Diagonal Sudoku (Mini Sudoku Series #17) - Free Printable

Let’s solve this Diagonal Sudoku puzzle step by step.

In Diagonal Sudoku, you must fill the 9x9 grid so that:
- Each row has numbers 1 to 9 (no repeats).
- Each column has numbers 1 to 9 (no repeats).
- Each of the nine 3x3 boxes has numbers 1 to 9 (no repeats).
- PLUS — both main diagonals (from top-left to bottom-right and top-right to bottom-left) also have numbers 1 to 9 with no repeats.

We’ll start by looking at what we already have and filling in cells where only one number can go.

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First, let’s label rows 1 to 9 from top to bottom, and columns A to I from left to right.

Given numbers:

Row 1: D=3, E=4
Row 2: C=2, H=6
Row 3: B=1, I=3
Row 4: A=1, G=6
Row 5: A=3, F=1
Row 6: B=6, C=5

Also, remember the two diagonals:

Main diagonal (top-left to bottom-right): A1, B2, C3, D4, E5, F6, G7, H8, I9
Anti-diagonal (top-right to bottom-left): I1, H2, G3, F4, E5, D6, C7, B8, A9

Note: E5 is on BOTH diagonals — it’s the center cell.

Let’s look for easy placements first.

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Look at Row 5: We have A5=3, F5=1. So missing: 2,4,5,6,7,8,9

But also check Column A: A1=?, A2=?, A3=?, A4=1, A5=3, A6=?, A7=?, A8=?, A9=?

Column A has 1 and 3 already. Also, A9 is on anti-diagonal.

Wait — let’s try a different approach. Look at Box 5 (center box: rows 4-6, cols D-F).

Box 5 currently has:
D4=?, E4=?, F4=?
D5=?, E5=?, F5=1
D6=?, E6=?, F6=?

We know F5=1.

Also, E5 is center — part of both diagonals.

What do we know about E5? Let’s see what’s missing in its row, column, box, and diagonals.

Row 5: A5=3, F5=1 → so E5 cannot be 1 or 3.

Column E: E1=4, E2=?, E3=?, E4=?, E5=?, E6=?, E7=?, E8=?, E9=?

E1=4 → so E5 ≠ 4.

Box 5: F5=1 → E5 ≠ 1.

Diagonals: E5 is on both. Main diagonal: A1,B2,C3,D4,E5,F6,G7,H8,I9 — none filled yet except maybe later. Anti-diagonal same.

Not enough info yet.

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Let’s look at Cell C3. It’s in Row 3, Column C, Box 1, and on main diagonal.

Row 3: B3=1, I3=3 → so C3 ≠ 1,3

Column C: C2=2, C6=5 → so C3 ≠ 2,5

Box 1 (rows 1-3, cols A-C): has B3=1, C2=2 → so C3 ≠ 1,2

Main diagonal: A1,B2,C3,... — nothing known yet.

Possible for C3: 4,6,7,8,9

Still too many.

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Try Cell G7. It’s on main diagonal. Row 7, Column G, Box 9.

Row 7: all empty except possibly others? Wait, given: Row 7 has nothing given? Wait no — let’s list all givens again clearly.

Givens:

R1C4=3, R1C5=4
R2C3=2, R2C8=6
R3C2=1, R3C9=3
R4C1=1, R4C7=6
R5C1=3, R5C6=1
R6C2=6, R6C3=5

So Row 7: all blank? No — wait, R7C2=6? No, R6C2=6. R7C2 is blank.

Actually, R7C2 is not given. Only R6C2=6.

Let me write the grid as is:

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

And diagonals must have 1-9 unique.

Now, let’s consider the anti-diagonal: positions are:

I1, H2, G3, F4, E5, D6, C7, B8, A9

We know some values:

H2 = ? — but R2C8=6 → that’s H2? Columns: A B C D E F G H I → so C8 is H? Yes, column 8 is H.

So R2C8 = H2 = 6 → so anti-diagonal has H2=6.

Similarly, G3: R3C7? No — G is column 7. R3C7 is G3? Row 3, column 7 → yes, G3.

But R3C7 is not given. Given R3C9=3 → that’s I3.

F4: R4C6 → not given. R4C7=6 → that’s G4.

E5: center — unknown.

D6: R6C4 → unknown.

C7: R7C3 → unknown.

B8: R8C2 → unknown.

A9: R9C1 → unknown.

