Pips answers for Friday, May 29, 2026

Start with the 20 region at the bottom: the only way to make it with these dominoes is 5 + 5 + 5 + 5.

• the “<2” region at the top-left: 0, 0
• the “=” region at the top: 6, 6
• the “>2” region at the top-right: 1, 2
• the 20 region at the bottom: 5, 5, 5, 5

1. The “<2” region at the top-left must total less than 2 — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–6 domino horizontally on row 1, column 1 and row 1, column 2.
2. Row 1, column 3 has only one open neighbor, so its domino must run to row 1, column 4; the “=” region at the top needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 2–6 domino horizontally on row 1, column 3 and row 1, column 4.
3. The 20 region at the bottom needs a total of exactly 20 — nothing else fits. Place the 5–5 domino horizontally on row 3, column 2 and row 3, column 3.
4. Row 2, column 4 has only one open neighbor, so its domino must run to row 3, column 4; the “>2” region at the top-right must total more than 2 (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.
5. Row 2, column 1 has only one open neighbor, so its domino must run to row 3, column 1; the “<2” region at the top-left must total less than 2 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–5 domino vertically on row 2, column 1 and row 3, column 1.

Start with the 3 region at the center: only 0 + 3, or 1 + 2 can make it with these dominoes.

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

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

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

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

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