Pips answers for Wednesday, July 8, 2026

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

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

1. The 11 region at the top needs a total of exactly 11 — nothing else fits. Place the 3–6 domino horizontally on row 1, column 2 and row 1, column 3.
2. The 11 region at the top needs a total of exactly 11 (1 of its 2 cells is already filled) — nothing else fits. Place the 4–5 domino horizontally on row 2, column 1 and row 2, column 2.
3. Row 3, column 2 has only one open neighbor, so its domino must run to row 4, column 2; the “=” region at the bottom needs every pip to match — nothing else fits. Place the 2–2 domino vertically on row 3, column 2 and row 4, column 2.
4. Row 4, column 3 has only one open neighbor, so its domino must run to row 4, column 4; the “=” region at the bottom needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 0–2 domino horizontally on row 4, column 3 and row 4, column 4.
5. Row 2, column 4 has only one open neighbor, so its domino must run to row 3, column 4; the single-cell “<6” region at the right must total less than 6 — nothing else fits. Place the 0–1 domino vertically on row 2, column 4 and row 3, column 4.

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

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

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

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

Start with the 1 region at the left: the only way to make it with these dominoes is 0 + 1.

• the 4 region at the top-left: 2, 2
• the “=” region at the top-left: 0, 0, 0
• the 1 region at the left: 0, 1
• the 3 region at the center: 1, 2
• the “=” region at the right: 6, 6
• the 10 region at the bottom-right: 5, 5
• the single-cell 1 region at the right: 1
• the 4 region at the bottom-left: 1, 3
• the single-cell “>0” region at the bottom: 4
• the 10 region at the bottom-right: 4, 6
• the 5 region at the bottom-left: 2, 3
• the single-cell “>4” region at the bottom: 6
• the “=” region at the bottom-right: 0, 0

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