Free Printable Medium Sudoku with the Answer #5270 - Free Printable

Let’s solve this Sudoku puzzle step by step.

Sudoku rules:
- Each row must have numbers 1 to 9, no repeats.
- Each column must have numbers 1 to 9, no repeats.
- Each 3x3 box (there are 9 of them) must have numbers 1 to 9, no repeats.

We’ll go box by box, row by row, and look for places where only one number can fit.

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Start with Row 6:
It has: 9, 2, 4, _, 7, _, _, 1, _
Missing numbers: 3, 5, 6, 8

Look at Column 4 (the 4th column from left):
Already has: 5 (row2), 9 (row4), 2 (row5), 1 (row8) → so missing in col4: 3,4,6,7,8 — but we’re looking at row6, col4.

In row6, col4: what can go?
Row6 needs: 3,5,6,8
Col4 already has: 5,9,2,1 → so 5 is taken → possible: 3,6,8
Box (middle-left 3x3 box: rows 4-6, cols 1-3) already has: 8,9,2,4 → so 3,5,6,7 missing in that box. But row6 col4 is NOT in that box — it’s in middle-center box (rows 4-6, cols 4-6).

Middle-center box (rows 4-6, cols 4-6):
Has: 9,4,3 (row4); 2,_ ,_ (row5); _ ,7,_ (row6) → so far: 2,3,4,7,9 → missing: 1,5,6,8

So for row6, col4: must be in {3,6,8} (from row) AND in {1,5,6,8} (from box) → common: 6,8

Still two options. Let’s try another cell.

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Look at Row 9:
_, _, 6, _, _, 2, _, 4, 1
Missing: 3,5,7,8,9

Column 1: has 8,9,5 → so row9 col1 cannot be 5,8,9 → possible: 3,7

Column 2: has 7,6,2,4 → so row9 col2 cannot be 2,4,6,7 → possible: 3,5,8,9 — but row9 missing 3,5,7,8,9 → so col2 allows 3,5,8,9

But let’s look at bottom-left box (rows 7-9, cols 1-3):
Has: 5,4 (row8); 6 (row9 col3) → also row7 col1-3: all empty; row8 col1-3: 5,4,_; row9 col1-3: _,_,6

So bottom-left box has: 4,5,6 → missing: 1,2,3,7,8,9

Row9 col1: can’t be 5,8,9 (col1 has them) → so 3 or 7
Row9 col2: can’t be 2,4,6,7 (col2 has them) → so 3,5,8,9 — but row9 missing 3,5,7,8,9 → so 3,5,8,9 ok

Wait — let’s try a different approach. Look for “naked singles” — cells where only one number fits.

Check Row 7:
_, _, _, _, _, 4, _, _, 5
Missing: 1,2,3,6,7,8,9

Too many. Try Column 9:

Column 9:
Row1: ?
Row2: ?
Row3: ?
Row4: ?
Row5: 9
Row6: ?
Row7: 5
Row8: ?
Row9: 1

So col9 has: 9,5,1 → missing: 2,3,4,6,7,8

Now look at Row 5:
8, _, _, 2, _, _, _, _, 9
Missing: 1,3,4,5,6,7

Col2 for row5: col2 has 7,6,2,4 → so row5 col2 can’t be 2,4,6,7 → possible: 1,3,5,8,9 — but row5 missing 1,3,4,5,6,7 → so 1,3,5

Box (middle-left: rows4-6, cols1-3): has 8,9,2,4 → missing 1,3,5,6,7

So row5 col2: possible 1,3,5 — still multiple.

Let’s try Row 3:
_, 6, _, _, _, _, 7, _, _
Missing: 1,2,3,4,5,8,9

Col1: has 8,9,5 → so row3 col1 can’t be 5,8,9 → possible: 1,2,3,4

Box top-left (rows1-3, cols1-3): has 5 (r1c3), 7 (r2c2), 6 (r3c2) → so missing: 1,2,3,4,8,9

Row3 col1: possible 1,2,3,4 — matches.

Not helping yet.

Let’s look at Box 1 (top-left: rows1-3, cols1-3):

Cells:
r1c1: ?
r1c2: ?
r1c3: 5
r2c1: ?
r2c2: 7
r2c3: ?
r3c1: ?
r3c2: 6
r3c3: ?

Numbers present: 5,7,6 → missing: 1,2,3,4,8,9

Now look at Row 1:
_, _, 5, _, _, _, 1, _, _
Missing: 2,3,4,6,7,8,9

Col1 for r1c1: col1 has 8,9,5 → so r1c1 can’t be 5,8,9 → possible: 2,3,4,6,7

But in box1, missing 1,2,3,4,8,9 — so r1c1 can be 2,3,4 (since 6,7 not in box missing list? Wait no — box missing includes 1,2,3,4,8,9 — so 6 and 7 are already in box? No — box has 5,7,6 — so 6 and 7 are present → so r1c1 cannot be 6 or 7 anyway.

So r1c1: possible 2,3,4

Similarly, r1c2: col2 has 7,6,2,4 → so r1c2 can’t be 2,4,6,7 → possible: 1,3,5,8,9 — but row1 missing 2,3,4,6,7,8,9 — so 3,8,9

And box1 missing 1,2,3,4,8,9 — so r1c2 can be 3,8,9

Still messy.

Let’s try a different strategy: look for numbers that can only go in one place in a row/column/box.

Take number 1 in Row 1:
Row1 has 5 and 1 already? Wait — row1: positions: c3=5, c7=1 → so 1 is already in row1! So no need to place 1 in row1.

Row2: has 7,5,1,3,2 → so missing 4,6,8,9

Where can 4 go in row2?
Col1: has 8,9,5 → ok for 4
Col3: ?
Col5: ?
Col6: ?
Col8: ?
Col9: ?

Col3: let's see what's in col3:
r1:5, r2:?, r3:?, r4:?, r5:?, r6:4, r7:?, r8:?, r9:6 → so col3 has 5,4,6 → so r2c3 can't be 4,5,6 → so if we want to put 4 in row2, not in c3.

Col5: r1:?, r2:?, r3:?, r4:4, r5:?, r6:?, r7:?, r8:?, r9:? → r4c5=4 → so col5 has 4 → so r2c5 can't be 4.

Col6: r1:?, r2:1, r3:?, r4:3, r5:?, r6:?, r7:4, r8:9, r9:2 → so col6 has 1,3,4,9,2 → so r2c6 can't be 4.

Col8: r1:?, r2:2, r3:?, r4:?, r5:?, r6:1, r7:?, r8:?, r9:4 → so col8 has 2,1,4 → so r2c8 can't be 4.

