Daily Sudoku Challenge
Today's Skudo daily Sudoku puzzle, who finished it first, and the full step-by-step solution to yesterday's puzzle. Free, printable, no sign-up.
Today's puzzle — 2 August 2026
| · | · | 5 | · | · | · | · | 7 | 3 |
| 2 | 4 | · | 7 | · | 6 | · | 5 | · |
| 6 | · | · | · | · | 1 | · | · | · |
| · | · | · | 4 | · | 5 | · | 3 | · |
| 9 | · | 3 | 8 | · | · | · | · | 2 |
| · | · | · | 3 | 6 | 9 | · | 8 | · |
| 3 | · | 1 | 2 | · | · | 4 | 6 | 9 |
| · | · | 2 | 6 | · | · | 3 | · | 7 |
| · | 9 | · | · | 7 | · | · | · | 5 |
First to finish
Ranked by when each player finished, not by how long they took — so looking up a solution never moves anyone up this board.
Open the app to see who finished first.
Yesterday's puzzle — 1 August 2026
| · | 1 | · | 8 | · | · | · | 6 | · |
| · | · | 6 | 5 | · | 3 | 8 | 4 | 2 |
| · | 2 | 4 | · | · | 6 | · | 3 | 1 |
| 1 | · | · | · | · | · | · | · | 3 |
| 2 | 6 | · | · | · | 9 | 1 | 5 | 4 |
| 7 | · | · | 1 | · | 5 | 2 | · | 6 |
| 6 | 9 | · | · | · | 7 | · | 1 | 8 |
| 4 | · | 1 | · | 9 | · | · | · | 5 |
| 3 | · | · | 4 | 8 | 1 | · | · | · |
| 5 | 1 | 3 | 8 | 2 | 4 | 9 | 6 | 7 |
| 9 | 7 | 6 | 5 | 1 | 3 | 8 | 4 | 2 |
| 8 | 2 | 4 | 9 | 7 | 6 | 5 | 3 | 1 |
| 1 | 4 | 5 | 2 | 6 | 8 | 7 | 9 | 3 |
| 2 | 6 | 8 | 7 | 3 | 9 | 1 | 5 | 4 |
| 7 | 3 | 9 | 1 | 4 | 5 | 2 | 8 | 6 |
| 6 | 9 | 2 | 3 | 5 | 7 | 4 | 1 | 8 |
| 4 | 8 | 1 | 6 | 9 | 2 | 3 | 7 | 5 |
| 3 | 5 | 7 | 4 | 8 | 1 | 6 | 2 | 9 |
Solve it in the app to get on the board
Step-by-step solution
Every cell below is filled by the same hint engine the app uses, in the order it can be deduced — 41 steps using 2 techniques.
Techniques used
- Hidden Single — inside one row, column or box this digit fits in only one cell, so it must go there
- Naked Single — every other digit is already used in this cell's row, column or box, so only one candidate is left
- R1C3 = 3 · Hidden SingleIn the top-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 3, R1C1 cannot hold 3.- Because row 2 already has a 3, R2C1 and R2C2 cannot hold 3.- Because row 3 already has a 3, R3C1 cannot hold 3.So 3 must go in R1C3.
- R2C2 = 7 · Hidden SingleIn the top-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 7, R1C1, R2C1 and R3C1 cannot hold 7.So 7 must go in R2C2.
- R3C1 = 8 · Hidden SingleIn the top-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 8, R1C1 cannot hold 8.- Because row 2 already has a 8, R2C1 cannot hold 8.So 8 must go in R3C1.
- R1C1 = 5 · Hidden SingleIn the top-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 5, R2C1 cannot hold 5.So 5 must go in R1C1.
- R2C1 = 9 · Naked SingleLooking at R2C1:- Row 2 already has 2, 3, 4, 5, 6, 7 and 8 → ruled out.- Column 1 already has 1 → ruled out.So this cell can only be 9.
- R2C5 = 1 · Naked SingleLooking at R2C5:- Row 2 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
- R3C4 = 9 · Hidden SingleIn the top-centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 9, R1C5 and R3C5 cannot hold 9.- Because column 6 already has a 9, R1C6 cannot hold 9.So 9 must go in R3C4.
- R3C7 = 5 · Hidden SingleIn the top-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 5, R1C7 and R1C9 cannot hold 5.So 5 must go in R3C7.
- R3C5 = 7 · Naked SingleLooking at R3C5:- Row 3 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
- R6C2 = 3 · Hidden SingleIn the middle-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 3, R4C2 and R4C3 cannot hold 3.- Because column 3 already has a 3, R5C3 and R6C3 cannot hold 3.So 3 must go in R6C2.
- R4C2 = 4 · Hidden SingleIn the middle-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 4, R4C3 and R6C3 cannot hold 4.- Because row 5 already has a 4, R5C3 cannot hold 4.So 4 must go in R4C2.
- R4C3 = 5 · Hidden SingleIn the middle-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 5, R5C3 cannot hold 5.- Because row 6 already has a 5, R6C3 cannot hold 5.So 5 must go in R4C3.
