Pips answers for Sunday, January 4, 2026

Start with the “=” region at the top: all 2 cells must show the same number, and only 4 or 5 or 6 appear on enough dominoes to fill it.

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

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

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

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

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

Start with the 12 region at the top-right: the only way to make it with these dominoes is 6 + 6.

• the single-cell “>4” region at the top-left: 6
• the “=” region at the top-right: 3, 3, 3, 3
• the “=” region at the top-left: 5, 5, 5, 5, 5, 5
• the 12 region at the top-right: 6, 6
• the single-cell “<4” region at the left: 3
• the “=” region at the bottom-left: 1, 1, 1, 1, 1
• the 4 region at the right: 1, 3
• the single-cell 4 region at the right: 4
• the single-cell “<4” region at the bottom-left: 2
• the 4 region at the bottom: 2, 2
• the 12 region at the bottom-left: 6, 6
• the single-cell 4 region at the bottom: 4

1. The 4 region at the right needs a total of exactly 4 — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–4 domino horizontally on row 5, column 5 and row 5, column 6.
2. The 4 region at the bottom needs a total of exactly 4 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 2–4 domino vertically on row 7, column 4 and row 8, column 4.
3. Row 5, column 4 has only one open neighbor, so its domino must run to row 6, column 4; the 4 region at the right needs a total of exactly 4 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–3 domino vertically on row 5, column 4 and row 6, column 4.
4. The 12 region at the bottom-left needs a total of exactly 12 — 5 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–6 domino horizontally on row 8, column 1 and row 8, column 2.
5. The 12 region at the bottom-left needs a total of exactly 12 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–6 domino vertically on row 6, column 1 and row 7, column 1.
6. The “=” region at the bottom-left needs every pip to match (1 of its 5 cells is already filled) — nothing else fits. Place the 1–1 domino vertically on row 6, column 2 and row 7, column 2.
7. The “=” region at the bottom-left needs every pip to match (3 of its 5 cells are already filled) — nothing else fits. Place the 1–3 domino vertically on row 4, column 2 and row 5, column 2.
8. The “=” region at the top-left needs every pip to match — nothing else fits. Place the 1–5 domino vertically on row 4, column 1 and row 5, column 1.
9. The “=” region at the top-left needs every pip to match (1 of its 6 cells is already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 5–5 domino horizontally on row 3, column 1 and row 3, column 2.
10. The “=” region at the top-left needs every pip to match (3 of its 6 cells are already filled) — nothing else fits. Place the 5–6 domino vertically on row 1, column 2 and row 2, column 2.
11. Row 2, column 1 has only one open neighbor, so its domino must run to row 1, column 1; the “=” region at the top-left needs every pip to match (4 of its 6 cells are already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–5 domino vertically on row 1, column 1 and row 2, column 1.
12. The “=” region at the top-left needs every pip to match (5 of its 6 cells are already filled) — nothing else fits. Place the 3–5 domino vertically on row 1, column 3 and row 2, column 3.
13. The “=” region at the top-right 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 4 and row 1, column 5.
14. Row 1, column 6 has only one open neighbor, so its domino must run to row 2, column 6; the “=” region at the top-right needs every pip to match (3 of its 4 cells are already filled) — nothing else fits. Place the 3–4 domino vertically on row 1, column 6 and row 2, column 6.
15. Row 2, column 4 has only one open neighbor, so its domino must run to row 2, column 5; the 12 region at the top-right needs a total of exactly 12 — nothing else fits. Place the 6–6 domino horizontally on row 2, column 4 and row 2, column 5.

This board has more than one valid layout; the steps follow the official answer.