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Daily Sudoku — 29 June 2026

The puzzle

  • 29 June 2026
  • Hard
  • 30 clues
Puzzle
···64·75·
··1··2···
·4·1·72··
1·9·····5
·2··8·1··
··5·7·3··
·1···4···
··87·····
5··219438
Solution
293648751
751392846
846157293
179423685
324586179
685971324
912834567
438765912
567219438

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. R2C2 = 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, R1C1, R1C2 and R1C3 cannot hold 5.- Because column 1 already has a 5, R2C1 and R3C1 cannot hold 5.- Because column 3 already has a 5, R3C3 cannot hold 5.So 5 must go in R2C2.
  2. R2C1 = 7 · Hidden SingleIn the top-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 7, R1C1, R1C2 and R1C3 cannot hold 7.- Because row 3 already has a 7, R3C1 and R3C3 cannot hold 7.So 7 must go in R2C1.
  3. R3C5 = 5 · Hidden SingleIn the top-centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 5, R1C6 cannot hold 5.- Because row 2 already has a 5, R2C4 and R2C5 cannot hold 5.So 5 must go in R3C5.
  4. R1C9 = 1 · Hidden SingleIn the top-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 1, R2C7, R2C8 and R2C9 cannot hold 1.- Because row 3 already has a 1, R3C8 and R3C9 cannot hold 1.So 1 must go in R1C9.
  5. R6C6 = 1 · Hidden SingleIn the centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 1, R4C4, R4C5 and R4C6 cannot hold 1.- Because row 5 already has a 1, R5C4 and R5C6 cannot hold 1.- Because column 4 already has a 1, R6C4 cannot hold 1.So 1 must go in R6C6.
  6. R4C5 = 2 · Hidden SingleIn the centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 2, R4C4 and R6C4 cannot hold 2.- Because column 6 already has a 2, R4C6 cannot hold 2.- Because row 5 already has a 2, R5C4 and R5C6 cannot hold 2.So 2 must go in R4C5.
  7. R8C1 = 4 · Hidden SingleIn the bottom-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 4, R7C1 and R7C3 cannot hold 4.- Because column 2 already has a 4, R8C2 cannot hold 4.- Because row 9 already has a 4, R9C2 and R9C3 cannot hold 4.So 4 must go in R8C1.
  8. R5C3 = 4 · Hidden SingleIn the middle-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 4, R4C2 and R6C2 cannot hold 4.- Because column 1 already has a 4, R5C1 and R6C1 cannot hold 4.So 4 must go in R5C3.
  9. R4C2 = 7 · Hidden SingleIn the middle-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 7, R5C1 cannot hold 7.- Because row 6 already has a 7, R6C1 and R6C2 cannot hold 7.So 7 must go in R4C2.
  10. R5C1 = 3 · Hidden SingleIn the middle-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 6 already has a 3, R6C1 and R6C2 cannot hold 3.So 3 must go in R5C1.
  11. R7C4 = 8 · Hidden SingleIn the bottom-centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 8, R7C5 cannot hold 8.- Because row 8 already has a 8, R8C5 and R8C6 cannot hold 8.So 8 must go in R7C4.
  12. R1C6 = 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.- Because column 5 already has a 8, R2C5 cannot hold 8.So 8 must go in R1C6.
  13. 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, R1C2 and R1C3 cannot hold 8.- Because column 3 already has a 8, R3C3 cannot hold 8.So 8 must go in R3C1.
  14. R3C3 = 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, R1C1, R1C2 and R1C3 cannot hold 6.So 6 must go in R3C3.
  15. R6C2 = 8 · Hidden SingleIn the middle-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 8, R6C1 cannot hold 8.So 8 must go in R6C2.
  16. R6C1 = 6 · Naked SingleLooking at R6C1:- Row 6 already has 1, 3, 5, 7 and 8 → ruled out.- Column 1 already has 4 → ruled out.- The middle-left 3×3 box already has 2 and 9 → ruled out.So this cell can only be 6.
  17. R8C6 = 5 · Hidden SingleIn the bottom-centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 5, R7C5 and R8C5 cannot hold 5.So 5 must go in R8C6.
  18. R5C4 = 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, R4C4 and R4C6 cannot hold 5.- Because column 6 already has a 5, R5C6 cannot hold 5.- Because row 6 already has a 5, R6C4 cannot hold 5.So 5 must go in R5C4.
  19. R6C4 = 9 · Hidden SingleIn the centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 9, R4C4 and R4C6 cannot hold 9.- Because column 6 already has a 9, R5C6 cannot hold 9.So 9 must go in R6C4.
  20. R2C5 = 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, R2C4 cannot hold 9.So 9 must go in R2C5.
  21. R2C4 = 3 · Naked SingleLooking at R2C4:- Row 2 already has 1, 2, 5, 7 and 9 → ruled out.- Column 4 already has 6 and 8 → ruled out.- The top-centre 3×3 box already has 4 → ruled out.So this cell can only be 3.