- R6C3 = 9 · Hidden SingleIn the middle-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 9, R5C3 cannot hold 9.So 9 must go in R6C3.
- R5C3 = 8 · Naked SingleLooking at R5C3:- Row 5 already has 1, 2, 4, 5, 6 and 9 → ruled out.- Column 3 already has 3 → ruled out.- The middle-left 3×3 box already has 7 → ruled out.So this cell can only be 8.
- R6C5 = 4 · Hidden SingleIn the centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 4, R4C4, R4C5 and R4C6 cannot hold 4.- Because row 5 already has a 4, R5C4 and R5C5 cannot hold 4.So 4 must go in R6C5.
- R6C8 = 8 · Naked SingleLooking at R6C8:- Row 6 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
- R1C6 = 4 · Hidden SingleIn the top-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 4, R1C5 cannot hold 4.So 4 must go in R1C6.
- R1C5 = 2 · Naked SingleLooking at R1C5:- Row 1 already has 1, 3, 4, 5, 6 and 8 → ruled out.- Column 5 already has 7 and 9 → ruled out.So this cell can only be 2.
- R4C6 = 8 · Hidden SingleIn the centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 8, R4C4 cannot hold 8.- Because column 5 already has a 8, R4C5 cannot hold 8.- Because row 5 already has a 8, R5C4 and R5C5 cannot hold 8.So 8 must go in R4C6.
- R8C6 = 2 · Naked SingleLooking at R8C6:- Row 8 already has 1, 4, 5 and 9 → ruled out.- Column 6 already has 3, 6, 7 and 8 → ruled out.So this cell can only be 2.
- R4C4 = 2 · Hidden SingleIn the centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 2, R4C5 cannot hold 2.- Because row 5 already has a 2, R5C4 and R5C5 cannot hold 2.So 2 must go in R4C4.
- R4C5 = 6 · Hidden SingleIn the centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 6, R5C4 and R5C5 cannot hold 6.So 6 must go in R4C5.
- R5C4 = 7 · Hidden SingleIn the centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 7, R5C5 cannot hold 7.So 7 must go in R5C4.
- R5C5 = 3 · Naked SingleLooking at R5C5:- Row 5 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
- R7C5 = 5 · Naked SingleLooking at R7C5:- Row 7 already has 1, 6, 7, 8 and 9 → ruled out.- Column 5 already has 2, 3 and 4 → ruled out.So this cell can only be 5.
- R9C2 = 5 · Hidden SingleIn the bottom-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 5, R7C3 cannot hold 5.- Because row 8 already has a 5, R8C2 cannot hold 5.- Because column 3 already has a 5, R9C3 cannot hold 5.So 5 must go in R9C2.
- R8C2 = 8 · Naked SingleLooking at R8C2:- Row 8 already has 1, 2, 4, 5 and 9 → ruled out.- Column 2 already has 3, 6 and 7 → ruled out.So this cell can only be 8.
- R9C3 = 7 · Hidden SingleIn the bottom-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 7, R7C3 cannot hold 7.So 7 must go in R9C3.
- R7C3 = 2 · Naked SingleLooking at R7C3:- Row 7 already has 1, 5, 6, 7, 8 and 9 → ruled out.- Column 3 already has 3 and 4 → ruled out.So this cell can only be 2.
- R8C4 = 6 · Hidden SingleIn the bottom-centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 6, R7C4 cannot hold 6.So 6 must go in R8C4.
- R7C4 = 3 · Naked SingleLooking at R7C4:- Row 7 already has 1, 2, 5, 6, 7, 8 and 9 → ruled out.- Column 4 already has 4 → ruled out.So this cell can only be 3.
- R7C7 = 4 · Naked SingleLooking at R7C7:- Row 7 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
- R9C8 = 2 · Hidden SingleIn the bottom-right 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 2, R8C7 and R8C8 cannot hold 2.- Because column 7 already has a 2, R9C7 cannot hold 2.- Because column 9 already has a 2, R9C9 cannot hold 2.So 2 must go in R9C8.
- R8C7 = 3 · Hidden SingleIn the bottom-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 3, R8C8 cannot hold 3.- Because row 9 already has a 3, R9C7 and R9C9 cannot hold 3.So 3 must go in R8C7.
- R8C8 = 7 · Naked SingleLooking at R8C8:- Row 8 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
- R4C8 = 9 · Naked SingleLooking at R4C8:- Row 4 already has 1, 2, 3, 4, 5, 6 and 8 → ruled out.- Column 8 already has 7 → ruled out.So this cell can only be 9.
- R4C7 = 7 · Naked SingleLooking at R4C7:- Row 4 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
- R1C9 = 7 · Hidden SingleIn the top-right 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 7, R1C7 cannot hold 7.So 7 must go in R1C9.
- R1C7 = 9 · Naked SingleLooking at R1C7:- Row 1 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
- R9C7 = 6 · Naked SingleLooking at R9C7:- Row 9 already has 1, 2, 3, 4, 5, 7 and 8 → ruled out.- Column 7 already has 9 → ruled out.So this cell can only be 6.
- R9C9 = 9 · Naked SingleLooking at R9C9:- Row 9 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.