Pips answers for Friday, August 29, 2025

Start with the “=” region at the bottom: all 4 cells must show the same number, and only 5s appear on enough dominoes to fill it.

• the single-cell 3 region at the right: 3
• the “=” region at the bottom: 5, 5, 5, 5
• the single-cell 4 region at the bottom-right: 4

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

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

• the 12 region at the top: 6, 6
• the 4 region at the right: 0, 2, 2
• the 8 region at the bottom-left: 4, 4
• the 4 region at the center: 2, 2
• the 0 region at the bottom: 0, 0

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

Start with the 2 region at the center: the only way to make it with these dominoes is 0 + 0 + 1 + 1.

• the 9 region at the top-left: 4, 5
• the “=” region at the top: 1, 1
• the “=” region at the left: 2, 2
• the 12 region at the right: 6, 6
• the 2 region at the center: 0, 0, 1, 1
• the 10 region at the bottom: 5, 5
• the single-cell 3 region at the bottom-left: 3
• the 18 region at the bottom-right: 6, 6, 6

1. The 18 region at the bottom-right needs a total of exactly 18 — nothing else fits. Place the 6–6 domino horizontally on row 10, column 1 and row 10, column 2.
2. The 18 region at the bottom-right needs a total of exactly 18 (2 of its 3 cells are already filled) — nothing else fits. Place the 5–6 domino vertically on row 8, column 2 and row 9, column 2.
3. The 10 region at the bottom needs a total of exactly 10 (1 of its 2 cells is already filled) — nothing else fits. Place the 3–5 domino vertically on row 8, column 1 and row 9, column 1.
4. The 12 region at the right needs a total of exactly 12 — 4 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 4, column 1 and row 4, column 2.
5. 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 0–6 domino vertically on row 5, column 2 and row 6, column 2.
6. The 2 region at the center needs a total of exactly 2 (1 of its 4 cells is already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–2 domino vertically on row 5, column 1 and row 6, column 1.
7. Row 7, column 1 has only one open neighbor, so its domino must run to row 7, column 2; the 2 region at the center needs a total of exactly 2 (2 of its 4 cells are already filled) — nothing else fits. Place the 1–1 domino horizontally on row 7, column 1 and row 7, column 2.
8. The 9 region at the top-left 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 3–4 domino horizontally on row 1, column 1 and row 1, column 2.
9. The “=” region at the top needs every pip to match — nothing else fits. Place the 1–5 domino vertically on row 2, column 1 and row 3, column 1.
10. Row 3, 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 (1 of its 2 cells is already filled) — nothing else fits. Place the 1–2 domino vertically on row 2, column 2 and row 3, column 2.