Pips answers for Friday, March 13, 2026

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

• the 11 region at the top-left: 5, 6
• the “=” region at the top-right: 3, 3
• the 8 region at the bottom-left: 2, 6
• the 6 region at the bottom: 2, 4

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

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

• the 1 region at the top-left: 0, 1
• the 9 region at the top: 4, 5
• the 6 region at the right: 3, 3
• the 5 region at the bottom-left: 0, 2, 3
• the 9 region at the bottom-right: 4, 5
• the 5 region at the bottom: 1, 4

1. The 1 region at the top-left needs a total of exactly 1 — 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.
2. Row 2, column 1 has only one open neighbor, so its domino must run to row 3, column 1; the 1 region at the top-left needs a total of exactly 1 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–0 domino vertically on row 2, column 1 and row 3, column 1.
3. Row 1, column 3 has only one open neighbor, so its domino must run to row 2, column 3; the 9 region at the top needs a total of exactly 9 (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 3–5 domino vertically on row 1, column 3 and row 2, column 3.
4. Row 2, column 4 has only one open neighbor, so its domino must run to row 3, column 4; the 6 region at the right needs a total of exactly 6 (1 of its 2 cells is already filled) — nothing else fits. Place the 3–4 domino vertically on row 2, column 4 and row 3, column 4.
5. The 5 region at the bottom needs a total of exactly 5 — nothing else fits. Place the 4–5 domino horizontally on row 4, column 3 and row 4, column 4.
6. Row 4, column 1 has only one open neighbor, so its domino must run to row 4, column 2; the 5 region at the bottom-left needs a total of exactly 5 (1 of its 3 cells is already filled) — nothing else fits. Place the 1–3 domino horizontally on row 4, column 1 and row 4, column 2.
7. Row 3, column 2 has only one open neighbor, so its domino must run to row 3, column 3; the 5 region at the bottom-left needs a total of exactly 5 (2 of its 3 cells are already filled) — nothing else fits. Place the 2–6 domino horizontally on row 3, column 2 and row 3, column 3.

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

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

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