Anti Diagonal Sudoku 9x9 - Easy to Extreme - Volume 1 - 276 ... - Free Printable

To solve this Anti-Diagonal Sudoku puzzle, we need to fill the $9 \times 9$ grid with numbers from 1 to 9.

The Rules:
1. Standard Sudoku: Each row, each column, and each of the nine $3 \times 3$ boxes must contain the digits 1 through 9 exactly once.
2. Anti-Diagonal Rule: The two main diagonals (marked with dashed pink lines) must NOT contain any repeated digits. This means the 9 cells on the top-left to bottom-right diagonal must all be different, and the 9 cells on the top-right to bottom-left diagonal must all be different.

Let's solve it step-by-step by finding the "easy" numbers first—cells where only one number can possibly fit.

Step 1: Fill in obvious single candidates

Row 5:
The numbers currently in Row 5 are: `5, 1, [blank], [blank], 8, 6, [blank], [blank], [blank]`.
Missing numbers: 2, 3, 4, 7, 9.
Let's look at Cell R5C3 (Row 5, Column 3).
- Column 3 already has: 6, 4, 9, 7, 8, 1. Missing in Col 3: 2, 3, 5. But Row 5 already has a 5. So R5C3 is either 2 or 3.
- Let's look at Cell R5C4. Column 4 has 2, 8, 5, 9, 3, 6. Missing: 1, 4, 7. Row 5 has 1. Box 5 (center) needs 2, 3, 5, 7, 9? No, let's look closer at Box 5.
- Box 5 contains: 5, 4, 8, 6, 9, 1, 3. Wait, looking at the image:
- R4: `_, _, 9 | 5, 4, _ | 7, _, _`
- R5: `5, 1, _ | _, 8, 6 | _, _, _`
- R6: `4, 8, 7 | 9, 1, 3 | _, 6, 5`
- Box 5 (center) currently has: 5, 4, 8, 6, 9, 1, 3. The missing numbers in Box 5 are 2 and 7.
- The empty cells in Box 5 are R5C4 and R4C6.
- Look at Column 4: It contains 2, 8, 5, 9, 3, 6. It already has a 2. Therefore, R5C4 cannot be 2. It must be 7.
- Consequently, R4C6 must be 2.

Update Box 5:
- R5C4 = 7
- R4C6 = 2

Now Row 5 is: `5, 1, _, 7, 8, 6, _, _, _`. Missing: 2, 3, 4, 9.
Cells remaining in Row 5: C3, C7, C8, C9.
- R5C3: Column 3 has 6, 4, 9, 7, 8, 1. Missing in Col 3: 2, 3, 5. Row 5 has 5. So R5C3 is 2 or 3.
- Look at Box 4 (middle left). It has 5, 1, 4, 8, 7, 9. Missing: 2, 3, 6.
- Cells in Box 4: R4C1, R4C2, R5C3.
- R4C1 and R4C2 are in Row 4. Row 4 currently: `_, _, 9, 5, 4, 2, 7, _, _`. Missing in Row 4: 1, 3, 6, 8.
- In Box 4, the missing numbers are 2, 3, 6.
- R5C3 is in Box 4. So R5C3 is 2, 3, or 6. But Col 3 already has 6 (R1C3). So R5C3 is 2 or 3.
- Let's check Column 3 again. Existing: 6(R1), 4(R2), _(R3), 9(R4), _(R5), 7(R6), 8(R7), 1(R8), _(R9).
- Wait, R4C3 is 9. R6C3 is 7. R7C3 is 8. R8C3 is 1.
- Col 3 values so far: 6, 4, ?, 9, ?, 7, 8, 1, ?. Missing: 2, 3, 5.
- R5C3 cannot be 5 (Row 5 has 5). So R5C3 is 2 or 3.
- R3C3 and R9C3 are the other blanks in Col 3.

Let's look at Row 6: `4, 8, 7, 9, 1, 3, _, 6, 5`.
- Only one number missing: 2.
- So, R6C7 = 2.

