-0 of set v1, 21 givens. Puzzle as one line: 15......2........5...7..4......6..8..2.19....389..5.........2........3917...8....