Pips answers for Wednesday, December 31, 2025

Start with the 18 region at the bottom: the only way to make it with these dominoes is 6 + 6 + 6.

• the 7 region at the left: 2, 5
• the 8 region at the center: 4, 4
• the 18 region at the bottom: 6, 6, 6
• the single-cell 3 region at the right: 3
• the “=” region at the bottom-left: 3, 3

1. The 18 region at the bottom needs a total of exactly 18 — nothing else fits. Place the 3–6 domino horizontally on row 3, column 4 and row 3, column 5.
2. The 7 region at the left needs a total of exactly 7 — nothing else fits. Place the 2–6 domino horizontally on row 3, column 2 and row 3, column 3.
3. The 7 region at the left needs a total of exactly 7 (1 of its 2 cells is already filled) — nothing else fits. Place the 1–5 domino vertically on row 1, column 2 and row 2, column 2.
4. Row 2, column 3 has only one open neighbor, so its domino must run to row 2, column 4; the 8 region at the center needs a total of exactly 8 — nothing else fits. Place the 4–4 domino horizontally on row 2, column 3 and row 2, column 4.
5. Row 4, column 1 has only one open neighbor, so its domino must run to row 4, column 2; the “=” region at the bottom-left needs every pip to match — nothing else fits. Place the 3–3 domino horizontally on row 4, column 1 and row 4, column 2.
6. Row 4, column 3 has only one open neighbor, so its domino must run to row 4, column 4; the 18 region at the bottom needs a total of exactly 18 (2 of its 3 cells are already filled) — nothing else fits. Place the 0–6 domino horizontally on row 4, column 3 and row 4, column 4.

Start with the “=” region at the top: all 3 cells must show the same number, and only 4 or 6 appear on enough dominoes to fill it.

• the “=” region at the top: 4, 4, 4
• the “=” region at the right: 6, 6, 6
• the 11 region at the center: 3, 3, 5
• the 7 region at the right: 1, 6

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

Start with the “=” region at the center: all 5 cells must show the same number, and only 3s appear on enough dominoes to fill it.

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

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