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Daily Sudoku — 8 October 2026

The puzzle

  • 8 October 2026
  • Hard
  • 30 clues
Puzzle
···7···4·
·641····7
······39·
·523··9·8
·······1·
·418··5·6
91·······
5·6·3218·
···981···
Solution
139758642
264193857
785246391
652314978
897625413
341879526
918567234
576432189
423981765

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 4 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
  • Locked Candidates (pointing/claiming) — a digit inside a box is confined to one row or column, so it is eliminated from the rest of that row or column
  • XY-Chain — a chain of two-candidate cells forces a digit to sit at one of the two ends of the chain
  1. R1C7 = 6 · Hidden SingleIn the top-right 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 6, R1C9 and R3C9 cannot hold 6.- Because row 2 already has a 6, R2C7 and R2C8 cannot hold 6.So 6 must go in R1C7.
  2. R2C7 = 8 · Hidden SingleIn the top-right 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 9 already has an 8, R1C9 and R3C9 cannot hold 8.- Because column 8 already has an 8, R2C8 cannot hold 8.So 8 must go in R2C7.
  3. R4C5 = 1 · Hidden SingleIn the centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 1, R4C6 cannot hold 1.- Because row 5 already has a 1, R5C4, R5C5 and R5C6 cannot hold 1.- Because row 6 already has a 1, R6C5 and R6C6 cannot hold 1.So 1 must go in R4C5.
  4. R9C1 = 4 · Hidden SingleIn the bottom-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 4, R7C3 and R9C3 cannot hold 4.- Because column 2 already has a 4, R8C2 and R9C2 cannot hold 4.So 4 must go in R9C1.
  5. R9C2 = 2 · Hidden SingleIn the bottom-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 2, R7C3 and R9C3 cannot hold 2.- Because row 8 already has a 2, R8C2 cannot hold 2.So 2 must go in R9C2.
  6. R7C3 = 8 · Hidden SingleIn the bottom-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 8 already has an 8, R8C2 cannot hold 8.- Because row 9 already has an 8, R9C3 cannot hold 8.So 8 must go in R7C3.
  7. R9C3 = 3 · Hidden SingleIn the bottom-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 3, R8C2 cannot hold 3.So 3 must go in R9C3.
  8. R8C2 = 7 · Naked SingleLooking at R8C2:- Row 8 already has 1, 2, 3, 5, 6 and 8 → ruled out.- Column 2 already has 4 → ruled out.- The bottom-left 3×3 box already has 9 → ruled out.So this cell can only be 7.
  9. R8C9 = 9 · Hidden SingleIn the bottom-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 9, R7C7, R7C8 and R7C9 cannot hold 9.- Because row 9 already has a 9, R9C7, R9C8 and R9C9 cannot hold 9.So 9 must go in R8C9.
  10. R8C4 = 4 · Naked SingleLooking at R8C4:- Row 8 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  11. R4C6 = 4 · Hidden SingleIn row 4, 4 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 4, R4C1 cannot hold 4.- Because column 8 already has a 4, R4C8 cannot hold 4.So 4 must go in R4C6.
  12. R3C5 = 4 · Hidden SingleIn the top-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 4, R1C5 and R1C6 cannot hold 4.- Because row 2 already has a 4, R2C5 and R2C6 cannot hold 4.- Because column 4 already has a 4, R3C4 cannot hold 4.- Because column 6 already has a 4, R3C6 cannot hold 4.So 4 must go in R3C5.
  13. R4C1 = 6 · Hidden SingleIn row 4, 6 fits in only one cell. Looking at the other empty cells:- Because the middle-right 3×3 box already has a 6, R4C8 cannot hold 6.So 6 must go in R4C1.
  14. R4C8 = 7 · Naked SingleLooking at R4C8:- Row 4 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  15. R9C8 = 6 · Hidden SingleIn row 9, 6 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 6, R9C7 cannot hold 6.- Because column 9 already has a 6, R9C9 cannot hold 6.So 6 must go in R9C8.
  16. R9C9 = 5 · Hidden SingleIn row 9, 5 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 5, R9C7 cannot hold 5.So 5 must go in R9C9.
  17. R9C7 = 7 · Naked SingleLooking at R9C7:- Row 9 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  18. R2C8 = 5 · Hidden SingleIn the top-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 5, R1C9 and R3C9 cannot hold 5.So 5 must go in R2C8.
  19. R3C2 = 8 · Naked SingleLooking at R3C2:- Row 3 already has 3, 4 and 9 → ruled out.- Column 2 already has 1, 2, 5, 6 and 7 → ruled out.So this cell can only be 8.
  20. R1C6 = 8 · Hidden SingleIn the top-centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 5 already has an 8, R1C5 cannot hold 8.- Because row 2 already has an 8, R2C5 and R2C6 cannot hold 8.- Because row 3 already has an 8, R3C4 and R3C6 cannot hold 8.So 8 must go in R1C6.
  21. R2C6 = 3 · Hidden SingleIn the top-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 3, R1C5 and R2C5 cannot hold 3.- Because row 3 already has a 3, R3C4 and R3C6 cannot hold 3.So 3 must go in R2C6.
  22. R5C1 = 8 · Hidden SingleIn the middle-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 2 already has an 8, R5C2 cannot hold 8.- Because column 3 already has an 8, R5C3 cannot hold 8.- Because row 6 already has an 8, R6C1 cannot hold 8.So 8 must go in R5C1.
