Pips answers for Tuesday, March 3, 2026

Start with the 7 region at the left: the only way to make it with these dominoes is 3 + 4.

• the 7 region at the left: 3, 4
• the 4 region at the center: 1, 3
• the “=” region at the center: 0, 0
• the single-cell 3 region at the right: 3
• the single-cell 1 region at the bottom-left: 1

1. The 7 region at the left needs a total of exactly 7 — nothing else fits. Place the 1–3 domino vertically on row 3, column 2 and row 4, column 2.
2. The 7 region at the left needs a total of exactly 7 (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. 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 0–3 domino horizontally on row 3, column 4 and row 3, column 5.
4. The 4 region at the center needs a total of exactly 4 — nothing else fits. Place the 0–1 domino vertically on row 2, column 3 and row 3, column 3.
5. Row 2, column 4 has only one open neighbor, so its domino must run to row 1, column 4; the 4 region at the center needs a total of exactly 4 (1 of its 2 cells is already filled) — nothing else fits. Place the 3–5 domino vertically on row 1, column 4 and row 2, column 4.

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

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

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

Start with the 5 region at the bottom — it has the fewest pip combinations on the board, so the dominoes around it are the easiest to pin down.

• the single-cell 2 region at the top-left: 2
• the single-cell 4 region at the top-left: 4
• the single-cell 0 region at the top: 0
• the single-cell 2 region at the top: 2
• the single-cell 2 region at the top-right: 2
• the single-cell 3 region at the top-right: 3
• the single-cell 4 region at the top-left: 4
• the single-cell 3 region at the top-left: 3
• the single-cell 2 region at the top: 2
• the single-cell 4 region at the top-right: 4
• the single-cell 0 region at the left: 0
• the 5 region at the bottom: 1, 1, 1, 1, 1
• the single-cell 0 region at the right: 0
• the single-cell 0 region at the right: 0
• the single-cell 3 region at the bottom-left: 3
• the single-cell 0 region at the bottom-left: 0
• the single-cell 4 region at the bottom: 4
• the single-cell 3 region at the bottom-right: 3
• the single-cell 4 region at the bottom-right: 4
• the single-cell 3 region at the bottom-left: 3
• the single-cell 2 region at the bottom: 2

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