Pips answers for Wednesday, September 2, 2026

Start with the 4 region at the top-left: only 0 + 4, or 1 + 3 can make it with these dominoes.

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

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

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

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

• the 3 region at the top: 0, 3
• the “=” region at the center: 5, 5, 5
• the “=” region at the center: 4, 4
• the single-cell “>4” region at the right: 5
• the single-cell “>4” region at the left: 6
• the single-cell “<2” region at the right: 1
• the 9 region at the bottom: 3, 3, 3

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

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

• the “=” region at the top-left: 2, 2, 2
• the 12 region at the top: 6, 6
• the single-cell “<3” region at the top: 2
• the “=” region at the top-right: 4, 4, 4, 4, 4, 4
• the “=” region at the left: 5, 5, 5, 5
• the single-cell “>3” region at the top: 4
• the single-cell “>3” region at the top: 5
• the single-cell “<3” region at the center: 0
• the single-cell 1 region at the right: 1
• the 12 region at the bottom-left: 6, 6
• the 8 region at the bottom: 3, 5
• the “=” region at the bottom-right: 3, 3

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