Pips answers for Monday, May 18, 2026

Start with the 6 region at the top: the only way to make it with these dominoes is 2 + 2 + 2.

• the 6 region at the top: 2, 2, 2
• the single-cell “<5” region at the left: 2
• the 9 region at the bottom-left: 4, 5

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

Start with the “=” region at the top-right: all 3 cells must show the same number, and only 3 or 5 appear on enough dominoes to fill it.

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

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

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

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

• the single-cell “>1” region at the top-right: 2
• the 7 region at the top-right: 2, 5
• the single-cell “>2” region at the top-left: 4
• the 11 region at the top-left: 5, 6
• the “>11” region at the top-left: 6, 6
• the single-cell “>2” region at the top: 4
• the single-cell “<2” region at the top-right: 0
• the 7 region at the right: 1, 3, 3
• the single-cell 3 region at the left: 3
• the single-cell 3 region at the left: 3
• the single-cell 3 region at the left: 3
• the 14 region at the bottom: 4, 5, 5
• the “=” region at the bottom-right: 0, 0, 0
• the single-cell “>4” region at the bottom-left: 6
• the single-cell “<2” region at the bottom-left: 0

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