NYT Pips hints and answers, Thursday, August 6, 2026

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

That cell is a 4, and only one domino has a 4: the 4–4.

Place the 4–4 domino with the 4 at row 3, column 1 and the 4 at row 4, column 1.

Full solution (domino: first half → second half): 1–1: R1C1→R1C2; 3–3: R1C3→R2C3; 4–4: R3C1→R4C1; 5–5: R3C2→R3C3; 6–6: R2C1→R2C2.

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

That cell is a 0, and only one domino has a 0: the 0–2.

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

Full solution (domino: first half → second half): 4–2: R2C2→R2C1; 0–2: R3C2→R3C1; 4–4: R3C3→R4C3; 1–3: R1C4→R2C4; 6–2: R1C2→R1C1; 4–1: R2C3→R1C3; 5–4: R3C5→R3C4.

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

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

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

Full solution (domino: first half → second half): 2–0: R3C2→R2C2; 0–5: R8C1→R8C2; 1–1: R8C4→R8C5; 5–5: R1C5→R2C5; 1–3: R9C5→R10C5; 4–4: R5C5→R6C5; 0–0: R5C1→R6C1; 1–2: R4C5→R3C5; 0–4: R9C3→R10C3; 5–2: R2C4→R3C4; 6–0: R3C1→R4C1; 3–4: R1C1→R2C1; 0–1: R9C1→R10C1; 2–2: R3C3→R4C3.