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Daily Sudoku — 19 July 2026

The puzzle

  • 19 July 2026
  • Easy
  • 40 clues
Puzzle
3··1·2··7
6·1···8··
·7·8··162
9·532····
71·······
2·3·7·5·4
·5··1·7··
132·4·9·5
46·985213
Solution
398162457
621457839
574839162
945328671
716594328
283671594
859213746
132746985
467985213

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 — 41 steps using 2 techniques.

Techniques used

  • Naked Single — every other digit is already used in this cell's row, column or box, so only one candidate is left
  • Hidden Single — inside one row, column or box this digit fits in only one cell, so it must go there
  1. R9C3 = 7 · Naked SingleLooking at R9C3:- Row 9 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  2. R2C2 = 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, R1C2 and R1C3 cannot hold 2.- Because row 3 already has a 2, R3C1 and R3C3 cannot hold 2.So 2 must go in R2C2.
  3. R3C1 = 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 cannot hold 5.- Because column 3 already has a 5, R1C3 and R3C3 cannot hold 5.So 5 must go in R3C1.
  4. R7C1 = 8 · Naked SingleLooking at R7C1:- Row 7 already has 1, 5 and 7 → ruled out.- Column 1 already has 2, 3, 4, 6 and 9 → ruled out.So this cell can only be 8.
  5. R7C3 = 9 · Naked SingleLooking at R7C3:- Row 7 already has 1, 5, 7 and 8 → ruled out.- Column 3 already has 2 and 3 → ruled out.- The bottom-left 3×3 box already has 4 and 6 → ruled out.So this cell can only be 9.
  6. 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 and R3C3 cannot hold 9.So 9 must go in R1C2.
  7. R1C3 = 8 · Hidden SingleIn the top-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 3 already has a 8, R3C3 cannot hold 8.So 8 must go in R1C3.
  8. R3C3 = 4 · Naked SingleLooking at R3C3:- Row 3 already has 1, 2, 5, 6, 7 and 8 → ruled out.- Column 3 already has 3 and 9 → ruled out.So this cell can only be 4.
  9. R5C3 = 6 · Naked SingleLooking at R5C3:- Row 5 already has 1 and 7 → ruled out.- Column 3 already has 2, 3, 4, 5, 8 and 9 → ruled out.So this cell can only be 6.
  10. R1C5 = 6 · Hidden SingleIn the top-centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 6, R2C4, R2C5 and R2C6 cannot hold 6.- Because row 3 already has a 6, R3C5 and R3C6 cannot hold 6.So 6 must go in R1C5.
  11. R2C8 = 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 R1C8 cannot hold 3.- Because column 9 already has a 3, R2C9 cannot hold 3.So 3 must go in R2C8.
  12. R1C8 = 5 · Hidden SingleIn the top-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 5, R1C7 cannot hold 5.- Because column 9 already has a 5, R2C9 cannot hold 5.So 5 must go in R1C8.
  13. R1C7 = 4 · Naked SingleLooking at R1C7:- Row 1 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  14. R2C9 = 9 · Naked SingleLooking at R2C9:- Row 2 already has 1, 2, 3, 6 and 8 → ruled out.- Column 9 already has 4, 5 and 7 → ruled out.So this cell can only be 9.
  15. R4C2 = 4 · Hidden SingleIn the middle-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 6 already has a 4, R6C2 cannot hold 4.So 4 must go in R4C2.
  16. R6C2 = 8 · Naked SingleLooking at R6C2:- Row 6 already has 2, 3, 4, 5 and 7 → ruled out.- Column 2 already has 1, 6 and 9 → ruled out.So this cell can only be 8.
  17. R4C9 = 1 · Hidden SingleIn the middle-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 1, R4C7 cannot hold 1.- Because column 8 already has a 1, R4C8 and R6C8 cannot hold 1.- Because row 5 already has a 1, R5C7, R5C8 and R5C9 cannot hold 1.So 1 must go in R4C9.
  18. 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, R4C6 cannot hold 1.- Because row 5 already has a 1, R5C4, R5C5 and R5C6 cannot hold 1.- Because column 4 already has a 1, R6C4 cannot hold 1.So 1 must go in R6C6.
