NYT Pips hints and answers, Wednesday, August 12, 2026

Start with the cell that must be less than 2 at row 3, column 1. It has the fewest ways to be satisfied.

That cell is a 0, and only one domino has a 0: the 0–3.

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

Full solution (domino: first half → second half): 3–3: R1C4→R2C4; 2–6: R1C3→R1C2; 0–3: R3C1→R2C1; 2–2: R3C3→R3C4; 4–5: R3C2→R2C2.

Start with the cell that must equal 4 at row 3, column 2. It has the fewest ways to be satisfied.

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

Place the 4–2 domino with the 4 at row 3, column 2 and the 2 at row 3, column 3.

Full solution (domino: first half → second half): 4–2: R3C2→R3C3; 1–4: R1C3→R1C4; 3–6: R1C5→R2C5; 2–1: R2C2→R1C2; 5–4: R3C4→R3C5; 5–2: R2C4→R2C3; 3–1: R2C1→R1C1.

Start with the cell that must equal 3 at row 1, column 2. It has the fewest ways to be satisfied.

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

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

Full solution (domino: first half → second half): 0–0: R4C4→R4C5; 0–1: R4C2→R4C1; 0–2: R9C8→R8C8; 0–3: R2C2→R1C2; 0–5: R3C4→R3C5; 0–6: R3C2→R3C1; 1–2: R8C1→R7C1; 1–4: R6C4→R6C5; 1–6: R4C8→R4C7; 3–3: R6C6→R7C6; 3–4: R9C6→R9C5; 3–5: R8C6→R8C5; 4–4: R8C4→R9C4; 5–5: R2C7→R3C7; 5–6: R9C1→R9C2; 6–6: R7C8→R7C9.