Evil Sudoku Online #1600053 - Live Sudoku - 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 fill in what we can logically deduce.

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First, let’s label the grid for clarity:

Rows: R1 to R9 (top to bottom)
Columns: C1 to C9 (left to right)
Boxes: B1 to B9 (top-left to bottom-right)

Given clues:

R2C7=7, R2C8=2, R2C9=5
R3C2=2, R3C4=1, R3C9=8
R4C2=4, R4C3=8, R4C4=3, R4C6=6, R4C8=1
R5C3=2
R6C5=5, R6C6=8, R6C7=4, R6C9=6
R7C2=5, R7C4=9, R7C6=3, R7C9=1
R8C2=3, R8C3=7
R9C2=8, R9C3=1, R9C4=5, R9C5=4, R9C8=3

Now let’s start filling.

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Look at Box 1 (R1-R3, C1-C3):

Cells:
R1C1, R1C2, R1C3
R2C1, R2C2, R2C3
R3C1, R3C2=2, R3C3

Missing numbers in Box 1: 1,3,4,5,6,7,8,9 — but wait, only 2 is given. Actually, let’s list what’s missing per row/column/box as we go.

Better approach: Look for rows/columns/boxes with many filled cells.

Look at Row 9:
R9C2=8, R9C3=1, R9C4=5, R9C5=4, R9C8=3 → So missing: 2,6,7,9

Check columns for those positions:

R9C1: Column 1 has? Let’s see what’s in C1 so far:
R3C4=1 → not C1
Actually, let’s build column by column.

Maybe better to look at Box 7 (bottom-left: R7-R9, C1-C3)

Box 7 cells:
R7C1, R7C2=5, R7C3
R8C1, R8C2=3, R8C3=7
R9C1, R9C2=8, R9C3=1

So in Box 7, we have: 5,3,7,8,1 → missing: 2,4,6,9

Now check Row 7: R7C2=5, R7C4=9, R7C6=3, R7C9=1 → so R7C1 and R7C3 need to be from {2,4,6,7,8} but constrained by box.

In Box 7, available numbers: 2,4,6,9

Row 7 already has 5,9,3,1 → so R7C1 and R7C3 cannot be 9 (already in row), so they must be from {2,4,6}

Similarly, Row 8: R8C2=3, R8C3=7 → so R8C1 must be from Box 7 missing: 2,4,6,9 minus what’s used.

Also, Column 1: let’s see what’s there.

Actually, let’s try a different spot.

Look at Row 6:
R6C5=5, R6C6=8, R6C7=4, R6C9=6 → missing: 1,2,3,7,9

Now look at Box 5 (center: R4-R6, C4-C6)

Box 5:
R4C4=3, R4C5=?, R4C6=6
R5C4=?, R5C5=?, R5C6=?
R6C4=?, R6C5=5, R6C6=8

So in Box 5, we have: 3,6,5,8 → missing: 1,2,4,7,9

But Row 6 in Box 5: R6C4, R6C5=5, R6C6=8 → so R6C4 must be one of missing in row 6: 1,2,3,7,9 — but 3 is already in Box 5 (R4C4=3), so R6C4 ≠3. Also, 5 and 8 are taken.

Wait, perhaps focus on where only one number fits.

Let’s look at R5C3=2 (given). That’s in Box 4 (R4-R6, C1-C3)

Box 4:
R4C1, R4C2=4, R4C3=8
R5C1, R5C2, R5C3=2
R6C1, R6C2, R6C3

So in Box 4, we have: 4,8,2 → missing: 1,3,5,6,7,9

Now Row 5: only R5C3=2 is given. Not helpful yet.

Another idea: Look at Column 9.

Column 9:
R2C9=5
R3C9=8
R6C9=6
R7C9=1
So filled: 5,8,6,1 → missing: 2,3,4,7,9

Now Row 1 to 9 for C9:

R1C9, R4C9, R5C9, R8C9, R9C9 are empty.

But R9C8=3, so R9C9 cannot be 3.

Let’s try Box 3 (top-right: R1-R3, C7-C9)

Box 3:
R1C7, R1C8, R1C9
R2C7=7, R2C8=2, R2C9=5
R3C7, R3C8, R3C9=8

So in Box 3, we have: 7,2,5,8 → missing: 1,3,4,6,9

Now Row 3: R3C2=2, R3C4=1, R3C9=8 → so R3C7 and R3C8 must be from missing in row 3: 3,4,5,6,7,9 — but 5 and 7 are in Box 3 already? Wait, Box 3 has 7,2,5,8 — so 5 and 7 are used, so R3C7 and R3C8 cannot be 5 or 7.

Also, Row 3 missing: 3,4,5,6,7,9 minus what's placed: R3C2=2, R3C4=1, R3C9=8 → so missing: 3,4,5,6,7,9

But in Box 3, 5 and 7 are already present (R2C9=5, R2C7=7), so R3C7 and R3C8 cannot be 5 or 7. So possible for R3C7,R3C8: 3,4,6,9

Now look at Column 7:
R2C7=7
R6C7=4
So filled: 7,4 → missing: 1,2,3,5,6,8,9

R3C7 is in Col 7, so cannot be 7 or 4.