Only known on anti-diagonal so far is H2=6.

Main diagonal: A1, B2, C3, D4, E5, F6, G7, H8, I9

Known: none directly, but I9 is R9C9 — unknown.

Wait — R3C9=3 → that’s I3, not I9.

No direct hits.

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Let’s try to find a cell with only one possibility.

Look at Row 4: [1, ?, ?, ?, ?, ?, 6, ?, ?]

Missing: 2,3,4,5,7,8,9

Column 1: R4C1=1, R5C1=3 → so other cells in col 1 can’t be 1 or 3.

Box 4 (rows 4-6, cols A-C): has R4C1=1, R5C1=3, R6C2=6, R6C3=5 → so missing 2,4,7,8,9

Cell R4C2: row 4, col 2, box 4.

Col 2: R3C2=1, R6C2=6 → so R4C2 ≠ 1,6

Box 4: has 1,3,5,6 → so R4C2 ≠ 1,3,5,6

Row 4: has 1,6 → so R4C2 ≠ 1,6

Possible: 2,4,7,8,9

Still many.

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Try Cell R5C2. Row 5, Col 2, Box 4.

Row 5: R5C1=3, R5C6=1 → so R5C2 ≠ 1,3

Col 2: R3C2=1, R6C2=6 → so R5C2 ≠ 1,6

Box 4: has 1,3,5,6 → so R5C2 ≠ 1,3,5,6

Possible: 2,4,7,8,9

Same issue.

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Let’s use the diagonals more.

Consider E5 (center). It must be unique in row 5, col 5, box 5, and both diagonals.

Row 5: has 3 (C1), 1 (C6) → so E5 ≠ 1,3

Col 5: R1C5=4 → so E5 ≠ 4

Box 5: R5C6=1 → so E5 ≠ 1

Main diagonal: no conflicts yet.

Anti-diagonal: no conflicts yet.

So E5 cannot be 1,3,4.

Possible: 2,5,6,7,8,9

Still broad.

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Another idea: look at Box 1 (top-left 3x3: rows 1-3, cols A-C)

Cells:
R1C1, R1C2, R1C3
R2C1, R2C2, R2C3=2
R3C1, R3C2=1, R3C3

Given: R2C3=2, R3C2=1

So missing in box 1: 3,4,5,6,7,8,9

Now, Row 1 has R1C4=3, R1C5=4 → so in row 1, C1,C2,C3 cannot be 3 or 4.

So in Box 1, R1C1,R1C2,R1C3 ≠ 3,4

Also, Col 1: R4C1=1, R5C1=3 → so R1C1,R2C1,R3C1 ≠ 1,3

Col 2: R3C2=1, R6C2=6 → so R1C2,R2C2 ≠ 1,6

Col 3: R2C3=2, R6C3=5 → so R1C3,R3C3 ≠ 2,5

Let’s try R3C3.

Row 3: R3C2=1, R3C9=3 → so R3C3 ≠ 1,3

Col 3: R2C3=2, R6C3=5 → so R3C3 ≠ 2,5

Box 1: has 1,2 → so R3C3 ≠ 1,2

Also, R3C3 is on main diagonal.

Possible for R3C3: 4,6,7,8,9

Still many.

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Perhaps start with a number that appears often.

Number 1 is placed at: R3C2, R4C1, R5C6

Where else can 1 go?

In Box 1: already has R3C2=1 → so no other 1 in box 1.

Box 2 (rows 1-3, cols D-F): R1C4=3, R1C5=4 → no 1 yet. Can 1 go here?

Row 1: no 1 yet, but R1C4=3, R1C5=4, so 1 could be in R1C6 or R1C1-3 but box 1 has 1 already, so R1C1-3 no 1. So in row 1, 1 must be in C6,C7,C8,C9.

Similarly, col 6: R5C6=1 → so no other 1 in col 6.

So in Box 2, 1 cannot be in col 6, and row 1 can have 1 in C6 but col 6 has 1 already, so R1C6 cannot be 1.

So in Box 2, possible cells for 1: R2C4, R2C5, R2C6, R3C4, R3C5, R3C6

But row 2: R2C3=2, R2C8=6 → no 1 yet.

Col 4: R1C4=3 → no 1.

Col 5: R1C5=4 → no 1.

Col 6: R5C6=1 → so R2C6 and R3C6 cannot be 1.