  22. R4C4 = 4 · Naked SingleLooking at R4C4:- Row 4 already has 1, 2, 5, 7 and 9 → ruled out.- Column 4 already has 3, 6 and 8 → ruled out.So this cell can only be 4.
  23. R3C9 = 3 · Hidden SingleIn the top-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 3, R2C7, R2C8 and R2C9 cannot hold 3.- Because column 8 already has a 3, R3C8 cannot hold 3.So 3 must go in R3C9.
  24. R3C8 = 9 · Naked SingleLooking at R3C8:- Row 3 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  25. R4C6 = 3 · Hidden SingleIn the centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 3, R5C6 cannot hold 3.So 3 must go in R4C6.
  26. R5C6 = 6 · Naked SingleLooking at R5C6:- Row 5 already has 1, 2, 3, 4, 5 and 8 → ruled out.- Column 6 already has 7 and 9 → ruled out.So this cell can only be 6.
  27. R5C9 = 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 and R4C8 cannot hold 9.- Because column 8 already has a 9, R5C8 cannot hold 9.- Because row 6 already has a 9, R6C8 and R6C9 cannot hold 9.So 9 must go in R5C9.
  28. R5C8 = 7 · Naked SingleLooking at R5C8:- Row 5 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  29. R8C8 = 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 7 already has a 1, R8C7 cannot hold 1.- Because column 9 already has a 1, R8C9 cannot hold 1.So 1 must go in R8C8.
  30. R7C7 = 5 · Hidden SingleIn the bottom-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 5, R7C8 cannot hold 5.- Because column 9 already has a 5, R7C9 cannot hold 5.- Because row 8 already has a 5, R8C7 and R8C9 cannot hold 5.So 5 must go in R7C7.
  31. R7C9 = 7 · Hidden SingleIn the bottom-right 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 7, R7C8 cannot hold 7.- Because row 8 already has a 7, R8C7 and R8C9 cannot hold 7.So 7 must go in R7C9.
  32. 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, R7C1 and R7C3 cannot hold 7.- Because row 8 already has a 7, R8C2 cannot hold 7.- Because column 2 already has a 7, R9C2 cannot hold 7.So 7 must go in R9C3.
  33. 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.
  34. R8C7 = 9 · Hidden SingleIn the bottom-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 9, R7C8 cannot hold 9.- Because column 9 already has a 9, R8C9 cannot hold 9.So 9 must go in R8C7.
  35. R7C1 = 9 · Hidden SingleIn the bottom-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 9, R7C3 cannot hold 9.- Because row 8 already has a 9, R8C2 cannot hold 9.So 9 must go in R7C1.
  36. R1C1 = 2 · Naked SingleLooking at R1C1:- Row 1 already has 1, 4, 5, 6, 7 and 8 → ruled out.- Column 1 already has 3 and 9 → ruled out.So this cell can only be 2.
  37. R1C2 = 9 · Hidden SingleIn the top-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 9, R1C3 cannot hold 9.So 9 must go in R1C2.
  38. R1C3 = 3 · Naked SingleLooking at R1C3:- Row 1 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  39. R8C2 = 3 · Naked SingleLooking at R8C2:- Row 8 already has 1, 4, 5, 7, 8 and 9 → ruled out.- Column 2 already has 2 and 6 → ruled out.So this cell can only be 3.
  40. R7C3 = 2 · Naked SingleLooking at R7C3:- Row 7 already has 1, 4, 5, 7, 8 and 9 → ruled out.- Column 3 already has 3 and 6 → ruled out.So this cell can only be 2.
  41. R7C5 = 3 · Hidden SingleIn the bottom-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 3, R8C5 cannot hold 3.So 3 must go in R7C5.
  42. 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.
  43. R8C5 = 6 · Naked SingleLooking at R8C5:- Row 8 already has 1, 3, 4, 5, 7, 8 and 9 → ruled out.- Column 5 already has 2 → ruled out.So this cell can only be 6.
  44. R8C9 = 2 · Naked SingleLooking at R8C9:- Row 8 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  45. R6C8 = 2 · Hidden SingleIn the middle-right 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 2, R4C7 and R4C8 cannot hold 2.- Because column 9 already has a 2, R6C9 cannot hold 2.So 2 must go in R6C8.
  46. R6C9 = 4 · Naked SingleLooking at R6C9:- Row 6 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  47. R2C9 = 6 · Naked SingleLooking at R2C9:- Row 2 already has 1, 2, 3, 5, 7 and 9 → ruled out.- Column 9 already has 4 and 8 → ruled out.So this cell can only be 6.
  48. R2C8 = 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, R2C7 cannot hold 4.So 4 must go in R2C8.
  49. R2C7 = 8 · Naked SingleLooking at R2C7:- Row 2 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  50. R4C7 = 6 · Naked SingleLooking at R4C7:- Row 4 already has 1, 2, 3, 4, 5, 7 and 9 → ruled out.- Column 7 already has 8 → ruled out.So this cell can only be 6.
  51. 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.

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