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Daily Sudoku — 13 July 2026

The puzzle

  • 13 July 2026
  • Hard
  • 30 clues
Puzzle
·4···1·95
··3·9·7··
·····7·8·
·3·65····
···7·8·3·
8·7··9···
·7·····64
5·94·2··8
··4·73··9
Solution
742381695
183596742
695247183
931654827
456728931
827139456
378915264
519462378
264873519

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. R1C1 = 7 · Hidden SingleIn the top-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 7, R1C3 cannot hold 7.- Because row 2 already has a 7, R2C1 and R2C2 cannot hold 7.- Because row 3 already has a 7, R3C1, R3C2 and R3C3 cannot hold 7.So 7 must go in R1C1.
  2. R4C7 = 8 · Hidden SingleIn the middle-right 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 8, R4C8 cannot hold 8.- Because column 9 already has a 8, R4C9 cannot hold 8.- Because row 5 already has a 8, R5C7 and R5C9 cannot hold 8.- Because row 6 already has a 8, R6C7, R6C8 and R6C9 cannot hold 8.So 8 must go in R4C7.
  3. R5C7 = 9 · Hidden SingleIn the middle-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 9, R4C8 cannot hold 9.- Because column 9 already has a 9, R4C9 and R5C9 cannot hold 9.- Because row 6 already has a 9, R6C7, R6C8 and R6C9 cannot hold 9.So 9 must go in R5C7.
  4. R4C1 = 9 · Hidden SingleIn the middle-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 9, R4C3 cannot hold 9.- Because row 5 already has a 9, R5C1, R5C2 and R5C3 cannot hold 9.- Because row 6 already has a 9, R6C2 cannot hold 9.So 9 must go in R4C1.
  5. R3C2 = 9 · Hidden SingleIn the top-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 9, R1C3 cannot hold 9.- Because row 2 already has a 9, R2C1 and R2C2 cannot hold 9.- Because column 1 already has a 9, R3C1 cannot hold 9.- Because column 3 already has a 9, R3C3 cannot hold 9.So 9 must go in R3C2.
  6. R5C1 = 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 R5C3 cannot hold 4.- Because column 2 already has a 4, R5C2 and R6C2 cannot hold 4.So 4 must go in R5C1.
  7. R7C1 = 3 · Hidden SingleIn the bottom-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 3, R7C3 cannot hold 3.- Because column 2 already has a 3, R8C2 cannot hold 3.- Because row 9 already has a 3, R9C1 and R9C2 cannot hold 3.So 3 must go in R7C1.
  8. R8C5 = 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, R7C5 and R7C6 cannot hold 6.- Because column 4 already has a 6, R9C4 cannot hold 6.So 6 must go in R8C5.
  9. R2C6 = 6 · Hidden SingleIn the top-centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 6, R1C4, R2C4 and R3C4 cannot hold 6.- Because column 5 already has a 6, R1C5 and R3C5 cannot hold 6.So 6 must go in R2C6.
  10. R3C5 = 4 · Hidden SingleIn the top-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 4, R1C4 and R1C5 cannot hold 4.- Because column 4 already has a 4, R2C4 and R3C4 cannot hold 4.So 4 must go in R3C5.
  11. R2C8 = 4 · Hidden SingleIn the top-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 4, R1C7 cannot hold 4.- Because column 9 already has a 4, R2C9 cannot hold 4.- Because row 3 already has a 4, R3C7 and R3C9 cannot hold 4.So 4 must go in R2C8.
  12. R4C6 = 4 · Hidden SingleIn the centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 4, R5C5 cannot hold 4.- Because column 4 already has a 4, R6C4 cannot hold 4.- Because column 5 already has a 4, R6C5 cannot hold 4.So 4 must go in R4C6.
  13. R7C6 = 5 · Naked SingleLooking at R7C6:- Row 7 already has 3, 4, 6 and 7 → ruled out.- Column 6 already has 1, 2, 8 and 9 → ruled out.So this cell can only be 5.
  14. R6C7 = 4 · Hidden SingleIn the middle-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 4, R4C8 and R4C9 cannot hold 4.- Because row 5 already has a 4, R5C9 cannot hold 4.- Because column 8 already has a 4, R6C8 cannot hold 4.- Because column 9 already has a 4, R6C9 cannot hold 4.So 4 must go in R6C7.
  15. R6C8 = 5 · Hidden SingleIn the middle-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 5, R4C8 and R4C9 cannot hold 5.- Because column 9 already has a 5, R5C9 and R6C9 cannot hold 5.So 5 must go in R6C8.
  16. R7C4 = 9 · Hidden SingleIn the bottom-centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 9, R7C5 cannot hold 9.- Because row 9 already has a 9, R9C4 cannot hold 9.So 9 must go in R7C4.
  17. R8C7 = 3 · Hidden SingleIn the bottom-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 3, R7C7 cannot hold 3.- Because column 8 already has a 3, R8C8 cannot hold 3.- Because row 9 already has a 3, R9C7 and R9C8 cannot hold 3.So 3 must go in R8C7.
  18. R3C9 = 3 · Hidden SingleIn the top-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 3, R1C7 and R3C7 cannot hold 3.- Because row 2 already has a 3, R2C9 cannot hold 3.So 3 must go in R3C9.
  19. R9C7 = 5 · Hidden SingleIn the bottom-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 5, R7C7 cannot hold 5.- Because row 8 already has a 5, R8C8 cannot hold 5.- Because column 8 already has a 5, R9C8 cannot hold 5.So 5 must go in R9C7.
  20. R8C8 = 7 · Hidden SingleIn the bottom-right 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 7, R7C7 cannot hold 7.- Because row 9 already has a 7, R9C8 cannot hold 7.So 7 must go in R8C8.