So in Box 2, 1 can only be in R2C4, R2C5, R3C4, R3C5

Now, also, if we consider the diagonals, but 1 is not on any diagonal yet except possibly later.

This is taking too long. Let's try to fill in what we can with certainty.

Look at Cell R6C1. Row 6, Col 1, Box 4.

Row 6: R6C2=6, R6C3=5 → so R6C1 ≠ 5,6

Col 1: R4C1=1, R5C1=3 → so R6C1 ≠ 1,3

Box 4: has R4C1=1, R5C1=3, R6C2=6, R6C3=5 → so missing 2,4,7,8,9

So R6C1 can be 2,4,7,8,9

Still many.

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Perhaps use the fact that in each row, column, box, diagonal, numbers 1-9 appear once.

Let’s consider the main diagonal: A1,B2,C3,D4,E5,F6,G7,H8,I9

Suppose we assume a value for E5 and see if it works, but that might not be efficient.

Another idea: look at Row 9. All empty, but A9 is on anti-diagonal, I9 on main diagonal.

Similarly, Row 1 has R1C4=3, R1C5=4.

Let’s try to place number 6.

6 is at: R2C8, R4C7, R6C2

Where else can 6 go?

In Box 1: no 6 yet. Cells available: R1C1,R1C2,R1C3,R2C1,R2C2,R3C1,R3C3 (since R2C3=2, R3C2=1)

Row 1: has 3,4 → so 6 can be in C1,C2,C3,C6,C7,C8,C9

But Box 1: can have 6.

Col 1: has 1,3 → so 6 can be in R1C1,R2C1,R3C1,R6C1,R7C1,R8C1,R9C1 — but R6C1 is in box 4, which has R6C2=6, so R6C1 cannot be 6.

Similarly, col 2: R3C2=1, R6C2=6 → so no other 6 in col 2.

Col 3: R2C3=2, R6C3=5 → so 6 can be in R1C3,R3C3,R4C3,R5C3,R7C3,R8C3,R9C3

In Box 1, 6 can be in R1C1,R1C2,R1C3,R2C1,R2C2,R3C1,R3C3

But R1C2: col 2 has R6C2=6, so R1C2 cannot be 6.

R2C2: col 2 has 6, so cannot be 6.

R3C1: col 1 has no 6 yet, but let's see.

Perhaps R3C3 can be 6.

Assume R3C3=6.

Then main diagonal has C3=6.

Is that possible? Let's check conflicts.

Row 3: R3C2=1, R3C9=3, so 6 is ok.

Col 3: R2C3=2, R6C3=5, so 6 is ok.

Box 1: has 1,2, and now 6 — ok.

So possible.

But is it necessary? Not yet.

Let's look for a cell that must be a specific number.

Consider Cell R1C1.

Row 1: has 3,4 in C4,C5 → so R1C1 can be 1,2,5,6,7,8,9

Col 1: has 1,3 in R4,R5 → so R1C1 ≠ 1,3

Box 1: has 1,2 in R3C2,R2C3 → so R1C1 ≠ 1,2

Also, R1C1 is on main diagonal.

So R1C1 ≠ 1,2,3

Possible: 4,5,6,7,8,9

But row 1 has 4 in C5, so R1C1 ≠ 4

So R1C1 can be 5,6,7,8,9

Still many.

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I recall that in Sudoku, sometimes you can use "naked singles" or "hidden singles".

Let's try hidden single for number 2 in Box 1.

Box 1 needs 2. Where can 2 go?

Cells: R1C1,R1C2,R1C3,R2C1,R2C2,R3C1,R3C3 (R2C3=2 already? No, R2C3=2 is given, so 2 is already in Box 1 at R2C3.

Oh! I forgot — R2C3=2 is given, so Box 1 already has 2.

So no need to place 2 in Box 1.

Similarly, number 1 is at R3C2.

So Box 1 has 1 and 2.

Missing: 3,4,5,6,7,8,9

Now, in Row 1, for Box 1 cells, cannot have 3 or 4 because row 1 has them in C4,C5.

So R1C1,R1C2,R1C3 cannot be 3 or 4.

Also, col 1 has 1,3, so R1C1 cannot be 1,3 — already knew.

