Pips answers for Wednesday, December 3, 2025

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

• the 0 region at the top-left: 0, 0
• the single-cell 0 region at the top-left: 0
• the 7 region at the top: 2, 5
• the “=” region at the center: 1, 1, 1, 1

1. The 0 region at the top-left 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 0–0 domino vertically on row 1, column 1 and row 2, column 1.
2. The “=” region at the center 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 1–1 domino horizontally on row 2, column 2 and row 2, column 3.
3. The 7 region at the top needs a total of exactly 7 — nothing else fits. Place the 1–2 domino vertically on row 1, column 4 and row 2, column 4.
4. Row 1, column 2 has only one open neighbor, so its domino must run to row 1, column 3; the single-cell 0 region at the top-left needs a total of exactly 0 — nothing else fits. Place the 0–5 domino horizontally on row 1, column 2 and row 1, column 3.
5. 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 (3 of its 4 cells are already filled) — nothing else fits. Place the 0–1 domino horizontally on row 3, column 4 and row 3, column 5.

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

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

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

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

• the 4 region at the top-left: 0, 2, 2
• the “≠” region at the top-right: 2, 4
• the single-cell “>4” region at the top-left: 6
• the 3 region at the center: 1, 1, 1
• the “>5” region at the center: 3, 3
• the “=” region at the center: 4, 4, 4, 4
• the 15 region at the bottom-left: 5, 5, 5
• the single-cell “>4” region at the bottom-left: 6
• the 17 region at the bottom-right: 5, 6, 6

1. The 4 region at the top-left needs a total of exactly 4 — 2 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.
2. Row 1, column 1 has only one open neighbor, so its domino must run to row 1, column 2; the 4 region at the top-left needs a total of exactly 4 (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 1, column 1 and row 1, column 2.
3. The 17 region at the bottom-right needs a total of exactly 17 — 3 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 6, column 7 and row 6, column 8.
4. Row 5, column 7 has only one open neighbor, so its domino must run to row 5, column 6; the 17 region at the bottom-right needs a total of exactly 17 (2 of its 3 cells are already filled) — 3 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 6 and row 5, column 7.
5. The 15 region at the bottom-left needs a total of exactly 15 — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 5–6 domino horizontally on row 5, column 2 and row 5, column 3.
6. Row 6, column 1 has only one open neighbor, so its domino must run to row 6, column 2; the 15 region at the bottom-left needs a total of exactly 15 (1 of its 3 cells is already filled) — nothing else fits. Place the 5–5 domino horizontally on row 6, column 1 and row 6, column 2.
7. The “≠” region at the top-right needs every pip to be different — 7 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 1–2 domino horizontally on row 2, column 6 and row 2, column 7.
8. The 3 region at the center needs a total of exactly 3 (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 1–4 domino vertically on row 3, column 6 and row 4, column 6.
9. The 3 region at the center needs a total of exactly 3 (2 of its 3 cells are already filled) — nothing else fits. Place the 1–3 domino horizontally on row 3, column 4 and row 3, column 5.
10. The “=” region at the center needs every pip to match (1 of its 4 cells is already filled) — nothing else fits. Place the 4–4 domino horizontally on row 4, column 4 and row 4, column 5.
11. Row 3, column 3 has only one open neighbor, so its domino must run to row 4, column 3; the “>5” region at the center must total more than 5 (1 of its 2 cells is already filled) — nothing else fits. Place the 3–4 domino vertically on row 3, column 3 and row 4, column 3.
12. Row 1, column 7 has only one open neighbor, so its domino must run to row 1, column 8; the “≠” region at the top-right needs every pip to be different (1 of its 2 cells is already filled) — nothing else fits. Place the 2–4 domino horizontally on row 1, column 7 and row 1, column 8.