Pips answers for Saturday, December 13, 2025

Start with the 11 region at the bottom-left: the only way to make it with these dominoes is 5 + 6.

• the single-cell 0 region at the top: 0
• the 7 region at the top-right: 1, 6
• the 7 region at the right: 1, 2, 4
• the 11 region at the bottom-left: 5, 6

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

Start with the 2 region at the top: only 0 + 2, or 1 + 1 can make it with these dominoes.

• the 2 region at the top: 1, 1
• the 9 region at the top-right: 3, 6
• the 2 region at the right: 0, 2
• the single-cell 1 region at the left: 1
• the “=” region at the center: 3, 3
• the 3 region at the bottom-left: 0, 3
• the “=” region at the bottom-left: 4, 4
• the 8 region at the bottom-right: 3, 5
• the single-cell 1 region at the bottom-right: 1

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

Start with the “=” region at the bottom: all 4 cells must show the same number, and only 1 or 4 appear on enough dominoes to fill it.

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

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