Pips answers for Thursday, October 30, 2025

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

• the “=” region at the top: 4, 4, 4
• the “=” region at the center: 3, 3
• the 0 region at the bottom: 0, 0

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

Start with the “=” region at the center: all 5 cells must show the same number, and only 1s appear on enough dominoes to fill it.

• the 6 region at the top-left: 2, 2, 2
• the “=” region at the center: 1, 1, 1, 1, 1
• the 6 region at the right: 2, 4
• the 11 region at the bottom: 5, 6
• the 8 region at the bottom-right: 3, 5

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

Start with the 17 region at the left: the only way to make it with these dominoes is 5 + 6 + 6.

• the single-cell “>1” region at the top-left: 3
• the single-cell “<5” region at the top: 1
• the “=” region at the top-right: 5, 5, 5
• the 17 region at the left: 5, 6, 6
• the 25 region at the center: 2, 2, 4, 5, 6, 6
• the single-cell 0 region at the right: 0
• the single-cell 2 region at the bottom: 2
• the 1 region at the bottom-right: 0, 1

1. The 25 region at the center needs a total of exactly 25 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–2 domino horizontally on row 3, column 4 and row 3, column 5.
2. The single-cell 2 region at the bottom needs a total of exactly 2 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 2–4 domino vertically on row 4, column 3 and row 5, column 3.
3. Row 4, column 4 has only one open neighbor, so its domino must run to row 5, column 4; the 25 region at the center needs a total of exactly 25 (2 of its 6 cells are already filled) — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–5 domino vertically on row 4, column 4 and row 5, column 4.
4. Row 5, column 5 has only one open neighbor, so its domino must run to row 5, column 6; the 1 region at the bottom-right needs a total of exactly 1 (1 of its 2 cells is already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–1 domino horizontally on row 5, column 5 and row 5, column 6.
5. The “=” region at the top-right 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 5–5 domino horizontally on row 1, column 5 and row 1, column 6.
6. The “=” region at the top-right needs every pip to match (2 of its 3 cells are already filled) — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 5–6 domino vertically on row 1, column 4 and row 2, column 4.
7. The 25 region at the center needs a total of exactly 25 (4 of its 6 cells are already filled) — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–6 domino vertically on row 1, column 3 and row 2, column 3.
8. The 17 region at the left needs a total of exactly 17 — nothing else fits. Place the 2–6 domino horizontally on row 3, column 2 and row 3, column 3.
9. The 17 region at the left needs a total of exactly 17 (1 of its 3 cells is already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 3–5 domino vertically on row 1, column 1 and row 2, column 1.
10. Row 3, column 1 has only one open neighbor, so its domino must run to row 4, column 1; the 17 region at the left needs a total of exactly 17 (2 of its 3 cells are already filled) — nothing else fits. Place the 6–6 domino vertically on row 3, column 1 and row 4, column 1.

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