Pips answers for Wednesday, November 11, 2026

Start with the 10 region at the left: the only way to make it with these dominoes is 5 + 5.

• the “=” region at the bottom-right: 1, 1
• the 10 region at the left: 5, 5
• the “=” region at the top: 2, 2
• the single-cell “>4” region at the top: 6
• the single-cell “>1” region at the bottom-left: 3
• the single-cell “>0” region at the right: 5

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

Start with the “<2” region at the top-left: the only way to make it with these dominoes is 0 + 1.

• the “<2” region at the top-left: 0, 1
• the “<8” region at the bottom-left: 1, 3, 3
• the 10 region at the left: 4, 6
• the “≠” region at the bottom: 2, 3, 4
• the “=” region at the bottom-right: 6, 6, 6

1. The “=” region at the bottom-right needs every pip to match — nothing else fits. Place the 6–6 domino horizontally on row 3, column 5 and row 3, column 6.
2. The “≠” region at the bottom 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 4–6 domino horizontally on row 2, column 4 and row 2, column 5.
3. The 10 region at the left needs a total of exactly 10 — 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 1, column 3 and row 2, column 3.
4. The “<2” region at the top-left must total less than 2 — nothing else fits. Place the 1–4 domino vertically on row 1, column 2 and row 2, column 2.
5. Row 1, column 1 has only one open neighbor, so its domino must run to row 2, column 1; the “<2” region at the top-left must total less than 2 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–3 domino vertically on row 1, column 1 and row 2, column 1.
6. Row 3, column 1 has only one open neighbor, so its domino must run to row 3, column 2; the “<8” region at the bottom-left must total less than 8 (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 1–3 domino horizontally on row 3, column 1 and row 3, column 2.
7. 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 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 2–3 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 4 region at the bottom-left: only 0 + 1 + 1 + 2, or 1 + 1 + 1 + 1 can make it with these dominoes.

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

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