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Daily Sudoku — 1 October 2026

The puzzle

  • 1 October 2026
  • Hard
  • 30 clues
Puzzle
3·9··18·2
61···597·
·5····4··
·3···6···
··1·5···4
9···7··6·
····3··85
·8·7···4·
·43·6···7
Solution
379641852
614285973
258397416
437816529
861952734
925473168
792134685
586729341
143568297

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. R2C3 = 4 · Hidden SingleIn the top-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 4, R1C2 cannot hold 4.- Because row 3 already has a 4, R3C1 and R3C3 cannot hold 4.So 4 must go in R2C3.
  2. R3C6 = 7 · Hidden SingleIn the top-centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 7, R1C4 and R3C4 cannot hold 7.- Because column 5 already has a 7, R1C5 and R3C5 cannot hold 7.- Because row 2 already has a 7, R2C4 and R2C5 cannot hold 7.So 7 must go in R3C6.
  3. R1C2 = 7 · Hidden SingleIn the top-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 3 already has a 7, R3C1 and R3C3 cannot hold 7.So 7 must go in R1C2.
  4. R1C8 = 5 · Hidden SingleIn the top-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 5, R2C9 cannot hold 5.- Because row 3 already has a 5, R3C8 and R3C9 cannot hold 5.So 5 must go in R1C8.
  5. R3C9 = 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, R2C9 cannot hold 6.- Because column 8 already has a 6, R3C8 cannot hold 6.So 6 must go in R3C9.
  6. R1C4 = 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, R1C5 cannot hold 6.- Because row 2 already has a 6, R2C4 and R2C5 cannot hold 6.- Because row 3 already has a 6, R3C4 and R3C5 cannot hold 6.So 6 must go in R1C4.
  7. R1C5 = 4 · Naked SingleLooking at R1C5:- Row 1 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  8. R3C8 = 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, R2C9 cannot hold 1.So 1 must go in R3C8.
  9. R2C9 = 3 · Naked SingleLooking at R2C9:- Row 2 already has 1, 4, 5, 6, 7 and 9 → ruled out.- Column 9 already has 2 → ruled out.- The top-right 3×3 box already has 8 → ruled out.So this cell can only be 3.
  10. R3C4 = 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 and R2C5 cannot hold 3.- Because column 5 already has a 3, R3C5 cannot hold 3.So 3 must go in R3C4.
  11. R3C5 = 9 · Hidden SingleIn the top-centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 9, R2C4 and R2C5 cannot hold 9.So 9 must go in R3C5.
  12. R4C1 = 4 · Hidden SingleIn the middle-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 4, R4C3 and R6C3 cannot hold 4.- 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 R4C1.
  13. R5C2 = 6 · Hidden SingleIn the middle-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 6, R4C3 cannot hold 6.- Because column 1 already has a 6, R5C1 cannot hold 6.- Because row 6 already has a 6, R6C2 and R6C3 cannot hold 6.So 6 must go in R5C2.
  14. R7C2 = 9 · Hidden SingleIn the bottom-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 9, R7C1, R8C1 and R9C1 cannot hold 9.- Because column 3 already has a 9, R7C3 and R8C3 cannot hold 9.So 9 must go in R7C2.
  15. R6C2 = 2 · Naked SingleLooking at R6C2:- Row 6 already has 6, 7 and 9 → ruled out.- Column 2 already has 1, 3, 4, 5 and 8 → ruled out.So this cell can only be 2.
  16. R9C4 = 5 · Hidden SingleIn the bottom-centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 5, R7C4 and R7C6 cannot hold 5.- Because column 5 already has a 5, R8C5 cannot hold 5.- Because column 6 already has a 5, R8C6 and R9C6 cannot hold 5.So 5 must go in R9C4.
  17. R9C6 = 8 · Hidden SingleIn the bottom-centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 7 already has an 8, R7C4 and R7C6 cannot hold 8.- Because row 8 already has an 8, R8C5 and R8C6 cannot hold 8.So 8 must go in R9C6.
  18. R8C6 = 9 · Hidden SingleIn the bottom-centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 9, R7C4 and R7C6 cannot hold 9.- Because column 5 already has a 9, R8C5 cannot hold 9.So 9 must go in R8C6.
  19. R8C7 = 3 · Hidden SingleIn the bottom-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 3, R7C7 cannot hold 3.- Because column 9 already has a 3, R8C9 cannot hold 3.- Because row 9 already has a 3, R9C7 and R9C8 cannot hold 3.So 3 must go in R8C7.
  20. R5C8 = 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, R4C8 and R4C9 cannot hold 3.- Because column 7 already has a 3, R5C7 and R6C7 cannot hold 3.- Because column 9 already has a 3, R6C9 cannot hold 3.So 3 must go in R5C8.
  21. R6C6 = 3 · Hidden SingleIn the centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 3, R4C4 and R4C5 cannot hold 3.- Because row 5 already has a 3, R5C4 and R5C6 cannot hold 3.- Because column 4 already has a 3, R6C4 cannot hold 3.So 3 must go in R6C6.