This is getting messy. Let’s try to find a cell with only one possibility.

Look at R4C5. It’s in Row 4, which has: R4C2=4, R4C3=8, R4C4=3, R4C6=6, R4C8=1 → so missing in Row 4: 2,5,7,9

Column 5: let’s see what’s there.

Col 5:
R6C5=5
R7C5=?
R9C5=4
So filled: 5,4 → missing: 1,2,3,6,7,8,9

Box 5 (R4-R6,C4-C6): has R4C4=3, R4C6=6, R6C5=5, R6C6=8 → so missing: 1,2,4,7,9

For R4C5: must be in Row 4 missing: 2,5,7,9 — but 5 is in Box 5 (R6C5=5), so R4C5 cannot be 5. So possible: 2,7,9

Also, Col 5 has 5 and 4, so 2,7,9 are ok for now.

Not unique.

Let’s try R5C5.

Row 5: only R5C3=2 known.

Col 5: has R6C5=5, R9C5=4

Box 5: missing 1,2,4,7,9 — but 2 is in R5C3, which is not in Box 5, so 2 is still available for Box 5? No, R5C3 is in Box 4, so Box 5 doesn't have 2 yet.

Box 5 cells: R4C4=3, R4C5=?, R4C6=6, R5C4=?, R5C5=?, R5C6=?, R6C4=?, R6C5=5, R6C6=8

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

R5C5 is in Row 5, which has only 2 so far (at R5C3), so R5C5 can be 1,4,7,9 (since 2 is already in row)

Also, Col 5 has 5 and 4, so R5C5 cannot be 4 or 5. So possible: 1,7,9

Still not unique.

Perhaps look at Box 9 (bottom-right: R7-R9, C7-C9)

Box 9:
R7C7, R7C8, R7C9=1
R8C7, R8C8, R8C9
R9C7, R9C8=3, R9C9

Given: R7C9=1, R9C8=3

So missing in Box 9: 2,4,5,6,7,8,9

Row 7: R7C2=5, R7C4=9, R7C6=3, R7C9=1 → so R7C7 and R7C8 must be from {2,4,6,7,8} (missing in row)

But in Box 9, 1 and 3 are used, so R7C7 and R7C8 can be 2,4,6,7,8

Column 7: R2C7=7, R6C7=4 → so R7C7 cannot be 7 or 4. So for R7C7: possible 2,6,8

Similarly, Column 8: R2C8=2, R4C8=1, R9C8=3 → so filled: 2,1,3 → missing: 4,5,6,7,8,9

R7C8 is in Col 8, so cannot be 2,1,3 — which is fine.

But let’s see if we can find a number that must go somewhere.

Another approach: Look at number 1 in each box.

Box 1: where can 1 go?
Cells: R1C1, R1C2, R1C3, R2C1, R2C2, R2C3, R3C1, R3C2=2, R3C3

Row 3 has R3C4=1, so 1 is in Row 3, so in Box 1, 1 cannot be in R3C1 or R3C3. So 1 must be in R1 or R2 of Box 1.

Col 1: no 1 yet. Col 2: R3C2=2, R4C2=4, R7C2=5, R8C2=3, R9C2=8 — no 1. Col 3: R4C3=8, R5C3=2, R8C3=7, R9C3=1 — oh! R9C3=1, so Col 3 has 1. Therefore, in Box 1, Col 3 cannot have 1, so R1C3 and R2C3 cannot be 1.

So in Box 1, 1 can only be in R1C1, R1C2, R2C1, R2C2

But Row 2: R2C7=7, R2C8=2, R2C9=5 — no 1 yet, so possible.

Not helping.

Let’s try to fill in obvious ones.

Look at R3C1. Row 3: has 2,1,8 — so missing 3,4,5,6,7,9

Col 1: let's list all known in Col 1.

From given:

R3C4=1 — not C1

Actually, let's list all given values by position:

I think I need to write out the grid.

Let me create a 9x9 grid with given numbers:

Row 1: . . . | . . . | . . .
Row 2: . . . | . . . | 7 2 5
Row 3: . 2 . | 1 . . | . . 8
------+-------+------
Row 4: . 4 8 | 3 . 6 | . 1 .
Row 5: . . 2 | . . . | . . .
Row 6: . . . | . 5 8 | 4 . 6
------+-------+------
Row 7: . 5 . | 9 . 3 | . . 1
Row 8: . 3 7 | . . . | . . .
Row 9: . 8 1 | 5 4 . | . 3 .

Now, let's look at Box 7 (R7-R9, C1-C3):

Cells:
R7C1, R7C2=5, R7C3
R8C1, R8C2=3, R8C3=7
R9C1, R9C2=8, R9C3=1

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

Now, Row 7: has 5,9,3,1 → so R7C1 and R7C3 cannot be 9 (already in row), so they must be from {2,4,6}

Similarly, Row 8: has 3,7 → so R8C1 can be 2,4,6,9

Row 9: has 8,1,5,4,3 → so R9C1 can be 2,6,7,9 — but in Box 7, 7 is already present (R8C3=7), so R9C1 cannot be 7. Also, 1,3,4,5,8 are used in row or box, so for R9C1, possible: 2,6,9

But in Box 7, missing 2,4,6,9, and R9C1 is in it.

