Pips answers for Thursday, April 23, 2026

Start with the 2 region at the bottom-left: the only way to make it with these dominoes is 1 + 1.

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

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

Start with the 3 region at the bottom-left: the only way to make it with these dominoes is 1 + 1 + 1.

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

1. The 3 region at the bottom-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 1–1 domino vertically on row 5, column 1 and row 6, column 1.
2. Row 6, column 2 has only one open neighbor, so its domino must run to row 6, column 3; the 3 region at the bottom-left needs a total of exactly 3 (2 of its 3 cells are already filled) — nothing else fits. Place the 1–3 domino horizontally on row 6, column 2 and row 6, column 3.
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 4–4 domino vertically on row 4, column 3 and row 5, column 3.
4. The “=” region at the top-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 3–3 domino vertically on row 2, column 6 and row 3, column 6.
5. The single-cell “>4” region at the center must total more than 4 — nothing else fits. Place the 3–5 domino horizontally on row 3, column 4 and row 3, column 5.
6. The 10 region at the top-right needs a total of exactly 10 — nothing else fits. Place the 3–4 domino horizontally on row 1, column 5 and row 1, column 6.
7. Row 1, column 4 has only one open neighbor, so its domino must run to row 2, column 4; the 10 region at the top-right needs a total of exactly 10 (1 of its 2 cells is already filled) — nothing else fits. Place the 4–6 domino vertically on row 1, column 4 and row 2, column 4.
8. Row 4, column 2 has only one open neighbor, so its domino must run to row 4, column 1; the “=” region at the center needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 2–4 domino horizontally on row 4, column 1 and row 4, column 2.

Start with the 17 region at the bottom: the only way to make it with these dominoes is 5 + 6 + 6.

• the 2 region at the top-right: 0, 0, 2
• the single-cell “<5” region at the top-left: 3
• the 10 region at the top: 5, 5
• the “=” region at the left: 1, 1, 1
• the single-cell 4 region at the center: 4
• the single-cell “>2” region at the right: 5
• the single-cell “>3” region at the center: 6
• the “≠” region at the right: 4, 6
• the single-cell 5 region at the left: 5
• the single-cell “<4” region at the center: 3
• the single-cell “>3” region at the right: 6
• the 17 region at the bottom: 5, 6, 6
• the single-cell “<3” region at the bottom: 2
• the single-cell “<3” region at the bottom: 2

1. The single-cell 5 region at the left needs a total of exactly 5 — nothing else fits. Place the 5–6 domino vertically on row 5, column 2 and row 6, column 2.
2. The “=” region at the left 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 1–1 domino horizontally on row 4, column 2 and row 4, column 3.
3. The “=” region at the left needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 1–5 domino horizontally on row 3, column 1 and row 3, column 2.
4. Row 1, column 5 has only one open neighbor, so its domino must run to row 2, column 5; the 2 region at the top-right needs a total of exactly 2 — 2 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 1, column 5 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 2 region at the top-right needs a total of exactly 2 (2 of its 3 cells are already filled) — nothing else fits. Place the 0–5 domino vertically on row 2, column 6 and row 3, column 6.
6. The 17 region at the bottom needs a total of exactly 17 (1 of its 3 cells is already filled) — nothing else fits. Place the 2–5 domino vertically on row 6, column 3 and row 7, column 3.
7. Row 5, column 4 has only one open neighbor, so its domino must run to row 6, column 4; the single-cell “<4” region at the center must total less than 4 — nothing else fits. Place the 3–6 domino vertically on row 5, column 4 and row 6, column 4.
8. Row 6, column 5 has only one open neighbor, so its domino must run to row 6, column 6; the single-cell “<3” region at the bottom must total less than 3 — nothing else fits. Place the 2–3 domino horizontally on row 6, column 5 and row 6, column 6.
9. The 10 region at the top 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 4–5 domino vertically on row 2, column 4 and row 3, column 4.
10. Row 2, column 2 has only one open neighbor, so its domino must run to row 2, column 3; the single-cell “<5” region at the top-left must total less than 5 — nothing else fits. Place the 3–5 domino horizontally on row 2, column 2 and row 2, column 3.
11. The “≠” region at the 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 4–6 domino horizontally on row 4, column 5 and row 4, column 6.
12. Row 5, column 6 has only one open neighbor, so its domino must run to row 5, column 7; the “≠” region at the right needs every pip to be different (1 of its 2 cells is already filled) — nothing else fits. Place the 6–6 domino horizontally on row 5, column 6 and row 5, column 7.

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