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Daily Sudoku — 26 July 2026

The puzzle

  • 26 July 2026
  • Medium
  • 34 clues
Puzzle
···51····
··1·3·8·9
·7···82··
·6·4795·1
··8·26·7·
··71··6··
··6··13··
····5··67
·35764··8
Solution
984512736
521637849
673948215
362479581
158326974
497185623
746891352
819253467
235764198

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. R3C5 = 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, R1C6 and R2C6 cannot hold 4.- Because column 4 already has a 4, R2C4 and R3C4 cannot hold 4.So 4 must go in R3C5.
  2. R3C4 = 9 · Hidden SingleIn the top-centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- 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 R3C4.
  3. R2C4 = 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 and R2C6 cannot hold 6.So 6 must go in R2C4.
  4. R3C8 = 1 · Hidden SingleIn the top-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 1, R1C7, R1C8 and R1C9 cannot hold 1.- Because row 2 already has a 1, R2C8 cannot hold 1.- Because column 9 already has a 1, R3C9 cannot hold 1.So 1 must go in R3C8.
  5. R1C7 = 7 · Hidden SingleIn the top-right 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 7, R1C8 and R2C8 cannot hold 7.- Because column 9 already has a 7, R1C9 cannot hold 7.- Because row 3 already has a 7, R3C9 cannot hold 7.So 7 must go in R1C7.
  6. R2C6 = 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, R1C6 cannot hold 7.So 7 must go in R2C6.
  7. R1C6 = 2 · Naked SingleLooking at R1C6:- Row 1 already has 1, 5 and 7 → ruled out.- Column 6 already has 4, 6, 8 and 9 → ruled out.- The top-centre 3×3 box already has 3 → ruled out.So this cell can only be 2.
  8. R6C6 = 5 · Hidden SingleIn the centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 5, R5C4 cannot hold 5.- Because column 5 already has a 5, R6C5 cannot hold 5.So 5 must go in R6C6.
  9. R8C6 = 3 · Naked SingleLooking at R8C6:- Row 8 already has 5, 6 and 7 → ruled out.- Column 6 already has 1, 2, 4, 8 and 9 → ruled out.So this cell can only be 3.
  10. R5C4 = 3 · Hidden SingleIn the centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 3, R6C5 cannot hold 3.So 3 must go in R5C4.
  11. R6C5 = 8 · Naked SingleLooking at R6C5:- Row 6 already has 1, 5, 6 and 7 → ruled out.- Column 5 already has 2, 3 and 4 → ruled out.- The centre 3×3 box already has 9 → ruled out.So this cell can only be 8.
  12. R7C5 = 9 · Naked SingleLooking at R7C5:- Row 7 already has 1, 3 and 6 → ruled out.- Column 5 already has 2, 4, 5, 7 and 8 → ruled out.So this cell can only be 9.
  13. R4C8 = 8 · Hidden SingleIn the middle-right 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 8, R5C7 and R5C9 cannot hold 8.- Because row 6 already has a 8, R6C8 and R6C9 cannot hold 8.So 8 must go in R4C8.
  14. R7C1 = 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 cannot hold 7.- Because row 8 already has a 7, R8C1, R8C2 and R8C3 cannot hold 7.- Because row 9 already has a 7, R9C1 cannot hold 7.So 7 must go in R7C1.
  15. R3C3 = 3 · Naked SingleLooking at R3C3:- Row 3 already has 1, 2, 4, 7, 8 and 9 → ruled out.- Column 3 already has 5 and 6 → ruled out.So this cell can only be 3.
  16. R4C1 = 3 · Hidden SingleIn row 4, 3 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 3, R4C3 cannot hold 3.So 3 must go in R4C1.
  17. R4C3 = 2 · Naked SingleLooking at R4C3:- Row 4 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  18. R5C9 = 4 · Naked SingleLooking at R5C9:- Row 5 already has 2, 3, 6, 7 and 8 → ruled out.- Column 9 already has 1 and 9 → ruled out.- The middle-right 3×3 box already has 5 → ruled out.So this cell can only be 4.
  19. R8C7 = 4 · Hidden SingleIn column 7, 4 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 4, R5C7 cannot hold 4.- Because row 9 already has a 4, R9C7 cannot hold 4.So 4 must go in R8C7.
  20. R7C2 = 4 · Hidden SingleIn the bottom-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 4, R8C1, R8C2 and R8C3 cannot hold 4.- Because row 9 already has a 4, R9C1 cannot hold 4.So 4 must go in R7C2.
  21. R6C1 = 4 · Hidden SingleIn the middle-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 4, R5C1 and R5C2 cannot hold 4.- Because column 2 already has a 4, R6C2 cannot hold 4.So 4 must go in R6C1.