Now, Column 1: let's see what's there.

Col 1: R1C1, R2C1, R3C1, R4C1, R5C1, R6C1, R7C1, R8C1, R9C1

Known: none yet, except we know some constraints.

Notice that in Box 7, the number 4 must go somewhere. Where can 4 go in Box 7?

Possible cells: R7C1, R7C3, R8C1, R9C1

But Row 7 has no 4 yet, Row 8 has no 4, Row 9 has R9C5=4, so Row 9 already has 4, so R9C1 cannot be 4.

So 4 in Box 7 must be in R7C1, R7C3, or R8C1.

Now, Column 1: if we can find where 4 can go.

Look at Row 4: R4C2=4, so Col 2 has 4, but that's not Col 1.

Col 1 has no 4 yet.

But let's see if there's a row or column that forces it.

Another idea: Look at R6C4.

Row 6: R6C5=5, R6C6=8, R6C7=4, R6C9=6 → so missing: 1,2,3,7,9

Col 4: R3C4=1, R4C4=3, R7C4=9, R9C4=5 → so filled: 1,3,9,5 → missing: 2,4,6,7,8

Box 5: R4C4=3, R4C6=6, R6C5=5, R6C6=8 → so for R6C4, it must be in Row 6 missing: 1,2,3,7,9 — but Col 4 has 1,3,9,5, so R6C4 cannot be 1,3,9. Also, 5 is in row, so not 5. So possible: 2,7

Also, Box 5 has 3,6,5,8, so 2 and 7 are missing, so R6C4 can be 2 or 7.

Now, look at R4C5.

Row 4: R4C2=4, R4C3=8, R4C4=3, R4C6=6, R4C8=1 → missing: 2,5,7,9

Col 5: R6C5=5, R9C5=4 → so filled: 5,4 → missing: 1,2,3,6,7,8,9

Box 5: missing 1,2,4,7,9 — but 4 is in R9C5, which is not in Box 5, so 4 is still available for Box 5? No, R9C5 is in Box 8, so Box 5 doesn't have 4 yet.

Box 5 cells: R4C4=3, R4C5=?, R4C6=6, R5C4=?, R5C5=?, R5C6=?, R6C4=?, R6C5=5, R6C6=8

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

For R4C5: must be in Row 4 missing: 2,5,7,9 — but 5 is in Box 5 (R6C5=5), so R4C5 cannot be 5. So possible: 2,7,9

Also, Col 5 has 5 and 4, so 2,7,9 are ok.

Now, if we look at R5C5, it could be 1,7,9 as earlier.

Let's try to see if there's a cell with only one option.

Look at R1C5.

Row 1: all empty.

Col 5: has R6C5=5, R9C5=4

Box 2 (R1-R3, C4-C6):
R1C4, R1C5, R1C6
R2C4, R2C5, R2C6
R3C4=1, R3C5, R3C6

So in Box 2, only R3C4=1 is given.

Not helpful.

Another spot: Look at Box 6 (R4-R6, C7-C9)

Box 6:
R4C7, R4C8=1, R4C9
R5C7, R5C8, R5C9
R6C7=4, R6C8, R6C9=6

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

Row 4: has 4,8,3,6,1 — so R4C7 and R4C9 must be from 2,5,7,9 (missing in row)

But in Box 6, 1,4,6 are used, so R4C7 and R4C9 can be 2,5,7,9

Col 7: R2C7=7, R6C7=4 → so R4C7 cannot be 7 or 4. So for R4C7: possible 2,5,9

Similarly, Col 9: R2C9=5, R3C9=8, R6C9=6, R7C9=1 → so filled: 5,8,6,1 → missing: 2,3,4,7,9

R4C9 is in Col 9, so cannot be 5,8,6,1 — which is fine.

But let's see R6C8.

Row 6: missing 1,2,3,7,9 (since has 5,8,4,6)

Col 8: R2C8=2, R4C8=1, R9C8=3 → so filled: 2,1,3 → missing: 4,5,6,7,8,9

Box 6: missing 2,3,5,7,8,9 — but 2,3 are in Col 8, so for R6C8, it must be in Row 6 missing: 1,2,3,7,9 — but Col 8 has 2,1,3, so R6C8 cannot be 1,2,3. So possible: 7,9

Also, Box 6 has 1,4,6, so 7,9 are missing, so R6C8 can be 7 or 9.

Now, if we look at R5C8.

Row 5: only R5C3=2 known.

Col 8: has 2,1,3 — so R5C8 can be 4,5,6,7,8,9

Box 6: missing 2,3,5,7,8,9 — 2 and 3 are in Col 8, so for R5C8, possible 5,7,8,9

Not unique.

Let's try to use the fact that in Row 9, we have R9C2=8, R9C3=1, R9C4=5, R9C5=4, R9C8=3 — so missing: 2,6,7,9

Col 1: for R9C1, possible 2,6,7,9 — but in Box 7, 7 is already present (R8C3=7), so R9C1 cannot be 7. So R9C1: 2,6,9

Col 6: R4C6=6, R6C6=8, R7C6=3 — so filled: 6,8,3 → missing: 1,2,4,5,7,9

R9C6 is in Row 9, so must be from 2,6,7,9 — but Col 6 has 6, so R9C6 cannot be 6. Also, 3,8 are in col, but not in row missing. So R9C6: 2,7,9

But in Box 8 (R7-R9, C4-C6):
R7C4=9, R7C5=?, R7C6=3
R8C4=?, R8C5=?, R8C6=?
R9C4=5, R9C5=4, R9C6=?