Step 2: Use the new number to progress

Column 7:
Current values in Col 7:
- R1: ?
- R2: ?
- R3: 8
- R4: 7
- R5: ?
- R6: 2 (just found)
- R7: ?
- R8: 4
- R9: ?
Missing in Col 7: 1, 3, 5, 6, 9.

Let's look at Box 6 (middle right).
Cells in Box 6:
- R4: 7, _, _
- R5: _, _, _
- R6: 2, 6, 5
Numbers present in Box 6: 2, 5, 6, 7.
Missing in Box 6: 1, 3, 4, 8, 9.
Empty cells in Box 6: R4C8, R4C9, R5C7, R5C8, R5C9.

Look at Row 4: `_, _, 9, 5, 4, 2, 7, _, _`.
Missing in Row 4: 1, 3, 6, 8.
Cells: R4C1, R4C2, R4C8, R4C9.
- R4C8 and R4C9 are in Box 6.
- Box 6 needs 1, 3, 4, 8, 9.
- Row 4 needs 1, 3, 6, 8.
- Intersection for R4C8/R4C9: Must be from {1, 3, 8} (since 6 is not in Box 6's missing list? Wait. Box 6 has 2,5,6,7. So 6 IS in Box 6. So Box 6 missing is 1,3,4,8,9. Correct.)
- So R4C8 and R4C9 must be chosen from {1, 3, 8}. They cannot be 6.
- Therefore, the 6 in Row 4 must be in R4C1 or R4C2.

Let's look at Column 8.
Values: 1(R1), 2(R2), _(R3), _(R4), _(R5), 6(R6), 9(R7), 5(R8), _(R9).
Missing: 3, 4, 7, 8.
- R4C8 is in Row 4 (needs 1,3,6,8) and Col 8 (needs 3,4,7,8). Common: 3, 8.
- R5C8 is in Row 5 (needs 2,3,4,9) and Col 8 (needs 3,4,7,8). Common: 3, 4.
- R3C8 is in Row 3 and Col 8.
- R9C8 is in Row 9 and Col 8.

Let's go back to Box 4 (Middle Left).
Missing: 2, 3, 6.
Cells: R4C1, R4C2, R5C3.
We established R5C3 is 2 or 3.
If R5C3 is 2, then R4C1/R4C2 are 3,6.
If R5C3 is 3, then R4C1/R4C2 are 2,6.

Check Column 1.
Values: 8(R1), _(R2), 9(R3), _(R4), 5(R5), 4(R6), _(R7), _(R8), 2(R9).
Missing: 1, 3, 6, 7.
- R4C1 is in Row 4 (needs 1,3,6,8) and Col 1 (needs 1,3,6,7). Common: 1, 3, 6.
- We know R4C1 is either 3 or 6 (from Box 4 logic above, since R5C3 takes 2 or 3, leaving 6 for the other spot if R5C3!=6. Wait. Box 4 missing 2,3,6. R5C3 is 2 or 3. So R4C1 and R4C2 are the remaining two. One is 6, the other is 2 or 3.)
- Actually, simpler: R4C1 and R4C2 contain {6, X} where X is 2 or 3. And R5C3 is the other of 2 or 3.
- Since Col 1 needs 1,3,6,7, and R4C1 is 3 or 6...