Col9: r1:?, r2:?, r3:?, r4:?, r5:9, r6:?, r7:5, r8:?, r9:1 → so col9 has 9,5,1 → so r2c9 can be 4.

Col1: as above, can be 4.

So in row2, 4 can go in c1 or c9.

Now check box1 for row2 c1: box1 has 5,7,6 — no 4 yet — so ok.

Box3 (top-right: rows1-3, cols7-9): has r1c7=1, r2c7=3, r2c8=2, r3c7=7 — so has 1,2,3,7 — missing 4,5,6,8,9

So r2c9 is in box3 — can be 4.

So 4 can go in r2c1 or r2c9.

Not unique.

Let’s try number 8 in Row 3.

Row3: _,6,_,_,_,_,7,_,_ → missing 1,2,3,4,5,8,9

Where can 8 go?

Col1: has 8 (r5c1=8) → so r3c1 can't be 8.

Col3: has 5,4,6 → no 8 yet — possible.

Col4: has 5,9,2,1 → no 8 — possible.

Col5: has 4 — no 8 — possible.

Col6: has 1,3,4,9,2 — no 8 — possible.

Col8: has 2,1,4 — no 8 — possible.

Col9: has 9,5,1 — no 8 — possible.

Many options.

This is taking too long. Let me try to fill in obvious ones.

Look at Row 4:
_, _, _, 9, 4, 3, _, _, _
Missing: 1,2,5,6,7,8

Col1: has 8,9,5 → so r4c1 can't be 5,8,9 — possible: 1,2,6,7

Col2: has 7,6,2,4 — so r4c2 can't be 2,4,6,7 — possible: 1,3,5,8,9 — but row4 missing 1,2,5,6,7,8 — so 1,5,8

Col3: has 5,4,6 — so r4c3 can't be 4,5,6 — possible: 1,2,3,7,8,9 — row4 missing 1,2,5,6,7,8 — so 1,2,7,8

Box4 (middle-left: rows4-6, cols1-3): has r5c1=8, r6c1=9, r6c2=2, r6c3=4 — so has 8,9,2,4 — missing 1,3,5,6,7

So for r4c1: must be in {1,2,6,7} (from row and col) and in {1,3,5,6,7} (box) → common: 1,6,7

r4c2: {1,5,8} and box missing 1,3,5,6,7 → common: 1,5

r4c3: {1,2,7,8} and box missing 1,3,5,6,7 → common: 1,7

So r4c2 can be 1 or 5.

If I assume r4c2 = 1, then let's see.

But better to find a cell with only one possibility.

Let's look at Row 8:
5,4,_,1,_,9,_,_,_
Missing: 2,3,6,7,8

Col3: has 5,4,6 — so r8c3 can't be 4,5,6 — possible: 1,2,3,7,8,9 — row8 missing 2,3,6,7,8 — so 2,3,7,8

Col5: has 4 — no restriction yet.

Col7: has 1,3,7 — so r8c7 can't be 1,3,7 — possible: 2,4,5,6,8,9 — row8 missing 2,3,6,7,8 — so 2,6,8

Col8: has 2,1,4 — so r8c8 can't be 1,2,4 — possible: 3,5,6,7,8,9 — row8 missing 2,3,6,7,8 — so 3,6,7,8

Col9: has 9,5,1 — so r8c9 can't be 1,5,9 — possible: 2,3,4,6,7,8 — row8 missing 2,3,6,7,8 — so 2,3,6,7,8

Now, look at Box 7 (bottom-left: rows7-9, cols1-3):
Has r8c1=5, r8c2=4, r9c3=6 — so has 4,5,6 — missing 1,2,3,7,8,9

Row7 c1-3: all empty
Row8 c1-3: 5,4,_
Row9 c1-3: _,_,6

So for r8c3: in box7, can be 1,2,3,7,8,9 — but from earlier, for row8, r8c3 can be 2,3,7,8

Also, col3 has 5,4,6 — so r8c3 can be 2,3,7,8

No single choice.

Let's try a different number. Let's take number 3 in the entire grid.

Where can 3 go in Box 1 (top-left)?

Box1: cells r1c1, r1c2, r2c1, r2c3, r3c1, r3c3 — missing 1,2,3,4,8,9

Row1 has 5,1 — so r1c1 and r1c2 can't be 1 or 5 — but 1 is already in row1, so for 3, possible.

Col1 has 8,9,5 — so r1c1 can't be 8,9,5 — so for 3, ok.

Col2 has 7,6,2,4 — so r1c2 can't be 2,4,6,7 — so 3 is ok.

Row2 has 7,5,1,3,2 — oh! Row2 already has 3 in r2c7 — so in row2, 3 is already placed — so r2c1 and r2c3 can't be 3.

Row3 has 6,7 — no 3 yet.

So in box1, 3 can only go in r1c1, r1c2, or r3c1, r3c3.

But r3c1: col1 has 8,9,5 — no 3 — ok.

r3c3: col3 has 5,4,6 — no 3 — ok.

So still multiple.

Perhaps I should use a solver or think differently.

Let's look at Row 6 again: 9,2,4, _,7,_, _,1,_

Missing: 3,5,6,8

Now, col4: let's list all values in col4:

r1: ?
r2: 5
r3: ?
r4: 9
r5: 2
r6: ?
r7: ?
r8: 1
r9: ?

So col4 has: 5,9,2,1 — missing: 3,4,6,7,8

For r6c4: row6 missing 3,5,6,8 — col4 missing 3,4,6,7,8 — common: 3,6,8

Box5 (middle-center: rows4-6, cols4-6): has r4c4=9, r4c5=4, r4c6=3, r5c4=2, r6c5=7 — so has 2,3,4,7,9 — missing: 1,5,6,8

So r6c4 must be in {3,6,8} (row and col) and in {1,5,6,8} (box) — common: 6,8

So r6c4 is 6 or 8.

Similarly, r6c6: row6 missing 3,5,6,8 — col6: let's see col6 values:

r1: ?
r2: 1
r3: ?
r4: 3
r5: ?
r6: ?
r7: 4
r8: 9
r9: 2

So col6 has: 1,3,4,9,2 — missing: 5,6,7,8

For r6c6: row6 missing 3,5,6,8 — col6 missing 5,6,7,8 — common: 5,6,8

Box5 missing 1,5,6,8 — so r6c6 can be 5,6,8

So still multiple.

But notice that in box5, missing 1,5,6,8, and r6c4 and r6c6 are both in row6 and box5, and they can be 6,8 for c4 and 5,6,8 for c6.

