Pips answers for Wednesday, April 15, 2026

Start with the “=” region at the top-right: all 3 cells must show the same number, and only 0 or 1 or 3 appear on enough dominoes to fill it.

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

1. The “=” region at the top-right needs every pip to match — nothing else fits. Place the 0–5 domino horizontally on row 2, column 6 and row 2, column 7.
2. The “=” region at the top-right needs every pip to match (1 of its 3 cells is already filled) — nothing else fits. Place the 0–0 domino horizontally on row 1, column 5 and row 1, column 6.
3. The “=” region at the center 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 3–3 domino vertically on row 3, column 4 and row 4, column 4.
4. The “=” region at the top needs every pip to match — nothing else fits. Place the 2–3 domino vertically on row 1, column 4 and row 2, column 4.
5. The “=” region at the top-left needs every pip to match — nothing else fits. Place the 1–2 domino horizontally on row 1, column 2 and row 1, column 3.
6. Row 2, column 1 has only one open neighbor, so its domino must run to row 2, column 2; the single-cell “<2” region at the left must total less than 2 — nothing else fits. Place the 1–1 domino horizontally on row 2, column 1 and row 2, column 2.

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

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

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

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

Start with the 23 region at the top-right: the only way to make it with these dominoes is 5 + 6 + 6 + 6.

• the “=” region at the top-left: 4, 4, 4, 4, 4
• the 4 region at the top: 1, 3
• the 23 region at the top-right: 5, 6, 6, 6
• the single-cell “<5” region at the right: 0
• the “=” region at the left: 5, 5, 5
• the “=” region at the right: 5, 5, 5
• the 4 region at the bottom-left: 1, 3
• the 7 region at the bottom-right: 1, 2, 2, 2
• the “=” region at the bottom: 3, 3

1. The 23 region at the top-right needs a total of exactly 23 — 6 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 6–6 domino vertically on row 2, column 5 and row 3, column 5.
2. Row 1, column 5 has only one open neighbor, so its domino must run to row 1, column 4; the 23 region at the top-right needs a total of exactly 23 (2 of its 4 cells are already filled) — 4 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 3–5 domino horizontally on row 1, column 4 and row 1, column 5.
3. The 23 region at the top-right needs a total of exactly 23 (3 of its 4 cells are already filled) — nothing else fits. Place the 5–6 domino vertically on row 4, column 5 and row 5, column 5.
4. The “=” region at the top-left needs every pip to match — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 4–4 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 5 cells are already filled) — nothing else fits. Place the 1–4 domino horizontally on row 1, column 2 and row 1, column 3.
6. The “=” region at the top-left needs every pip to match (3 of its 5 cells are already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–4 domino vertically on row 3, column 1 and row 4, column 1.
7. Row 3, column 2 has only one open neighbor, so its domino must run to row 4, column 2; the “=” region at the top-left needs every pip to match (4 of its 5 cells are already filled) — nothing else fits. Place the 4–5 domino vertically on row 3, column 2 and row 4, column 2.
8. The “=” region at the right needs every pip to match (1 of its 3 cells is already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–5 domino vertically on row 3, column 4 and row 4, column 4.
9. The 7 region at the bottom-right needs a total of exactly 7 — 2 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 5, column 4 and row 6, column 4.
10. The “=” region at the left needs every pip to match (1 of its 3 cells is already filled) — 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 5, column 2 and row 6, column 2.
11. Row 6, column 5 has only one open neighbor, so its domino must run to row 7, column 5; the 7 region at the bottom-right needs a total of exactly 7 (1 of its 4 cells is already filled) — 7 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–2 domino vertically on row 6, column 5 and row 7, column 5.
12. The “=” region at the bottom needs every pip to match — nothing else fits. Place the 2–3 domino horizontally on row 7, column 3 and row 7, column 4.
13. The 4 region at the bottom-left needs a total of exactly 4 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 1–3 domino horizontally on row 7, column 1 and row 7, column 2.
14. Row 6, column 1 has only one open neighbor, so its domino must run to row 5, column 1; the 4 region at the bottom-left needs a total of exactly 4 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–3 domino vertically on row 5, column 1 and row 6, column 1.