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Daily Sudoku — 8 July 2026

The puzzle

  • 8 July 2026
  • Hard
  • 30 clues
Puzzle
9·82···5·
···8·7··1
·3·6·5···
····38···
···4·6·83
··9·5·4··
·····9·16
2··5617·8
·1····3··
Solution
978213654
456897231
132645879
764938125
521476983
389152467
847329516
293561748
615784392

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 — 51 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. R1C6 = 3 · Hidden SingleIn the top-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 3, R1C5 and R2C5 cannot hold 3.- Because row 3 already has a 3, R3C5 cannot hold 3.So 3 must go in R1C6.
  2. R2C8 = 3 · Hidden SingleIn the top-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 3, R1C7 and R1C9 cannot hold 3.- Because column 7 already has a 3, R2C7 cannot hold 3.- Because row 3 already has a 3, R3C7, R3C8 and R3C9 cannot hold 3.So 3 must go in R2C8.
  3. R3C7 = 8 · Hidden SingleIn the top-right 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 8, R1C7 and R1C9 cannot hold 8.- Because row 2 already has a 8, R2C7 cannot hold 8.- Because column 8 already has a 8, R3C8 cannot hold 8.- Because column 9 already has a 8, R3C9 cannot hold 8.So 8 must go in R3C7.
  4. R6C1 = 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, R4C1, R4C2 and R4C3 cannot hold 3.- Because row 5 already has a 3, R5C1, R5C2 and R5C3 cannot hold 3.- Because column 2 already has a 3, R6C2 cannot hold 3.So 3 must go in R6C1.
  5. R6C2 = 8 · Hidden SingleIn the middle-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 8, R4C1, R4C2 and R4C3 cannot hold 8.- Because row 5 already has a 8, R5C1, R5C2 and R5C3 cannot hold 8.So 8 must go in R6C2.
  6. R8C2 = 9 · Hidden SingleIn the bottom-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 9, R7C1, R7C2 and R7C3 cannot hold 9.- Because column 3 already has a 9, R8C3 and R9C3 cannot hold 9.- Because column 1 already has a 9, R9C1 cannot hold 9.So 9 must go in R8C2.
  7. R7C4 = 3 · Hidden SingleIn the bottom-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 3, R7C5 cannot hold 3.- Because row 9 already has a 3, R9C4, R9C5 and R9C6 cannot hold 3.So 3 must go in R7C4.
  8. R8C3 = 3 · Hidden SingleIn the bottom-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 3, R7C1, R7C2 and R7C3 cannot hold 3.- Because row 9 already has a 3, R9C1 and R9C3 cannot hold 3.So 3 must go in R8C3.
  9. R8C8 = 4 · Naked SingleLooking at R8C8:- Row 8 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  10. R1C5 = 1 · Hidden SingleIn row 1, 1 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 1, R1C2 cannot hold 1.- Because the top-right 3×3 box already has a 1, R1C7 cannot hold 1.- Because column 9 already has a 1, R1C9 cannot hold 1.So 1 must go in R1C5.
  11. R6C4 = 1 · Hidden SingleIn row 6, 1 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 1, R6C6 cannot hold 1.- Because column 8 already has a 1, R6C8 cannot hold 1.- Because column 9 already has a 1, R6C9 cannot hold 1.So 1 must go in R6C4.
  12. R6C8 = 6 · Hidden SingleIn row 6, 6 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 6, R6C6 cannot hold 6.- Because column 9 already has a 6, R6C9 cannot hold 6.So 6 must go in R6C8.
  13. R6C9 = 7 · Hidden SingleIn row 6, 7 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 7, R6C6 cannot hold 7.So 7 must go in R6C9.
  14. R6C6 = 2 · Naked SingleLooking at R6C6:- Row 6 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  15. R9C6 = 4 · Naked SingleLooking at R9C6:- Row 9 already has 1 and 3 → ruled out.- Column 6 already has 2, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  16. R3C8 = 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.- Because column 9 already has a 7, R1C9 and R3C9 cannot hold 7.- Because row 2 already has a 7, R2C7 cannot hold 7.So 7 must go in R3C8.
  17. R1C2 = 7 · Hidden SingleIn the top-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 7, R2C1, R2C2 and R2C3 cannot hold 7.- Because row 3 already has a 7, R3C1 and R3C3 cannot hold 7.So 7 must go in R1C2.
  18. R1C9 = 4 · Hidden SingleIn row 1, 4 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 4, R1C7 cannot hold 4.So 4 must go in R1C9.
  19. R1C7 = 6 · Naked SingleLooking at R1C7:- Row 1 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  20. R4C4 = 9 · Hidden SingleIn column 4, 9 fits in only one cell. Looking at the other empty cells:- Because the bottom-centre 3×3 box already has a 9, R9C4 cannot hold 9.So 9 must go in R4C4.
  21. R9C4 = 7 · Naked SingleLooking at R9C4:- Row 9 already has 1, 3 and 4 → ruled out.- Column 4 already has 2, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  22. R5C5 = 7 · Naked SingleLooking at R5C5:- Row 5 already has 3, 4, 6 and 8 → ruled out.- Column 5 already has 1 and 5 → ruled out.- The centre 3×3 box already has 2 and 9 → ruled out.So this cell can only be 7.
