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Daily Sudoku — 17 September 2026

The puzzle

  • 17 September 2026
  • Hard
  • 30 clues
Puzzle
··618···9
7··9····5
4··562·8·
···3····6
·37··952·
9···2····
·2·7·····
3·1··6··7
·7····69·
Solution
256187439
718934265
493562781
582371946
137649528
964825173
629758314
341296857
875413692

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 8 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
  • Naked Pair — two cells hold exactly the same two digits, so those digits are eliminated from every other cell in the unit
  • 2-String Kite — a row and a column each hold a digit in exactly two places, joined through one box; at least one free end holds the digit, so it is eliminated where the free ends cross
  • Turbot Fish — a strong–weak–strong chain on one digit forces at least one of the two chain ends to hold it
  • X-Wing — one digit forms a 2×2 rectangle across two rows and columns, so it is eliminated from the rest of those columns (or rows)
  • XY-Wing — a pivot with two pincers guarantees one digit lands in one of the pincers, so it is eliminated from any cell that sees both
  1. R1C6 = 7 · Hidden SingleIn the top-centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 7, R2C5 and R2C6 cannot hold 7.So 7 must go in R1C6.
  2. R2C8 = 6 · Hidden SingleIn the top-right 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 6, R1C7 and R1C8 cannot hold 6.- Because column 7 already has a 6, R2C7 cannot hold 6.- Because row 3 already has a 6, R3C7 and R3C9 cannot hold 6.So 6 must go in R2C8.
  3. R3C7 = 7 · Hidden SingleIn the top-right 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 7, R1C7 and R1C8 cannot hold 7.- Because row 2 already has a 7, R2C7 cannot hold 7.- Because column 9 already has a 7, R3C9 cannot hold 7.So 7 must go in R3C7.
  4. R4C5 = 7 · Hidden SingleIn the centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 7, R4C6 and R6C6 cannot hold 7.- Because row 5 already has a 7, R5C4 and R5C5 cannot hold 7.- Because column 4 already has a 7, R6C4 cannot hold 7.So 7 must go in R4C5.
  5. R6C8 = 7 · Hidden SingleIn the middle-right 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 7, R4C7 and R4C8 cannot hold 7.- Because row 5 already has a 7, R5C9 cannot hold 7.- Because column 7 already has a 7, R6C7 cannot hold 7.- Because column 9 already has a 7, R6C9 cannot hold 7.So 7 must go in R6C8.
  6. R4C7 = 9 · Hidden SingleIn the middle-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 9, R4C8 cannot hold 9.- Because row 5 already has a 9, R5C9 cannot hold 9.- Because row 6 already has a 9, R6C7 and R6C9 cannot hold 9.So 9 must go in R4C7.
  7. R7C1 = 6 · Hidden SingleIn the bottom-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 6, R7C3 cannot hold 6.- Because row 8 already has a 6, R8C2 cannot hold 6.- Because row 9 already has a 6, R9C1 and R9C3 cannot hold 6.So 6 must go in R7C1.
  8. R6C2 = 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, R4C2 and R4C3 cannot hold 6.- Because column 1 already has a 6, R5C1 cannot hold 6.- Because column 3 already has a 6, R6C3 cannot hold 6.So 6 must go in R6C2.
  9. R5C4 = 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 column 5 already has a 6, R5C5 cannot hold 6.- Because row 6 already has a 6, R6C4 and R6C6 cannot hold 6.So 6 must go in R5C4.
  10. R9C9 = 2 · Hidden SingleIn column 9, 2 fits in only one cell. Looking at the other empty cells:- Because row 3 already has a 2, R3C9 cannot hold 2.- Because row 5 already has a 2, R5C9 cannot hold 2.- Because row 6 already has a 2, R6C9 cannot hold 2.- Because row 7 already has a 2, R7C9 cannot hold 2.So 2 must go in R9C9.
  11. R8C4 = 2 · Hidden SingleIn the bottom-centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 2, R7C5 and R7C6 cannot hold 2.- Because column 5 already has a 2, R8C5 cannot hold 2.- Because row 9 already has a 2, R9C4, R9C5 and R9C6 cannot hold 2.So 2 must go in R8C4.