So in Box 8, we have: 9,3,5,4 → missing: 1,2,6,7,8

For R9C6: must be in Row 9 missing: 2,6,7,9 — but 9 is in Box 8 (R7C4=9), so R9C6 cannot be 9. Also, 6 is in Col 6 (R4C6=6), so R9C6 cannot be 6. So R9C6: 2,7

Similarly, for R9C1: 2,6,9 — but 6 and 9 may be restricted.

Let's look at R9C7.

Row 9: missing 2,6,7,9

Col 7: R2C7=7, R6C7=4 — so filled: 7,4 → missing: 1,2,3,5,6,8,9

Box 9: R7C9=1, R9C8=3 — so missing 2,4,5,6,7,8,9

For R9C7: must be in Row 9 missing: 2,6,7,9 — and Col 7 allows 2,6,9 (not 7 because R2C7=7), so R9C7: 2,6,9

But in Box 9, 7 is not yet placed, but R9C7 cannot be 7 because Col 7 has 7.

So R9C7: 2,6,9

Now, notice that in Row 9, the missing numbers are 2,6,7,9, and they must go in C1, C6, C7, C9

R9C9 is also empty.

Col 9: has R2C9=5, R3C9=8, R6C9=6, R7C9=1 — so filled: 5,8,6,1 → missing: 2,3,4,7,9

R9C9 must be from Row 9 missing: 2,6,7,9 — but Col 9 has 6, so R9C9 cannot be 6. Also, 5,8,1 are in col, but not in row missing. So R9C9: 2,7,9

But in Box 9, R9C8=3, R7C9=1, so for R9C9, it can be 2,7,9

Now, let's consider that in Row 9, 7 must go somewhere. Where can 7 go in Row 9?

Possible cells: R9C1, R9C6, R9C7, R9C9

But R9C1: in Box 7, which has R8C3=7, so R9C1 cannot be 7.

R9C6: in Box 8, which has R7C4=9, etc., but no 7 yet, and Col 6 has no 7, so possible.

R9C7: Col 7 has R2C7=7, so cannot be 7.

R9C9: Col 9 has no 7, Box 9 has no 7, so possible.

So 7 in Row 9 can be in R9C6 or R9C9.

Similarly, for 2,6,9.

This is taking too long. Let's try to look for a different strategy.

Let's look at number 1 in the entire grid.

Where can 1 go in Box 1?

As before, R3C4=1, so Row 3 has 1, so in Box 1, 1 cannot be in R3C1 or R3C3.

Col 3 has R9C3=1, so Col 3 has 1, so in Box 1, 1 cannot be in R1C3 or R2C3.

So in Box 1, 1 can only be in R1C1, R1C2, R2C1, R2C2

Now, Row 2: R2C7=7, R2C8=2, R2C9=5 — no 1, so possible.

But let's see if there's a column that has 1.

Col 1: no 1 yet.

Col 2: R3C2=2, R4C2=4, R7C2=5, R8C2=3, R9C2=8 — no 1.

So 1 can be in any of those four cells.

Not helpful.

Let's try Box 4 (R4-R6, C1-C3)

Box 4:
R4C1, R4C2=4, R4C3=8
R5C1, R5C2, R5C3=2
R6C1, R6C2, R6C3

Numbers: 4,8,2 → missing: 1,3,5,6,7,9

Row 4: has 4,8,3,6,1 — so R4C1 must be from 2,5,7,9 — but 2 is in Box 4 (R5C3=2), so R4C1 cannot be 2. Also, 3,6,1 are in row, so for R4C1: 5,7,9

Col 1: no constraints yet.

Similarly, R5C1: Row 5 has only 2, so can be 1,3,4,5,6,7,8,9 — but in Box 4, missing 1,3,5,6,7,9, so possible.

Let's look at R6C1.

Row 6: missing 1,2,3,7,9

Col 1: no info.

Box 4: missing 1,3,5,6,7,9

So R6C1 can be 1,3,7,9 (since 2 is in row, and 5,6 may be possible, but let's see)

Perhaps we can use the given that in R5C3=2, and see what else.

Another idea: Look at Column 3.

Col 3: R4C3=8, R5C3=2, R8C3=7, R9C3=1 — so filled: 8,2,7,1 → missing: 3,4,5,6,9

Cells: R1C3, R2C3, R3C3, R6C3, R7C3

Row 3: R3C2=2, R3C4=1, R3C9=8 — so R3C3 can be 3,4,5,6,7,9 — but Col 3 has 8,2,7,1, so R3C3 cannot be 7, so possible 3,4,5,6,9

But in Box 1, we have to place numbers.

Let's try to assume or look for a breakthrough.

Let's look at R4C5 again.

Perhaps I can search online or recall that this is a specific puzzle, but since I'm supposed to solve it, let's continue.

Let's consider Box 5 again.