Let's look at the Anti-Diagonals. This is often the key in these puzzles.
Diagonal 1 (Top-Left to Bottom-Right):
Cells: R1C1, R2C2, R3C3, R4C4, R5C5, R6C6, R7C7, R8C8, R9C9.
Values given:
- R1C1 = 8
- R4C4 = 5
- R5C5 = 8 ... WAIT.
- R1C1 is 8.
- R5C5 is 8.
- This violates the Anti-Diagonal rule if they are on the same diagonal.
- Let's check the coordinates.
- R1C1 is (1,1). R5C5 is (5,5). Yes, they are on the main diagonal.
- The problem states "Anti Diagonal Sudoku". Usually, this means the diagonals *cannot* have repeats.
- Let me re-read the board carefully.
- R1C1 = 8.
- R5C5 = 8.
- Is it possible I misidentified the diagonal?
- The pink dashed lines go from corner to corner.
- Line 1: R1C1 -> R9C9.
- Line 2: R1C9 -> R9C1.
- If R1C1=8 and R5C5=8, the puzzle is invalid under standard "Anti-Diagonal" rules (which usually mean "X-Sudoku" where diagonals MUST be unique).
- However, some variants called "Anti-Diagonal" might mean something else? No, standard terminology: "Diagonal Sudoku" requires unique diagonals. "Anti-Diagonal" isn't a standard distinct name from "Diagonal Sudoku" in most contexts, but sometimes "Anti-Knight" etc exist.
- Let's look at the cover text: "Anti Diagonal Sudoku". And the stars show difficulty.
- Let's re-examine the image. Maybe R1C1 isn't 8? It looks like an 8. Maybe R5C5 isn't 8? It looks like an 8.
- Ah, wait. Look at the dashed lines.
- One line goes through R1C2, R2C3...? No.
- The lines clearly go through the corners.
- Is it possible that "Anti Diagonal" means the diagonals *must* contain specific patterns or *cannot* contain 1-9 uniquely?
- Actually, there is a variant called "Anti-Diagonal Sudoku" where the two main diagonals do not contain the digits 1-9 uniquely? No, that would make it easier/irrelevant.
- Let's assume the standard rule: The two main diagonals must contain digits 1-9 exactly once.
- If R1C1=8 and R5C5=8, this is a contradiction.
- Let me look really closely at R1C1. It is an 8.
- Let me look really closely at R5C5. It is an 8.
- Is it possible the diagonal is offset? No, the lines are drawn through the center cells.
- Let's check the other diagonal: R1C9 to R9C1.
- R1C9 is blank. R9C1 is 2.
- Let's check for other duplicates on Diagonal 1.
- R4C4=5. R6C6=3. R2C2=?. R3C3=?. R7C7=?. R8C8=5.
- R4C4=5 and R8C8=5. Another duplicate!
- Okay, so the rule cannot be that the diagonals contain unique numbers 1-9.
- What does "Anti Diagonal Sudoku" mean in this specific book/publisher context (Nick Snels)?
- Researching "Nick Snels Anti Diagonal Sudoku":
- In many of these puzzle books, "Anti-Diagonal" actually refers to a constraint where no digit appears more than once on the two main diagonals.
- If my observation of duplicates (8 and 5) is correct, then either:
1. I am misreading the numbers.
2. The rule is different.

Let's re-read the numbers.
Row 1: 8, _, 6 | 2, 3, 4 | _, 1, _
Row 5: 5, 1, _ | _, 8, 6 | _, _, _

Is R1C1 definitely 8? Yes.
Is R5C5 definitely 8? Yes.

Is it possible the dashed lines indicate cells that must be different from their orthogonal neighbors? No.

Let's look at the title again. "Anti Diagonal Sudoku".
There is a variant called "Anti-Diagonal" where the constraint is: The digits on the two main diagonals are NOT 1-9 unique. i.e., they *must* have repeats? That seems unlikely for a logic puzzle as it adds little constraint.

Wait, could "Anti-Diagonal" mean that the sum of the diagonals is constant? No.

Let's look at similar puzzles online. "Anti-Diagonal Sudoku" often implies that the two main diagonals do not follow the standard sudoku rule (i.e., they are just normal cells), BUT there is a different constraint?

Actually, looking at Nick Snels' definitions:
"Anti-Diagonal Sudoku": The two main diagonals do not contain the digits 1-9. Instead, usually, this term is used incorrectly or refers to "Non-Consecutive" or something else?

Let's try a different interpretation. Maybe the dashed lines highlight cells that form a "King's Move" constraint? No.

Let's look at the visual cue. The dashed lines cross out the diagonals.
In some contexts, "Anti-Diagonal" means that digits cannot repeat on the diagonals.
If R1C1=8 and R5C5=8, the puzzle as printed is broken OR I am misinterpreting the grid.

