Pips answers for Friday, November 14, 2025

Start with the “=” region at the center: all 4 cells must show the same number, and only 6s appear on enough dominoes to fill it.

• the single-cell 5 region at the top-right: 5
• the “=” region at the center: 6, 6, 6, 6
• the “=” region at the bottom-left: 3, 3
• the single-cell 1 region at the bottom: 1

1. Row 1, column 4 has only one open neighbor, so its domino must run to row 2, column 4; the single-cell 5 region at the top-right needs a total of exactly 5 — nothing else fits. Place the 5–6 domino vertically on row 1, column 4 and row 2, column 4.
2. The “=” region at the center needs every pip to match (1 of its 4 cells is already filled) — nothing else fits. Place the 6–6 domino horizontally on row 2, column 2 and row 2, column 3.
3. Row 2, column 1 has only one open neighbor, so its domino must run to row 3, column 1; the “=” region at the center needs every pip to match (3 of its 4 cells are already filled) — nothing else fits. Place the 3–6 domino vertically on row 2, column 1 and row 3, column 1.
4. Row 3, column 2 has only one open neighbor, so its domino must run to row 3, column 3; the “=” region at the bottom-left needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 1–3 domino horizontally on row 3, column 2 and row 3, column 3.

Start with the 9 region at the bottom: the only way to make it with these dominoes is 4 + 5.

• the “=” region at the top: 0, 0
• the “=” region at the right: 1, 1, 1
• the single-cell “>2” region at the left: 6
• the “=” region at the left: 2, 2, 2
• the 9 region at the bottom: 4, 5
• the 9 region at the bottom-right: 4, 5

1. The “=” region at the left needs every pip to match — 5 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–6 domino horizontally on row 2, column 1 and row 2, column 2.
2. Row 3, column 2 has only one open neighbor, so its domino must run to row 3, column 3; the “=” region at the left needs every pip to match (1 of its 3 cells is already filled) — nothing else fits. Place the 2–4 domino horizontally on row 3, column 2 and row 3, column 3.
3. The “=” region at the left needs every pip to match (2 of its 3 cells are already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–2 domino horizontally on row 2, column 3 and row 2, column 4.
4. The “=” region at the top needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 0–1 domino horizontally on row 1, column 3 and row 1, column 4.
5. Row 1, column 5 has only one open neighbor, so its domino must run to row 2, column 5; the “=” region at the right needs every pip to match — nothing else fits. Place the 1–1 domino vertically on row 1, column 5 and row 2, column 5.
6. Row 2, column 6 has only one open neighbor, so its domino must run to row 3, column 6; the “=” region at the right needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 1–5 domino vertically on row 2, column 6 and row 3, column 6.
7. Row 3, column 4 has only one open neighbor, so its domino must run to row 3, column 5; the 9 region at the bottom needs a total of exactly 9 (1 of its 2 cells is already filled) — nothing else fits. Place the 4–5 domino horizontally on row 3, column 4 and row 3, column 5.

Start with the “=” region at the top-left: all 5 cells must show the same number, and only 0s appear on enough dominoes to fill it.

• the “=” region at the top-left: 0, 0, 0, 0, 0
• the single-cell 5 region at the top: 5
• the single-cell 5 region at the top-right: 5
• the 5 region at the top-right: 0, 1, 4
• the 5 region at the left: 2, 3
• the 5 region at the center: 1, 4
• the single-cell 5 region at the right: 5
• the 5 region at the bottom-left: 2, 3
• the 5 region at the bottom-left: 2, 3
• the 5 region at the bottom-right: 2, 3

1. The single-cell 5 region at the top needs a total of exactly 5 — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–5 domino horizontally on row 1, column 3 and row 1, column 4.
2. The “=” region at the top-left needs every pip to match (1 of its 5 cells is already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–0 domino horizontally on row 1, column 1 and row 1, column 2.
3. The 5 region at the top-right needs a total of exactly 5 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 4–5 domino horizontally on row 1, column 5 and row 1, column 6.
4. The 5 region at the center needs a total of exactly 5 — nothing else fits. Place the 1–5 domino horizontally on row 3, column 4 and row 3, column 5.
5. Row 4, column 4 has only one open neighbor, so its domino must run to row 5, column 4; the 5 region at the center needs a total of exactly 5 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–4 domino vertically on row 4, column 4 and row 5, column 4.
6. Row 2, column 6 has only one open neighbor, so its domino must run to row 3, column 6; the 5 region at the top-right needs a total of exactly 5 (1 of its 3 cells is already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–1 domino vertically on row 2, column 6 and row 3, column 6.
7. The “=” region at the top-left needs every pip to match (3 of its 5 cells are already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–3 domino vertically on row 2, column 1 and row 3, column 1.
8. Row 2, column 2 has only one open neighbor, so its domino must run to row 3, column 2; the “=” region at the top-left needs every pip to match (4 of its 5 cells are already filled) — nothing else fits. Place the 0–2 domino vertically on row 2, column 2 and row 3, column 2.
9. Row 5, column 5 has only one open neighbor, so its domino must run to row 5, column 6; the 5 region at the bottom-right needs a total of exactly 5 (1 of its 2 cells is already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–3 domino horizontally on row 5, column 5 and row 5, column 6.
10. The 5 region at the bottom-left needs a total of exactly 5 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 2–2 domino horizontally on row 4, column 1 and row 4, column 2.
11. Row 5, column 1 has only one open neighbor, so its domino must run to row 5, column 2; the 5 region at the bottom-left needs a total of exactly 5 (1 of its 2 cells is already filled) — nothing else fits. Place the 3–3 domino horizontally on row 5, column 1 and row 5, column 2.

This board has more than one valid layout; the steps follow the official answer.