Also, r5c5 and r5c6 are in box5.

Row5: 8, _, _, 2, _, _, _, _, 9 — so r5c5 and r5c6 are empty.

Col5: r1:?, r2:?, r3:?, r4:4, r5:?, r6:?, r7:?, r8:?, r9:? — only r4c5=4 known.

Col6: as above.

Perhaps for box5, the missing numbers 1,5,6,8 must go in r5c5, r5c6, r6c4, r6c6.

Row5 has 8,2,9 — so r5c5 and r5c6 can't be 8,2,9 — so for 1,5,6,8, they can be 1,5,6 (since 8 is in row5 already? Row5 has r5c1=8, so yes, 8 is in row5, so r5c5 and r5c6 can't be 8.

So in box5, 8 must go in r6c4 or r6c6.

From earlier, r6c4 can be 6 or 8, r6c6 can be 5,6,8 — so 8 can be in either.

But if 8 is in r6c4, then r6c6 can be 5 or 6, etc.

Let's look at Row 7: _, _, _, _, _, 4, _, _, 5

Missing: 1,2,3,6,7,8,9

Col6 has 4 in r7c6, so for other columns.

Notice that in Col 6, we have r2c6=1, r4c6=3, r7c6=4, r8c6=9, r9c6=2 — so missing 5,6,7,8

Row7 c6 is 4, so for r7c1 to r7c5 and r7c7 to r7c9.

Perhaps for number 6 in Col 6.

Col6 missing 5,6,7,8

Row1 c6: row1 has 5,1 — so can be 6? Yes.

Row3 c6: row3 has 6,7 — so can't be 6.

Row5 c6: row5 has 8,2,9 — so can be 6.

Row6 c6: can be 5,6,8 — so can be 6.

So 6 can be in r1c6, r5c6, r6c6.

Not unique.

I recall that in Sudoku, sometimes you can look for "hidden singles" — where a number can only go in one cell in a row/column/box.

Let's try number 1 in Box 2 (top-middle: rows1-3, cols4-6)

Box2: r1c4, r1c5, r1c6, r2c4=5, r2c5=?, r2c6=1, r3c4=?, r3c5=?, r3c6=?

So has 5,1 — missing 2,3,4,6,7,8,9

Row1 has 5,1 — so r1c4, r1c5, r1c6 can't be 1 or 5 — but 1 is already in box2 (r2c6=1), so for 1, it's already placed.

For number 2 in Box 2.

Where can 2 go?

Row1: has 5,1 — no 2 — so r1c4, r1c5, r1c6 can be 2.

Row2: has 7,5,1,3,2 — oh! r2c8=2, but that's not in this box — in box2, r2c4=5, r2c6=1, r2c5=? — and row2 has 2 in r2c8, which is in box3, so in row2, 2 is already used, so r2c5 can't be 2.

Row3: has 6,7 — no 2 — so r3c4, r3c5, r3c6 can be 2.

Col4: has 5,9,2,1 — r5c4=2, so col4 has 2 — so r1c4, r3c4 can't be 2.

Col5: has 4 — no 2 yet — so r1c5, r2c5, r3c5 can be 2 — but r2c5 can't be 2 because row2 has 2 already.

Col6: has 1,3,4,9,2 — r9c6=2, so col6 has 2 — so r1c6, r3c6 can't be 2.

So for box2, 2 can only go in r1c5 or r3c5.

Because:
- r1c4: col4 has 2 — no
- r1c6: col6 has 2 — no
- r2c5: row2 has 2 — no
- r3c4: col4 has 2 — no
- r3c6: col6 has 2 — no
- only r1c5 and r3c5 are possible.

So 2 is in r1c5 or r3c5.

Now, look at Row 1: if r1c5=2, or r3c5=2.

But let's see if we can eliminate one.

Col5: currently has r4c5=4 — and others unknown.

Row1: if r1c5=2, then ok.

Row3: if r3c5=2, then ok.

No conflict yet.

Perhaps later.

Let's try to fill in what we can.

Another idea: look at Row 9: _, _, 6, _, _, 2, _, 4, 1

Missing: 3,5,7,8,9

Col1: has 8,9,5 — so r9c1 can't be 5,8,9 — so must be 3 or 7

Col2: has 7,6,2,4 — so r9c2 can't be 2,4,6,7 — so must be 3,5,8,9 — but row9 missing 3,5,7,8,9 — so 3,5,8,9

But if r9c1 is 3 or 7, and r9c2 is 3,5,8,9, then if r9c1=3, r9c2 can't be 3, etc.

Also, Box 7 (bottom-left): has r8c1=5, r8c2=4, r9c3=6 — so missing 1,2,3,7,8,9

Row7 c1-3: all empty
Row8 c1-3: 5,4,_
Row9 c1-3: _,_,6

So for r9c1: can be 3,7 (from col1) and in box7 missing 1,2,3,7,8,9 — so 3,7

r9c2: can be 3,5,8,9 (from col2 and row) and in box7 missing 1,2,3,7,8,9 — so 3,8,9 (since 5 is in box7 already? r8c1=5, so 5 is in box7, so r9c2 can't be 5 — so 3,8,9

So r9c2: 3,8,9

Now, if r9c1=3, then r9c2 can't be 3, so 8 or 9

If r9c1=7, then r9c2 can be 3,8,9

Also, r8c3: in box7, can be 1,2,3,7,8,9 — but row8 has 5,4,1,9 — so missing 2,3,6,7,8 — so r8c3 can be 2,3,7,8

Col3 has 5,4,6 — so r8c3 can be 2,3,7,8

So no help.

Let's consider that in Box 7, the number 1 must go somewhere.

Box7 missing 1,2,3,7,8,9

Row7 c1-3: can be 1,2,3,7,8,9

Row8 c3: can be 2,3,7,8 (as above)

Row9 c1: 3,7; r9c2: 3,8,9

So 1 can only go in row7 c1, c2, or c3, because row8 and row9 have no room for 1 in this box? Row8 has r8c4=1, but that's not in this box — in box7, row8 has c1=5, c2=4, c3=? — so r8c3 could be 1? But row8 has 1 in r8c4, so row8 already has 1, so r8c3 can't be 1.

Row9 has r9c9=1, so row9 has 1, so r9c1 and r9c2 can't be 1.

Therefore, in box7, 1 can only go in row7 c1, c2, or c3.

So r7c1, r7c2, or r7c3 = 1.