Let's check R4C4 and R8C8 again.
R4: `_ _ 9 5 4 2 7 _ _` -> R4C4 is 5.
R8: `_ 9 1 3 _ _ 4 5 _` -> R8C8 is 5.

Okay, both 8 and 5 are duplicated on the main diagonal in the *given* clues.
Therefore, the rule cannot be "Diagonals must be unique".

What if the rule is: The diagonals must NOT contain any of the digits 1-9? Impossible.

What if "Anti-Diagonal" means that the cells on the diagonal cannot be equal to the cell symmetric to them? No.

Let's reconsider the standard definition. Is it possible that "Anti-Diagonal Sudoku" means that the two main diagonals are NOT constrained? i.e. It's just a regular Sudoku, and the "Anti" part is a red herring or refers to the fact that you *don't* apply the diagonal constraint?

If it's just a Regular Sudoku, I can solve it using standard rules. The dashed lines might just be decorative or indicating "Ignore these for diagonal constraints"?

However, usually, extra lines mean extra constraints.

Let's look for another variant: "Disjoint Groups"? No.

Let's assume it is a Standard Sudoku first. If I hit a contradiction, I'll revisit. The presence of duplicates in the givens on the diagonal proves that the "Unique Diagonal" rule does NOT apply.

Hypothesis: This is a Standard Sudoku Puzzle. The "Anti Diagonal" title might refer to a variant where the diagonals are explicitly *excluded* from being unique (which is the default in standard sudoku), or it's a poorly named "Diagonal Sudoku" where the publisher made an error in the givens. Given the duplicates, we must ignore the diagonal uniqueness constraint. We will solve it as a standard Sudoku.

Solving as Standard Sudoku

Recap Current State:
R1: `8 . 6 | 2 3 4 | . 1 .`
R2: `. . 4 | 8 9 5 | . 2 7`
R3: `9 2 . | . . 7 | 8 . 3`
R4: `. . 9 | 5 4 2 | 7 . .` (R4C6=2 derived earlier)
R5: `5 1 . | 7 8 6 | . . .` (R5C4=7 derived earlier)
R6: `4 8 7 | 9 1 3 | 2 6 5` (R6C7=2 derived earlier)
R7: `. 5 8 | . 7 1 | . 9 2`
R8: `. 9 1 | 3 . . | 4 5 .`
R9: `2 4 . | 6 5 9 | . . 8`

Step 3: Continue Filling Rows/Columns

Row 6 is complete.

Box 6 (Middle Right):
Cells: R4C7(7), R4C8(?), R4C9(?), R5C7(?), R5C8(?), R5C9(?), R6C7(2), R6C8(6), R6C9(5).
Missing in Box 6: 1, 3, 4, 8, 9.
Wait, R4C7 is 7. R6C7 is 2.
Numbers in Box 6 so far: 2, 5, 6, 7.
Missing: 1, 3, 4, 8, 9.
Empty cells: R4C8, R4C9, R5C7, R5C8, R5C9.

Row 4: `. . 9 5 4 2 7 . .`
Missing: 1, 3, 6, 8.
Cells: R4C1, R4C2, R4C8, R4C9.
In Box 6, R4C8 and R4C9 must be from {1, 3, 8, 9, 4} intersected with Row 4 missing {1, 3, 6, 8}.
Intersection: {1, 3, 8}.
So R4C8, R4C9 $\in \{1, 3, 8\}$.
This leaves 6 for R4C1 or R4C2.

Row 5: `5 1 . 7 8 6 . . .`
Missing: 2, 3, 4, 9.
Cells: R5C3, R5C7, R5C8, R5C9.
R5C3 is in Box 4. Box 4 missing: 2, 3, 6.
R5C3 $\in \{2, 3\}$ (since Col 3 has 6? No, Col 3 has 6 in R1. So R5C3 != 6).
So R5C3 is 2 or 3.
Remaining cells in Row 5: R5C7, R5C8, R5C9.
They must contain the remaining numbers from {2, 3, 4, 9} minus whatever is in R5C3.
Also, these cells are in Box 6.
Box 6 missing: 1, 3, 4, 8, 9.
R5C7, R5C8, R5C9 must be in Box 6's missing set.
Intersection of Row 5 missing {2,3,4,9} and Box 6 missing {1,3,4,8,9}:
Common: 3, 4, 9.
So R5C7, R5C8, R5C9 $\in \{3, 4, 9\}$.
This implies R5C3 must be 2. (Because 2 is in Row 5's missing list but NOT in the intersection for the Box 6 cells).
So, R5C3 = 2.

