NYT Pips hints and answers, Saturday, September 12, 2026

Start with the cell that must be less than 1 at row 1, column 2. It has the fewest ways to be satisfied.

That cell is a 0, and only one domino has a 0: the 0–5.

Place the 0–5 domino with the 0 at row 1, column 2 and the 5 at row 1, column 3.

Full solution (domino: first half → second half): 3–1: R1C1→R2C1; 6–6: R3C3→R3C4; 1–4: R2C2→R3C2; 4–2: R1C4→R2C4; 0–5: R1C2→R1C3.

Start with the cell that must equal 3 at row 3, column 5. It has the fewest ways to be satisfied.

That cell is a 3. Only 3 dominoes have a 3: 6–3, 3–3, 3–0.

Place the 6–3 domino with the 6 at row 3, column 4 and the 3 at row 3, column 5.

Full solution (domino: first half → second half): 1–1: R1C4→R2C4; 0–2: R1C2→R1C3; 0–0: R1C1→R2C1; 6–3: R3C4→R3C5; 3–3: R3C2→R3C3; 3–0: R2C3→R2C2; 0–6: R4C3→R4C4.

Start with the cell that must equal 4 at row 1, column 7. It has the fewest ways to be satisfied.

That cell is a 4. Only 5 dominoes have a 4: 0–4, 4–3, 4–1, 6–4, 2–4.

Place the 0–4 domino with the 0 at row 2, column 7 and the 4 at row 1, column 7.

Full solution (domino: first half → second half): 0–4: R2C7→R1C7; 5–6: R10C4→R10C5; 3–3: R3C8→R3C9; 1–6: R2C3→R1C3; 0–3: R7C7→R8C7; 1–1: R3C1→R3C2; 4–3: R8C1→R8C2; 2–5: R4C7→R5C7; 3–6: R9C7→R10C7; 4–1: R1C1→R2C1; 3–2: R8C9→R7C9; 6–4: R8C4→R8C5; 1–3: R3C3→R4C3; 6–6: R9C9→R10C9; 2–4: R3C5→R3C6; 3–5: R9C2→R10C2.