Pips answers for Saturday, November 1, 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 top-right: 6, 6
• the single-cell 4 region at the bottom: 4

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

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

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

1. The single-cell 0 region at the center needs a total of exactly 0 — nothing else fits. Place the 0–3 domino vertically on row 3, column 3 and row 4, column 3.
2. The single-cell 1 region at the right needs a total of exactly 1 — nothing else fits. Place the 1–2 domino horizontally on row 3, column 4 and row 3, column 5.
3. The single-cell 6 region at the right needs a total of exactly 6 — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 3–6 domino horizontally on row 2, column 5 and row 2, column 6.
4. The 12 region at the left needs a total of exactly 12 — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 6–6 domino vertically on row 2, column 1 and row 3, column 1.
5. 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 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 4–4 domino horizontally on row 1, column 1 and row 1, column 2.
6. 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; this one matches the official answer (this board has more than one valid layout). Place the 4–6 domino vertically on row 1, column 3 and row 2, column 3.
7. Row 2, column 2 has only one open neighbor, so its domino must run to row 3, column 2; the 12 region at the left needs a total of exactly 12 (1 of its 3 cells is already filled) — nothing else fits. Place the 3–3 domino vertically on row 2, column 2 and row 3, column 2.
8. Row 2, column 4 has only one open neighbor, so its domino must run to row 1, column 4; the 5 region at the center needs a total of exactly 5 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–3 domino vertically on row 1, column 4 and row 2, column 4.

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

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

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

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