Now, look at Row 7: _, _, _, _, _, 4, _, _, 5

So r7c6=4, r7c9=5

Col1: has 8,9,5 — so r7c1 can't be 5,8,9 — so for 1, ok.

Col2: has 7,6,2,4 — so r7c2 can't be 2,4,6,7 — so for 1, ok.

Col3: has 5,4,6 — so r7c3 can't be 4,5,6 — so for 1, ok.

So 1 can be in any of r7c1, r7c2, r7c3.

Not helpful yet.

Perhaps I need to guess or use a different method.

Let's look at the answer or think of a standard way.

I recall that in Sudoku, you can use the process of elimination.

Let me try to fill in r6c4.

Earlier, r6c4 can be 6 or 8.

Suppose r6c4 = 6.

Then in box5, 6 is placed, so missing 1,5,8 for r5c5, r5c6, r6c6.

Row6: if r6c4=6, then missing 3,5,8 for r6c6, r6c7, r6c9 (since r6c8=1)

Row6: 9,2,4,6,7,_, _,1,_ so missing 3,5,8 for c6,c7,c9.

Col6: missing 5,6,7,8 — but if r6c4=6, then for r6c6, can be 5,8 (since 6 is used in row6)

Box5: if r6c4=6, then missing 1,5,8 for r5c5, r5c6, r6c6.

Row5: 8, _, _, 2, _, _, _, _, 9 — so r5c5 and r5c6 can't be 8,2,9 — so can be 1,5,6 — but 6 is used in box5, so 1,5

So r5c5 and r5c6 are 1 and 5 in some order.

Then r6c6 must be 8 (since box5 missing 1,5,8, and r5c5 and r5c6 take 1 and 5, so r6c6=8)

Then row6: r6c6=8, so missing 3,5 for r6c7 and r6c9.

Col7: has r1c7=1, r2c7=3, r3c7=7, r6c7=?, r7c7=?, r8c7=?, r9c7=? — so has 1,3,7 — missing 2,4,5,6,8,9

For r6c7: can be 3 or 5 — but col7 has 3 already (r2c7=3), so r6c7 can't be 3 — so must be 5.

Then r6c9 = 3.

So if r6c4=6, then r6c6=8, r6c7=5, r6c9=3.

Now check if this works.

Row6: 9,2,4,6,7,8,5,1,3 — good, all unique.

Now, in box5, r6c4=6, r6c6=8, so missing 1,5 for r5c5 and r5c6.

Row5: 8, _, _, 2, _, _, _, _, 9 — so r5c5 and r5c6 are 1 and 5.

Col5: has r4c5=4 — so r5c5 can be 1 or 5.

Col6: has r2c6=1, r4c6=3, r7c6=4, r8c6=9, r9c6=2 — so has 1,3,4,9,2 — so r5c6 can't be 1 — so r5c6 must be 5, then r5c5=1.

So r5c5=1, r5c6=5.

Row5: 8, _, _, 2, 1, 5, _, _, 9

Missing for row5: 3,4,6,7 for c2,c3,c7,c8.

Col2: has 7,6,2,4 — so r5c2 can't be 2,4,6,7 — so must be 3 or 5 or 8 or 9 — but row5 missing 3,4,6,7 — so only 3 is common — so r5c2=3.

Then row5: 8,3, _, 2,1,5, _, _, 9

Missing 4,6,7 for c3,c7,c8.

Col3: has 5,4,6 — so r5c3 can't be 4,5,6 — so must be 7 (since 4 and 6 are forbidden, and 7 is in missing)

So r5c3=7.

Then row5: 8,3,7,2,1,5, _, _, 9

Missing 4,6 for c7,c8.

Col7: has 1,3,7 — so r5c7 can be 4 or 6.

Col8: has 2,1,4 — so r5c8 can be 6 or 4 — but col8 has 4? r9c8=4, so col8 has 4 — so r5c8 can't be 4 — so r5c8=6, then r5c7=4.

So row5: 8,3,7,2,1,5,4,6,9

Good.

Now, back to row6: we have 9,2,4,6,7,8,5,1,3

Now, let's do row4: _, _, _, 9,4,3, _, _, _

Missing: 1,2,5,6,7,8

But in box4 (middle-left), we have r5c1=8, r5c2=3, r5c3=7, r6c1=9, r6c2=2, r6c3=4 — so box4 has 8,3,7,9,2,4 — missing 1,5,6

So r4c1, r4c2, r4c3 must be 1,5,6 in some order.

Row4: r4c4=9, r4c5=4, r4c6=3 — so for c1,c2,c3: 1,5,6

Col1: has r5c1=8, r6c1=9, r8c1=5, r9c1=? — so has 8,9,5 — so r4c1 can't be 5,8,9 — so can be 1 or 6

Col2: has r5c2=3, r6c2=2, r8c2=4, r9c2=? — so has 3,2,4 — so r4c2 can be 1,5,6

Col3: has r5c3=7, r6c3=4, r8c3=?, r9c3=6 — so has 7,4,6 — so r4c3 can't be 4,6,7 — so can be 1 or 5

So for r4c1: 1 or 6

r4c2: 1,5,6

r4c3: 1 or 5

And they must be 1,5,6.

If r4c1=1, then r4c3 can be 5, r4c2=6

If r4c1=6, then r4c3 can be 1 or 5, r4c2=1 or 5.

But also, row4 missing 1,2,5,6,7,8 — but in c1,c2,c3 we have 1,5,6, so for c7,c8,c9: 2,7,8

Col7: has r1c7=1, r2c7=3, r3c7=7, r5c7=4, r6c7=5 — so has 1,3,7,4,5 — missing 2,6,8,9

For r4c7: can be 2,7,8 — but col7 has 7 already, so can be 2 or 8

Similarly, col8: has r2c8=2, r6c8=1, r9c8=4 — so has 2,1,4 — missing 3,5,6,7,8,9

For r4c8: can be 2,7,8 — but col8 has 2, so can be 7 or 8

Col9: has r5c9=9, r6c9=3, r7c9=5, r9c9=1 — so has 9,3,5,1 — missing 2,4,6,7,8

For r4c9: can be 2,7,8 — all possible.

Now, let's assume r4c1=1.

Then r4c3=5 (since if r4c1=1, r4c3 can be 5, and r4c2=6)

Then row4: 1,6,5,9,4,3, _, _, _

Missing 2,7,8 for c7,c8,c9.

Col7: can be 2 or 8 (since 7 is in col7 already)

Col8: can be 7 or 8

Col9: can be 2,7,8

If r4c7=2, then r4c8 and r4c9 are 7 and 8.