  12. R1C2 = 5 · Naked SingleLooking at R1C2:- Row 1 already has 1, 6, 7, 8 and 9 → ruled out.- Column 2 already has 2 and 3 → ruled out.- The top-left 3×3 box already has 4 → ruled out.So this cell can only be 5.
  13. R1C1 = 2 · Naked SingleLooking at R1C1:- Row 1 already has 1, 5, 6, 7, 8 and 9 → ruled out.- Column 1 already has 3 and 4 → ruled out.So this cell can only be 2.
  14. R2C7 = 2 · Hidden SingleIn the top-right 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 2, R1C7 and R1C8 cannot hold 2.- Because row 3 already has a 2, R3C9 cannot hold 2.So 2 must go in R2C7.
  15. R3C9 = 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 and R1C8 cannot hold 1.So 1 must go in R3C9.
  16. R2C2 = 1 · Hidden SingleIn the top-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 1, R2C3 cannot hold 1.- Because row 3 already has a 1, R3C2 and R3C3 cannot hold 1.So 1 must go in R2C2.
  17. R2C3 = 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 an 8, R3C2 and R3C3 cannot hold 8.So 8 must go in R2C3.
  18. R3C3 = 3 · Hidden SingleIn the top-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 3, R3C2 cannot hold 3.So 3 must go in R3C3.
  19. R3C2 = 9 · Naked SingleLooking at R3C2:- Row 3 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  20. R4C3 = 2 · Hidden SingleIn the middle-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 2, R4C1 cannot hold 2.- Because column 2 already has a 2, R4C2 cannot hold 2.- Because row 5 already has a 2, R5C1 cannot hold 2.- Because row 6 already has a 2, R6C3 cannot hold 2.So 2 must go in R4C3.
  21. R7C3 = 9 · Hidden SingleIn the bottom-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 9, R8C2 cannot hold 9.- Because row 9 already has a 9, R9C1 and R9C3 cannot hold 9.So 9 must go in R7C3.
  22. R8C5 = 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, R7C5 and R7C6 cannot hold 9.- Because row 9 already has a 9, R9C4, R9C5 and R9C6 cannot hold 9.So 9 must go in R8C5.
  23. R8C8 = 5 · Hidden SingleIn row 8, 5 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 5, R8C2 cannot hold 5.- Because column 7 already has a 5, R8C7 cannot hold 5.So 5 must go in R8C8.
  24. R7C5 = 5 · Hidden SingleStep 1 · Pick a digit and a box, and find where it can goThe centre 3×3 box has no 5 yet; 5 can only go in R4C6 and R6C6.Step 2 · Places on one column: point along it (pointing)All of them lie on column 6, so the box's 5 is on column 6. The rest of column 6 loses 5: R7C6 and R9C6.Step 3 · Pick a digit and a box, and find where it can goThe bottom-left 3×3 box has no 5 yet; 5 can only go in R9C1 and R9C3.Step 4 · Places on one row: point along it (pointing)All of them lie on row 9, so the box's 5 is on row 9. The rest of row 9 loses 5: R9C5.Step 5 · Collect what the eliminations unlockIn the bottom-centre 3×3 box, 5 has one place left, R7C5 (lost 5: R7C6 in step 2, R9C5 in step 4 and R9C6 in step 2) → place 5 in R7C5.
  25. R5C1 = 1 · Hidden SingleStep 1 · Find a cell with exactly two candidatesPivot: R4C2 holds only {4, 8}.Step 2 · Find two pincers that share the same ZR4C8 {1, 4} and R5C1 {1, 8} see the pivot and each hold one of its digits; their other digit is 1, so Z = 1.Step 3 · Follow both cases: one pincer is always ZIf the pivot is 4, R4C8 is 1; if it is 8, R5C1 is 1.Step 4 · Remove Z from every cell that sees both pincersThe cells that see both pincers lose 1: R4C1.Step 5 · Collect what the elimination unlocksIn the middle-left 3×3 box, 1 has one place left, R5C1 (lost 1: R4C1 in step 4) → place 1 in R5C1.
