Pips answers for Sunday, January 11, 2026

Start with the 4 region at the center: the only way to make it with these dominoes is 1 + 3.

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

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

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

• the “=” region at the top: 3, 3, 3, 3
• the single-cell “<1” region at the right: 0
• the “<2” region at the left: 0, 0
• the 8 region at the right: 3, 5
• the single-cell 2 region at the right: 2
• the “=” region at the bottom-left: 4, 4
• the single-cell “>5” region at the bottom: 6
• the single-cell 2 region at the bottom-right: 2

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

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 0–2 domino vertically on row 1, column 1 and row 2, column 1.
2. The single-cell 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 3–4 domino horizontally on row 1, column 2 and row 1, column 3.
3. The single-cell 3 region at the top-right needs a total of exactly 3 — nothing else fits. Place the 1–3 domino horizontally on row 1, column 5 and row 1, column 6.
4. Row 1, column 4 has only one open neighbor, so its domino must run to row 2, column 4; the single-cell 2 region at the top needs a total of exactly 2 — nothing else fits. Place the 2–3 domino vertically on row 1, column 4 and row 2, column 4.
5. The single-cell 2 region at the top-left needs a total of exactly 2 — nothing else fits. Place the 1–2 domino horizontally on row 2, column 2 and row 2, column 3.
6. The single-cell 3 region at the bottom-left needs a total of exactly 3 — nothing else fits. Place the 3–3 domino vertically on row 3, column 2 and row 4, column 2.
7. Row 3, column 1 has only one open neighbor, so its domino must run to row 4, column 1; the single-cell 0 region at the left needs a total of exactly 0 — nothing else fits. Place the 0–4 domino vertically on row 3, column 1 and row 4, column 1.
8. The single-cell 4 region at the top-right needs a total of exactly 4 — nothing else fits. Place the 4–4 domino horizontally on row 2, column 5 and row 2, column 6.
9. Row 5, column 1 has only one open neighbor, so its domino must run to row 5, column 2; the single-cell 0 region at the bottom-left needs a total of exactly 0 — nothing else fits. Place the 0–1 domino horizontally on row 5, column 1 and row 5, column 2.
10. The single-cell 0 region at the right needs a total of exactly 0 — nothing else fits. Place the 0–0 domino horizontally on row 3, column 5 and row 3, column 6.
11. The single-cell 1 region at the bottom needs a total of exactly 1 — nothing else fits. Place the 1–1 domino vertically on row 3, column 4 and row 4, column 4.
12. Row 3, column 3 has only one open neighbor, so its domino must run to row 4, column 3; the single-cell 4 region at the center needs a total of exactly 4 — nothing else fits. Place the 2–4 domino vertically on row 3, column 3 and row 4, column 3.
13. The single-cell 0 region at the bottom-right needs a total of exactly 0 — nothing else fits. Place the 0–3 domino vertically on row 4, column 5 and row 5, column 5.
14. Row 4, column 6 has only one open neighbor, so its domino must run to row 5, column 6; the single-cell 1 region at the bottom-right needs a total of exactly 1 — nothing else fits. Place the 1–4 domino vertically on row 4, column 6 and row 5, column 6.
15. Row 5, column 3 has only one open neighbor, so its domino must run to row 5, column 4; the single-cell 2 region at the bottom needs a total of exactly 2 — nothing else fits. Place the 2–2 domino horizontally on row 5, column 3 and row 5, column 4.