Skudo
ENVI
Play on web

Daily Sudoku — 22 July 2026

The puzzle

  • 22 July 2026
  • Hard
  • 30 clues
Puzzle
··5312···
4···8··2·
·1·····5·
6·8·71···
·3·64·5··
··12··6··
·24······
1·39····6
·6·15··4·
Solution
785312469
496785321
312496758
648571293
239648517
571239684
924867135
153924876
867153942

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. R3C1 = 3 · Hidden SingleIn the top-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 3, R1C1 and R1C2 cannot hold 3.- Because column 2 already has a 3, R2C2 cannot hold 3.- Because column 3 already has a 3, R2C3 and R3C3 cannot hold 3.So 3 must go in R3C1.
  2. R3C3 = 2 · Hidden SingleIn the top-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 2, R1C1 and R1C2 cannot hold 2.- Because row 2 already has a 2, R2C2 and R2C3 cannot hold 2.So 2 must go in R3C3.
  3. 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, R1C1 cannot hold 6.- Because column 2 already has a 6, R1C2 and R2C2 cannot hold 6.So 6 must go in R2C3.
  4. R1C8 = 6 · Hidden SingleIn the top-right 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 6, R1C7 and R3C7 cannot hold 6.- Because column 9 already has a 6, R1C9 and R3C9 cannot hold 6.- Because row 2 already has a 6, R2C7 and R2C9 cannot hold 6.So 6 must go in R1C8.
  5. R5C1 = 2 · Hidden SingleIn the middle-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 2, R4C2 cannot hold 2.- Because column 3 already has a 2, R5C3 cannot hold 2.- Because row 6 already has a 2, R6C1 and R6C2 cannot hold 2.So 2 must go in R5C1.
  6. R8C5 = 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, R7C4, R7C5 and R7C6 cannot hold 2.- Because column 6 already has a 2, R8C6 and R9C6 cannot hold 2.So 2 must go in R8C5.
  7. R8C6 = 4 · Hidden SingleIn the bottom-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 4, R7C4, R7C5 and R7C6 cannot hold 4.- Because row 9 already has a 4, R9C6 cannot hold 4.So 4 must go in R8C6.
  8. R3C4 = 4 · Hidden SingleIn the top-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 4, R2C4 and R2C6 cannot hold 4.- Because column 5 already has a 4, R3C5 cannot hold 4.- Because column 6 already has a 4, R3C6 cannot hold 4.So 4 must go in R3C4.
  9. R7C9 = 5 · Hidden SingleIn the bottom-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 5, R7C7 and R8C7 cannot hold 5.- Because column 8 already has a 5, R7C8 and R8C8 cannot hold 5.- Because row 9 already has a 5, R9C7 and R9C9 cannot hold 5.So 5 must go in R7C9.
  10. R8C2 = 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 cannot hold 5.- Because row 9 already has a 5, R9C1 and R9C3 cannot hold 5.So 5 must go in R8C2.
  11. R6C1 = 5 · Hidden SingleIn the middle-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 5, R4C2 and R6C2 cannot hold 5.- Because row 5 already has a 5, R5C3 cannot hold 5.So 5 must go in R6C1.
  12. 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, R5C6 cannot hold 5.- Because row 6 already has a 5, R6C5 and R6C6 cannot hold 5.So 5 must go in R4C4.
  13. R2C6 = 5 · Hidden SingleIn the top-centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 5, R2C4 cannot hold 5.- Because row 3 already has a 5, R3C5 and R3C6 cannot hold 5.So 5 must go in R2C6.
  14. R1C2 = 8 · Hidden SingleIn column 2, 8 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 8, R2C2 cannot hold 8.- Because row 4 already has a 8, R4C2 cannot hold 8.- Because the middle-left 3×3 box already has a 8, R6C2 cannot hold 8.So 8 must go in R1C2.
  15. R7C4 = 8 · Hidden SingleIn column 4, 8 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 8, R2C4 cannot hold 8.So 8 must go in R7C4.
  16. R2C4 = 7 · Naked SingleLooking at R2C4:- Row 2 already has 2, 4, 5, 6 and 8 → ruled out.- Column 4 already has 1, 3 and 9 → ruled out.So this cell can only be 7.
  17. R1C1 = 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.So 7 must go in R1C1.
  18. R2C2 = 9 · Naked SingleLooking at R2C2:- Row 2 already has 2, 4, 5, 6, 7 and 8 → ruled out.- Column 2 already has 1 and 3 → ruled out.So this cell can only be 9.
  19. 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.So 9 must go in R5C3.
  20. R9C3 = 7 · Naked SingleLooking at R9C3:- Row 9 already has 1, 4, 5 and 6 → ruled out.- Column 3 already has 2, 3, 8 and 9 → ruled out.So this cell can only be 7.
  21. R6C2 = 7 · Hidden SingleIn the middle-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 7, R4C2 cannot hold 7.So 7 must go in R6C2.
  22. R4C2 = 4 · Naked SingleLooking at R4C2:- Row 4 already has 1, 5, 6, 7 and 8 → ruled out.- Column 2 already has 2, 3 and 9 → ruled out.So this cell can only be 4.
  23. R6C9 = 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, R4C7, R4C8 and R4C9 cannot hold 4.- Because row 5 already has a 4, R5C8 and R5C9 cannot hold 4.- Because column 8 already has a 4, R6C8 cannot hold 4.So 4 must go in R6C9.