  21. R8C2 = 1 · Naked SingleLooking at R8C2:- Row 8 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  22. R4C9 = 7 · Hidden SingleIn the middle-right 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 7, R4C8 cannot hold 7.- Because row 5 already has a 7, R5C9 cannot hold 7.- Because row 6 already has a 7, R6C9 cannot hold 7.So 7 must go in R4C9.
  23. R7C7 = 2 · Hidden SingleIn row 7, 2 fits in only one cell. Looking at the other empty cells:- Because the bottom-centre 3×3 box already has a 2, R7C5 cannot hold 2.- Applying XY-Wing (a pivot with two pincers guarantees one digit lands in one of the pincers, so it is eliminated from any cell that sees both) means R7C3 cannot hold 2 either.So 2 must go in R7C7.
  24. R9C8 = 1 · Naked SingleLooking at R9C8:- Row 9 already has 3, 4, 5, 7 and 9 → ruled out.- Column 8 already has 6 and 8 → ruled out.- The bottom-right 3×3 box already has 2 → ruled out.So this cell can only be 1.
  25. R4C8 = 2 · Naked SingleLooking at R4C8:- Row 4 already has 3, 4, 5, 6, 7, 8 and 9 → ruled out.- Column 8 already has 1 → ruled out.So this cell can only be 2.
  26. R4C3 = 1 · Naked SingleLooking at R4C3:- Row 4 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  27. R2C9 = 2 · Hidden SingleIn the top-right 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 2, R1C7 and R3C7 cannot hold 2.So 2 must go in R2C9.
  28. R3C7 = 1 · Hidden SingleIn the top-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 1, R1C7 cannot hold 1.So 1 must go in R3C7.
  29. R1C7 = 6 · Naked SingleLooking at R1C7:- Row 1 already has 1, 4, 5, 7 and 9 → ruled out.- Column 7 already has 2, 3 and 8 → ruled out.So this cell can only be 6.
  30. R2C1 = 1 · Hidden SingleIn the top-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 1, R1C3 cannot hold 1.- Because column 2 already has a 1, R2C2 cannot hold 1.- Because row 3 already has a 1, R3C1 and R3C3 cannot hold 1.So 1 must go in R2C1.
  31. R7C5 = 1 · Hidden SingleIn the bottom-centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 9 already has a 1, R9C4 cannot hold 1.So 1 must go in R7C5.
  32. R7C3 = 8 · Naked SingleLooking at R7C3:- Row 7 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  33. R9C4 = 8 · Naked SingleLooking at R9C4:- Row 9 already has 1, 3, 4, 5, 7 and 9 → ruled out.- Column 4 already has 6 → ruled out.- The bottom-centre 3×3 box already has 2 → ruled out.So this cell can only be 8.
  34. R2C2 = 8 · Hidden SingleIn the top-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 8, R1C3 cannot hold 8.- Because row 3 already has a 8, R3C1 and R3C3 cannot hold 8.So 8 must go in R2C2.
  35. R2C4 = 5 · Naked SingleLooking at R2C4:- Row 2 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  36. R3C3 = 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, R1C3 cannot hold 5.- Because column 1 already has a 5, R3C1 cannot hold 5.So 5 must go in R3C3.
  37. R3C1 = 6 · Hidden SingleIn the top-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 6, R1C3 cannot hold 6.So 6 must go in R3C1.
  38. R3C4 = 2 · Naked SingleLooking at R3C4:- Row 3 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  39. R9C1 = 2 · Naked SingleLooking at R9C1:- Row 9 already has 1, 3, 4, 5, 7, 8 and 9 → ruled out.- Column 1 already has 6 → ruled out.So this cell can only be 2.
  40. R9C2 = 6 · Naked SingleLooking at R9C2:- Row 9 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  41. R1C3 = 2 · Naked SingleLooking at R1C3:- Row 1 already has 1, 4, 5, 6, 7 and 9 → ruled out.- Column 3 already has 3 and 8 → ruled out.So this cell can only be 2.
  42. R5C3 = 6 · Naked SingleLooking at R5C3:- Row 5 already has 3, 4, 7, 8 and 9 → ruled out.- Column 3 already has 1, 2 and 5 → ruled out.So this cell can only be 6.
  43. 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.So 8 must go in R1C5.
  44. R1C4 = 3 · Naked SingleLooking at R1C4:- Row 1 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  45. R6C4 = 1 · Naked SingleLooking at R6C4:- Row 6 already has 4, 5, 7, 8 and 9 → ruled out.- Column 4 already has 2, 3 and 6 → ruled out.So this cell can only be 1.
  46. R5C2 = 5 · Hidden SingleIn the middle-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 6 already has a 5, R6C2 cannot hold 5.So 5 must go in R5C2.
  47. R6C2 = 2 · Naked SingleLooking at R6C2:- Row 6 already has 1, 4, 5, 7, 8 and 9 → ruled out.- Column 2 already has 3 and 6 → ruled out.So this cell can only be 2.
  48. R5C5 = 2 · Hidden SingleIn the centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 6 already has a 2, R6C5 cannot hold 2.So 2 must go in R5C5.
  49. R5C9 = 1 · Naked SingleLooking at R5C9:- Row 5 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  50. R6C5 = 3 · Naked SingleLooking at R6C5:- Row 6 already has 1, 2, 4, 5, 7, 8 and 9 → ruled out.- Column 5 already has 6 → ruled out.So this cell can only be 3.
  51. R6C9 = 6 · Naked SingleLooking at R6C9:- Row 6 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.

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