Now, Box 4 (Middle Left):
Missing were 2, 3, 6.
R5C3 = 2.
Remaining cells R4C1, R4C2 must be 3 and 6.
Row 4 missing: 1, 3, 6, 8.
R4C1, R4C2 are 3, 6.
So R4C8, R4C9 must be 1, 8. (From earlier intersection {1,3,8}, removing 3 which is used in C1/C2? Wait. If R4C1/C2 are 3/6, then 3 is used in Row 4. So R4C8/C9 cannot be 3. They must be 1 and 8.)
So R4C8, R4C9 $\in \{1, 8\}$.

Back to Row 5:
R5C3 = 2.
Remaining missing in Row 5: 3, 4, 9.
Cells: R5C7, R5C8, R5C9.
These are in Box 6.
Box 6 missing now: 1, 3, 4, 8, 9.
R4C8, R4C9 take 1, 8.
So remaining cells in Box 6 (R5C7, R5C8, R5C9) must take 3, 4, 9.
This matches perfectly.

Let's determine R5C7, R5C8, R5C9 specifically.
Look at Column 7.
Values: R1(?), R2(?), R3(8), R4(7), R5(?), R6(2), R7(?), R8(4), R9(?).
Missing: 1, 3, 5, 6, 9.
R5C7 $\in \{3, 4, 9\}$.
Col 7 already has 4 (R8). So R5C7 $\neq 4$.
So R5C7 $\in \{3, 9\}$.

Look at Column 8.
Values: R1(1), R2(2), R3(?), R4(?), R5(?), R6(6), R7(9), R8(5), R9(?).
Missing: 3, 4, 7, 8.
R5C8 $\in \{3, 4, 9\}$.
Col 8 has 9 (R7). So R5C8 $\neq 9$.
Col 8 missing 3, 4, 7, 8.
So R5C8 $\in \{3, 4\}$.

Look at Column 9.
Values: R1(?), R2(7), R3(3), R4(?), R5(?), R6(5), R7(2), R8(?), R9(8).
Missing: 1, 4, 6, 9.
R5C9 $\in \{3, 4, 9\}$.
Col 9 has 3 (R3). So R5C9 $\neq 3$.
So R5C9 $\in \{4, 9\}$.

Summary for Row 5 ends:
R5C7 $\in \{3, 9\}$
R5C8 $\in \{3, 4\}$
R5C9 $\in \{4, 9\}$

Let's look at Column 8 again.
Missing: 3, 4, 7, 8.
Cells: R3C8, R4C8, R5C8, R9C8.
We know R4C8 $\in \{1, 8\}$.
But Col 8 missing does not include 1. Col 8 has 1 (R1).
So R4C8 cannot be 1.
Therefore, R4C8 = 8.
And consequently, R4C9 = 1.

Now update Column 8:
Missing: 3, 4, 7. (8 is placed).
Cells: R3C8, R5C8, R9C8.
R5C8 $\in \{3, 4\}$.

Update Row 4:
R4C8=8, R4C9=1.
Row 4 was `. . 9 5 4 2 7 8 1`.
Missing: 3, 6.
Cells: R4C1, R4C2.
So R4C1, R4C2 $\in \{3, 6\}$.

Step 4: Solve Box 4 and Column 1/2

Box 4 (Middle Left):
Cells:
R4C1, R4C2 (3, 6)
R5C1=5, R5C2=1, R5C3=2
R6C1=4, R6C2=8, R6C3=7
Box 4 is complete except for R4C1, R4C2.

Look at Column 1:
Values: 8(R1), _(R2), 9(R3), _(R4), 5(R5), 4(R6), _(R7), _(R8), 2(R9).
Missing: 1, 3, 6, 7.
R4C1 $\in \{3, 6\}$.

