Pips answers for Friday, December 19, 2025

Start with the 12 region at the bottom-right: the only way to make it with these dominoes is 6 + 6.

• the single-cell 5 region at the top: 5
• the single-cell 0 region at the left: 0
• the “=” region at the center: 3, 3
• the single-cell “<5” region at the right: 4
• the 12 region at the bottom-right: 6, 6

1. Row 1, column 2 has only one open neighbor, so its domino must run to row 2, column 2; the single-cell 5 region at the top needs a total of exactly 5 — nothing else fits. Place the 3–5 domino vertically on row 1, column 2 and row 2, column 2.
2. The “=” region at the center needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 0–3 domino horizontally on row 3, column 2 and row 3, column 3.
3. Row 2, column 1 has only one open neighbor, so its domino must run to row 3, column 1; the single-cell 0 region at the left needs a total of exactly 0 — nothing else fits. Place the 0–6 domino vertically on row 2, column 1 and row 3, column 1.
4. The single-cell “<5” region at the right must total less than 5 — nothing else fits. Place the 4–6 domino horizontally on row 2, column 3 and row 2, column 4.
5. Row 3, column 4 has only one open neighbor, so its domino must run to row 4, column 4; the 12 region at the bottom-right needs a total of exactly 12 — nothing else fits. Place the 6–6 domino vertically on row 3, column 4 and row 4, column 4.

Start with the “=” region at the center: all 4 cells must show the same number, and only 0s appear on enough dominoes to fill it.

• the “=” region at the top-left: 1, 1
• the single-cell 5 region at the top: 5
• the “=” region at the center: 0, 0, 0, 0
• the “=” region at the right: 4, 4
• the single-cell 3 region at the center: 3
• the 9 region at the bottom: 4, 5

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

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

Start with the 1 region at the bottom-right: the only way to make it with these dominoes is 0 + 1.

• the 3 region at the top-left: 1, 2
• the single-cell 4 region at the top: 4
• the single-cell 6 region at the top-right: 6
• the 9 region at the left: 3, 3, 3
• the 6 region at the center: 2, 2, 2
• the 7 region at the right: 3, 4
• the 10 region at the bottom: 5, 5
• the 9 region at the bottom-right: 4, 5
• the 2 region at the bottom: 1, 1
• the 1 region at the bottom-right: 0, 1

1. The 1 region at the bottom-right needs a total of exactly 1 — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–4 domino vertically on row 5, column 6 and row 6, column 6.
2. The 1 region at the bottom-right needs a total of exactly 1 (1 of its 2 cells is already filled) — nothing else fits. Place the 1–1 domino horizontally on row 6, column 4 and row 6, column 5.
3. The 2 region at the bottom needs a total of exactly 2 (1 of its 2 cells is already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 1–3 domino horizontally on row 6, column 2 and row 6, column 3.
4. The single-cell 6 region at the top-right needs a total of exactly 6 — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 5–6 domino horizontally on row 2, column 4 and row 2, column 5.
5. The single-cell 4 region at the top needs a total of exactly 4 — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 1–4 domino horizontally on row 2, column 2 and row 2, column 3.
6. Row 2, column 1 has only one open neighbor, so its domino must run to row 1, column 1; the 3 region at the top-left needs a total of exactly 3 (1 of its 2 cells is already filled) — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–6 domino vertically on row 1, column 1 and row 2, column 1.
7. The 9 region at the bottom-right needs a total of exactly 9 (1 of its 2 cells is already filled) — 4 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 4–5 domino vertically on row 4, column 5 and row 5, column 5.
8. The 7 region at the right needs a total of exactly 7 (1 of its 2 cells is already filled) — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 3–6 domino horizontally on row 3, column 5 and row 3, column 6.
9. The 10 region at the bottom needs a total of exactly 10 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 5–5 domino vertically on row 4, column 3 and row 5, column 3.
10. The 9 region at the left needs a total of exactly 9 — 4 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 3–3 domino vertically on row 3, column 2 and row 4, column 2.
11. Row 3, column 3 has only one open neighbor, so its domino must run to row 3, column 4; the 9 region at the left needs a total of exactly 9 (2 of its 3 cells are already filled) — nothing else fits. Place the 2–3 domino horizontally on row 3, column 3 and row 3, column 4.
12. Row 4, column 4 has only one open neighbor, so its domino must run to row 5, column 4; the 6 region at the center needs a total of exactly 6 (1 of its 3 cells is already filled) — nothing else fits. Place the 2–2 domino vertically on row 4, column 4 and row 5, column 4.