  26. R5C9 = 8 · Hidden SingleIn row 5, 8 fits in only one cell. Looking at the other empty cells:- Because column 5 already has an 8, R5C5 cannot hold 8.So 8 must go in R5C9.
  27. R5C5 = 4 · Naked SingleLooking at R5C5:- Row 5 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  28. R2C6 = 4 · Hidden SingleIn the top-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 4, R2C5 cannot hold 4.So 4 must go in R2C6.
  29. R2C5 = 3 · Naked SingleLooking at R2C5:- Row 2 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  30. R9C5 = 1 · Naked SingleLooking at R9C5:- Row 9 already has 2, 6, 7 and 9 → ruled out.- Column 5 already has 3, 4, 5 and 8 → ruled out.So this cell can only be 1.
  31. R9C4 = 4 · Hidden SingleIn the bottom-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 4, R7C6 and R9C6 cannot hold 4.So 4 must go in R9C4.
  32. R6C4 = 8 · Naked SingleLooking at R6C4:- Row 6 already has 2, 6, 7 and 9 → ruled out.- Column 4 already has 1, 3, 4 and 5 → ruled out.So this cell can only be 8.
  33. R8C2 = 4 · Hidden SingleIn the bottom-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 9 already has a 4, R9C1 and R9C3 cannot hold 4.So 4 must go in R8C2.
  34. R8C7 = 8 · Naked SingleLooking at R8C7:- Row 8 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  35. R4C2 = 8 · Naked SingleLooking at R4C2:- Row 4 already has 2, 3, 6, 7 and 9 → ruled out.- Column 2 already has 1, 4 and 5 → ruled out.So this cell can only be 8.
  36. R6C3 = 4 · Hidden SingleIn the middle-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 4, R4C1 cannot hold 4.So 4 must go in R6C3.
  37. R9C3 = 5 · Naked SingleLooking at R9C3:- Row 9 already has 1, 2, 4, 6, 7 and 9 → ruled out.- Column 3 already has 3 and 8 → ruled out.So this cell can only be 5.
  38. R4C1 = 5 · Naked SingleLooking at R4C1:- Row 4 already has 2, 3, 6, 7, 8 and 9 → ruled out.- Column 1 already has 1 and 4 → ruled out.So this cell can only be 5.
  39. R9C1 = 8 · Naked SingleLooking at R9C1:- Row 9 already has 1, 2, 4, 5, 6, 7 and 9 → ruled out.- Column 1 already has 3 → ruled out.So this cell can only be 8.
  40. R9C6 = 3 · Naked SingleLooking at R9C6:- Row 9 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  41. R7C6 = 8 · Naked SingleLooking at R7C6:- Row 7 already has 2, 5, 6, 7 and 9 → ruled out.- Column 6 already has 3 and 4 → ruled out.- The bottom-centre 3×3 box already has 1 → ruled out.So this cell can only be 8.
  42. R6C6 = 5 · Hidden SingleIn the centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 5, R4C6 cannot hold 5.So 5 must go in R6C6.
  43. R4C6 = 1 · Naked SingleLooking at R4C6:- Row 4 already has 2, 3, 5, 6, 7, 8 and 9 → ruled out.- Column 6 already has 4 → ruled out.So this cell can only be 1.
  44. R4C8 = 4 · Naked SingleLooking at R4C8:- Row 4 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  45. R1C7 = 4 · Hidden SingleIn the top-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 4, R1C8 cannot hold 4.So 4 must go in R1C7.
  46. R1C8 = 3 · Naked SingleLooking at R1C8:- Row 1 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  47. R7C8 = 1 · Naked SingleLooking at R7C8:- Row 7 already has 2, 5, 6, 7, 8 and 9 → ruled out.- Column 8 already has 3 and 4 → ruled out.So this cell can only be 1.
  48. R6C7 = 1 · Hidden SingleIn the middle-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 1, R6C9 cannot hold 1.So 1 must go in R6C7.
  49. R6C9 = 3 · Naked SingleLooking at R6C9:- Row 6 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  50. R7C7 = 3 · Naked SingleLooking at R7C7:- Row 7 already has 1, 2, 5, 6, 7, 8 and 9 → ruled out.- Column 7 already has 4 → ruled out.So this cell can only be 3.
  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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