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Daily Sudoku — 11 July 2026

The puzzle

  • 11 July 2026
  • Medium
  • 34 clues
Puzzle
··2······
9·6·4···7
·8·257·9·
··9·1·5·2
82·9·6·13
1·45···76
·········
··18···29
35·7·9·6·
Solution
572691348
916348257
483257691
769413582
825976413
134582976
297164835
641835729
358729164

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. R1C1 = 5 · Hidden SingleIn the top-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 5, R1C2 and R2C2 cannot hold 5.- Because row 3 already has a 5, R3C1 and R3C3 cannot hold 5.So 5 must go in R1C1.
  2. 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, R2C2 cannot hold 7.- Because row 3 already has a 7, R3C1 and R3C3 cannot hold 7.So 7 must go in R1C2.
  3. R2C2 = 1 · Hidden SingleIn the top-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 1, R3C1 cannot hold 1.- Because column 3 already has a 1, R3C3 cannot hold 1.So 1 must go in R2C2.
  4. R3C3 = 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, R3C1 cannot hold 3.So 3 must go in R3C3.
  5. R3C1 = 4 · Naked SingleLooking at R3C1:- Row 3 already has 2, 3, 5, 7, 8 and 9 → ruled out.- Column 1 already has 1 → ruled out.- The top-left 3×3 box already has 6 → ruled out.So this cell can only be 4.
  6. R1C5 = 9 · Hidden SingleIn the top-centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 9, R1C4 cannot hold 9.- Because column 6 already has a 9, R1C6 cannot hold 9.- Because row 2 already has a 9, R2C4 and R2C6 cannot hold 9.So 9 must go in R1C5.
  7. R1C4 = 6 · Hidden SingleIn the top-centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 6, R1C6 cannot hold 6.- Because row 2 already has a 6, R2C4 and R2C6 cannot hold 6.So 6 must go in R1C4.
  8. R1C6 = 1 · Hidden SingleIn the top-centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 1, R2C4 and R2C6 cannot hold 1.So 1 must go in R1C6.
  9. R2C6 = 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, R2C4 cannot hold 8.So 8 must go in R2C6.
  10. R2C4 = 3 · Naked SingleLooking at R2C4:- Row 2 already has 1, 4, 6, 7, 8 and 9 → ruled out.- Column 4 already has 2 and 5 → ruled out.So this cell can only be 3.
  11. 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 row 3 already has a 2, R3C7 and R3C9 cannot hold 2.So 2 must go in R2C7.
  12. R2C8 = 5 · Naked SingleLooking at R2C8:- Row 2 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  13. R3C7 = 6 · Hidden SingleIn the top-right 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 6, R1C7, R1C8 and R1C9 cannot hold 6.- Because column 9 already has a 6, R3C9 cannot hold 6.So 6 must go in R3C7.
  14. R3C9 = 1 · Naked SingleLooking at R3C9:- Row 3 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  15. R5C3 = 5 · Hidden SingleIn the middle-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 5, R4C1 and R4C2 cannot hold 5.- Because row 6 already has a 5, R6C2 cannot hold 5.So 5 must go in R5C3.
  16. R4C1 = 7 · Hidden SingleIn the middle-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 7, R4C2 cannot hold 7.- Because row 6 already has a 7, R6C2 cannot hold 7.So 7 must go in R4C1.
  17. R4C2 = 6 · Hidden SingleIn the middle-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 6 already has a 6, R6C2 cannot hold 6.So 6 must go in R4C2.
  18. R6C2 = 3 · Naked SingleLooking at R6C2:- Row 6 already has 1, 4, 5, 6 and 7 → ruled out.- Column 2 already has 2 and 8 → ruled out.- The middle-left 3×3 box already has 9 → ruled out.So this cell can only be 3.
  19. R4C6 = 3 · Hidden SingleIn the centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 3, R4C4 cannot hold 3.- Because row 5 already has a 3, R5C5 cannot hold 3.- Because row 6 already has a 3, R6C5 and R6C6 cannot hold 3.So 3 must go in R4C6.
  20. R4C4 = 4 · Hidden SingleIn the centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 4, R5C5 cannot hold 4.- Because row 6 already has a 4, R6C5 and R6C6 cannot hold 4.So 4 must go in R4C4.
