Pips answers for Thursday, May 28, 2026

Start with the “>6” region at the bottom-right: only 3 + 4, or 4 + 4 can make it with these dominoes.

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

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

Start with the “≠” region at the center: the only way to make it with these dominoes is 2 + 3 + 4 + 5 + 6.

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

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

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

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

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