Pips answers for Wednesday, December 24, 2025

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

• the “=” region at the left: 2, 2
• the single-cell “>3” region at the center: 4
• the 12 region at the right: 6, 6
• the single-cell 6 region at the bottom-left: 6
• the single-cell 1 region at the bottom-left: 1
• the “=” region at the bottom-right: 3, 3

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

Start with the 3 region at the top: the only way to make it with these dominoes is 0 + 3.

• the “=” region at the top-left: 1, 1
• the 3 region at the top: 0, 3
• the single-cell 4 region at the top-right: 4
• the “=” region at the left: 0, 0
• the single-cell 4 region at the left: 4
• the 4 region at the bottom-right: 1, 3
• the “>7” region at the bottom: 4, 5

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

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

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

• the 12 region at the top-left: 3, 3, 6
• the single-cell 6 region at the top-right: 6
• the 12 region at the top-right: 4, 4, 4
• the single-cell 4 region at the top-left: 4
• the single-cell 5 region at the top-left: 5
• the “=” region at the left: 2, 2
• the 0 region at the right: 0, 0
• the 11 region at the center: 3, 3, 5
• the 2 region at the center: 1, 1
• the single-cell 0 region at the bottom: 0
• the 11 region at the bottom: 5, 6

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