Pips answers for Sunday, March 8, 2026

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

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

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

Start with the “<2” region at the left: the only way to make it with these dominoes is 0 + 0.

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

1. The “<2” region at the left must total less than 2 — nothing else fits. Place the 0–0 domino horizontally on row 3, column 2 and row 3, column 3.
2. Row 4, column 4 has only one open neighbor, so its domino must run to row 4, column 5; the single-cell “>5” region at the bottom must total more than 5 — nothing else fits. Place the 0–6 domino horizontally on row 4, column 4 and row 4, column 5.
3. Row 3, column 5 has only one open neighbor, so its domino must run to row 3, column 6; the “<2” region at the center must total less than 2 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–5 domino horizontally on row 3, column 5 and row 3, column 6.
4. Row 5, column 5 has only one open neighbor, so its domino must run to row 5, column 6; the “=” region at the bottom-right needs every pip to match (1 of its 4 cells is already filled) — nothing else fits. Place the 5–5 domino horizontally on row 5, column 5 and row 5, column 6.
5. The “=” region at the top-left needs every pip to match — nothing else fits. Place the 3–3 domino horizontally on row 1, column 2 and row 1, column 3.
6. Row 2, column 1 has only one open neighbor, so its domino must run to row 2, column 2; the single-cell 3 region at the top-left needs a total of exactly 3 — nothing else fits. Place the 3–6 domino horizontally on row 2, column 1 and row 2, column 2.
7. Row 2, column 3 has only one open neighbor, so its domino must run to row 2, column 4; the “=” region at the top-left needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 5–6 domino horizontally on row 2, column 3 and row 2, column 4.
8. Row 4, column 6 has only one open neighbor, so its domino must run to row 4, column 7; the “=” region at the bottom-right needs every pip to match (3 of its 4 cells are already filled) — nothing else fits. Place the 4–5 domino horizontally on row 4, column 6 and row 4, column 7.

Start with the 0 region at the bottom-right: the only way to make it with these dominoes is 0 + 0.

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

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