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Daily Sudoku — 30 June 2026

The puzzle

  • 30 June 2026
  • Hard
  • 30 clues
Puzzle
7·3·8··5·
9······81
·8·9···2·
8·4·3·2··
···2·7···
··7·6··3·
···6·8··5
69··7··42
3····2·9·
Solution
723481956
956723481
481956723
864539217
539217864
217864539
142698375
698375142
375142698

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. R2C5 = 2 · Hidden SingleIn the top-centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 2, R1C4 and R2C4 cannot hold 2.- Because column 6 already has a 2, R1C6 and R2C6 cannot hold 2.- Because row 3 already has a 2, R3C5 and R3C6 cannot hold 2.So 2 must go in R2C5.
  2. R1C2 = 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, R2C2 and R2C3 cannot hold 2.- Because row 3 already has a 2, R3C1 and R3C3 cannot hold 2.So 2 must go in R1C2.
  3. R2C4 = 7 · Hidden SingleIn the top-centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 7, R1C4 and R1C6 cannot hold 7.- Because column 6 already has a 7, R2C6 and R3C6 cannot hold 7.- Because column 5 already has a 7, R3C5 cannot hold 7.So 7 must go in R2C4.
  4. R6C1 = 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, R4C2 cannot hold 2.- Because row 5 already has a 2, R5C1, R5C2 and R5C3 cannot hold 2.- Because column 2 already has a 2, R6C2 cannot hold 2.So 2 must go in R6C1.
  5. R5C2 = 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, R4C2 cannot hold 3.- Because column 1 already has a 3, R5C1 cannot hold 3.- Because column 3 already has a 3, R5C3 cannot hold 3.- Because row 6 already has a 3, R6C2 cannot hold 3.So 3 must go in R5C2.
  6. R5C3 = 9 · Hidden SingleIn the middle-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 9, R4C2 and R6C2 cannot hold 9.- Because column 1 already has a 9, R5C1 cannot hold 9.So 9 must go in R5C3.
  7. 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, R5C1 cannot hold 6.- Because row 6 already has a 6, R6C2 cannot hold 6.So 6 must go in R4C2.
  8. 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, R4C4 and R4C6 cannot hold 8.- Because column 5 already has a 8, R5C5 cannot hold 8.- Because column 6 already has a 8, R6C6 cannot hold 8.So 8 must go in R6C4.
  9. R7C3 = 2 · Hidden SingleIn the bottom-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 2, R7C1 cannot hold 2.- Because column 2 already has a 2, R7C2 cannot hold 2.- Because row 8 already has a 2, R8C3 cannot hold 2.- Because row 9 already has a 2, R9C2 and R9C3 cannot hold 2.So 2 must go in R7C3.
  10. R7C5 = 9 · Hidden SingleIn the bottom-centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 9, R8C4 and R8C6 cannot hold 9.- Because row 9 already has a 9, R9C4 and R9C5 cannot hold 9.So 9 must go in R7C5.
  11. R7C7 = 3 · Hidden SingleIn row 7, 3 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 3, R7C1 cannot hold 3.- Because column 2 already has a 3, R7C2 cannot hold 3.- Because column 8 already has a 3, R7C8 cannot hold 3.So 3 must go in R7C7.
  12. R3C9 = 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 and R3C7 cannot hold 3.So 3 must go in R3C9.
  13. R2C6 = 3 · Hidden SingleIn the top-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 3, R1C4 and R1C6 cannot hold 3.- Because row 3 already has a 3, R3C5 and R3C6 cannot hold 3.So 3 must go in R2C6.
  14. R3C7 = 7 · Hidden SingleIn the top-right 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 7, R1C7 and R1C9 cannot hold 7.- Because row 2 already has a 7, R2C7 cannot hold 7.So 7 must go in R3C7.
  15. R8C4 = 3 · Hidden SingleIn the bottom-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 3, R8C6 cannot hold 3.- Because row 9 already has a 3, R9C4 and R9C5 cannot hold 3.So 3 must go in R8C4.
  16. R5C8 = 6 · Hidden SingleIn column 8, 6 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 6, R4C8 cannot hold 6.- Because row 7 already has a 6, R7C8 cannot hold 6.So 6 must go in R5C8.
  17. R5C1 = 5 · Hidden SingleIn column 1, 5 fits in only one cell. Looking at the other empty cells:- Because 5 inside the top-centre 3×3 box is confined to row 3 (locked candidates), R3C1 cannot hold 5.- Because row 7 already has a 5, R7C1 cannot hold 5.So 5 must go in R5C1.
  18. R6C2 = 1 · Naked SingleLooking at R6C2:- Row 6 already has 2, 3, 6, 7 and 8 → ruled out.- Column 2 already has 9 → ruled out.- The middle-left 3×3 box already has 4 and 5 → ruled out.So this cell can only be 1.
  19. R6C7 = 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.- Because column 9 already has a 5, R4C9 and R6C9 cannot hold 5.- Because row 5 already has a 5, R5C7 and R5C9 cannot hold 5.So 5 must go in R6C7.
  20. R1C7 = 9 · Hidden SingleIn column 7, 9 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 9, R2C7 cannot hold 9.- Because row 5 already has a 9, R5C7 cannot hold 9.- Because row 8 already has a 9, R8C7 cannot hold 9.- Because row 9 already has a 9, R9C7 cannot hold 9.So 9 must go in R1C7.
