Pips answers for Tuesday, August 11, 2026

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

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

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

Start with the “<3” region at the top-left: only 0 + 0 + 0, or 0 + 0 + 1, or 0 + 0 + 2 can make it with these dominoes.

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

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

Start with the 17 region at the right: the only way to make it with these dominoes is 5 + 6 + 6.

• the 5 region at the top-left: 1, 4
• the single-cell 2 region at the top-left: 2
• the single-cell 3 region at the top: 3
• the 0 region at the top-right: 0, 0
• the 16 region at the top: 5, 5, 6
• the 17 region at the right: 5, 6, 6
• the “=” region at the left: 0, 0, 0, 0, 0
• the single-cell 2 region at the left: 2
• the single-cell 3 region at the center: 3
• the “=” region at the right: 3, 3
• the single-cell 5 region at the right: 5
• the single-cell 0 region at the right: 0
• the 10 region at the bottom-left: 4, 6
• the 9 region at the bottom-right: 3, 6
• the single-cell 1 region at the bottom-left: 1
• the single-cell 2 region at the bottom: 2
• the single-cell 3 region at the bottom-right: 3

1. The single-cell 2 region at the bottom needs a total of exactly 2 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 2–6 domino horizontally on row 7, column 2 and row 7, column 3.
2. The “=” region at the left needs every pip to match — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–1 domino vertically on row 6, column 1 and row 7, column 1.
3. The “=” region at the left needs every pip to match (1 of its 5 cells is already filled) — nothing else fits. Place the 0–0 domino vertically on row 4, column 1 and row 5, column 1.
4. Row 5, column 2 has only one open neighbor, so its domino must run to row 6, column 2; the “=” region at the left needs every pip to match (3 of its 5 cells are already filled) — nothing else fits. Place the 0–4 domino vertically on row 5, column 2 and row 6, column 2.
5. The 0 region at the top-right needs a total of exactly 0 — 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 5 and row 2, column 5.
6. The 0 region at the top-right needs a total of exactly 0 (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 0–6 domino vertically on row 1, column 4 and row 2, column 4.
7. The single-cell 0 region at the right needs a total of exactly 0 — nothing else fits. Place the 0–3 domino vertically on row 5, column 5 and row 6, column 5.
8. The 17 region at the right needs a total of exactly 17 (1 of its 3 cells is already filled) — nothing else fits. Place the 6–6 domino vertically on row 3, column 5 and row 4, column 5.
9. Row 7, column 4 has only one open neighbor, so its domino must run to row 7, column 5; the single-cell 3 region at the bottom-right needs a total of exactly 3 — nothing else fits. Place the 3–6 domino horizontally on row 7, column 4 and row 7, column 5.
10. The “=” region at the left needs every pip to match (4 of its 5 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 0–2 domino horizontally on row 3, column 1 and row 3, column 2.
11. The “=” region at the right needs every pip to match — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 3–5 domino vertically on row 4, column 4 and row 5, column 4.
12. The single-cell 3 region at the top needs a total of exactly 3 — nothing else fits. Place the 2–3 domino horizontally on row 1, column 2 and row 1, column 3.
13. The 16 region at the top needs a total of exactly 16 (1 of its 3 cells is already filled) — nothing else fits. Place the 5–5 domino horizontally on row 2, column 2 and row 2, column 3.
14. Row 3, column 3 has only one open neighbor, so its domino must run to row 3, column 4; the single-cell 3 region at the center needs a total of exactly 3 — nothing else fits. Place the 3–3 domino horizontally on row 3, column 3 and row 3, column 4.
15. Row 1, column 1 has only one open neighbor, so its domino must run to row 2, column 1; the 5 region at the top-left needs a total of exactly 5 — 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 vertically on row 1, column 1 and row 2, column 1.

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