Col 2 has 1,6, so R1C2 cannot be 1,6

Col 3 has 2,5, so R1C3 cannot be 2,5

So for R1C1: cannot be 1,2,3,4 (from above) — so 5,6,7,8,9

R1C2: cannot be 1,2,3,4,6 (row has 3,4; col has 1,6; box has 1,2) — so cannot be 1,2,3,4,6 — possible 5,7,8,9

R1C3: cannot be 1,2,3,4,5 (row has 3,4; col has 2,5; box has 1,2) — so cannot be 1,2,3,4,5 — possible 6,7,8,9

Now, let's look at R2C1.

Row 2: R2C3=2, R2C8=6 → so R2C1 ≠ 2,6

Col 1: R4C1=1, R5C1=3 → so R2C1 ≠ 1,3

Box 1: has 1,2 → so R2C1 ≠ 1,2

So R2C1 can be 4,5,7,8,9

Similarly, R2C2: row 2 has 2,6; col 2 has 1,6; box has 1,2 — so cannot be 1,2,6 — possible 3,4,5,7,8,9

But row 2 has no 3,4,5, etc yet.

This is not helping.

Let's try a different strategy. Let's look at the anti-diagonal.

Anti-diagonal: I1, H2, G3, F4, E5, D6, C7, B8, A9

We know H2 = R2C8 = 6

So anti-diagonal has 6 at position 2.

Now, what about I1? R1C9

Row 1: has R1C4=3, R1C5=4 → so R1C9 can be 1,2,5,6,7,8,9

Col 9: R3C9=3 → so R1C9 ≠ 3

Box 3 (rows 1-3, cols G-I): has R1C4=3? No, C4 is D, not in box 3. Box 3 is cols G,H,I i.e. 7,8,9.

R1C7,R1C8,R1C9
R2C7,R2C8=6,R2C9
R3C7,R3C8,R3C9=3

So box 3 has R2C8=6, R3C9=3

So for R1C9, in box 3, cannot be 3,6

Row 1 has 3,4, so R1C9 ≠ 3,4

Col 9 has R3C9=3, so R1C9 ≠ 3

So R1C9 can be 1,2,5,7,8,9

And it's on anti-diagonal.

Similarly, A9 = R9C1, on anti-diagonal.

Col 1: R4C1=1, R5C1=3 → so R9C1 ≠ 1,3

Row 9: all empty.

Box 7 (rows 7-9, cols A-C): no givens yet.

So R9C1 can be 2,4,5,6,7,8,9

But on anti-diagonal, and H2=6, so if A9=6, then duplicate on anti-diagonal, so A9 ≠ 6

Similarly, other constraints.

Perhaps set E5 to a value.

Let's assume E5 = 2, and see if it works, but that might not be reliable.

I recall that in some Sudoku puzzles, the center cell can be deduced from the diagonals.

Another idea: count how many times each number appears in the given clues.

Given numbers:

1: R3C2, R4C1, R5C6 — three 1's

2: R2C3 — one 2

3: R1C4, R3C9, R5C1 — three 3's

4: R1C5 — one 4

5: R6C3 — one 5

6: R2C8, R4C7, R6C2 — three 6's

So missing: 7,8,9 not present at all, and 2,4,5 only once.

So likely, 7,8,9 will be placed in various places.

But not helpful yet.

Let's try to fill in Box 5 (center box: rows 4-6, cols D-F)

Cells:
R4D, R4E, R4F
R5D, R5E, R5F=1
R6D, R6E, R6F

Given: R5F=1

Also, row 4: R4C1=1, R4C7=6 — so for R4D,R4E,R4F, cannot be 1,6

Row 5: R5C1=3, R5C6=1 — so R5D,R5E cannot be 1,3

Row 6: R6C2=6, R6C3=5 — so R6D,R6E,R6F cannot be 5,6

Col D (col 4): R1C4=3 — so R4D,R5D,R6D ≠ 3

Col E (col 5): R1C5=4 — so R4E,R5E,R6E ≠ 4

Col F (col 6): R5C6=1 — so R4F,R6F ≠ 1

Also, Box 5 must have 1-9, has R5F=1, so missing 2,3,4,5,6,7,8,9

Now, let's see possible for R5E (center of box, and E5).

As before, R5E cannot be 1 (row 5 has it), cannot be 3 (row 5 has R5C1=3), cannot be 4 (col 5 has R1C5=4), cannot be 1 (box has it).

So R5E ≠ 1,3,4

Possible: 2,5,6,7,8,9

Now, also, R5E is on both diagonals.

Suppose we look at what is missing in row 5.

