Pips answers for Tuesday, May 12, 2026

Start with the “=” region at the center: all 3 cells must show the same number, and only 1 or 3 appear on enough dominoes to fill it.

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

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

Start with the 9 region at the left: only 3 + 6, or 4 + 5 can make it with these dominoes.

• the 12 region at the top: 1, 3, 4, 4
• the 9 region at the left: 3, 6
• the “=” region at the right: 3, 3
• the “<2” region at the center: 0, 0
• the “=” region at the bottom-left: 2, 2
• the “>5” region at the bottom-right: 4, 5

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

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

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

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

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