Col8 has 2,1,4 — so r4c8 can be 7 or 8.

Col9 has 9,3,5,1 — so r4c9 can be 2,7,8 — but if r4c7=2, then r4c9 can be 7 or 8.

Suppose r4c7=2, r4c8=7, r4c9=8.

Check col7: r4c7=2 — ok, not duplicate.

Col8: r4c8=7 — ok.

Col9: r4c9=8 — ok.

So possible.

If r4c1=6, then r4c2 and r4c3 are 1 and 5.

Say r4c2=1, r4c3=5.

Then row4: 6,1,5,9,4,3, _, _, _

Missing 2,7,8 for c7,c8,c9.

Same as above.

Or r4c2=5, r4c3=1.

Then row4: 6,5,1,9,4,3, _, _, _

Still missing 2,7,8.

So in both cases, c7,c8,c9 are 2,7,8.

Now, to decide, let's look at col1.

If r4c1=1, then col1 has r4c1=1, r5c1=8, r6c1=9, r8c1=5, r9c1=? — so has 1,8,9,5 — missing 2,3,4,6,7

Row9 c1: can be 3 or 7 (as earlier)

If r4c1=6, then col1 has 6,8,9,5 — missing 1,2,3,4,7 — and r9c1 can be 3 or 7.

Both possible.

Perhaps look at box4.

Box4 has r4c1, r4c2, r4c3 = 1,5,6 in some order, and we have r5c1=8, r5c2=3, r5c3=7, r6c1=9, r6c2=2, r6c3=4 — so all filled except r4c1,c2,c3.

So no issue.

Let's move to row3.

Row3: _,6,_,_,_,_,7,_,_

With our assumption, we have r6c4=6, etc.

Perhaps continue.

Another cell: let's do r3c7=7, already given.

Col7: has r1c7=1, r2c7=3, r3c7=7, r4c7=2 (assume), r5c7=4, r6c7=5 — so has 1,3,7,2,4,5 — missing 6,8,9

Row7 c7: row7 has r7c6=4, r7c9=5 — so missing 1,2,3,6,7,8,9 — so r7c7 can be 6,8,9

Similarly, r8c7, r9c7.

But let's see box6 (middle-right: rows4-6, cols7-9)

Box6: r4c7, r4c8, r4c9, r5c7=4, r5c8=6, r5c9=9, r6c7=5, r6c8=1, r6c9=3

So has 4,6,9,5,1,3 — missing 2,7,8

And r4c7, r4c8, r4c9 are 2,7,8 — perfect.

So whatever order, it's fine for now.

Assume r4c7=2, r4c8=7, r4c9=8.

Then box6 is complete.

Now, row4: 1,6,5,9,4,3,2,7,8 — if we chose r4c1=1, r4c2=6, r4c3=5.

Is this consistent?

Col1: r4c1=1, r5c1=8, r6c1=9, r8c1=5, r9c1=? — so has 1,8,9,5 — good.

Col2: r4c2=6, r5c2=3, r6c2=2, r8c2=4, r9c2=? — has 6,3,2,4 — good.

Col3: r4c3=5, r5c3=7, r6c3=4, r8c3=?, r9c3=6 — has 5,7,4,6 — good.

Now, let's do row3.

Row3: _,6,_,_,_,_,7,_,_

Missing: 1,2,3,4,5,8,9

Col1: has r4c1=1, r5c1=8, r6c1=9, r8c1=5 — so has 1,8,9,5 — so r3c1 can't be 1,5,8,9 — so can be 2,3,4

Col2: has r3c2=6, r4c2=6? Oh! r4c2=6, but r3c2=6 — same column! Conflict!

Oh no! I assumed r4c2=6, but r3c2=6 already — so col2 has two 6s — impossible.

So my assumption that r4c2=6 is wrong.

Therefore, in row4, r4c2 cannot be 6.

Earlier, for r4c1, r4c2, r4c3 = 1,5,6

And r4c2 cannot be 6 because col2 has r3c2=6.

So r4c2 must be 1 or 5.

Also, r4c1 can be 1 or 6, r4c3 can be 1 or 5.

But if r4c2=1, then r4c1 and r4c3 are 5 and 6.

If r4c2=5, then r4c1 and r4c3 are 1 and 6.

Now, col2 has r3c2=6, r4c2=?, r5c2=3, r6c2=2, r8c2=4 — so if r4c2=1 or 5, both ok since 1 and 5 not in col2 yet.

But if r4c2=1, then col2 has 6,1,3,2,4 — good.

If r4c2=5, then col2 has 6,5,3,2,4 — good.

Now, col1: if r4c1=6, then col1 has r4c1=6, r5c1=8, r6c1=9, r8c1=5 — so has 6,8,9,5 — good.

If r4c1=1, then has 1,8,9,5 — good.

Col3: if r4c3=5, then col3 has r4c3=5, r5c3=7, r6c3=4, r9c3=6 — so has 5,7,4,6 — good.

If r4c3=1, then has 1,7,4,6 — good.

But we have the constraint that r4c2 cannot be 6, which we already used.

Now, also, in row4, for c7,c8,c9: 2,7,8

And col7 has r3c7=7, so r4c7 cannot be 7 — so r4c7 must be 2 or 8.

Similarly, col8 has r2c8=2, so r4c8 cannot be 2 — so r4c8 must be 7 or 8.

Col9 has no restriction yet for 2,7,8.

So for r4c7: 2 or 8

r4c8: 7 or 8

r4c9: 2,7,8

But they must be 2,7,8 all different.

So possible combinations:

- r4c7=2, r4c8=7, r4c9=8

- r4c7=2, r4c8=8, r4c9=7

- r4c7=8, r4c8=7, r4c9=2

- r4c7=8, r4c8=8 — invalid

So three options.

Now, back to r4c1,c2,c3.

Suppose we set r4c2=1.

Then r4c1 and r4c3 are 5 and 6.

If r4c1=5, but col1 has r8c1=5 — so r4c1 can't be 5 — so r4c1 must be 6, r4c3=5.

So r4c1=6, r4c2=1, r4c3=5.

Then row4: 6,1,5,9,4,3, _, _, _

Then for c7,c8,c9: 2,7,8

With r4c7=2 or 8, r4c8=7 or 8, etc.

Also, col1: r4c1=6, r5c1=8, r6c1=9, r8c1=5 — so has 6,8,9,5 — good.

Col2: r4c2=1, r3c2=6, r5c2=3, r6c2=2, r8c2=4 — so has 1,6,3,2,4 — good.

Col3: r4c3=5, r5c3=7, r6c3=4, r9c3=6 — so has 5,7,4,6 — good.

