Pips answers for Friday, November 28, 2025

Start with the “=” region at the top: all 3 cells must show the same number, and only 1 or 3 appear on enough dominoes to fill it.

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

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

Start with the 4 region at the right: the only way to make it with these dominoes is 1 + 1 + 2.

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

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

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

• the 10 region at the top-left: 4, 6
• the 4 region at the top: 1, 3
• the 12 region at the top-right: 2, 5, 5
• the single-cell 0 region at the left: 0
• the 11 region at the center: 5, 6
• the 12 region at the bottom-left: 0, 6, 6
• the 12 region at the bottom-right: 4, 4, 4

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