Row 5: R5C1=3, R5C6=1, so missing 2,4,5,6,7,8,9 for C2,C3,C4,C5,C7,C8,C9

But C5 is R5E.

Similarly, col 5: R1C5=4, so missing 1,2,3,5,6,7,8,9 for R2C5 to R9C5, but R5C5 is R5E.

Perhaps consider that in col 5, number 1 is not placed yet, but R5C6=1 is in col 6, not col 5.

Col 5 has only R1C5=4 given, so 1 can be in col 5.

But R5E cannot be 1 because row 5 has R5C6=1.

So in col 5, 1 can be in R2C5, R3C5, R4C5, R6C5, R7C5, R8C5, R9C5

Similarly for other numbers.

Let's try to place number 5 in Box 5.

Box 5 needs 5. Where can it go?

Cells: R4D,R4E,R4F, R5D,R5E, R6D,R6E,R6F (R5F=1)

Row 4: has 1,6 — so R4D,R4E,R4F ≠ 1,6

Row 5: has 3,1 — so R5D,R5E ≠ 1,3

Row 6: has 6,5 — R6C2=6, R6C3=5, so R6D,R6E,R6F ≠ 5,6

So in Box 5, for 5: cannot be in row 6 (because R6C3=5, so no other 5 in row 6), cannot be in row 4 or 5 if they have 5, but they don't have 5 yet.

Row 4 has no 5, row 5 has no 5, row 6 has 5 in C3, so in box 5, row 6 cells cannot have 5.

So 5 can be in R4D,R4E,R4F, R5D,R5E

Col D: R1C4=3 — no 5 yet

Col E: R1C5=4 — no 5

Col F: R5C6=1 — no 5

So all possible.

But also, if we consider the diagonals, but 5 is not on diagonal yet.

Perhaps R6E cannot be 5 because row 6 has 5 in C3, so yes, already excluded.

So 5 can be in R4D,R4E,R4F, R5D,R5E

Now, let's see if there's a constraint from elsewhere.

For example, in col E, if we can eliminate some.

But hard.

Let's look at Cell R4C4 (D4).

Row 4: R4C1=1, R4C7=6 — so R4C4 ≠ 1,6

Col 4: R1C4=3 — so R4C4 ≠ 3

Box 5: R5F=1 — so R4C4 ≠ 1

Also, R4C4 is on main diagonal.

So R4C4 ≠ 1,3,6

Possible: 2,4,5,7,8,9

Similarly, R6C4 (D6) is on anti-diagonal.

Row 6: R6C2=6, R6C3=5 — so R6C4 ≠ 5,6

Col 4: R1C4=3 — so R6C4 ≠ 3

Box 5: R5F=1 — so R6C4 ≠ 1

So R6C4 ≠ 1,3,5,6

Possible: 2,4,7,8,9

And on anti-diagonal.

Now, perhaps we can consider that in the anti-diagonal, we have H2=6, and other cells.

Let's list the anti-diagonal cells and what we know:

Position 1: I1 = R1C9 — can be 1,2,5,7,8,9 (as earlier)

Position 2: H2 = R2C8 = 6 — fixed

Position 3: G3 = R3C7 — row 3: R3C2=1, R3C9=3 — so R3C7 ≠ 1,3; col 7: R4C7=6 — so R3C7 ≠ 6; box 3: has R2C8=6, R3C9=3 — so R3C7 can be 2,4,5,7,8,9

Position 4: F4 = R4C6 — row 4: R4C1=1, R4C7=6 — so R4C6 ≠ 1,6; col 6: R5C6=1 — so R4C6 ≠ 1; box 5: R5F=1 — so R4C6 ≠ 1; so R4C6 ≠ 1,6; possible 2,3,4,5,7,8,9

Position 5: E5 = R5C5 — as before, ≠ 1,3,4; possible 2,5,6,7,8,9

Position 6: D6 = R6C4 — as above, ≠ 1,3,5,6; possible 2,4,7,8,9

Position 7: C7 = R7C3 — row 7: all empty; col 3: R2C3=2, R6C3=5 — so R7C3 ≠ 2,5; box 7: no givens; so possible 1,3,4,6,7,8,9

Position 8: B8 = R8C2 — row 8: empty; col 2: R3C2=1, R6C2=6 — so R8C2 ≠ 1,6; box 8: no givens; so possible 2,3,4,5,7,8,9

Position 9: A9 = R9C1 — row 9: empty; col 1: R4C1=1, R5C1=3 — so R9C1 ≠ 1,3; box 7: no givens; so possible 2,4,5,6,7,8,9

And anti-diagonal must have 1-9 unique, with H2=6.

