NYT Pips hints and answers, Sunday, October 11, 2026

Start with the cell that must be greater than 4 at row 2, column 5. It has the fewest ways to be satisfied.

That cell is a 6, and only one domino has a 6: the 6–4.

Place the 6–4 domino with the 6 at row 2, column 5 and the 4 at row 2, column 4.

Full solution (domino: first half → second half): 6–4: R2C5→R2C4; 2–2: R1C4→R1C5; 4–0: R3C1→R3C2; 2–4: R1C3→R2C3; 3–4: R2C1→R2C2.

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

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

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

Full solution (domino: first half → second half): 1–0: R1C1→R2C1; 4–2: R1C3→R1C2; 2–2: R1C4→R2C4; 5–1: R4C4→R3C4; 0–5: R4C2→R4C3; 3–4: R4C1→R3C1; 2–0: R2C2→R2C3.

Start with the cell that must equal 5 at row 1, column 1. It has the fewest ways to be satisfied.

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

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

Full solution (domino: first half → second half): 0–4: R2C7→R1C7; 6–1: R1C4→R2C4; 4–2: R4C6→R5C6; 6–3: R5C7→R4C7; 5–5: R2C2→R3C2; 4–4: R3C1→R4C1; 2–5: R2C1→R1C1; 0–6: R4C4→R4C3; 2–2: R6C6→R6C7; 0–0: R5C4→R6C4; 3–1: R1C6→R2C6; 3–2: R5C1→R6C1; 4–5: R3C3→R3C4; 2–6: R3C7→R3C6.