Now, for r4c7, r4c8, r4c9.

Suppose r4c7=2, r4c8=7, r4c9=8.

Then col7: r4c7=2, and r3c7=7, r5c7=4, r6c7=5, r1c7=1, r2c7=3 — so has 2,7,4,5,1,3 — good.

Col8: r4c8=7, r2c8=2, r6c8=1, r9c8=4 — so has 7,2,1,4 — good.

Col9: r4c9=8, r5c9=9, r6c9=3, r7c9=5, r9c9=1 — so has 8,9,3,5,1 — good.

So this works.

So row4: 6,1,5,9,4,3,2,7,8

Now, let's do row3.

Row3: _,6,_,_,_,_,7,_,_

Missing: 1,2,3,4,5,8,9

Col1: has r4c1=6, r5c1=8, r6c1=9, r8c1=5 — so has 6,8,9,5 — so r3c1 can't be 5,6,8,9 — so can be 1,2,3,4

Col2: has r3c2=6, r4c2=1, r5c2=3, r6c2=2, r8c2=4 — so has 6,1,3,2,4 — so r3c2 is already 6, so for other cells, but r3c1 is in col1, not col2.

For r3c1: can be 1,2,3,4

Col3: has r4c3=5, r5c3=7, r6c3=4, r9c3=6 — so has 5,7,4,6 — so r3c3 can't be 4,5,6,7 — so can be 1,2,3,8,9

But row3 missing 1,2,3,4,5,8,9 — so for r3c3: 1,2,3,8,9

Now, box1 (top-left): has r1c3=5, r2c2=7, r3c2=6, and r4c1=6? No, r4c1 is in box4, not box1.

Box1: rows1-3, cols1-3: r1c1, r1c2, r1c3=5, r2c1, r2c2=7, r2c3, r3c1, r3c2=6, r3c3

So has 5,7,6 — missing 1,2,3,4,8,9

Row2: has r2c2=7, r2c4=5, r2c6=1, r2c7=3, r2c8=2 — so missing 4,6,8,9 for r2c1, r2c3, r2c5, r2c9

But in box1, for r2c1 and r2c3, can be 4,8,9 (since 6 is in r3c2, but 6 is already in box1? r3c2=6, so 6 is in box1, so r2c1 and r2c3 can't be 6 — so for row2 missing 4,6,8,9, but 6 is in box1, so r2c1 and r2c3 can be 4,8,9

Similarly, r1c1, r1c2, r3c1, r3c3 can be 1,2,3,4,8,9 minus what's used.

Also, col1: r3c1 can be 1,2,3,4

Col3: r3c3 can be 1,2,3,8,9

Now, let's look at number 4 in box1.

Where can 4 go?

Row1: has 5,1 — no 4 — so r1c1, r1c2 can be 4.

Row2: has 7,5,1,3,2 — no 4 — so r2c1, r2c3 can be 4.

Row3: has 6,7 — no 4 — so r3c1, r3c3 can be 4.

Col1: has r4c1=6, r5c1=8, r6c1=9, r8c1=5 — so no 4 yet — so r1c1, r2c1, r3c1 can be 4.

Col2: has r3c2=6, r4c2=1, r5c2=3, r6c2=2, r8c2=4 — oh! r8c2=4, so col2 has 4 — so r1c2, r2c2, r3c2 can't be 4 — but r2c2=7, r3c2=6, so for r1c2, can't be 4.

So in col2, 4 is already in r8c2, so r1c2 cannot be 4.

Therefore, in box1, 4 can only go in r1c1, r2c1, r2c3, r3c1, r3c3 — but not r1c2.

Also, col3: has r4c3=5, r5c3=7, r6c3=4 — oh! r6c3=4, so col3 has 4 — so r1c3, r2c3, r3c3 can't be 4 — but r1c3=5, so for r2c3 and r3c3, can't be 4.

So in box1, 4 can only go in r1c1, r2c1, or r3c1.

Because:
- r1c2: col2 has 4 — no
- r2c3: col3 has 4 — no
- r3c3: col3 has 4 — no
- only r1c1, r2c1, r3c1 are possible.

So 4 is in col1, in rows 1,2, or 3.

Now, col1 has r4c1=6, r5c1=8, r6c1=9, r8c1=5 — so missing 1,2,3,4,7

So r1c1, r2c1, r3c1 can be 1,2,3,4,7

But for 4, it must be in one of them.

Now, let's look at row1.

Row1: _, _, 5, _, _, _, 1, _, _

Missing: 2,3,4,6,7,8,9

Col1: can be 1,2,3,4,7 — but row1 missing 2,3,4,6,7,8,9 — so for r1c1: 2,3,4,7

Similarly, r1c2: col2 has r3c2=6, r4c2=1, r5c2=3, r6c2=2, r8c2=4 — so has 6,1,3,2,4 — so r1c2 can't be 1,2,3,4,6 — so can be 5,7,8,9 — but row1 has 5,1 — so can be 7,8,9

And in box1, for r1c2, can be 7,8,9 — but 7 is in r2c2, so in box1, 7 is already there, so r1c2 can't be 7 — so 8 or 9.

So r1c2 = 8 or 9.

Now, perhaps for number 8 in row1.

But let's try to fill r3c1.

Suppose we set r3c1 = 4.

Then in box1, 4 is placed.

Then col1 has r3c1=4, r4c1=6, r5c1=8, r6c1=9, r8c1=5 — so has 4,6,8,9,5 — good.

Then for row3: r3c1=4, r3c2=6, r3c7=7 — so missing 1,2,3,5,8,9 for c3,c4,c5,c6,c8,c9

Col3: has r4c3=5, r5c3=7, r6c3=4, r9c3=6 — so has 5,7,4,6 — so r3c3 can't be 4,5,6,7 — so can be 1,2,3,8,9

So ok.

Now, let's look at box2 (top-middle): r1c4, r1c5, r1c6, r2c4=5, r2c5=?, r2c6=1, r3c4=?, r3c5=?, r3c6=?

Has 5,1 — missing 2,3,4,6,7,8,9

Row1: r1c4, r1c5, r1c6 can be 2,3,4,6,7,8,9 minus what's in col.

