Pips answers for Tuesday, May 19, 2026

Start with the “<4” region at the top-left: only 1 + 1, or 1 + 2 can make it with these dominoes.

• the “<4” region at the top-left: 1, 2
• the 5 region at the left: 1, 4
• the “=” region at the right: 3, 3
• the “>9” region at the bottom-left: 4, 6

1. The 5 region at the left needs a total of exactly 5 — nothing else fits. Place the 4–4 domino vertically on row 3, column 1 and row 4, column 1.
2. The “<4” region at the top-left must total less than 4 — nothing else fits. Place the 1–1 domino vertically on row 1, column 1 and row 2, column 1.
3. Row 1, column 2 has only one open neighbor, so its domino must run to row 1, column 3; the “<4” region at the top-left must total less than 4 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–2 domino horizontally on row 1, column 2 and row 1, column 3.
4. Row 4, column 2 has only one open neighbor, so its domino must run to row 4, column 3; the “>9” region at the bottom-left must total more than 9 (1 of its 2 cells is already filled) — nothing else fits. Place the 6–6 domino horizontally on row 4, column 2 and row 4, column 3.
5. Row 3, column 4 has only one open neighbor, so its domino must run to row 3, column 5; the “=” region at the right needs every pip to match — nothing else fits. Place the 3–3 domino horizontally on row 3, column 4 and row 3, column 5.

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

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

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

Start with the 2 region at the center: only 0 + 2, or 1 + 1 can make it with these dominoes.

• the “=” region at the top: 1, 1, 1, 1
• the single-cell “>0” region at the top-right: 2
• the 9 region at the center: 3, 3, 3
• the 2 region at the center: 1, 1
• the 9 region at the bottom-right: 4, 5
• the 8 region at the bottom-left: 4, 4
• the 9 region at the bottom: 4, 5
• the “=” region at the bottom-right: 6, 6, 6, 6
• the 9 region at the bottom-left: 2, 2, 5
• the 8 region at the bottom: 3, 5

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

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