So the other 8 cells must be 1,2,3,4,5,7,8,9

Now, notice that in position 1 (I1=R1C9), it cannot be 3,4,6 (from earlier), and 6 is already used, so can be 1,2,5,7,8,9

Similarly, position 3 (G3=R3C7) cannot be 1,3,6 — 6 used, so can be 2,4,5,7,8,9

Position 4 (F4=R4C6) cannot be 1,6 — 6 used, so can be 2,3,4,5,7,8,9

Position 5 (E5) cannot be 1,3,4 — so can be 2,5,6,7,8,9 but 6 used, so 2,5,7,8,9

Position 6 (D6=R6C4) cannot be 1,3,5,6 — 6 used, so can be 2,4,7,8,9

Position 7 (C7=R7C3) can be 1,3,4,6,7,8,9 but 6 used, so 1,3,4,7,8,9

Position 8 (B8=R8C2) can be 2,3,4,5,7,8,9

Position 9 (A9=R9C1) can be 2,4,5,6,7,8,9 but 6 used, so 2,4,5,7,8,9

Now, number 1 must be in one of these positions for the anti-diagonal.

Where can 1 go in anti-diagonal?

Position 1: R1C9 — can be 1 (earlier said can be 1,2,5,7,8,9)

Position 3: R3C7 — cannot be 1 (row 3 has R3C2=1)

Position 4: R4C6 — cannot be 1 (row 4 has R4C1=1)

Position 5: R5C5 — cannot be 1 (row 5 has R5C6=1)

Position 6: R6C4 — cannot be 1 (row 6 has no 1 yet, but col 4 has R1C4=3, no 1, but row 6: R6C2=6, R6C3=5, so 1 is possible? Wait, row 6 has no 1 given, so R6C4 can be 1? But earlier I said for R6C4, in box 5, R5F=1, so R6C4 cannot be 1 because same box? Box 5 is rows 4-6, cols D-F, so R6C4 is in box 5, and R5F=1 is in box 5, so yes, R6C4 cannot be 1.

Similarly, position 7: R7C3 — can be 1? Row 7 no 1, col 3 has R2C3=2, R6C3=5, no 1, box 7 no 1, so yes, can be 1.

Position 8: R8C2 — col 2 has R3C2=1, so cannot be 1.

Position 9: R9C1 — col 1 has R4C1=1, R5C1=3, so cannot be 1.

So for anti-diagonal, 1 can only be in position 1 (R1C9) or position 7 (R7C3)

Similarly, number 3: where can it go in anti-diagonal?

Position 1: R1C9 — cannot be 3 (row 1 has R1C4=3)

Position 3: R3C7 — cannot be 3 (row 3 has R3C9=3)

Position 4: R4C6 — can be 3? Row 4 has R4C1=1, R4C7=6, no 3; col 6 has R5C6=1, no 3; box 5 has no 3 yet; so yes.

Position 5: R5C5 — cannot be 3 (row 5 has R5C1=3)

Position 6: R6C4 — cannot be 3 (col 4 has R1C4=3)

Position 7: R7C3 — can be 3? Row 7 no 3, col 3 has R2C3=2, R6C3=5, no 3, box 7 no 3, so yes.

Position 8: R8C2 — can be 3? Row 8 no 3, col 2 has R3C2=1, R6C2=6, no 3, box 8 no 3, so yes.

Position 9: R9C1 — cannot be 3 (col 1 has R5C1=3)

So 3 can be in position 4,7,8

Number 4: where in anti-diagonal?

Position 1: R1C9 — cannot be 4 (row 1 has R1C5=4)

Position 3: R3C7 — can be 4? Row 3 has 1,3, no 4; col 7 has R4C7=6, no 4; box 3 has no 4; so yes.

Position 4: R4C6 — can be 4? Row 4 has 1,6, no 4; col 6 has 1, no 4; box 5 has no 4; so yes.

Position 5: R5C5 — cannot be 4 (col 5 has R1C5=4)

Position 6: R6C4 — can be 4? Row 6 has 5,6, no 4; col 4 has 3, no 4; box 5 has no 4; so yes.

Position 7: R7C3 — can be 4? Yes.

Position 8: R8C2 — can be 4? Yes.

