Pips answers for Friday, September 26, 2025

Start with the “=” region at the top: all 5 cells must show the same number, and only 6s appear on enough dominoes to fill it.

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

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

• the single-cell 6 region at the top-left: 6
• the 15 region at the top-right: 5, 5, 5
• the 9 region at the center: 4, 5
• the 3 region at the right: 1, 2
• the 10 region at the center: 4, 6
• the 1 region at the bottom: 0, 1
• the single-cell 0 region at the bottom-right: 0

1. The single-cell 6 region at the top-left needs a total of exactly 6 — nothing else fits. Place the 5–6 domino horizontally on row 1, column 2 and row 1, column 3.
2. The 15 region at the top-right 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 1, column 4 and row 1, column 5.
3. The 10 region at the center needs a total of exactly 10 — 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 2 and row 4, column 2.
4. The 1 region at the bottom 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 4, column 3 and row 4, column 4.
5. The 3 region at the right needs a total of exactly 3 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–4 domino horizontally on row 3, column 3 and row 3, column 4.
6. The 9 region at the center needs a total of exactly 9 — nothing else fits. Place the 1–5 domino horizontally on row 2, column 3 and row 2, column 4.
7. Row 2, column 2 has only one open neighbor, so its domino must run to row 2, column 1; the 9 region at the center needs a total of exactly 9 (1 of its 2 cells is already filled) — nothing else fits. Place the 3–4 domino horizontally on row 2, column 1 and row 2, column 2.

Start with the 22 region at the bottom-right: the only way to make it with these dominoes is 5 + 5 + 6 + 6.

• the single-cell “<2” region at the top: 1
• the “=” region at the top-right: 0, 0, 0, 0
• the single-cell “<2” region at the top: 0
• the single-cell 2 region at the left: 2
• the “≠” region at the left: 3, 5
• the single-cell “<2” region at the center: 1
• the 22 region at the bottom-right: 5, 5, 6, 6
• the “=” region at the bottom-left: 4, 4, 4
• the single-cell 2 region at the bottom-left: 2

1. The 22 region at the bottom-right needs a total of exactly 22 — 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 5, column 4 and row 5, column 5.
2. 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 0–0 domino horizontally on row 2, column 4 and row 2, column 5.
3. The single-cell “<2” region at the top must total less than 2 — nothing else fits. Place the 0–1 domino horizontally on row 1, column 3 and row 1, column 4.
4. The 22 region at the bottom-right needs a total of exactly 22 (2 of its 4 cells are already filled) — nothing else fits. Place the 0–6 domino vertically on row 3, column 5 and row 4, column 5.
5. The 22 region at the bottom-right needs a total of exactly 22 (3 of its 4 cells are already filled) — nothing else fits. Place the 2–6 domino vertically on row 3, column 4 and row 4, column 4.
6. 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 1–4 domino vertically on row 4, column 3 and row 5, column 3.
7. Row 2, column 3 has only one open neighbor, so its domino must run to row 3, column 3; the single-cell “<2” region at the top must total less than 2 — nothing else fits. Place the 0–2 domino vertically on row 2, column 3 and row 3, column 3.
8. The “=” region at the bottom-left needs every pip to match (1 of its 3 cells is already filled) — nothing else fits. Place the 2–4 domino vertically on row 5, column 2 and row 6, column 2.
9. The single-cell 2 region at the left needs a total of exactly 2 — nothing else fits. Place the 2–5 domino vertically on row 2, column 1 and row 3, column 1.
10. The “≠” region at the left needs every pip to be different — nothing else fits. Place the 3–4 domino vertically on row 4, column 1 and row 5, column 1.
11. Row 4, 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 be different (1 of its 2 cells is already filled) — nothing else fits. Place the 3–5 domino vertically on row 3, column 2 and row 4, column 2.