  24. R1C7 = 4 · Hidden SingleIn the top-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 4, R1C9 cannot hold 4.- Because row 2 already has a 4, R2C7 and R2C9 cannot hold 4.- Because row 3 already has a 4, R3C7 and R3C9 cannot hold 4.So 4 must go in R1C7.
  25. R1C9 = 9 · Naked SingleLooking at R1C9:- Row 1 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  26. R9C1 = 8 · Hidden SingleIn the bottom-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 8, R7C1 cannot hold 8.So 8 must go in R9C1.
  27. R7C1 = 9 · Naked SingleLooking at R7C1:- Row 7 already has 2, 4, 5 and 8 → ruled out.- Column 1 already has 1, 3, 6 and 7 → ruled out.So this cell can only be 9.
  28. R7C6 = 7 · Hidden SingleIn the bottom-centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 7, R7C5 cannot hold 7.- Because row 9 already has a 7, R9C6 cannot hold 7.So 7 must go in R7C6.
  29. R7C5 = 6 · Hidden SingleIn the bottom-centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 9 already has a 6, R9C6 cannot hold 6.So 6 must go in R7C5.
  30. R9C6 = 3 · Naked SingleLooking at R9C6:- Row 9 already has 1, 4, 5, 6, 7 and 8 → ruled out.- Column 6 already has 2 → ruled out.- The bottom-centre 3×3 box already has 9 → ruled out.So this cell can only be 3.
  31. R3C6 = 6 · Hidden SingleIn the top-centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 6, R3C5 cannot hold 6.So 6 must go in R3C6.
  32. R3C5 = 9 · Naked SingleLooking at R3C5:- Row 3 already has 1, 2, 3, 4, 5 and 6 → ruled out.- Column 5 already has 7 and 8 → ruled out.So this cell can only be 9.
  33. R6C5 = 3 · Naked SingleLooking at R6C5:- Row 6 already has 1, 2, 4, 5, 6 and 7 → ruled out.- Column 5 already has 8 and 9 → ruled out.So this cell can only be 3.
  34. R6C6 = 9 · Hidden SingleIn the centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 9, R5C6 cannot hold 9.So 9 must go in R6C6.
  35. R6C8 = 8 · Naked SingleLooking at R6C8:- Row 6 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  36. R5C6 = 8 · Naked SingleLooking at R5C6:- Row 5 already has 2, 3, 4, 5, 6 and 9 → ruled out.- Column 6 already has 1 and 7 → ruled out.So this cell can only be 8.
  37. R8C7 = 8 · Hidden SingleIn the bottom-right 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 8, R7C7 and R7C8 cannot hold 8.- Because column 8 already has a 8, R8C8 cannot hold 8.- Because row 9 already has a 8, R9C7 and R9C9 cannot hold 8.So 8 must go in R8C7.
  38. R8C8 = 7 · Naked SingleLooking at R8C8:- Row 8 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  39. R3C9 = 8 · Hidden SingleIn the top-right 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 8, R2C7 and R2C9 cannot hold 8.- Because column 7 already has a 8, R3C7 cannot hold 8.So 8 must go in R3C9.
  40. R3C7 = 7 · Naked SingleLooking at R3C7:- Row 3 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  41. R5C9 = 7 · Hidden SingleIn the middle-right 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 7, R4C7, R4C8 and R4C9 cannot hold 7.- Because column 8 already has a 7, R5C8 cannot hold 7.So 7 must go in R5C9.
  42. R5C8 = 1 · Naked SingleLooking at R5C8:- Row 5 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  43. R7C7 = 1 · Hidden SingleIn the bottom-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 1, R7C8 cannot hold 1.- Because row 9 already has a 1, R9C7 and R9C9 cannot hold 1.So 1 must go in R7C7.
  44. R7C8 = 3 · Naked SingleLooking at R7C8:- Row 7 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  45. R4C8 = 9 · Naked SingleLooking at R4C8:- Row 4 already has 1, 4, 5, 6, 7 and 8 → ruled out.- Column 8 already has 2 and 3 → ruled out.So this cell can only be 9.
  46. R2C9 = 1 · Hidden SingleIn the top-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 1, R2C7 cannot hold 1.So 1 must go in R2C9.
  47. R2C7 = 3 · Naked SingleLooking at R2C7:- Row 2 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  48. R4C9 = 3 · Hidden SingleIn the middle-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 3, R4C7 cannot hold 3.So 3 must go in R4C9.
  49. R4C7 = 2 · Naked SingleLooking at R4C7:- Row 4 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  50. R9C7 = 9 · Naked SingleLooking at R9C7:- Row 9 already has 1, 3, 4, 5, 6, 7 and 8 → ruled out.- Column 7 already has 2 → ruled out.So this cell can only be 9.
  51. R9C9 = 2 · Naked SingleLooking at R9C9:- Row 9 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.

First to finish

Ranked by when each player finished, not by how long they took — so looking up a solution never moves anyone up this board.

Open the app to see who finished first.

Recent days

Full archive →