  23. R2C5 = 9 · Hidden SingleIn row 2, 9 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 9, R2C1 cannot hold 9.So 9 must go in R2C5.
  24. R2C1 = 2 · Naked SingleLooking at R2C1:- Row 2 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  25. R6C6 = 9 · Hidden SingleIn row 6, 9 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 9, R6C1 cannot hold 9.- Because column 5 already has a 9, R6C5 cannot hold 9.- Because column 8 already has a 9, R6C8 cannot hold 9.So 9 must go in R6C6.
  26. R1C2 = 3 · Hidden SingleStep 1 · List the bivalue cells and pick a startStart from the bivalue cell R1C2 {3, 9} with Z = 3: if it is not 3, it is 9.Step 2 · Walk the chain one link at a timeEach cell sees the one before and loses the digit it just took: R1C3 {5, 9} → 5; R1C5 {2, 5} → 2; R6C5 {2, 7} → 7.Step 3 · Stop at a cell that is forced to ZR6C1 {3, 7} sees R6C5, so it is not 7 — it is 3. A chain of 5 cells: if R1C2 is not 3, R6C1 is 3.Step 4 · Either end is Z: clear Z from cells that see bothThere is always a 3 in R1C2 or R6C1, so the cells that see both ends lose 3: R1C1 and R5C2.Step 5 · Collect what the elimination unlocksIn the top-left 3×3 box, 3 has one place left, R1C2 (lost 3: R1C1 in step 4) → place 3 in R1C2.
  27. R5C2 = 9 · Naked SingleLooking at R5C2:- Row 5 already has 1 and 8 → ruled out.- Column 2 already has 2, 3, 4, 5, 6 and 7 → ruled out.So this cell can only be 9.
  28. R1C3 = 9 · Hidden SingleIn the top-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 9, R1C1 cannot hold 9.- Because row 3 already has a 9, R3C1 and R3C3 cannot hold 9.So 9 must go in R1C3.
  29. R3C3 = 5 · Hidden SingleIn the top-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 5, R1C1 and R3C1 cannot hold 5.So 5 must go in R3C3.
  30. R5C3 = 7 · Naked SingleLooking at R5C3:- Row 5 already has 1, 8 and 9 → ruled out.- Column 3 already has 2, 3, 4, 5 and 6 → ruled out.So this cell can only be 7.
  31. R6C1 = 3 · Naked SingleLooking at R6C1:- Row 6 already has 1, 4, 5, 6, 8 and 9 → ruled out.- Column 1 already has 2 → ruled out.- The middle-left 3×3 box already has 7 → ruled out.So this cell can only be 3.
  32. R3C1 = 7 · Hidden SingleIn the top-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 7, R1C1 cannot hold 7.So 7 must go in R3C1.
  33. R1C1 = 1 · Naked SingleLooking at R1C1:- Row 1 already has 3, 4, 6, 7, 8 and 9 → ruled out.- Column 1 already has 2 and 5 → ruled out.So this cell can only be 1.
  34. R1C5 = 5 · Hidden SingleIn the top-centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 3 already has a 5, R3C4 and R3C6 cannot hold 5.So 5 must go in R1C5.
  35. R1C9 = 2 · Naked SingleLooking at R1C9:- Row 1 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  36. R3C9 = 1 · Naked SingleLooking at R3C9:- Row 3 already has 3, 4, 5, 7, 8 and 9 → ruled out.- Column 9 already has 2 and 6 → ruled out.So this cell can only be 1.
  37. R3C4 = 2 · Hidden SingleIn the top-centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 2, R3C6 cannot hold 2.So 2 must go in R3C4.
  38. R3C6 = 6 · Naked SingleLooking at R3C6:- Row 3 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  39. R6C5 = 7 · Hidden SingleIn the centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 7, R5C4, R5C5 and R5C6 cannot hold 7.So 7 must go in R6C5.
  40. R6C8 = 2 · Naked SingleLooking at R6C8:- Row 6 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  41. R7C8 = 3 · Naked SingleLooking at R7C8:- Row 7 already has 1, 8 and 9 → ruled out.- Column 8 already has 2, 4, 5, 6 and 7 → ruled out.So this cell can only be 3.
  42. R5C5 = 2 · Hidden SingleIn the centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 2, R5C4 cannot hold 2.- Because column 6 already has a 2, R5C6 cannot hold 2.So 2 must go in R5C5.
  43. R7C5 = 6 · Naked SingleLooking at R7C5:- Row 7 already has 1, 3, 8 and 9 → ruled out.- Column 5 already has 2, 4, 5 and 7 → ruled out.So this cell can only be 6.
  44. R5C4 = 6 · Hidden SingleIn the centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 6, R5C6 cannot hold 6.So 6 must go in R5C4.
  45. R7C4 = 5 · Naked SingleLooking at R7C4:- Row 7 already has 1, 3, 6, 8 and 9 → ruled out.- Column 4 already has 2, 4 and 7 → ruled out.So this cell can only be 5.
  46. R5C6 = 5 · Naked SingleLooking at R5C6:- Row 5 already has 1, 2, 6, 7, 8 and 9 → ruled out.- Column 6 already has 3 and 4 → ruled out.So this cell can only be 5.
  47. R7C6 = 7 · Naked SingleLooking at R7C6:- Row 7 already has 1, 3, 5, 6, 8 and 9 → ruled out.- Column 6 already has 2 and 4 → ruled out.So this cell can only be 7.
  48. R5C9 = 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, R5C7 cannot hold 3.So 3 must go in R5C9.
  49. R5C7 = 4 · Naked SingleLooking at R5C7:- Row 5 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  50. R7C7 = 2 · Naked SingleLooking at R7C7:- Row 7 already has 1, 3, 5, 6, 7, 8 and 9 → ruled out.- Column 7 already has 4 → ruled out.So this cell can only be 2.
  51. R7C9 = 4 · Naked SingleLooking at R7C9:- Row 7 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.

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