Look at Column 2:
Values: _(R1), _(R2), 2(R3), _(R4), 1(R5), 8(R6), 5(R7), 9(R8), 4(R9).
Missing: 3, 6, 7.
R4C2 $\in \{3, 6\}$.

Let's look at Row 1: `8 . 6 2 3 4 . 1 .`
Missing: 5, 7, 9.
Cells: R1C2, R1C7, R1C9.
Col 2 missing: 3, 6, 7.
R1C2 is in Col 2. So R1C2 $\in \{3, 6, 7\}$.
Row 1 missing: 5, 7, 9.
Intersection for R1C2: 7.
So, R1C2 = 7.

Now Row 1 missing: 5, 9.
Cells: R1C7, R1C9.
Col 7 missing: 1, 3, 5, 6, 9.
Col 9 missing: 1, 4, 6, 9.
R1C7 $\in \{5, 9\}$.
R1C9 $\in \{5, 9\}$.

Update Column 2:
Values: 7(R1), _(R2), 2(R3), _(R4), 1(R5), 8(R6), 5(R7), 9(R8), 4(R9).
Missing: 3, 6.
Cells: R2C2, R4C2.
We know R4C2 $\in \{3, 6\}$.
So R2C2 $\in \{3, 6\}$.

Look at Row 2: `. . 4 8 9 5 . 2 7`
Missing: 1, 3, 6.
Cells: R2C1, R2C2, R2C7.
R2C2 $\in \{3, 6\}$.
So R2C1, R2C7 must contain the remaining numbers.
Col 1 missing: 1, 3, 6, 7.
R2C1 is in Col 1.
Row 2 missing 1, 3, 6.
If R2C2 is 3 or 6, then R2C1 is 1 or the other of 3/6.

Let's look at Box 1 (Top Left).
Cells:
R1: 8, 7, 6
R2: R2C1, R2C2, 4
R3: 9, 2, R3C3
Numbers present: 2, 4, 6, 7, 8, 9.
Missing: 1, 3, 5.
Cells: R2C1, R2C2, R3C3.
So R2C1, R2C2, R3C3 $\in \{1, 3, 5\}$.
We know R2C2 $\in \{3, 6\}$. But Box 1 missing doesn't have 6.
Contradiction?
Wait. Col 2 missing was 3, 6.
R1C2=7. R3C2=2. R5C2=1. R6C2=8. R7C2=5. R8C2=9. R9C2=4.
Col 2 values: 7, ?, 2, ?, 1, 8, 5, 9, 4.
Missing: 3, 6.
So R2C2 and R4C2 are 3 and 6.
But R2C2 is in Box 1.
Box 1 missing: 1, 3, 5.
So R2C2 MUST be 3. (It can't be 6 because 6 is already in Box 1 at R1C3).
Therefore, R2C2 = 3.
And consequently, R4C2 = 6.
And since R4C1, R4C2 were {3,6}, R4C1 = 3.

Now Box 1 missing: 1, 5.
Cells: R2C1, R3C3.
Row 2 missing: 1, 6. (Since R2C2=3, Row 2 `. 3 4 8 9 5 . 2 7`. Missing 1, 6).
Cells: R2C1, R2C7.
R2C1 is in Box 1. Box 1 missing 1, 5.
So R2C1 must be 1 or 5.
Row 2 missing 1, 6.
Intersection: R2C1 = 1.
Therefore, R3C3 = 5. (Last spot in Box 1).
And R2C7 = 6. (Last spot in Row 2).

Step 5: Fill remaining easy spots

Column 1:
Values: 8, 1, 9, 3, 5, 4, ?, ?, 2.
Missing: 6, 7.
Cells: R7C1, R8C1.
Row 7: `. 5 8 . 7 1 . 9 2`. Missing: 3, 4, 6.
R7C1 $\in \{6, 7\}$. Row 7 missing doesn't have 7.
So R7C1 = 6.
Therefore, R8C1 = 7.

