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Daily Sudoku — 12 July 2026

The puzzle

  • 12 July 2026
  • Medium
  • 34 clues
Puzzle
735·8···9
1··9·2·5·
9·6·5··8·
·9·····2·
·1····7·8
··4621···
4····86·3
··13··57·
369·····2
Solution
735184269
148962357
926753481
593847126
612539748
874621935
457218693
281396574
369475812

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. R3C2 = 2 · Hidden SingleIn the top-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 2, R2C2 and R2C3 cannot hold 2.So 2 must go in R3C2.
  2. R2C2 = 4 · Hidden SingleIn the top-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 4, R2C3 cannot hold 4.So 4 must go in R2C2.
  3. R2C3 = 8 · Naked SingleLooking at R2C3:- Row 2 already has 1, 2, 4, 5 and 9 → ruled out.- Column 3 already has 6 → ruled out.- The top-left 3×3 box already has 3 and 7 → ruled out.So this cell can only be 8.
  4. R1C7 = 2 · Hidden SingleIn the top-right 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 2, R1C8 cannot hold 2.- Because row 2 already has a 2, R2C7 and R2C9 cannot hold 2.- Because row 3 already has a 2, R3C7 and R3C9 cannot hold 2.So 2 must go in R1C7.
  5. R4C4 = 8 · Hidden SingleIn the centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 8, R4C5 cannot hold 8.- Because column 6 already has a 8, R4C6 cannot hold 8.- Because row 5 already has a 8, R5C4, R5C5 and R5C6 cannot hold 8.So 8 must go in R4C4.
  6. R7C2 = 5 · Hidden SingleIn the bottom-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 5, R7C3 cannot hold 5.- Because row 8 already has a 5, R8C1 and R8C2 cannot hold 5.So 5 must go in R7C2.
  7. R7C3 = 7 · Hidden SingleIn the bottom-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 7, R8C1 and R8C2 cannot hold 7.So 7 must go in R7C3.
  8. R6C2 = 7 · Hidden SingleIn the middle-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 7, R4C1 and R6C1 cannot hold 7.- Because column 3 already has a 7, R4C3 cannot hold 7.- Because row 5 already has a 7, R5C1 and R5C3 cannot hold 7.So 7 must go in R6C2.
  9. R8C2 = 8 · Naked SingleLooking at R8C2:- Row 8 already has 1, 3, 5 and 7 → ruled out.- Column 2 already has 2, 4, 6 and 9 → ruled out.So this cell can only be 8.
  10. R8C1 = 2 · Naked SingleLooking at R8C1:- Row 8 already has 1, 3, 5, 7 and 8 → ruled out.- Column 1 already has 4 and 9 → ruled out.- The bottom-left 3×3 box already has 6 → ruled out.So this cell can only be 2.
  11. R5C3 = 2 · Hidden SingleIn the middle-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 2, R4C1 and R4C3 cannot hold 2.- Because column 1 already has a 2, R5C1 cannot hold 2.- Because row 6 already has a 2, R6C1 cannot hold 2.So 2 must go in R5C3.
  12. R4C3 = 3 · Naked SingleLooking at R4C3:- Row 4 already has 2, 8 and 9 → ruled out.- Column 3 already has 1, 4, 5, 6 and 7 → ruled out.So this cell can only be 3.
  13. R6C1 = 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 a 8, R4C1 cannot hold 8.- Because row 5 already has a 8, R5C1 cannot hold 8.So 8 must go in R6C1.
  14. R7C4 = 2 · Hidden SingleIn the bottom-centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 2, R7C5 cannot hold 2.- Because row 8 already has a 2, R8C5 and R8C6 cannot hold 2.- Because row 9 already has a 2, R9C4, R9C5 and R9C6 cannot hold 2.So 2 must go in R7C4.
  15. R9C7 = 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, R7C8 cannot hold 8.- Because row 8 already has a 8, R8C9 cannot hold 8.- Because column 8 already has a 8, R9C8 cannot hold 8.So 8 must go in R9C7.
  16. R7C8 = 9 · Hidden SingleIn the bottom-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 9, R8C9 cannot hold 9.- Because row 9 already has a 9, R9C8 cannot hold 9.So 9 must go in R7C8.
  17. R7C5 = 1 · Naked SingleLooking at R7C5:- Row 7 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  18. R6C7 = 9 · Hidden SingleIn the middle-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 9, R4C7 and R4C9 cannot hold 9.- Because column 8 already has a 9, R5C8 and R6C8 cannot hold 9.- Because column 9 already has a 9, R6C9 cannot hold 9.So 9 must go in R6C7.
  19. R9C8 = 1 · Hidden SingleIn the bottom-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 1, R8C9 cannot hold 1.So 1 must go in R9C8.
