Common Sudoku Techniques Explained

Why it works

1. Look at the middle-right 3×3 box: rows 4–6 and columns 7–9. Six of its cells are still empty.
2. Row 6 already has a 5 at r6c6, so the whole bottom row of the box is ruled out.
3. Column 9 already has a 5 at r3c9, so r4c9 and r5c9 are ruled out too.
4. One empty cell survives, so r5c8 must be the 5.

Why it works

Three units meet at r7c4, and between them they use up eight digits: row 7 already holds 1, 2, 3, 4, 5, column 4 holds 2, 3, 6, and the box holds 3, 7, 9. Only 8 is left, so that is the answer — no pattern-hunting needed, just the three lines that cross on the cell.

Why it works

Row 6 has four cells left. Three of them — r6c6, r6c7 and r6c9 — cannot take a 1 at all, so the row's 1 has nowhere to go but r6c5. That cell still holds four candidates {1, 2, 8, 9}, so it is nowhere near a naked single; the 1 is hidden among them. Looking at the row as a whole, rather than at one cell, is what finds it.

Why it works

Inside the top-right box, candidate 8 appears only at r1c7 and r1c8. Both cells are in row 1, so the 8 for that box must be somewhere in row 1. Therefore, candidate 8 can be removed from the other cells of row 1 outside the box: remove 8 from r1c1 and r1c2.

Why it works

In row 5, candidate 6 appears only in r5c1 and r5c2, and both cells are inside the middle-left box. Therefore, the 6 for row 5 must be placed inside that box. Candidate 6 can then be removed from the rest of the same box: remove 6 from r6c2.

Why it works

Cells r3c3 and r9c3 are both exactly {2, 8}. Whichever way round they go, the two of them use up the 2 and the 8 of column 3 between them — so no other cell in that column can be either digit.

Remove 2 from r2c3 and 8 from r6c3. r1c3 keeps both its candidates: it never held a 2 or an 8 to begin with.

Why it works

In column 9, the digits 3 and 9 appear in only two cells: r5c9 and r8c9. Two digits needing two cells fills both of them, so nothing else can live there.

Strip the extras: remove 5 from r5c9, and 4 and 5 from r8c9. Both cells become {3, 9} — a naked pair that was there all along, just buried under spare candidates.

Why it works

The three highlighted cells in row 9 use only the digits 3, 4, and 5. Those three digits must occupy those three cells. Therefore, in the rest of row 9, only the overlapping candidates can be removed: remove 4 from r9c8.

Why it works

In row 2, the digits 2, 3, and 6 appear only in r2c3, r2c8, and r2c9. This is a proper hidden triple: 2 appears in r2c3/r2c8, 3 appears in r2c8/r2c9, and 6 appears in r2c3/r2c9. Those three cells must contain 2, 3, and 6 in some order. Therefore, remove the extra candidates: remove 8 and 9 from r2c8, and remove 9 from r2c9.

Why it works

In row 2, digit 5 fits only in r2c2 or r2c7, and in row 8 only in r8c2 or r8c7 — the same two columns. The four corners form a rectangle, and the two 5s must sit on opposite corners of it. Either way, one of them lands in column 2 and the other in column 7.

Those two columns are therefore spoken for. Every other cell in them loses the 5: remove it from r1c7, r3c2, r7c2 and r7c7.

Why it works

In rows 1, 4, and 7, candidate 6 is confined to columns 7, 8, and 9. This forms a row-based Swordfish. Those three rows must place their 6s somewhere inside columns 7, 8, and 9, so candidate 6 can be removed from the other cells in those same columns: remove 6 from r2c7 and r9c9.

Why it works

In row 5, digit 3 fits only in r5c2 or r5c6. In row 7, only in r7c2 or r7c5. Both rows include column 2, and a column holds each digit once — so r5c2 and r7c2 cannot both be 3. At least one of the two roof cells, r5c6 or r7c5, must therefore be the 3. Any cell that sees both roofs can be cleared: remove 3 from r4c5 and r9c6.

Why it works

In row 9, digit 2 fits only in r9c3 or r9c6. In column 5, only in r2c5 or r8c5. The cells r9c6 and r8c5 sit in the same box, so they cannot both be 2. That forces the 2 to the far end of one string or the other: r9c3 or r2c5. r2c3 sees both far ends, so remove 2 from it.

Why it works

In the bottom-middle box, digit 2 fits only in r9c4, r9c6 and r7c5 — all of them on row 9 or column 5, which cross at r9c5. Now look at row 4: the 2 there is either at r4c5 or r4c7.

1. If r4c5 is 2, column 5 is used up, so the box's 2 must sit on row 9 — and then r9c7, on row 9 outside the box, cannot be 2.
2. If r4c7 is 2, then r9c7 cannot be 2 either, because they share column 7.

Both branches agree, so remove 2 from r9c7.

Why it works

The pivot cell r5c6 has candidates {1, 4}. One wing is r3c6 with {3, 4}, and the other wing is r6c5 with {1, 3}. No matter whether the pivot becomes 1 or 4, one of the wings must become 3. Therefore, any cell that sees both wings cannot contain 3: remove 3 from r6c6.

Why it works

The pivot r8c7 holds {2, 3, 8}. Its wings are r4c7 with {2, 3} (same column) and r9c8 with {2, 8} (same box). Run through the pivot's three options: if it is 3 the first wing must be 2; if it is 8 the second wing must be 2; if it is 2, it is the 2 itself. So one of these three cells is always the 2. Unlike an XY-Wing, the target must see all three cells, pivot included: r7c7 does, so remove 2 from it.

Why it works

r1c3 and r2c6 both hold exactly {4, 6}, and they do not see each other. In row 3, digit 4 has only two places left: r3c1 and r3c6 — one of them is the 4. r3c1 shares a box with r1c3, and r3c6 shares column 6 with r2c6.

1. If r3c1 is 4, then r1c3 cannot be 4, so it is 6.
2. If r3c6 is 4, then r2c6 cannot be 4, so it is 6.

Either way, one of the two matching cells is the 6. Any cell seeing both can be cleared: remove 6 from r1c5 and r2c3.