Pips answers for Saturday, April 18, 2026

Start with the 11 region at the top-left: the only way to make it with these dominoes is 5 + 6.

• the 11 region at the top-left: 5, 6
• the 9 region at the center: 2, 2, 5
• the “=” region at the bottom-right: 3, 3
• the “=” region at the bottom: 0, 0

1. The 11 region at the top-left needs a total of exactly 11 — 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 vertically on row 1, column 1 and row 2, column 1.
2. The 11 region at the top-left needs a total of exactly 11 (1 of its 2 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 5–5 domino horizontally on row 1, column 2 and row 1, column 3.
3. The 9 region at the center needs a total of exactly 9 (1 of its 3 cells is already filled) — 6 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.
4. The 9 region at the center needs a total of exactly 9 (2 of its 3 cells are already filled) — nothing else fits. Place the 2–3 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 “=” region at the bottom needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 0–3 domino horizontally on row 3, column 3 and row 3, column 4.

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

Start with the 12 region at the bottom: the only way to make it with these dominoes is 6 + 6.

• the “<7” region at the top-left: 0, 0
• the single-cell “<3” region at the left: 2
• the 6 region at the center: 0, 6
• the 8 region at the bottom-left: 4, 4
• the 4 region at the center: 2, 2
• the 6 region at the bottom-right: 2, 4
• the 12 region at the bottom: 6, 6

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

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

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

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

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

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