Box 5: R4C4=3, R4C6=6, R6C5=5, R6C6=8 — so missing 1,2,4,7,9

Now, Row 5: R5C3=2, so in Row 5, 2 is used, so for R5C4, R5C5, R5C6, they cannot be 2.

So in Box 5, 2 must be in R4C5 or R6C4.

R4C5: Row 4 has no 2 yet, Col 5 has no 2, so possible.

R6C4: Row 6 has no 2, Col 4 has R3C4=1, R4C4=3, R7C4=9, R9C4=5 — no 2, so possible.

So 2 in Box 5 is in R4C5 or R6C4.

Now, if we look at Row 4, missing 2,5,7,9 — so R4C5 can be 2,5,7,9

But if 2 is in R4C5, then good.

Otherwise, in R6C4.

But let's see if there's a conflict.

Suppose R4C5 = 2.

Then Row 4 has 2, so missing 5,7,9 for other cells.

Col 5 has 2, so ok.

Box 5 has 2, so missing 1,4,7,9

Then R6C4 must be from Row 6 missing: 1,2,3,7,9 — but 2 is used in Box 5, so R6C4 cannot be 2, so 1,3,7,9 — but Col 4 has 1,3,9,5, so R6C4 cannot be 1,3,9, so only 7 left.

So if R4C5=2, then R6C4=7.

Is that possible? Let's check.

If R6C4=7, then Row 6 has 7, so missing 1,2,3,9 for other cells.

Col 4 has 7, so ok.

Box 5 has 7, so missing 1,4,9

Then for R5C4, R5C5, R5C6, they must be 1,4,9 in some order.

Row 5 has only 2, so ok.

Col 4 has R3C4=1, R4C4=3, R6C4=7, R7C4=9, R9C4=5 — so filled: 1,3,7,9,5 — missing: 2,4,6,8

So for R5C4, it must be in Col 4 missing: 2,4,6,8 — but in Box 5, for R5C4, it must be 1,4,9 — intersection is 4.

So R5C4 = 4.

Then in Box 5, R5C4=4, so missing 1,9 for R5C5 and R5C6.

Col 5: has R6C5=5, R9C5=4, and if R4C5=2, then Col 5 has 2,5,4 — so missing 1,3,6,7,8,9

R5C5 must be 1 or 9.

Similarly, Col 6: R4C6=6, R6C6=8, R7C6=3 — so filled: 6,8,3 — missing: 1,2,4,5,7,9

R5C6 must be 1 or 9.

So possible.

Now, let's see if this works with other constraints.

So assume R4C5=2, R6C4=7, R5C4=4.

Then in Box 5, R5C5 and R5C6 are 1 and 9.

Now, Row 5: R5C3=2, R5C4=4, so missing 1,3,5,6,7,8,9 — but R5C5 and R5C6 are 1 and 9, so ok.

Now, look at R5C5.

If R5C5=1, then Col 5 has 1,2,4,5 — ok.

If R5C5=9, Col 5 has 9,2,4,5 — ok.

Similarly for R5C6.

But let's see Box 2 or other.

Perhaps we can proceed.

Another cell: R4C7.

Row 4: if R4C5=2, then missing 5,7,9 for R4C1, R4C7, R4C9

Col 7: R2C7=7, R6C7=4 — so R4C7 cannot be 7, so must be 5 or 9.

Similarly, Col 9: R2C9=5, R3C9=8, R6C9=6, R7C9=1 — so R4C9 cannot be 5, so must be 7 or 9 — but if R4C5=2, then R4C9 can be 7 or 9, but Col 9 has no 7 or 9 yet, so ok.

But if R4C7 is 5 or 9, and R4C9 is 7 or 9, and R4C1 is the remaining.

Also, in Box 6, we have R4C8=1, R6C7=4, R6C9=6, so for R4C7 and R4C9, they are in Box 6.

Box 6 missing 2,3,5,7,8,9 — but 2 is in R4C5, which is not in Box 6, so 2 is still missing in Box 6? No, R4C5 is in Box 5, so Box 6 doesn't have 2 yet.

Box 6: R4C7, R4C8=1, R4C9, R5C7, R5C8, R5C9, R6C7=4, R6C8, R6C9=6

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

If R4C7=5 or 9, R4C9=7 or 9, etc.

Suppose R4C7=5, then Col 7 has 5,7,4 — ok.

Then R4C9 must be 7 or 9, but if R4C7=5, then for Row 4, missing 7,9 for R4C1 and R4C9.

Col 9 has no 7 or 9, so ok.

But let's see if we can find a contradiction or confirmation.

Perhaps look at R6C8.

Earlier, R6C8 can be 7 or 9.

If R6C4=7, then Row 6 has 7, so R6C8 cannot be 7, so must be 9.

So R6C8=9.

Then in Box 6, R6C8=9, so missing 2,3,5,7,8

Row 6: has R6C4=7, R6C5=5, R6C6=8, R6C7=4, R6C8=9, R6C9=6 — so missing 1,2,3 for R6C1, R6C2, R6C3

Col 8: R2C8=2, R4C8=1, R6C8=9, R9C8=3 — so filled: 2,1,9,3 — missing: 4,5,6,7,8

R6C8=9, so ok.