  21. R4C8 = 8 · Naked SingleLooking at R4C8:- Row 4 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  22. R7C4 = 1 · Naked SingleLooking at R7C4:- Column 4 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  23. R5C5 = 7 · Hidden SingleIn the centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 6 already has a 7, R6C5 and R6C6 cannot hold 7.So 7 must go in R5C5.
  24. R5C7 = 4 · Naked SingleLooking at R5C7:- Row 5 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  25. R6C7 = 9 · Naked SingleLooking at R6C7:- Row 6 already has 1, 3, 4, 5, 6 and 7 → ruled out.- Column 7 already has 2 → ruled out.- The middle-right 3×3 box already has 8 → ruled out.So this cell can only be 9.
  26. R6C5 = 8 · Hidden SingleIn the centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 8, R6C6 cannot hold 8.So 8 must go in R6C5.
  27. 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.
  28. R7C1 = 2 · Hidden SingleIn the bottom-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 2, R7C2 cannot hold 2.- Because column 3 already has a 2, R7C3 and R9C3 cannot hold 2.- Because row 8 already has a 2, R8C1 and R8C2 cannot hold 2.So 2 must go in R7C1.
  29. R8C1 = 6 · Naked SingleLooking at R8C1:- Row 8 already has 1, 2, 8 and 9 → ruled out.- Column 1 already has 3, 4, 5 and 7 → ruled out.So this cell can only be 6.
  30. R7C3 = 7 · Hidden SingleIn the bottom-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 7, R7C2 and R8C2 cannot hold 7.- Because row 9 already has a 7, R9C3 cannot hold 7.So 7 must go in R7C3.
  31. R9C3 = 8 · Naked SingleLooking at R9C3:- Row 9 already has 3, 5, 6, 7 and 9 → ruled out.- Column 3 already has 1, 2 and 4 → ruled out.So this cell can only be 8.
  32. R7C2 = 9 · Hidden SingleIn the bottom-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 9, R8C2 cannot hold 9.So 9 must go in R7C2.
  33. R8C2 = 4 · Naked SingleLooking at R8C2:- Row 8 already has 1, 2, 6, 8 and 9 → ruled out.- Column 2 already has 3, 5 and 7 → ruled out.So this cell can only be 4.
  34. R9C5 = 2 · Hidden SingleIn the bottom-centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 2, R7C5 and R7C6 cannot hold 2.- Because row 8 already has a 2, R8C5 and R8C6 cannot hold 2.So 2 must go in R9C5.
  35. R7C6 = 4 · Hidden SingleIn the bottom-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 4, R7C5 cannot hold 4.- Because row 8 already has a 4, R8C5 and R8C6 cannot hold 4.So 4 must go in R7C6.
  36. R8C6 = 5 · Naked SingleLooking at R8C6:- Row 8 already has 1, 2, 4, 6, 8 and 9 → ruled out.- Column 6 already has 3 and 7 → ruled out.So this cell can only be 5.
  37. R7C5 = 6 · Hidden SingleIn the bottom-centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 6, R8C5 cannot hold 6.So 6 must go in R7C5.
  38. R8C5 = 3 · Naked SingleLooking at R8C5:- Row 8 already has 1, 2, 4, 5, 6, 8 and 9 → ruled out.- Column 5 already has 7 → ruled out.So this cell can only be 3.
  39. R8C7 = 7 · Naked SingleLooking at R8C7:- Row 8 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  40. R9C7 = 1 · Hidden SingleIn the bottom-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 1, R7C7, R7C8 and R7C9 cannot hold 1.- Because column 9 already has a 1, R9C9 cannot hold 1.So 1 must go in R9C7.
  41. R9C9 = 4 · Naked SingleLooking at R9C9:- Row 9 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  42. R1C8 = 4 · Hidden SingleIn the top-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 4, R1C7 cannot hold 4.- Because column 9 already has a 4, R1C9 cannot hold 4.So 4 must go in R1C8.
  43. R7C8 = 3 · Naked SingleLooking at R7C8:- Row 7 already has 1, 2, 4, 6, 7 and 9 → ruled out.- Column 8 already has 5 and 8 → ruled out.So this cell can only be 3.
  44. R1C7 = 3 · Hidden SingleIn the top-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 3, R1C9 cannot hold 3.So 3 must go in R1C7.
  45. R1C9 = 8 · Naked SingleLooking at R1C9:- Row 1 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  46. R7C7 = 8 · Naked SingleLooking at R7C7:- Row 7 already has 1, 2, 3, 4, 6, 7 and 9 → ruled out.- Column 7 already has 5 → ruled out.So this cell can only be 8.
  47. R7C9 = 5 · Naked SingleLooking at R7C9:- Row 7 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.

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