Col4: has r2c4=5, r4c4=9, r5c4=2, r6c4=6 — so has 5,9,2,6 — missing 1,3,4,7,8

For r1c4: row1 missing 2,3,4,6,7,8,9 — col4 missing 1,3,4,7,8 — common: 3,4,7,8

Similarly, r1c5: col5 has r4c5=4, r5c5=1, r6c5=7 — so has 4,1,7 — missing 2,3,5,6,8,9

Row1 missing 2,3,4,6,7,8,9 — common: 2,3,6,8,9

r1c6: col6 has r2c6=1, r4c6=3, r5c6=5, r6c6=8, r7c6=4, r8c6=9, r9c6=2 — so has 1,3,5,8,4,9,2 — missing 6,7

So r1c6 can be 6 or 7.

Row1 missing 2,3,4,6,7,8,9 — so 6 or 7.

So r1c6 = 6 or 7.

Now, in box2, for r1c6, can be 6 or 7.

Also, r2c5: row2 missing 4,6,8,9 (since has 7,5,1,3,2) — col5 has 4,1,7 — so r2c5 can be 6,8,9 (since 4 is in col5)

Box2 missing 2,3,4,6,7,8,9 — so r2c5 can be 6,8,9

Similarly, r3c4, r3c5, r3c6.

But let's assume r1c6 = 6.

Then row1: r1c6=6, so missing 2,3,4,7,8,9 for c1,c2,c4,c5,c8,c9

Col6 has r1c6=6, and has 1,3,5,8,4,9,2 — so all except 6,7 — but we put 6, so now col6 has 1,2,3,4,5,6,8,9 — missing 7 — so r3c6 must be 7.

Then row3: r3c6=7, but r3c7=7 — same row! Conflict.

So r1c6 cannot be 6.

Therefore, r1c6 = 7.

Then col6 has r1c6=7, and has 1,3,5,8,4,9,2 — so missing 6 — so r3c6 must be 6.

Then row3: r3c6=6, but r3c2=6 — same row! Conflict again.

What's wrong?

Col6: before placing, has r2c6=1, r4c6=3, r5c6=5, r6c6=8, r7c6=4, r8c6=9, r9c6=2 — so values: 1,3,5,8,4,9,2 — so missing 6,7

So r1c6 and r3c6 must be 6 and 7.

If r1c6=6, then r3c6=7, but row3 has r3c7=7, so r3c6=7 would be duplicate in row3.

If r1c6=7, then r3c6=6, but row3 has r3c2=6, so duplicate in row3.

Oh no! Contradiction.

So my initial assumption that r6c4=6 must be wrong.

Therefore, r6c4 cannot be 6; it must be 8.

Let's start over with r6c4=8.

So back to row6: 9,2,4, _,7,_, _,1,_

With r6c4=8.

Then in box5, r6c4=8, so missing 1,5,6 for r5c5, r5c6, r6c6.

Row6: with r6c4=8, missing 3,5,6 for r6c6, r6c7, r6c9 (since r6c8=1)

Col6: missing 5,6,7,8 — but r6c4=8, so for r6c6, can be 5,6 (since 8 is used in row6)

Box5: missing 1,5,6 for r5c5, r5c6, r6c6.

Row5: 8, _, _, 2, _, _, _, _, 9 — so r5c5 and r5c6 can't be 8,2,9 — so can be 1,5,6

So r5c5, r5c6, r6c6 are 1,5,6 in some order.

Row6: r6c6 can be 5 or 6 (from above)

Also, col6: has r2c6=1, r4c6=3, r7c6=4, r8c6=9, r9c6=2 — so has 1,3,4,9,2 — missing 5,6,7,8

So r6c6 can be 5 or 6.

Suppose r6c6 = 5.

Then in box5, r6c6=5, so missing 1,6 for r5c5, r5c6.

Row5: r5c5 and r5c6 are 1 and 6.

Col5: has r4c5=4 — so r5c5 can be 1 or 6.

Col6: has r6c6=5, and has 1,3,4,9,2 — so has 1,2,3,4,5,9 — missing 6,7,8

So r5c6 can be 6 (since 1 is in col6 already? r2c6=1, so col6 has 1, so r5c6 can't be 1 — so r5c6=6, then r5c5=1.

So r5c5=1, r5c6=6.

Then row5: 8, _, _, 2, 1, 6, _, _, 9

Missing 3,4,5,7 for c2,c3,c7,c8.

Col2: has 7,6,2,4 — so r5c2 can't be 2,4,6,7 — so must be 3 or 5 or 8 or 9 — but row5 missing 3,4,5,7 — so 3 or 5

Col3: has 5,4,6 — so r5c3 can't be 4,5,6 — so must be 3 or 7 or 8 or 9 — row5 missing 3,4,5,7 — so 3 or 7

So r5c2 = 3 or 5, r5c3 = 3 or 7

If r5c2=3, then r5c3=7 (since 3 is used)

If r5c2=5, then r5c3=3 or 7.

But also, box4: has r5c1=8, r6c1=9, r6c2=2, r6c3=4 — so has 8,9,2,4 — missing 1,3,5,6,7

So r5c2 and r5c3 must be from 1,3,5,6,7 — but row5 has r5c5=1, r5c6=6, so 1 and 6 are used, so for r5c2 and r5c3, can be 3,5,7

So r5c2 = 3 or 5, r5c3 = 3 or 7, and they must be different.

So possible: r5c2=3, r5c3=7 or r5c2=5, r5c3=3 or r5c2=5, r5c3=7

Now, col2: if r5c2=3, then col2 has r5c2=3, r3c2=6, r6c2=2, r8c2=4 — so has 3,6,2,4 — good.

If r5c2=5, then has 5,6,2,4 — good.

Col3: if r5c3=7, then has r5c3=7, r6c3=4, r9c3=6 — so has 7,4,6 — good.

If r5c3=3, then has 3,4,6 — good.

So all possible.

Let's choose r5c2=3, r5c3=7.

Then row5: 8,3,7,2,1,6, _, _, 9

Missing 4,5 for c7,c8.

Col7: has r1c7=1, r2c7=3, r3c7=7, r6c7=? — so has 1,3,7 — missing 2,4,5,6,8,9

For r5c7: can be 4 or 5.

Col8: has r2c8=2, r6c8=1, r9c8=4 — so has 2,1,4 — missing 3,5,6,7,8,9

For r5c8: can be 4 or 5 — but col8 has 4, so r5c8 can't be 4 — so r5c8=5, then r5c7=4.

So row5: 8,3,7,2,1,6,4,5,9

Good.

Now, row6: with r6c4=8, r6c6=5, so missing 3,6 for r6c7, r6c9 (since r6c8=1)

Col7: has r5c7=4, r3c7=7, r2c7=3, r1c7=1 — so has 4,7,3,1 — missing 2,5,6,8,9

For r6c7: can be 3 or 6 — but col7 has 3 already, so r6c7 can't be 3 — so must be 6.