  21. R4C4 = 5 · Hidden SingleIn the centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 5, R5C5 cannot hold 5.- Because row 6 already has a 5, R6C6 cannot hold 5.- Applying W-Wing (two cells with the same pair, linked by a strong link, guarantee one of them is a; a is eliminated from cells seeing both) means R4C6 cannot hold 5 either.So 5 must go in R4C4.
  22. R2C3 = 6 · Hidden SingleIn the top-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 6, R2C2 cannot hold 6.- Because column 1 already has a 6, R3C1 cannot hold 6.- Applying Finned X-Wing (an X-Wing with one extra cell (the fin) confined to a box; the digit is eliminated only where the pattern and the fin both see it) means R3C3 cannot hold 6 either.So 6 must go in R2C3.
  23. R1C9 = 6 · Hidden SingleIn the top-right 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 6, R2C7 cannot hold 6.So 6 must go in R1C9.
  24. R2C7 = 4 · Naked SingleLooking at R2C7:- Row 2 already has 1, 2, 3, 6, 7, 8 and 9 → ruled out.- Column 7 already has 5 → ruled out.So this cell can only be 4.
  25. R2C2 = 5 · Naked SingleLooking at R2C2:- Row 2 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  26. R3C1 = 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 R3C1.
  27. R7C1 = 1 · Naked SingleLooking at R7C1:- Row 7 already has 2, 3, 5, 6, 8 and 9 → ruled out.- Column 1 already has 4 and 7 → ruled out.So this cell can only be 1.
  28. R3C3 = 1 · Naked SingleLooking at R3C3:- Row 3 already has 2, 3, 4, 7, 8 and 9 → ruled out.- Column 3 already has 6 → ruled out.- The top-left 3×3 box already has 5 → ruled out.So this cell can only be 1.
  29. R3C6 = 6 · Hidden SingleIn the top-centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 6, R1C4 and R1C6 cannot hold 6.- Because column 5 already has a 6, R3C5 cannot hold 6.So 6 must go in R3C6.
  30. R3C5 = 5 · Naked SingleLooking at R3C5:- Row 3 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  31. R8C6 = 5 · Hidden SingleIn the bottom-centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 5, R9C4 cannot hold 5.- Because column 5 already has a 5, R9C5 cannot hold 5.So 5 must go in R8C6.
  32. 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, R7C2 cannot hold 5.- Because row 8 already has a 5, R8C3 cannot hold 5.- Because column 2 already has a 5, R9C2 cannot hold 5.So 5 must go in R9C3.
  33. R8C3 = 8 · Naked SingleLooking at R8C3:- Row 8 already has 2, 3, 4, 5, 6, 7 and 9 → ruled out.- Column 3 already has 1 → ruled out.So this cell can only be 8.
  34. R8C7 = 1 · Naked SingleLooking at R8C7:- Row 8 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  35. R4C8 = 1 · Hidden SingleIn the middle-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 1, R4C9 and R5C9 cannot hold 1.- Because column 7 already has a 1, R5C7 cannot hold 1.- Because row 6 already has a 1, R6C9 cannot hold 1.So 1 must go in R4C8.
  36. R7C8 = 7 · Naked SingleLooking at R7C8:- Row 7 already has 1, 2, 3, 5, 6, 8 and 9 → ruled out.- Column 8 already has 4 → ruled out.So this cell can only be 7.
  37. R7C2 = 4 · Naked SingleLooking at R7C2:- Row 7 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  38. R9C2 = 7 · Naked SingleLooking at R9C2:- Row 9 already has 2, 3, 5 and 9 → ruled out.- Column 2 already has 1, 4, 6 and 8 → ruled out.So this cell can only be 7.
  39. R5C5 = 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, R4C6 cannot hold 1.- Because row 6 already has a 1, R6C6 cannot hold 1.So 1 must go in R5C5.
  40. R9C5 = 4 · Naked SingleLooking at R9C5:- Row 9 already has 2, 3, 5, 7 and 9 → ruled out.- Column 5 already has 1, 6 and 8 → ruled out.So this cell can only be 4.
  41. R9C4 = 1 · Naked SingleLooking at R9C4:- Row 9 already has 2, 3, 4, 5, 7 and 9 → ruled out.- Column 4 already has 6 and 8 → ruled out.So this cell can only be 1.
  42. R1C4 = 4 · Naked SingleLooking at R1C4:- Row 1 already has 2, 3, 5, 6, 7, 8 and 9 → ruled out.- Column 4 already has 1 → ruled out.So this cell can only be 4.
  43. R1C6 = 1 · Naked SingleLooking at R1C6:- Row 1 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  44. R6C6 = 4 · Hidden SingleIn the centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 4, R4C6 cannot hold 4.So 4 must go in R6C6.
  45. R6C9 = 9 · Naked SingleLooking at R6C9:- Row 6 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  46. R4C6 = 9 · Naked SingleLooking at R4C6:- Row 4 already has 1, 2, 3, 4, 5, 6 and 8 → ruled out.- Column 6 already has 7 → ruled out.So this cell can only be 9.
  47. R4C9 = 7 · Naked SingleLooking at R4C9:- Row 4 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  48. R5C9 = 4 · Hidden SingleIn the middle-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 4, R5C7 cannot hold 4.So 4 must go in R5C9.
  49. R5C7 = 8 · Naked SingleLooking at R5C7:- Row 5 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  50. R9C7 = 6 · Naked SingleLooking at R9C7:- Row 9 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. R9C9 = 8 · Naked SingleLooking at R9C9:- Row 9 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.

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