Pips answers for Saturday, February 7, 2026

Start with the 30 region at the center: the only way to make it with these dominoes is 6 + 6 + 6 + 6 + 6.

• the 30 region at the center: 6, 6, 6, 6, 6
• the single-cell “>3” region at the top-right: 4
• the single-cell “<1” region at the left: 0
• the single-cell “<3” region at the right: 2

1. Row 2, column 1 has only one open neighbor, so its domino must run to row 2, column 2; the single-cell “<1” region at the left must total less than 1 — nothing else fits. Place the 0–6 domino horizontally on row 2, column 1 and row 2, column 2.
2. Row 3, column 2 has only one open neighbor, so its domino must run to row 3, column 3; the 30 region at the center needs a total of exactly 30 (1 of its 5 cells is already filled) — nothing else fits. Place the 6–6 domino horizontally on row 3, column 2 and row 3, column 3.
3. The 30 region at the center needs a total of exactly 30 (3 of its 5 cells are already filled) — nothing else fits. Place the 4–6 domino horizontally on row 1, column 3 and row 1, column 4.
4. Row 2, column 3 has only one open neighbor, so its domino must run to row 2, column 4; the 30 region at the center needs a total of exactly 30 (4 of its 5 cells are already filled) — nothing else fits. Place the 2–6 domino horizontally on row 2, column 3 and row 2, column 4.

Start with the “<2” region at the right: the only way to make it with these dominoes is 0 + 1.

• the single-cell 6 region at the top-left: 6
• the “=” region at the top: 1, 1
• the single-cell 3 region at the top-right: 3
• the 7 region at the bottom-left: 1, 6
• the 7 region at the bottom-left: 3, 4
• the single-cell 2 region at the center: 2
• the “=” region at the bottom-right: 4, 4, 4, 4
• the “<2” region at the right: 0, 1

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

Start with the 3 region at the top: the only way to make it with these dominoes is 1 + 2.

• the 5 region at the top: 1, 4
• the 3 region at the top: 1, 2
• the 8 region at the top-left: 4, 4
• the 9 region at the top-right: 4, 5
• the 7 region at the left: 3, 4
• the 2 region at the right: 1, 1
• the 4 region at the bottom-left: 1, 3
• the 11 region at the bottom-right: 5, 6
• the 6 region at the bottom: 2, 4
• the 10 region at the bottom: 5, 5

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