Position 9: R9C1 — can be 4? Yes.

So 4 can be in 3,4,6,7,8,9

This is still messy.

Perhaps start with a cell that has few possibilities.

Let's consider Cell R3C1.

Row 3: R3C2=1, R3C9=3 — so R3C1 ≠ 1,3

Col 1: R4C1=1, R5C1=3 — so R3C1 ≠ 1,3

Box 1: has R2C3=2, R3C2=1 — so R3C1 ≠ 1,2

Also, R3C1 is not on diagonal.

So R3C1 can be 4,5,6,7,8,9

Same as before.

I think I need to look for a different approach or accept that this requires systematic solving.

Upon second thought, let's try to fill in the grid using a standard Sudoku solver logic, but since I'm doing it manually, let's focus on one area.

Let's look at Box 6 (rows 4-6, cols G-I)

Cells:
R4G=6 (given), R4H, R4I
R5G, R5H, R5I
R6G, R6H, R6I

Given: R4G=6

Row 4: R4C1=1, R4C7=6 — so for R4H,R4I, cannot be 1,6

Row 5: R5C1=3, R5C6=1 — so R5G,R5H,R5I ≠ 1,3

Row 6: R6C2=6, R6C3=5 — so R6G,R6H,R6I ≠ 5,6

Col G (col 7): R4C7=6 — so R5G,R6G ≠ 6

Col H (col 8): R2C8=6 — so R4H,R5H,R6H ≠ 6

Col I (col 9): R3C9=3 — so R4I,R5I,R6I ≠ 3

Also, Box 6 must have 1-9, has R4G=6, so missing 1,2,3,4,5,7,8,9

Now, number 1: where can it go in Box 6?

Row 4 has R4C1=1, so R4H,R4I ≠ 1

Row 5 has R5C6=1, so R5G,R5H,R5I ≠ 1

Row 6 has no 1 yet, so R6G,R6H,R6I can be 1, but col G has R4G=6, no 1; col H has R2C8=6, no 1; col I has R3C9=3, no 1; so yes, 1 can be in R6G,R6H,R6I

But also, in row 6, R6C2=6, R6C3=5, so 1 is possible.

So 1 can be in R6G,R6H,R6I

Similarly, number 3: row 4 has no 3, but col I has R3C9=3, so R4I,R5I,R6I ≠ 3

Row 5 has R5C1=3, so R5G,R5H,R5I ≠ 3

Row 6 has no 3, so R6G,R6H,R6I can be 3, but col I has 3, so R6I ≠ 3, so R6G or R6H can be 3.

Col G: no 3 yet, col H: no 3 yet.

So 3 can be in R6G or R6H

Number 5: row 6 has R6C3=5, so R6G,R6H,R6I ≠ 5

Row 4 has no 5, row 5 has no 5, so R4H,R4I, R5G,R5H,R5I can be 5, but col G has R4G=6, no 5; col H has R2C8=6, no 5; col I has R3C9=3, no 5; so yes.

But row 4: R4H,R4I can be 5; row 5: R5G,R5H,R5I can be 5.

So many places.

Perhaps for number 1 in Box 6, it must be in row 6, as above.

So R6G, R6H, or R6I = 1

Now, also, in row 6, we have R6C2=6, R6C3=5, so the 1 must be in C4 to C9, but C4,C5,C6 are in box 5, C7,C8,C9 in box 6.

So for row 6, 1 can be in R6C4,R6C5,R6C6,R6C7,R6C8,R6C9

But in box 5, R5F=1, so R6C4,R6C5,R6C6 cannot be 1 (same box).

So in row 6, 1 can only be in R6C7,R6C8,R6C9 — which are in Box 6.

So yes, 1 is in R6G,R6H, or R6I.

Now, similarly, for other numbers.

But let's consider that in col I, R3C9=3, so for R6I, it can be 1,2,4,5,6,7,8,9 but row 6 has 5,6, so R6I ≠ 5,6, and col I has 3, so R6I ≠ 3, so R6I can be 1,2,4,7,8,9

Similarly for others.

Perhaps assume that R6I = 1, and see.

But let's try to use the diagonal constraint for E5.

I recall that in some Diagonal Sudoku, the center cell can be found by process of elimination.

Let's list all cells that are on the diagonals and see what is missing.

Main diagonal: A1,B2,C3,D4,E5,F6,G7,H8,I9

Anti-diagonal: I1,H2=6,G3,F4,E5,D6,C7,B8,A9

E5 is common.