  19. R5C8 = 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 7 already has a 2, R5C7 cannot hold 2.- Because column 9 already has a 2, R5C9 cannot hold 2.- Because row 6 already has a 2, R6C8 cannot hold 2.So 2 must go in R5C8.
  20. R5C7 = 3 · Hidden SingleIn the middle-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 3, R4C7 and R4C8 cannot hold 3.- Because column 9 already has a 3, R5C9 cannot hold 3.- Because row 6 already has a 3, R6C8 cannot hold 3.So 3 must go in R5C7.
  21. R4C7 = 6 · Naked SingleLooking at R4C7:- Row 4 already has 1, 2, 3, 4, 5 and 9 → ruled out.- Column 7 already has 7 and 8 → ruled out.So this cell can only be 6.
  22. R6C4 = 6 · Hidden SingleIn the centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 6, R4C6 cannot hold 6.- Because row 5 already has a 6, R5C4, R5C5 and R5C6 cannot hold 6.So 6 must go in R6C4.
  23. R6C8 = 9 · Naked SingleLooking at R6C8:- Row 6 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  24. R4C8 = 7 · Hidden SingleIn the middle-right 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 7, R5C9 cannot hold 7.So 7 must go in R4C8.
  25. R4C6 = 8 · Naked SingleLooking at R4C6:- Row 4 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  26. R5C9 = 8 · Naked SingleLooking at R5C9:- Row 5 already has 1, 2, 3, 6 and 7 → ruled out.- Column 9 already has 4, 5 and 9 → ruled out.So this cell can only be 8.
  27. R7C9 = 6 · Naked SingleLooking at R7C9:- Row 7 already has 1, 5, 7, 8 and 9 → ruled out.- Column 9 already has 2, 3 and 4 → ruled out.So this cell can only be 6.
  28. R7C4 = 2 · Hidden SingleIn the bottom-centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 2, R7C6 cannot hold 2.- Because row 8 already has a 2, R8C4 and R8C6 cannot hold 2.So 2 must go in R7C4.
  29. R7C6 = 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, R8C4 and R8C6 cannot hold 3.So 3 must go in R7C6.
  30. R7C8 = 4 · Naked SingleLooking at R7C8:- Row 7 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  31. R8C8 = 8 · Naked SingleLooking at R8C8:- Row 8 already has 1, 2, 3, 4, 5 and 9 → ruled out.- Column 8 already has 6 and 7 → ruled out.So this cell can only be 8.
  32. R3C5 = 3 · Hidden SingleIn the top-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 3, R2C4, R2C5 and R2C6 cannot hold 3.- Because column 6 already has a 3, R3C6 cannot hold 3.So 3 must go in R3C5.
  33. R3C6 = 9 · Naked SingleLooking at R3C6:- Row 3 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  34. R5C5 = 9 · Hidden SingleIn the centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 9, R5C4 cannot hold 9.- Because column 6 already has a 9, R5C6 cannot hold 9.So 9 must go in R5C5.
  35. R2C5 = 5 · Naked SingleLooking at R2C5:- Row 2 already has 1, 2, 3, 6, 8 and 9 → ruled out.- Column 5 already has 4 and 7 → ruled out.So this cell can only be 5.
  36. R5C4 = 5 · Hidden SingleIn the centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 5, R5C6 cannot hold 5.So 5 must go in R5C4.
  37. R5C6 = 4 · Naked SingleLooking at R5C6:- Row 5 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  38. R2C4 = 4 · Hidden SingleIn the top-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 4, R2C6 cannot hold 4.So 4 must go in R2C4.
  39. R2C6 = 7 · Naked SingleLooking at R2C6:- Row 2 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  40. R8C4 = 7 · Naked SingleLooking at R8C4:- Row 8 already has 1, 2, 3, 4, 5, 8 and 9 → ruled out.- Column 4 already has 6 → ruled out.So this cell can only be 7.
  41. R8C6 = 6 · Naked SingleLooking at R8C6:- Row 8 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.

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