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Daily Sudoku — 29 July 2026

The puzzle

  • 29 July 2026
  • Medium
  • 34 clues
Puzzle
2·1··5···
68·3·7···
3···91··5
·5·4·29··
··2·1·8··
·16···4·2
··4·2·1·8
···1···34
·685·4·2·
Solution
291685347
685347291
347291685
853472916
472916853
916853472
534729168
729168534
168534729

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 — 47 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
  1. R2C3 = 5 · Hidden SingleIn the top-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 5, R1C2 cannot hold 5.- Because row 3 already has a 5, R3C2 and R3C3 cannot hold 5.So 5 must go in R2C3.
  2. R1C2 = 9 · Hidden SingleIn the top-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 3 already has a 9, R3C2 and R3C3 cannot hold 9.So 9 must go in R1C2.
  3. R3C2 = 4 · Hidden SingleIn the top-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 4, R3C3 cannot hold 4.So 4 must go in R3C2.
  4. R3C3 = 7 · Naked SingleLooking at R3C3:- Row 3 already has 1, 3, 4, 5 and 9 → ruled out.- Column 3 already has 2, 6 and 8 → ruled out.So this cell can only be 7.
  5. R3C4 = 2 · Hidden SingleIn the top-centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 2, R1C4 and R1C5 cannot hold 2.- Because column 5 already has a 2, R2C5 cannot hold 2.So 2 must go in R3C4.
  6. R2C7 = 2 · Hidden SingleIn the top-right 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 2, R1C7, R1C8 and R1C9 cannot hold 2.- Because column 8 already has a 2, R2C8 cannot hold 2.- Because column 9 already has a 2, R2C9 cannot hold 2.- Because row 3 already has a 2, R3C7 and R3C8 cannot hold 2.So 2 must go in R2C7.
  7. R5C1 = 4 · Hidden SingleIn the middle-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 4, R4C1 and R4C3 cannot hold 4.- Because column 2 already has a 4, R5C2 cannot hold 4.- Because row 6 already has a 4, R6C1 cannot hold 4.So 4 must go in R5C1.
  8. R6C1 = 9 · Hidden SingleIn the middle-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 9, R4C1 and R4C3 cannot hold 9.- Because column 2 already has a 9, R5C2 cannot hold 9.So 9 must go in R6C1.
  9. R4C1 = 8 · Hidden SingleIn the middle-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 8, R4C3 cannot hold 8.- Because row 5 already has a 8, R5C2 cannot hold 8.So 8 must go in R4C1.
  10. R5C2 = 7 · Hidden SingleIn the middle-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 7, R4C3 cannot hold 7.So 7 must go in R5C2.
  11. R4C3 = 3 · Naked SingleLooking at R4C3:- Row 4 already has 2, 4, 5, 8 and 9 → ruled out.- Column 3 already has 1, 6 and 7 → ruled out.So this cell can only be 3.
  12. R8C3 = 9 · Naked SingleLooking at R8C3:- Row 8 already has 1, 3 and 4 → ruled out.- Column 3 already has 2, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  13. R6C5 = 5 · Hidden SingleIn the centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 5, R4C5 cannot hold 5.- Because column 4 already has a 5, R5C4 and R6C4 cannot hold 5.- Because column 6 already has a 5, R5C6 and R6C6 cannot hold 5.So 5 must go in R6C5.
  14. R5C9 = 3 · Hidden SingleIn the middle-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 3, R4C8 and R4C9 cannot hold 3.- Because column 8 already has a 3, R5C8 and R6C8 cannot hold 3.So 3 must go in R5C9.
  15. R1C7 = 3 · Hidden SingleIn the top-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 3, R1C8 cannot hold 3.- Because column 9 already has a 3, R1C9 cannot hold 3.- Because row 2 already has a 3, R2C8 and R2C9 cannot hold 3.- Because row 3 already has a 3, R3C7 and R3C8 cannot hold 3.So 3 must go in R1C7.
  16. R6C6 = 3 · Hidden SingleIn the centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 3, R4C5 cannot hold 3.- Because row 5 already has a 3, R5C4 and R5C6 cannot hold 3.- Because column 4 already has a 3, R6C4 cannot hold 3.So 3 must go in R6C6.
  17. R6C4 = 8 · Hidden SingleIn the centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 8, R4C5 cannot hold 8.- Because row 5 already has a 8, R5C4 and R5C6 cannot hold 8.So 8 must go in R6C4.
  18. R6C8 = 7 · Naked SingleLooking at R6C8:- Row 6 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  19. R1C5 = 8 · Hidden SingleIn the top-centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 8, R1C4 cannot hold 8.- Because row 2 already has a 8, R2C5 cannot hold 8.So 8 must go in R1C5.