  20. R8C9 = 4 · Naked SingleLooking at R8C9:- Row 8 already has 1, 2, 3, 5, 7 and 8 → ruled out.- Column 9 already has 9 → ruled out.- The bottom-right 3×3 box already has 6 → ruled out.So this cell can only be 4.
  21. R1C4 = 1 · Hidden SingleIn row 1, 1 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 1, R1C6 cannot hold 1.- Because column 8 already has a 1, R1C8 cannot hold 1.So 1 must go in R1C4.
  22. R6C8 = 3 · Hidden SingleIn row 6, 3 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 3, R6C9 cannot hold 3.So 3 must go in R6C8.
  23. R6C9 = 5 · Naked SingleLooking at R6C9:- Row 6 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  24. R2C7 = 3 · Naked SingleLooking at R2C7:- Row 2 already has 1, 2, 4, 5, 8 and 9 → ruled out.- Column 7 already has 6 and 7 → ruled out.So this cell can only be 3.
  25. R3C6 = 3 · Hidden SingleIn the top-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 3, R1C6 cannot hold 3.- Because row 2 already has a 3, R2C5 cannot hold 3.- Because column 4 already has a 3, R3C4 cannot hold 3.So 3 must go in R3C6.
  26. R5C5 = 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, R4C5 and R4C6 cannot hold 3.- Because column 4 already has a 3, R5C4 cannot hold 3.- Because column 6 already has a 3, R5C6 cannot hold 3.So 3 must go in R5C5.
  27. R5C6 = 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, R4C5 and R4C6 cannot hold 9.- Because column 4 already has a 9, R5C4 cannot hold 9.So 9 must go in R5C6.
  28. R8C5 = 9 · Hidden SingleIn the bottom-centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 9, R8C6 cannot hold 9.- Because row 9 already has a 9, R9C4, R9C5 and R9C6 cannot hold 9.So 9 must go in R8C5.
  29. 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.
  30. R2C5 = 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 cannot hold 6.- Because row 3 already has a 6, R3C4 cannot hold 6.So 6 must go in R2C5.
  31. R2C9 = 7 · Naked SingleLooking at R2C9:- Row 2 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  32. R3C4 = 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 R3C4.
  33. R1C6 = 4 · Naked SingleLooking at R1C6:- Row 1 already has 1, 2, 3, 5, 7, 8 and 9 → ruled out.- Column 6 already has 6 → ruled out.So this cell can only be 4.
  34. R1C8 = 6 · Naked SingleLooking at R1C8:- Row 1 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  35. R5C8 = 4 · Naked SingleLooking at R5C8:- Row 5 already has 1, 2, 3, 7, 8 and 9 → ruled out.- Column 8 already has 5 and 6 → ruled out.So this cell can only be 4.
  36. R3C7 = 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, R3C9 cannot hold 4.So 4 must go in R3C7.
  37. R3C9 = 1 · Naked SingleLooking at R3C9:- Row 3 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  38. R4C7 = 1 · Naked SingleLooking at R4C7:- Row 4 already has 2, 3, 8 and 9 → ruled out.- Column 7 already has 4, 5, 6 and 7 → ruled out.So this cell can only be 1.
  39. R4C9 = 6 · Naked SingleLooking at R4C9:- Row 4 already has 1, 2, 3, 8 and 9 → ruled out.- Column 9 already has 4, 5 and 7 → ruled out.So this cell can only be 6.
  40. R5C1 = 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, R4C1 cannot hold 6.So 6 must go in R5C1.
  41. R5C4 = 5 · Naked SingleLooking at R5C4:- Row 5 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  42. R4C1 = 5 · Naked SingleLooking at R4C1:- Row 4 already has 1, 2, 3, 6, 8 and 9 → ruled out.- Column 1 already has 4 and 7 → ruled out.So this cell can only be 5.
  43. R9C4 = 4 · Naked SingleLooking at R9C4:- Row 9 already has 1, 2, 3, 6, 8 and 9 → ruled out.- Column 4 already has 5 and 7 → ruled out.So this cell can only be 4.
  44. R4C5 = 4 · Hidden SingleIn the centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 4, R4C6 cannot hold 4.So 4 must go in R4C5.
  45. R4C6 = 7 · Naked SingleLooking at R4C6:- Row 4 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  46. R9C5 = 7 · Naked SingleLooking at R9C5:- Row 9 already has 1, 2, 3, 4, 6, 8 and 9 → ruled out.- Column 5 already has 5 → ruled out.So this cell can only be 7.
  47. R9C6 = 5 · Naked SingleLooking at R9C6:- Row 9 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.

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