Pips answers for Friday, September 19, 2025

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

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

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

Start with the 3 region at the top-right: only 0 + 3, or 1 + 2 can make it with these dominoes.

• the single-cell “>3” region at the top-left: 4
• the 3 region at the top-right: 1, 2
• the 3 region at the right: 0, 3
• the single-cell “>3” region at the bottom-right: 4
• the 3 region at the bottom-right: 1, 2

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

Start with the 4 region at the center: the only way to make it with these dominoes is 1 + 1 + 1 + 1.

• the single-cell “<3” region at the top-left: 1
• the 17 region at the top: 4, 4, 4, 5
• the “=” region at the top-right: 3, 3, 3, 3, 3
• the 10 region at the left: 4, 6
• the single-cell “<3” region at the right: 2
• the 4 region at the center: 1, 1, 1, 1
• the 7 region at the bottom: 3, 4
• the single-cell 5 region at the bottom-right: 5

1. The 7 region at the bottom needs a total of exactly 7 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 4–5 domino horizontally on row 8, column 3 and row 8, column 4.
2. The 4 region at the center 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 1–1 domino vertically on row 5, column 4 and row 6, column 4.
3. The 4 region at the center needs a total of exactly 4 (2 of its 4 cells are 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 vertically on row 6, column 3 and row 7, column 3.
4. The 10 region at the left 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 1–6 domino horizontally on row 5, column 2 and row 5, column 3.
5. The 17 region at the top needs a total of exactly 17 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 1–4 domino vertically on row 1, column 1 and row 2, column 1.
6. The “=” region at the top-right 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 3–4 domino horizontally on row 4, column 2 and row 4, column 3.
7. The “=” region at the top-right needs every pip to match (1 of its 5 cells is already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–3 domino horizontally on row 4, column 4 and row 4, column 5.
8. The “=” region at the top-right needs every pip to match (2 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 3–3 domino horizontally on row 3, column 4 and row 3, column 5.
9. The 17 region at the top needs a total of exactly 17 (1 of its 4 cells is already filled) — nothing else fits. Place the 3–5 domino horizontally on row 2, column 4 and row 2, column 5.
10. Row 2, column 2 has only one open neighbor, so its domino must run to row 2, column 3; the 17 region at the top needs a total of exactly 17 (2 of its 4 cells are already filled) — nothing else fits. Place the 4–4 domino horizontally on row 2, column 2 and row 2, column 3.