Now, for R6C1, R6C2, R6C3: must be 1,2,3

Box 4: R4C2=4, R4C3=8, R5C3=2, so in Box 4, missing 1,3,5,6,7,9 — but R6C1, R6C2, R6C3 are in Box 4, and must be 1,2,3 — but 2 is already in Box 4 (R5C3=2), so R6C1, R6C2, R6C3 cannot be 2. Contradiction!

Oh! So if R6C4=7, then R6C1, R6C2, R6C3 must be 1,2,3, but 2 is already in Box 4, so impossible.

Therefore, our assumption that R4C5=2 is wrong.

So R4C5 cannot be 2.

Therefore, in Box 5, 2 must be in R6C4.

So R6C4=2.

Then, since R6C4=2, Row 6 has 2, so missing 1,3,7,9 for other cells.

Col 4 has R3C4=1, R4C4=3, R6C4=2, R7C4=9, R9C4=5 — so filled: 1,3,2,9,5 — missing: 4,6,7,8

So for R6C4=2, ok.

Now, in Box 5, R6C4=2, so missing 1,4,7,9 for R4C5, R5C4, R5C5, R5C6

Row 4: missing 2,5,7,9 — but 2 is now in R6C4, not in Row 4, so Row 4 still missing 2,5,7,9 for R4C1, R4C5, R4C7, R4C9

But R4C5 is in Box 5, and must be from 1,4,7,9 — but Row 4 missing 2,5,7,9, so intersection is 7,9.

So R4C5 = 7 or 9.

Similarly, Col 5: has R6C5=5, R9C5=4 — so if R4C5=7 or 9, ok.

Now, let's see R5C4.

Col 4: missing 4,6,7,8 — and R5C4 is in Col 4, so must be 4,6,7,8

But in Box 5, R5C4 must be from 1,4,7,9 — intersection is 4,7.

So R5C4 = 4 or 7.

Similarly, R5C5 and R5C6 must be from 1,4,7,9 minus what's used.

Now, let's look at Row 5: R5C3=2, so missing 1,3,4,5,6,7,8,9

But in Box 5, R5C4, R5C5, R5C6 are three cells, must be three of 1,4,7,9.

So possible.

Now, let's consider R4C5.

If R4C5=7, then Row 4 has 7, so missing 2,5,9 for R4C1, R4C7, R4C9

Col 5 has 7,5,4 — ok.

If R4C5=9, then Row 4 has 9, missing 2,5,7 for others.

Now, let's see if we can find which one.

Look at Col 5.

Col 5: R4C5=?, R5C5=?, R6C5=5, R7C5=?, R8C5=?, R9C5=4

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

R4C5 is 7 or 9.

Suppose R4C5=7.

Then Col 5 has 7,5,4 — missing 1,2,3,6,8,9

R5C5 must be from Box 5 missing: 1,4,9 (since R4C5=7, R6C4=2, so Box 5 has 3,6,5,8,2,7 — so missing 1,4,9)

So R5C5 = 1,4,9

But Col 5 has 7,5,4, so if R5C5=4, conflict with R9C5=4? No, R9C5=4, so Col 5 has 4, so R5C5 cannot be 4.

So R5C5 = 1 or 9.

Similarly, if R4C5=9, then Box 5 has 3,6,5,8,2,9 — missing 1,4,7

R5C5 = 1,4,7 — but Col 5 has 9,5,4, so if R5C5=4, conflict, so R5C5 = 1 or 7.

Now, let's look at R7C5.

Row 7: R7C2=5, R7C4=9, R7C6=3, R7C9=1 — so missing 2,4,6,7,8

Col 5: has R6C5=5, R9C5=4, and R4C5=7 or 9, so if R4C5=7, then Col 5 has 7,5,4 — so R7C5 cannot be 4,5,7 — so from Row 7 missing: 2,6,8 (since 4,7 are in col or row)

Row 7 missing 2,4,6,7,8 — Col 5 has 5,4, and say 7, so R7C5 cannot be 4,5,7 — so possible 2,6,8

Similarly, if R4C5=9, Col 5 has 9,5,4, so R7C5 cannot be 4,5,9 — so from Row 7 missing: 2,6,7,8 — so possible 2,6,7,8

Not unique.

Perhaps look at Box 8.

Box 8: R7C4=9, R7C5=?, R7C6=3, R8C4=?, R8C5=?, R8C6=?, R9C4=5, R9C5=4, R9C6=?

Numbers: 9,3,5,4 → missing: 1,2,6,7,8

Row 7: missing 2,4,6,7,8 — but 4 is in R9C5, not in Row 7, so for R7C5, it can be 2,6,7,8

But in Box 8, missing 1,2,6,7,8, so R7C5 can be 2,6,7,8

Col 5: as above.

Let's try to set R4C5=9.

Then in Box 5, R4C5=9, R6C4=2, so missing 1,4,7 for R5C4, R5C5, R5C6

Row 4: has R4C5=9, so missing 2,5,7 for R4C1, R4C7, R4C9

Col 5: has R4C5=9, R6C5=5, R9C5=4 — so filled: 9,5,4 — missing: 1,2,3,6,7,8

R5C5 must be from 1,4,7 — but Col 5 has 4, so R5C5 cannot be 4, so R5C5 = 1 or 7.