Row 7: `6 5 8 . 7 1 . 9 2`.
Missing: 3, 4.
Cells: R7C4, R7C7.
Col 4 values: 2, 8, ?, 5, 7, 9, ?, 3, 6.
Missing in Col 4: 1, 4.
R7C4 $\in \{3, 4\}$. Col 4 missing 1, 4.
So R7C4 = 4.
Therefore, R7C7 = 3.

Col 4:
Missing: 1.
Cell: R3C4.
So R3C4 = 1.

Row 3: `9 2 5 1 . 7 8 . 3`.
Missing: 4, 6.
Cells: R3C5, R3C8.
Col 5 values: 3, 9, ?, 4, 8, 1, 7, ?, 5.
Missing: 2, 6.
R3C5 $\in \{4, 6\}$. Col 5 missing 2, 6.
So R3C5 = 6.
Therefore, R3C8 = 4.

Col 5:
Missing: 2.
Cell: R8C5.
So R8C5 = 2.

Row 8: `7 9 1 3 2 . 4 5 .`.
Missing: 6, 8.
Cells: R8C6, R8C9.
Col 6 values: 4, 5, 7, 2, 6, 3, 1, ?, 9.
Missing: 8.
So R8C6 = 8.
Therefore, R8C9 = 6.

Row 9: `2 4 . 6 5 9 . . 8`.
Missing: 1, 3, 7.
Cells: R9C3, R9C7, R9C8.
Col 3 values: 6, 4, 5, 9, 2, 7, 8, 1, ?.
Missing: 3.
So R9C3 = 3.

Row 9 remaining: 1, 7.
Cells: R9C7, R9C8.
Col 7 values: ?, 6, 8, 7, ?, 2, 3, 4, ?.
Missing: 1, 5, 9.
R9C7 $\in \{1, 7\}$. Col 7 missing 1, 5, 9.
So R9C7 = 1.
Therefore, R9C8 = 7.

Col 7:
Missing: 5, 9.
Cells: R1C7, R5C7.
Row 1 missing: 5, 9.
Row 5 missing: 3, 4, 9? Let's check Row 5.
Row 5: `5 1 2 7 8 6 . . .`
We determined R5C7, R5C8, R5C9 $\in \{3, 4, 9\}$.
R5C7 is in Col 7. Col 7 missing 5, 9.
So R5C7 must be 9. (Since 5 is not in Row 5's options for those cells).
So R5C7 = 9.
Therefore, R1C7 = 5.

Row 1:
Last cell R1C9.
Row 1: `8 7 6 2 3 4 5 1 .`
Missing: 9.
So R1C9 = 9.

Row 5:
Remaining cells R5C8, R5C9.
Missing: 3, 4.
Col 8 values: 1, 3, 4, 8, ?, 6, 9, 5, 7.
Missing: 2. Wait.
Col 8:
R1:1, R2:2, R3:4, R4:8, R5:?, R6:6, R7:9, R8:5, R9:7.
Missing: 3.
So R5C8 = 3.
Therefore, R5C9 = 4.

Final Check of the Grid:

R1: 8 7 6 2 3 4 5 1 9
R2: 1 3 4 8 9 5 6 2 7
R3: 9 2 5 1 6 7 8 4 3
R4: 3 6 9 5 4 2 7 8 1
R5: 5 1 2 7 8 6 9 3 4
R6: 4 8 7 9 1 3 2 6 5
R7: 6 5 8 4 7 1 3 9 2
R8: 7 9 1 3 2 8 4 5 6
R9: 2 4 3 6 5 9 1 7 8

All rows, columns, and boxes contain 1-9 uniquely.

Final Answer:
The solved grid is:
8 7 6 2 3 4 5 1 9
1 3 4 8 9 5 6 2 7
9 2 5 1 6 7 8 4 3
3 6 9 5 4 2 7 8 1
5 1 2 7 8 6 9 3 4
4 8 7 9 1 3 2 6 5
6 5 8 4 7 1 3 9 2
7 9 1 3 2 8 4 5 6
2 4 3 6 5 9 1 7 8