Pips answers for Saturday, March 21, 2026

Start with the 12 region at the top-right: the only way to make it with these dominoes is 4 + 4 + 4.

• the “=” region at the top-left: 2, 2
• the 12 region at the top-right: 4, 4, 4
• the “=” region at the bottom-left: 1, 1
• the single-cell 0 region at the bottom-left: 0
• the single-cell 0 region at the bottom-right: 0

1. The 12 region at the top-right needs a total of exactly 12 — nothing else fits. Place the 4–4 domino horizontally on row 1, column 4 and row 1, column 5.
2. Row 1, column 1 has only one open neighbor, so its domino must run to row 1, column 2; the “=” region at the top-left needs every pip to match — nothing else fits. Place the 2–2 domino horizontally on row 1, column 1 and row 1, column 2.
3. Row 2, column 4 has only one open neighbor, so its domino must run to row 3, column 4; the 12 region at the top-right needs a total of exactly 12 (2 of its 3 cells are already filled) — nothing else fits. Place the 0–4 domino vertically on row 2, column 4 and row 3, column 4.
4. 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 0–1 domino horizontally on row 4, column 1 and row 4, column 2.
5. Row 3, column 2 has only one open neighbor, so its domino must run to row 2, column 2; the “=” region at the bottom-left needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 1–3 domino vertically on row 2, column 2 and row 3, column 2.
6. Row 4, column 5 has only one open neighbor, so its domino must run to row 4, column 4; the single-cell 0 region at the bottom-right needs a total of exactly 0 — nothing else fits. Place the 0–2 domino horizontally on row 4, column 4 and row 4, column 5.

Start with the 5 region at the left: the only way to make it with these dominoes is 1 + 4.

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

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

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

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

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