Pips answers for Friday, January 30, 2026

Start with the “=” region at the top-left: all 3 cells must show the same number, and only 1 or 2 appear on enough dominoes to fill it.

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

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

Start with the “≠” region at the center: the only way to make it with these dominoes is 0 + 1 + 2 + 3 + 4 + 5.

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

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

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

Start with the “=” region at the top: all 5 cells must show the same number, and only 3 or 5 appear on enough dominoes to fill it.

• the single-cell 5 region at the top: 5
• the single-cell 4 region at the top: 4
• the “=” region at the top: 6, 6
• the “=” region at the top: 3, 3, 3, 3, 3
• the single-cell “>4” region at the top-right: 5
• the 12 region at the right: 4, 4, 4
• the “=” region at the left: 2, 2, 2
• the single-cell 5 region at the center: 5
• the “=” region at the right: 0, 0
• the single-cell 2 region at the center: 2
• the “=” region at the bottom: 6, 6
• the single-cell “<2” region at the bottom-left: 1
• the “=” region at the bottom: 0, 0
• the “=” region at the bottom-left: 5, 5

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