Similarly, R5C4 must be from 1,4,7 — and Col 4 has R3C4=1, R4C4=3, R6C4=2, R7C4=9, R9C4=5 — so filled: 1,3,2,9,5 — missing: 4,6,7,8

So R5C4 = 4 or 7 (from Box 5 options 1,4,7, but 1 is in Col 4, so R5C4 cannot be 1, so 4 or 7)

If R5C4=4, then Col 4 has 4, ok.

If R5C4=7, ok.

Now, suppose R5C4=4.

Then in Box 5, R5C4=4, so missing 1,7 for R5C5 and R5C6.

R5C5 = 1 or 7.

Col 5 has 9,5,4, so if R5C5=1, ok; if 7, ok.

Say R5C5=1, then R5C6=7.

Or vice versa.

Now, Row 5: R5C3=2, R5C4=4, R5C5=1, R5C6=7 — so missing 3,5,6,8,9 for other cells.

Col 6: R4C6=6, R6C6=8, R7C6=3 — so filled: 6,8,3 — missing: 1,2,4,5,7,9

R5C6=7, so ok.

Now, let's look at R4C7.

Row 4: missing 2,5,7 — but R4C5=9, so missing 2,5,7 for R4C1, R4C7, R4C9

Col 7: R2C7=7, R6C7=4 — so R4C7 cannot be 7, so must be 2 or 5.

Similarly, Col 9: R2C9=5, R3C9=8, R6C9=6, R7C9=1 — so R4C9 cannot be 5, so must be 2 or 7 — but if R4C7 is 2 or 5, and R4C9 is 2 or 7, and R4C1 is the remaining.

Also, in Box 6, R4C7 and R4C9 are in it.

Box 6: R4C7, R4C8=1, R4C9, R5C7, R5C8, R5C9, R6C7=4, R6C8, R6C9=6

Numbers: 1,4,6 — missing: 2,3,5,7,8,9

If R4C7=2 or 5, R4C9=2 or 7.

Suppose R4C7=5, then Col 7 has 5,7,4 — ok.

Then R4C9 must be 2 or 7, but if R4C7=5, then for Row 4, missing 2,7 for R4C1 and R4C9.

Col 9 has no 2 or 7, so ok.

But if R4C9=2, then Col 9 has 2,5,8,6,1 — ok.

If R4C9=7, ok.

Now, let's see R6C8.

Row 6: R6C4=2, R6C5=5, R6C6=8, R6C7=4, R6C9=6 — so missing 1,3,7,9 for R6C1, R6C2, R6C3, R6C8

Col 8: R2C8=2, R4C8=1, R9C8=3 — so filled: 2,1,3 — missing: 4,5,6,7,8,9

R6C8 must be from Row 6 missing: 1,3,7,9 — but Col 8 has 1,2,3, so R6C8 cannot be 1,3, so must be 7 or 9.

Also, in Box 6, missing 2,3,5,7,8,9 — 2,3 are in Col 8, so for R6C8, possible 7,8,9 — but from row, 7 or 9, so 7 or 9.

Now, if we can find which.

Look at R5C8.

Row 5: if R5C4=4, R5C5=1, R5C6=7, R5C3=2, so missing 3,5,6,8,9 for R5C1, R5C2, R5C7, R5C8, R5C9

Col 8: has 2,1,3 — so R5C8 can be 4,5,6,7,8,9 — but in Box 6, missing 2,3,5,7,8,9 — 2,3 are in Col 8, so for R5C8, possible 5,7,8,9

But Row 5 has 7 in R5C6, so R5C8 cannot be 7, so 5,8,9

Similarly, etc.

This is still complicated.

Perhaps the intended solution is to recognize that this is a standard puzzle, but since I need to solve it, let's try to look for the answer or use a different method.

After thinking, I recall that in Sudoku, sometimes you can use the process of elimination with pencil marks, but for this format, perhaps I can provide the solved grid.

Since this is a "Wicked Puzzle" from LiveSudoku.com, and the number is #1600053, I can assume that the solution is unique, and after careful solving, the completed grid is:

Let me fill it based on logical deduction or known solution.

Upon solving or recalling, the completed Sudoku grid is:

Row 1: 6 9 3 | 7 8 2 | 5 4 1
Row 2: 8 1 4 | 6 3 9 | 7 2 5
Row 3: 5 2 7 | 1 4 5 | 3 6 8 -- wait, duplicate 5, mistake.

Let me correct.

Standard solution for this puzzle:

After research or proper solving, the grid is:

R1: 6 9 3 | 7 8 2 | 5 4 1
R2: 8 1 4 | 6 3 9 | 7 2 5
R3: 5 2 7 | 1 4 5 | 3 6 8 -- again duplicate.

I think I have a error.

Let's start over with a fresh approach.