Then r6c9 = 3.

So row6: 9,2,4,8,7,5,6,1,3

Good.

Now, box5: r4c4=9, r4c5=4, r4c6=3, r5c4=2, r5c5=1, r5c6=6, r6c4=8, r6c5=7, r6c6=5 — so has 9,4,3,2,1,6,8,7,5 — all good.

Now, row4: _, _, _, 9,4,3, _, _, _

Missing: 1,2,5,6,7,8

Box4: has r5c1=8, r5c2=3, r5c3=7, r6c1=9, r6c2=2, r6c3=4 — so has 8,3,7,9,2,4 — missing 1,5,6

So r4c1, r4c2, r4c3 = 1,5,6

Col1: has r5c1=8, r6c1=9, r8c1=5 — so has 8,9,5 — so r4c1 can't be 5,8,9 — so can be 1 or 6

Col2: has r5c2=3, r6c2=2, r8c2=4, r3c2=6 — so has 3,2,4,6 — so r4c2 can't be 2,3,4,6 — so can be 1,5,7,8,9 — but for box4, must be 1,5,6 — so 1 or 5

Col3: has r5c3=7, r6c3=4, r9c3=6 — so has 7,4,6 — so r4c3 can't be 4,6,7 — so can be 1,2,3,5,8,9 — for box4, 1,5,6 — so 1 or 5

So r4c1: 1 or 6

r4c2: 1 or 5

r4c3: 1 or 5

And they must be 1,5,6 all different.

So if r4c1=6, then r4c2 and r4c3 are 1 and 5.

If r4c1=1, then r4c2 and r4c3 are 5 and 6, but r4c2 can't be 6 (because col2 has r3c2=6), and r4c3 can't be 6 (col3 has r9c3=6), so if r4c1=1, then r4c2 and r4c3 must be 5 and 6, but neither can be 6, contradiction.

Therefore, r4c1 cannot be 1; must be 6.

Then r4c2 and r4c3 are 1 and 5.

Now, r4c2 can be 1 or 5, r4c3 can be 1 or 5.

But col2 has r3c2=6, r5c2=3, r6c2=2, r8c2=4 — so no 1 or 5 yet, so both ok.

Col3 has r5c3=7, r6c3=4, r9c3=6 — so no 1 or 5 yet, so both ok.

So say r4c2=1, r4c3=5.

Then row4: 6,1,5,9,4,3, _, _, _

Missing 2,7,8 for c7,c8,c9.

As before, r4c7 can be 2 or 8 (since col7 has r3c7=7), r4c8 can be 7 or 8 (col8 has r2c8=2), r4c9 can be 2,7,8.

And must be 2,7,8.

So possible: r4c7=2, r4c8=7, r4c9=8 or r4c7=2, r4c8=8, r4c9=7 or r4c7=8, r4c8=7, r4c9=2

Now, col7: has r1c7=1, r2c7=3, r3c7=7, r5c7=4, r6c7=6 — so has 1,3,7,4,6 — missing 2,5,8,9

So r4c7 can be 2 or 8.

Col8: has r2c8=2, r5c8=5, r6c8=1, r9c8=4 — so has 2,5,1,4 — missing 3,6,7,8,9

So r4c8 can be 7 or 8.

Col9: has r5c9=9, r6c9=3, r7c9=5, r9c9=1 — so has 9,3,5,1 — missing 2,4,6,7,8

So r4c9 can be 2,7,8.

Now, suppose r4c7=2, r4c8=7, r4c9=8.

Then col7: r4c7=2 — ok.

Col8: r4c8=7 — ok.

Col9: r4c9=8 — ok.

So row4: 6,1,5,9,4,3,2,7,8

Good.

Now, let's do row3.

Row3: _,6,_,_,_,_,7,_,_

Missing: 1,2,3,4,5,8,9

Col1: has r4c1=6, r5c1=8, r6c1=9, r8c1=5 — so has 6,8,9,5 — so r3c1 can't be 5,6,8,9 — so can be 1,2,3,4

Col2: has r3c2=6, r4c2=1, r5c2=3, r6c2=2, r8c2=4 — so has 6,1,3,2,4 — so r3c2 is 6, so for other, but r3c1 is in col1.

For r3c1: 1,2,3,4

Col3: has r4c3=5, r5c3=7, r6c3=4, r9c3=6 — so has 5,7,4,6 — so r3c3 can't be 4,5,6,7 — so can be 1,2,3,8,9

Box1: has r1c3=5, r2c2=7, r3c2=6, r4c1=6? No, r4c1 is in box4.

Box1: r1c1, r1c2, r1c3=5, r2c1, r2c2=7, r2c3, r3c1, r3c2=6, r3c3

Has 5,7,6 — missing 1,2,3,4,8,9

Row2: has r2c2=7, r2c4=5, r2c6=1, r2c7=3, r2c8=2 — so missing 4,6,8,9 for r2c1, r2c3, r2c5, r2c9

But in box1, for r2c1 and r2c3, can be 4,8,9 (since 6 is in r3c2, so 6 is in box1, so r2c1 and r2c3 can't be 6 — so 4,8,9

Similarly, r1c1, r1c2, r3c1, r3c3 can be 1,2,3,4,8,9

Now, col1: r3c1 can be 1,2,3,4

Suppose r3c1 = 4.

Then col1 has r3c1=4, r4c1=6, r5c1=8, r6c1=9, r8c1=5 — so has 4,6,8,9,5 — good.

Then for row3: r3c1=4, r3c2=6, r3c7=7 — so missing 1,2,3,5,8,9 for c3,c4,c5,c6,c8,c9

Col3: r3c3 can be 1,2,3,8,9

Now, box2: r1c4, r1c5, r1c6, r2c4=5, r2c5=?, r2c6=1, r3c4=?, r3c5=?, r3c6=?

Has 5,1 — missing 2,3,4,6,7,8,9

Col4: has r2c4=5, r4c4=9, r5c4=2, r6c4=8 — so has 5,9,2,8 — missing 1,3,4,6,7

For r1c4: row1 missing 2,3,4,6,7,8,9 — col4 missing 1,3,4,6,7 — common: 3,4,6,7

r1c5: col5 has r4c5=4, r5c5=1, r6c5=7 — so has 4,1,7 — missing 2,3,5,6,8,9

Row1 missing 2,3,4,6,7,8,9 — common: 2,3,6,8,9

r1c6: col6 has r2c6=1, r4c6=3, r5c6=6, r6c6=5, r7c6=4,