  22. R6C4 = 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, R4C4 and R4C5 cannot hold 4.- Because row 5 already has a 4, R5C4 and R5C6 cannot hold 4.So 4 must go in R6C4.
  23. R7C6 = 4 · Hidden SingleIn the bottom-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 4, R7C4 cannot hold 4.- Because row 8 already has a 4, R8C5 cannot hold 4.So 4 must go in R7C6.
  24. R5C6 = 2 · Naked SingleLooking at R5C6:- Row 5 already has 1, 3, 4, 5 and 6 → ruled out.- Column 6 already has 7, 8 and 9 → ruled out.So this cell can only be 2.
  25. R7C7 = 6 · Hidden SingleIn the bottom-right 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 6, R8C9 cannot hold 6.- Because row 9 already has a 6, R9C7 and R9C8 cannot hold 6.So 6 must go in R7C7.
  26. R8C3 = 6 · Hidden SingleIn the bottom-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 6, R7C1 and R7C3 cannot hold 6.- Because column 1 already has a 6, R8C1 cannot hold 6.- Because row 9 already has a 6, R9C1 cannot hold 6.So 6 must go in R8C3.
  27. R8C1 = 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 and R7C3 cannot hold 5.- Because row 9 already has a 5, R9C1 cannot hold 5.So 5 must go in R8C1.
  28. R9C8 = 9 · Hidden SingleIn the bottom-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 9, R8C9 cannot hold 9.- Because column 7 already has a 9, R9C7 cannot hold 9.So 9 must go in R9C8.
  29. R4C8 = 2 · Naked SingleLooking at R4C8:- Row 4 already has 3, 4 and 6 → ruled out.- Column 8 already has 1, 5, 7, 8 and 9 → ruled out.So this cell can only be 2.
  30. R4C9 = 9 · Hidden SingleIn the middle-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 9, R4C7 and R5C7 cannot hold 9.- Because row 6 already has a 9, R6C7 and R6C9 cannot hold 9.So 9 must go in R4C9.
  31. R5C4 = 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 R4C5 cannot hold 9.So 9 must go in R5C4.
  32. R6C9 = 8 · Hidden SingleIn the middle-right 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 7 already has an 8, R4C7, R5C7 and R6C7 cannot hold 8.So 8 must go in R6C9.
  33. R8C9 = 1 · Naked SingleLooking at R8C9:- Row 8 already has 3, 4, 5, 6, 7, 8 and 9 → ruled out.- Column 9 already has 2 → ruled out.So this cell can only be 1.
  34. R8C5 = 2 · Naked SingleLooking at R8C5:- Row 8 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  35. R7C4 = 1 · Naked SingleLooking at R7C4:- Row 7 already has 3, 4, 5, 6, 8 and 9 → ruled out.- Column 4 already has 7 → ruled out.- The bottom-centre 3×3 box already has 2 → ruled out.So this cell can only be 1.
  36. R9C7 = 2 · Naked SingleLooking at R9C7:- Row 9 already has 3, 4, 5, 6, 7, 8 and 9 → ruled out.- The bottom-right 3×3 box already has 1 → ruled out.So this cell can only be 2.
  37. R9C1 = 1 · Naked SingleLooking at R9C1:- Row 9 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  38. R2C4 = 2 · Hidden SingleIn the top-centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 2, R2C5 cannot hold 2.So 2 must go in R2C4.
  39. R2C5 = 8 · Naked SingleLooking at R2C5:- Row 2 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  40. R4C4 = 8 · Naked SingleLooking at R4C4:- Row 4 already has 2, 3, 4, 6 and 9 → ruled out.- Column 4 already has 1, 5 and 7 → ruled out.So this cell can only be 8.
  41. R4C5 = 1 · Naked SingleLooking at R4C5:- Row 4 already has 2, 3, 4, 6, 8 and 9 → ruled out.- Column 5 already has 5 and 7 → ruled out.So this cell can only be 1.
  42. R5C1 = 8 · Hidden SingleIn the middle-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 4 already has an 8, R4C3 cannot hold 8.- Because row 6 already has an 8, R6C3 cannot hold 8.So 8 must go in R5C1.
  43. R5C7 = 7 · Naked SingleLooking at R5C7:- Row 5 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  44. R3C3 = 8 · Hidden SingleIn the top-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 1 already has an 8, R3C1 cannot hold 8.So 8 must go in R3C3.
  45. R3C1 = 2 · Naked SingleLooking at R3C1:- Row 3 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  46. R7C1 = 7 · Naked SingleLooking at R7C1:- Row 7 already has 1, 3, 4, 5, 6, 8 and 9 → ruled out.- Column 1 already has 2 → ruled out.So this cell can only be 7.
  47. R7C3 = 2 · Naked SingleLooking at R7C3:- Row 7 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  48. R4C3 = 7 · Hidden SingleIn the middle-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 6 already has a 7, R6C3 cannot hold 7.So 7 must go in R4C3.
  49. R4C7 = 5 · Naked SingleLooking at R4C7:- Row 4 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  50. R6C3 = 5 · Naked SingleLooking at R6C3:- Row 6 already has 2, 3, 4, 6, 7, 8 and 9 → ruled out.- Column 3 already has 1 → ruled out.So this cell can only be 5.
  51. R6C7 = 1 · Naked SingleLooking at R6C7:- Row 6 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.

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