The two diagonals together have 17 cells, but E5 is shared, so 17 unique cells, but in 9x9, there are 81 cells, so many overlap with rows/columns/boxes.

Perhaps calculate what numbers are missing in the diagonals.

But we have only H2=6 given on anti-diagonal, and no given on main diagonal.

So not helpful.

After struggling, I decide to look for a solution online or use a solver, but since this is a reasoning exercise, I should continue.

Let's try Cell R1C6.

Row 1: R1C4=3, R1C5=4 — so R1C6 ≠ 3,4

Col 6: R5C6=1 — so R1C6 ≠ 1

Box 2 (rows 1-3, cols D-F): has R1C4=3, R1C5=4, R2C3=2 but C3 is not in box 2, box 2 is D,E,F i.e. 4,5,6.

So cells: R1D=3, R1E=4, R1F=?
R2D, R2E, R2F
R3D, R3E, R3F

Given: R1D=3, R1E=4

So for R1F, in row 1, cannot be 3,4; in col 6, cannot be 1; in box 2, cannot be 3,4; so R1F can be 2,5,6,7,8,9

Also, not on diagonal.

Now, number 2 in Box 2: where can it go?

Row 1 has no 2, but R1F can be 2.

Row 2 has R2C3=2, but C3 is not in box 2, so R2D,R2E,R2F can be 2.

Row 3 has no 2 yet.

Col D: R1D=3 — no 2

Col E: R1E=4 — no 2

Col F: R5C6=1 — no 2

So 2 can be in R1F, R2D, R2E, R2F, R3D, R3E, R3F

But if we consider that in row 2, R2C3=2, so no other 2 in row 2, so R2D,R2E,R2F cannot be 2.

So in Box 2, 2 can be in R1F, R3D, R3E, R3F

Row 3: R3C2=1, R3C9=3 — so R3D,R3E,R3F can be 2.

So possible.

Now, if R1F = 2, is that possible? Let's see.

Row 1: has 3,4, and if R1F=2, then ok.

Col 6: has R5C6=1, so 2 is ok.

Box 2: has 3,4,2 — ok.

So possible.

But not necessary.

Perhaps for number 5 in Box 2.

Box 2 needs 5.

Row 1: no 5, so R1F can be 5.

Row 2: no 5, so R2D,R2E,R2F can be 5.

Row 3: no 5, so R3D,R3E,R3F can be 5.

Col D: no 5

Col E: no 5

Col F: no 5

So many places.

I think I need to give up and provide the solved grid, but since the user wants step-by-step, and I'm stuck, let's try a different tactic.

Let's consider the cell R5C5 (E5).

Suppose we assume E5 = 5.

Then check if it works.

Row 5: has 3,1, and 5 — ok.

Col 5: has 4, and 5 — ok.

Box 5: has 1, and 5 — ok.

Main diagonal: has 5 — ok.

Anti-diagonal: has 5 — ok.

So possible.

Then, in Box 5, with E5=5, and R5F=1, so missing 2,3,4,6,7,8,9 for other cells.

Row 4: R4C1=1, R4C7=6 — so for R4D,R4E,R4F, cannot be 1,6

Row 5: R5C1=3, R5C6=1, R5C5=5 — so R5D,R5E cannot be 1,3,5

Row 6: R6C2=6, R6C3=5 — so R6D,R6E,R6F cannot be 5,6

Col D: R1C4=3 — so R4D,R5D,R6D ≠ 3

Col E: R1C5=4 — so R4E,R5E,R6E ≠ 4

Col F: R5C6=1 — so R4F,R6F ≠ 1

Also, R4D is on main diagonal, R6D on anti-diagonal, etc.

With E5=5, then for R4E, cannot be 4 (col E), cannot be 1,6 (row 4), cannot be 5 (box), so can be 2,3,7,8,9

But col E has R1C5=4, so R4E ≠ 4, already.

Similarly, R5E = E5 = 5, so done.

R6E: cannot be 4 (col E), cannot be 5,6 (row 6), cannot be 1 (box? R5F=1, so R6E ≠ 1), so R6E ≠ 1,4,5,6 — possible 2,3,7,8,9

Now, let's see if we can place number 2 in Box 5.

Box 5 needs 2.

Cells: R4D,R4E,R4F, R5D,R5E=5, R6D,R6E,R6F

Row 4: can