  23. R5C7 = 9 · Hidden SingleIn the middle-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 9, R4C7, R4C8 and R4C9 cannot hold 9.So 9 must go in R5C7.
  24. R3C9 = 9 · Hidden SingleIn the top-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 9, R2C7 cannot hold 9.So 9 must go in R3C9.
  25. R2C7 = 2 · Naked SingleLooking at R2C7:- Row 2 already has 1, 3, 7 and 8 → ruled out.- Column 7 already has 4, 6 and 9 → ruled out.- The top-right 3×3 box already has 5 → ruled out.So this cell can only be 2.
  26. R3C3 = 2 · Hidden SingleIn the top-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 2, R2C1, R2C2 and R2C3 cannot hold 2.- Because column 1 already has a 2, R3C1 cannot hold 2.So 2 must go in R3C3.
  27. R3C1 = 1 · Hidden SingleIn the top-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 1, R2C1, R2C2 and R2C3 cannot hold 1.So 1 must go in R3C1.
  28. R3C5 = 4 · Naked SingleLooking at R3C5:- Row 3 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  29. R2C5 = 9 · Naked SingleLooking at R2C5:- Row 2 already has 1, 2, 3, 7 and 8 → ruled out.- Column 5 already has 4, 5 and 6 → ruled out.So this cell can only be 9.
  30. R4C7 = 1 · Hidden SingleIn the middle-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 1, R4C8 cannot hold 1.- Because column 9 already has a 1, R4C9 cannot hold 1.So 1 must go in R4C7.
  31. R7C7 = 5 · Naked SingleLooking at R7C7:- Row 7 already has 1, 3, 6 and 9 → ruled out.- Column 7 already has 2, 4, 7 and 8 → ruled out.So this cell can only be 5.
  32. R5C3 = 1 · Hidden SingleIn the middle-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 1, R4C1, R4C2 and R4C3 cannot hold 1.- Because column 1 already has a 1, R5C1 cannot hold 1.- Because column 2 already has a 1, R5C2 cannot hold 1.So 1 must go in R5C3.
  33. R4C9 = 5 · Hidden SingleIn the middle-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 5, R4C8 cannot hold 5.So 5 must go in R4C9.
  34. R9C9 = 2 · Naked SingleLooking at R9C9:- Row 9 already has 1, 3, 4 and 7 → ruled out.- Column 9 already has 5, 6, 8 and 9 → ruled out.So this cell can only be 2.
  35. R4C8 = 2 · Naked SingleLooking at R4C8:- Row 4 already has 1, 3, 5, 8 and 9 → ruled out.- Column 8 already has 4, 6 and 7 → ruled out.So this cell can only be 2.
  36. R9C8 = 9 · Naked SingleLooking at R9C8:- Row 9 already has 1, 2, 3, 4 and 7 → ruled out.- Column 8 already has 5, 6 and 8 → ruled out.So this cell can only be 9.
  37. R5C2 = 2 · Hidden SingleIn the middle-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 2, R4C1, R4C2 and R4C3 cannot hold 2.- Because column 1 already has a 2, R5C1 cannot hold 2.So 2 must go in R5C2.
  38. R5C1 = 5 · Naked SingleLooking at R5C1:- Row 5 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  39. R9C3 = 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, R7C1, R7C2 and R7C3 cannot hold 5.- Because column 1 already has a 5, R9C1 cannot hold 5.So 5 must go in R9C3.
  40. R2C2 = 5 · Hidden SingleIn the top-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 5, R2C1 cannot hold 5.- Because column 3 already has a 5, R2C3 cannot hold 5.So 5 must go in R2C2.
  41. R9C1 = 6 · Hidden SingleIn the bottom-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 6, R7C1, R7C2 and R7C3 cannot hold 6.So 6 must go in R9C1.
  42. R9C5 = 8 · Naked SingleLooking at R9C5:- Row 9 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  43. R7C5 = 2 · Naked SingleLooking at R7C5:- Row 7 already has 1, 3, 5, 6 and 9 → ruled out.- Column 5 already has 4, 7 and 8 → ruled out.So this cell can only be 2.
  44. R2C3 = 6 · Hidden SingleIn the top-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 6, R2C1 cannot hold 6.So 6 must go in R2C3.
  45. R2C1 = 4 · Naked SingleLooking at R2C1:- Row 2 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  46. R4C2 = 6 · Hidden SingleIn the middle-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 6, R4C1 cannot hold 6.- Because column 3 already has a 6, R4C3 cannot hold 6.So 6 must go in R4C2.
  47. R7C2 = 4 · Naked SingleLooking at R7C2:- Row 7 already has 1, 2, 3, 5, 6 and 9 → ruled out.- Column 2 already has 7 and 8 → ruled out.So this cell can only be 4.
  48. R4C3 = 4 · Hidden SingleIn the middle-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 4, R4C1 cannot hold 4.So 4 must go in R4C3.
  49. R4C1 = 7 · Naked SingleLooking at R4C1:- Row 4 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  50. R7C1 = 8 · Naked SingleLooking at R7C1:- Row 7 already has 1, 2, 3, 4, 5, 6 and 9 → ruled out.- Column 1 already has 7 → ruled out.So this cell can only be 8.
  51. R7C3 = 7 · Naked SingleLooking at R7C3:- Row 7 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.

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