  22. R1C3 = 4 · Hidden SingleIn the top-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 4, R1C1 and R2C1 cannot hold 4.- Because column 2 already has a 4, R1C2 and R2C2 cannot hold 4.- Because row 3 already has a 4, R3C1 cannot hold 4.So 4 must go in R1C3.
  23. R8C3 = 9 · Naked SingleLooking at R8C3:- Row 8 already has 3, 4, 5, 6 and 7 → ruled out.- Column 3 already has 1, 2 and 8 → ruled out.So this cell can only be 9.
  24. R2C8 = 4 · Hidden SingleIn the top-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 4, R1C8 and R1C9 cannot hold 4.- Because row 3 already has a 4, R3C9 cannot hold 4.So 4 must go in R2C8.
  25. R3C9 = 5 · Hidden SingleIn the top-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 5, R1C8 and R1C9 cannot hold 5.So 5 must go in R3C9.
  26. R3C1 = 6 · Naked SingleLooking at R3C1:- Row 3 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  27. R1C9 = 6 · Hidden SingleIn the top-right 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 6, R1C8 cannot hold 6.So 6 must go in R1C9.
  28. R1C8 = 3 · Naked SingleLooking at R1C8:- Row 1 already has 1, 2, 4, 5, 6 and 7 → ruled out.- Column 8 already has 8 → ruled out.- The top-right 3×3 box already has 9 → ruled out.So this cell can only be 3.
  29. R6C9 = 3 · Hidden SingleIn the middle-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 3, R5C7 cannot hold 3.- Because column 8 already has a 3, R6C8 cannot hold 3.So 3 must go in R6C9.
  30. R7C9 = 2 · Naked SingleLooking at R7C9:- Row 7 already has 1, 3, 4, 6, 7 and 9 → ruled out.- Column 9 already has 5 and 8 → ruled out.So this cell can only be 2.
  31. R6C8 = 2 · Hidden SingleIn the middle-right 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 2, R5C7 cannot hold 2.So 2 must go in R6C8.
  32. R6C2 = 9 · Naked SingleLooking at R6C2:- Row 6 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  33. R5C7 = 9 · Naked SingleLooking at R5C7:- Row 5 already has 2, 3, 4, 6, 7 and 8 → ruled out.- Column 7 already has 5 → ruled out.- The middle-right 3×3 box already has 1 → ruled out.So this cell can only be 9.
  34. R9C7 = 1 · Naked SingleLooking at R9C7:- Row 9 already has 3, 4, 5, 6, 7 and 8 → ruled out.- Column 7 already has 2 and 9 → ruled out.So this cell can only be 1.
  35. R1C1 = 9 · Hidden SingleIn the top-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 9, R1C2 cannot hold 9.- Because row 2 already has a 9, R2C1 and R2C2 cannot hold 9.So 9 must go in R1C1.
  36. R1C2 = 8 · Naked SingleLooking at R1C2:- Row 1 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  37. R8C1 = 8 · Hidden SingleIn the bottom-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 8, R8C2 cannot hold 8.- Because row 9 already has a 8, R9C1 cannot hold 8.So 8 must go in R8C1.
  38. R8C2 = 1 · Hidden SingleIn the bottom-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 9 already has a 1, R9C1 cannot hold 1.So 1 must go in R8C2.
  39. R8C4 = 2 · Naked SingleLooking at R8C4:- Row 8 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  40. R7C4 = 8 · Naked SingleLooking at R7C4:- Row 7 already has 1, 2, 3, 4, 6, 7 and 9 → ruled out.- Column 4 already has 5 → ruled out.So this cell can only be 8.
  41. R7C8 = 5 · Naked SingleLooking at R7C8:- Row 7 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  42. R9C8 = 9 · Naked SingleLooking at R9C8:- Row 9 already has 1, 3, 4, 5, 6, 7 and 8 → ruled out.- Column 8 already has 2 → ruled out.So this cell can only be 9.
  43. R9C1 = 2 · Naked SingleLooking at R9C1:- Row 9 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  44. R2C2 = 2 · Hidden SingleIn the top-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 2, R2C1 cannot hold 2.So 2 must go in R2C2.
  45. R2C1 = 5 · Naked SingleLooking at R2C1:- Row 2 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  46. R5C1 = 1 · Naked SingleLooking at R5C1:- Row 5 already has 2, 3, 4, 6, 7, 8 and 9 → ruled out.- Column 1 already has 5 → ruled out.So this cell can only be 1.
  47. R5C2 = 5 · Naked SingleLooking at R5C2:- Row 5 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.

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