Let me list the given again:

Positions with numbers:

(2,7)=7, (2,8)=2, (2,9)=5
(3,2)=2, (3,4)=1, (3,9)=8
(4,2)=4, (4,3)=8, (4,4)=3, (4,6)=6, (4,8)=1
(5,3)=2
(6,5)=5, (6,6)=8, (6,7)=4, (6,9)=6
(7,2)=5, (7,4)=9, (7,6)=3, (7,9)=1
(8,2)=3, (8,3)=7
(9,2)=8, (9,3)=1, (9,4)=5, (9,5)=4, (9,8)=3

Now, let's look at Box 3 (R1-3, C7-9):

Cells: R1C7, R1C8, R1C9
R2C7=7, R2C8=2, R2C9=5
R3C7, R3C8, R3C9=8

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

Row 3: has R3C2=2, R3C4=1, R3C9=8 — so missing: 3,4,5,6,7,9

But in Box 3, 5 and 7 are already present, so R3C7 and R3C8 cannot be 5 or 7, so must be from 3,4,6,9

Col 7: R2C7=7, R6C7=4 — so R3C7 cannot be 7 or 4, so from 3,6,9

Col 8: R2C8=2, R4C8=1, R9C8=3 — so R3C8 cannot be 2,1,3, so from 4,6,9 (since 3 is in col)

So for R3C8: possible 4,6,9

For R3C7: 3,6,9

Now, if R3C7=3, then Col 7 has 3,7,4 — ok.

If R3C7=6, ok.

etc.

Perhaps look at R1C7.

But let's consider that in Row 3, the number 5 must be placed. Where can 5 go in Row 3?

Possible cells: R3C1, R3C3, R3C5, R3C6, R3C7, R3C8

But in Box 1, R3C1 and R3C3 are in it, and Box 1 has no 5 yet.

Col 1: no 5.

Col 3: R4C3=8, R5C3=2, R8C3=7, R9C3=1 — no 5, so possible.

But in Box 3, 5 is already in R2C9, so R3C7 and R3C8 cannot be 5.

So 5 in Row 3 can be in R3C1, R3C3, R3C5, R3C6

Now, Box 2 (R1-3, C4-6): R3C4=1, so R3C5 and R3C6 are in it.

Box 2 has only R3C4=1, so 5 can be there.

So possible.

Not helping.

Let's try to fill R9C6.

Row 9: missing 2,6,7,9

Col 6: R4C6=6, R6C6=8, R7C6=3 — so R9C6 cannot be 6, so must be 2,7,9

Box 8: R7C4=9, R7C6=3, R9C4=5, R9C5=4 — so for R9C6, it can be 2,7 (since 9 is in Box 8)

So R9C6 = 2 or 7

Similarly, R9C1 = 2,6,9 (as before)

R9C7 = 2,6,9 (but Col 7 has 7, so not 7, so 2,6,9)

R9C9 = 2,7,9 (Col 9 has 6, so not 6, so 2,7,9)

Now, if R9C6=2, then Row 9 has 2, so missing 6,7,9 for C1,C7,C9

Then R9C1 = 6,9 (since 2 is used)

R9C7 = 6,9

R9C9 = 7,9

But in Box 9, R9C8=3, R7C9=1, so for R9C9, it can be 7,9

Also, Col 9 has R2C9=5, R3C9=8, R6C9=6, R7C9=1 — so if R9C9=7 or 9, ok.

Now, if R9C6=2, then in Box 8, R9C6=2, so missing 1,6,7,8 for other cells.

Row 8: R8C2=3, R8C3=7 — so missing 1,2,4,5,6,8,9

But R9C6=2, so for R8C6, it can be from Box 8 missing: 1,6,7,8 — but 7 is in Row 8 (R8C3=7), so R8C6 cannot be 7, so 1,6,8

Col 6: has R4C6=6, R6C6=8, R7C6=3, R9C6=2 — so filled: 6,8,3,2 — missing: 1,4,5,7,9

So R8C6 = 1,4,5,7,9 — but from above, 1,6,8 — intersection 1.

So R8C6 = 1.

Then in Box 8, R8C6=1, so missing 6,7,8 for R7C5, R8C4, R8C5

Row 7: missing 2,4,6,7,8 — but R9C6=2, so for R7C5, it can be 4,6,7,8

But in Box 8, R7C5 must be 6,7,8 (since 1 is placed)

Col 5: has R6C5=5, R9C5=4 — so R7C5 can be 6,7,8

So possible.

Then R8C4 and R8C5 must be the remaining two of 6,7,8.

Row 8: has R8C2=3, R8C3=7, R8C6=1 — so missing 2,4,5,6,8,9

So R8C4 and R8C5 can be 6,8 (since 7 is in row)

So if R8C4 and R8C5 are 6 and 8, then good.

Col 4: has R3C4=1, R4C4=3, R6C4=2, R7C4=9, R9C4=5 — so filled: 1,3,2,9,5 — missing: 4,6,7,8

So R8C4 = 6,7,8 — but from above, 6 or 8.

Similarly, Col 5: has R6C5=5, R9C5=4 — so R8C5 = 6,7,8

So possible.

Say R8C4=6, R8C5=8, then R7C5=7.

Or other combinations.

Then R7C5=7.

Then Row 7: R7C2=5, R7C4=9, R7C5=7, R7C6=3, R7C9=1 — so missing 2,4,6,8 for R7C1, R7C3, R7C7, R7C8

But earlier, R9C6=2, so Row 7 missing 4,6,8 for those cells.

Col 1: for R7C1, etc.

This is working, but for the sake of time, I'll provide the final