Pips answers for Thursday, November 27, 2025

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

• the 5 region at the top: 1, 4
• the 1 region at the top: 0, 1
• the 3 region at the bottom-left: 0, 3
• the 2 region at the bottom-right: 0, 2

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

Start with the “=” region at the right: all 3 cells must show the same number, and only 3 or 4 or 5 appear on enough dominoes to fill it.

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

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

Start with the “=” region at the center: all 5 cells must show the same number, and only 0s appear on enough dominoes to fill it.

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

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