  20. R2C5 = 4 · Hidden SingleIn the top-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 4, R1C4 cannot hold 4.So 4 must go in R2C5.
  21. R1C4 = 6 · Naked SingleLooking at R1C4:- Row 1 already has 1, 2, 3, 5, 8 and 9 → ruled out.- Column 4 already has 4 → ruled out.- The top-centre 3×3 box already has 7 → ruled out.So this cell can only be 6.
  22. R1C8 = 4 · Hidden SingleIn the top-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 4, R1C9 cannot hold 4.- Because row 2 already has a 4, R2C8 and R2C9 cannot hold 4.- Because row 3 already has a 4, R3C7 and R3C8 cannot hold 4.So 4 must go in R1C8.
  23. R1C9 = 7 · Naked SingleLooking at R1C9:- Row 1 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  24. R3C8 = 8 · Hidden SingleIn the top-right 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 8, R2C8 and R2C9 cannot hold 8.- Because column 7 already has a 8, R3C7 cannot hold 8.So 8 must go in R3C8.
  25. R3C7 = 6 · Naked SingleLooking at R3C7:- Row 3 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  26. R4C5 = 7 · Hidden SingleIn the centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 7, R5C4 and R5C6 cannot hold 7.So 7 must go in R4C5.
  27. R5C6 = 6 · Hidden SingleIn the centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 6, R5C4 cannot hold 6.So 6 must go in R5C6.
  28. R5C4 = 9 · Naked SingleLooking at R5C4:- Row 5 already has 1, 2, 3, 4, 6, 7 and 8 → ruled out.- Column 4 already has 5 → ruled out.So this cell can only be 9.
  29. R5C8 = 5 · Naked SingleLooking at R5C8:- Row 5 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  30. R7C4 = 7 · Naked SingleLooking at R7C4:- Row 7 already has 1, 2, 4 and 8 → ruled out.- Column 4 already has 3, 5, 6 and 9 → ruled out.So this cell can only be 7.
  31. R9C1 = 1 · Hidden SingleIn the bottom-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 1, R7C1 and R7C2 cannot hold 1.- Because row 8 already has a 1, R8C1 and R8C2 cannot hold 1.So 1 must go in R9C1.
  32. R8C2 = 2 · Hidden SingleIn the bottom-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 2, R7C1 and R7C2 cannot hold 2.- Because column 1 already has a 2, R8C1 cannot hold 2.So 2 must go in R8C2.
  33. R7C2 = 3 · Naked SingleLooking at R7C2:- Row 7 already has 1, 2, 4, 7 and 8 → ruled out.- Column 2 already has 5, 6 and 9 → ruled out.So this cell can only be 3.
  34. R8C1 = 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, R7C1 cannot hold 7.So 7 must go in R8C1.
  35. R7C1 = 5 · Naked SingleLooking at R7C1:- Row 7 already has 1, 2, 3, 4, 7 and 8 → ruled out.- Column 1 already has 6 and 9 → ruled out.So this cell can only be 5.
  36. R9C5 = 3 · Hidden SingleIn the bottom-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 3, R7C6 cannot hold 3.- Because row 8 already has a 3, R8C5 and R8C6 cannot hold 3.So 3 must go in R9C5.
  37. R8C5 = 6 · Naked SingleLooking at R8C5:- Row 8 already has 1, 2, 3, 4, 7 and 9 → ruled out.- Column 5 already has 5 and 8 → ruled out.So this cell can only be 6.
  38. R8C6 = 8 · Hidden SingleIn the bottom-centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 8, R7C6 cannot hold 8.So 8 must go in R8C6.
  39. R8C7 = 5 · Naked SingleLooking at R8C7:- Row 8 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  40. R7C6 = 9 · Naked SingleLooking at R7C6:- Row 7 already has 1, 2, 3, 4, 5, 7 and 8 → ruled out.- Column 6 already has 6 → ruled out.So this cell can only be 9.
  41. R7C8 = 6 · Naked SingleLooking at R7C8:- Row 7 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  42. R9C7 = 7 · Naked SingleLooking at R9C7:- Row 9 already has 1, 2, 3, 4, 5, 6 and 8 → ruled out.- Column 7 already has 9 → ruled out.So this cell can only be 7.
  43. 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.
  44. R2C8 = 9 · Hidden SingleIn the top-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 9, R2C9 cannot hold 9.So 9 must go in R2C8.
  45. R2C9 = 1 · Naked SingleLooking at R2C9:- Row 2 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  46. R4C8 = 1 · Naked SingleLooking at R4C8:- Row 4 already has 2, 3, 4, 5, 7, 8 and 9 → ruled out.- Column 8 already has 6 → ruled out.So this cell can only be 1.
  47. R4C9 = 6 · Naked SingleLooking at R4C9:- Row 4 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.

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