Pips answers for Tuesday, January 13, 2026

Start with the “=” region at the center: all 2 cells must show the same number, and only 1 or 6 appear on enough dominoes to fill it.

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

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

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

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

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

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

• the single-cell 0 region at the top-left: 0
• the “=” region at the top: 1, 1, 1, 1, 1
• the “=” region at the top-right: 3, 3
• the 4 region at the top-left: 2, 2
• the single-cell 6 region at the top-left: 6
• the single-cell “<2” region at the top-left: 0
• the single-cell “>4” region at the top: 6
• the 8 region at the top: 2, 6
• the 0 region at the top-right: 0, 0, 0
• the 11 region at the left: 5, 6
• the single-cell 4 region at the left: 4
• the 7 region at the bottom: 3, 4
• the “=” region at the bottom-right: 3, 3, 3
• the single-cell “>4” region at the bottom-right: 5

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

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