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NYT Pips answers — Saturday, April 18, 2026

Pips #244 · Edited by Ian Livengood · Easy by Ian Livengood · Medium by Rodolfo Kurchan · Hard by Rodolfo Kurchan

Stuck on today's New York Times Pips? Below are three progressive hints for each difficulty, then the full solution as a tap-to-reveal board. Hints first — the answer only when you want it.

Jump to: Easy · Medium · Hard

Easy Pips answer — April 18, 2026

3×4 board · 5 dominoes · 5 regions · by Ian Livengood

5 dominoes, 5 regions, a 3×4 grid — that's the easy puzzle for April 18, 2026. You're working with 2 sum regions, 2 "all equal", 1 free region. The two-cell region at row 1, column 1 adds to 11: the only split is 5 + 6.. After that, the "=" pair at row 2, column 4 needs two identical pips — a double, or matching halves from two dominoes.. Rack check: 1 double, 2 blank halves, 1 six.

Hints

Hint 1 — where to start

The two-cell region at row 1, column 1 adds to 11: the only split is 5 + 6.

Hint 2 — first domino

Place the 3–6 domino with its 3 at row 2, column 1 and its 6 at row 1, column 1.

Hint 3 — second domino

Next, the 2–3 goes at row 2, column 3 → row 2, column 4. From there the rest should fall into place.

Solution

Tap any cell to reveal its pips, or reveal the whole board. Region rules: 11 must add up to 11; 9 must add up to 9; = must all be the same number.

11
9
=
=
Solution as text
  1. 0–2 — 0 at row 3, column 2, 2 at row 2, column 2
  2. 3–0 — 3 at row 3, column 4, 0 at row 3, column 3
  3. 3–6 — 3 at row 2, column 1, 6 at row 1, column 1
  4. 5–5 — 5 at row 1, column 2, 5 at row 1, column 3
  5. 2–3 — 2 at row 2, column 3, 3 at row 2, column 4

Medium Pips answer — April 18, 2026

4×5 board · 7 dominoes · 8 regions · by Rodolfo Kurchan

7 dominoes, 8 regions, a 4×5 grid — that's the medium puzzle for April 18, 2026. You're working with 5 sum regions, 2 "less than", 1 free region. The way in: the two-cell region at row 4, column 2 adds to 12: both cells must be 6. After that, the pair at row 2, column 3 adds to 6; list the ways to make 6 with the rack you have.. Rack check: 2 doubles, 3 blank halves, 3 sixes.

Hints

Hint 1 — where to start

The two-cell region at row 4, column 2 adds to 12: both cells must be 6. That's the 6–6 domino, or two 6s from different dominoes.

Hint 2 — first domino

Place the 6–4 domino with its 6 at row 4, column 2 and its 4 at row 4, column 1.

Hint 3 — second domino

Next, the 6–0 goes at row 2, column 3 → row 2, column 4. From there the rest should fall into place.

Solution

Tap any cell to reveal its pips, or reveal the whole board. Region rules: <7 must add up to less than 7; <3 must add up to less than 3; 6 must add up to 6; 8 must add up to 8; 4 must add up to 4.

<7
<3
6
8
4
6
12
Solution as text
  1. 6–4 — 6 at row 4, column 2, 4 at row 4, column 1
  2. 0–0 — 0 at row 1, column 2, 0 at row 2, column 2
  3. 4–2 — 4 at row 3, column 1, 2 at row 2, column 1
  4. 2–2 — 2 at row 3, column 2, 2 at row 3, column 3
  5. 6–0 — 6 at row 2, column 3, 0 at row 2, column 4
  6. 3–4 — 3 at row 3, column 5, 4 at row 3, column 4
  7. 2–6 — 2 at row 4, column 4, 6 at row 4, column 3

Hard Pips answer — April 18, 2026

8×6 board · 14 dominoes · 15 regions · by Rodolfo Kurchan

Hard today is a 8×6 board with 14 dominoes and 15 regions. Rules on the board: 7 sum regions, 4 "all equal", 1 "less than", 1 "greater than", 2 free regions. The way in: the single cell at row 1, column 3 is marked 3, so it can only hold a 3. After that, the single cell at row 3, column 2 is marked 0, so it can only hold a 0. Rack check: 4 doubles, 5 blank halves, 2 sixes.

Hints

Hint 1 — where to start

The single cell at row 1, column 3 is marked 3, so it can only hold a 3. Find the domino with a 3 that fits there.

Hint 2 — first domino

Place the 3–0 domino with its 3 at row 1, column 3 and its 0 at row 1, column 4.

Hint 3 — second domino

Next, the 1–0 goes at row 3, column 1 → row 3, column 2. From there the rest should fall into place.

Solution

Tap any cell to reveal its pips, or reveal the whole board. Region rules: 3 must add up to 3; 0 must add up to 0; = must all be the same number; >0 must add up to more than 0; <5 must add up to less than 5.

3
0
=
=
>0
0
3
=
<5
0
8
=
8
Solution as text
  1. 4–2 — 4 at row 5, column 1, 2 at row 5, column 2
  2. 5–3 — 5 at row 8, column 5, 3 at row 8, column 6
  3. 6–1 — 6 at row 8, column 3, 1 at row 8, column 4
  4. 0–0 — 0 at row 5, column 5, 0 at row 6, column 5
  5. 5–2 — 5 at row 4, column 3, 2 at row 5, column 3
  6. 1–0 — 1 at row 3, column 1, 0 at row 3, column 2
  7. 5–5 — 5 at row 2, column 3, 5 at row 3, column 3
  8. 2–1 — 2 at row 7, column 3, 1 at row 7, column 4
  9. 4–5 — 4 at row 3, column 5, 5 at row 3, column 4
  10. 6–3 — 6 at row 4, column 4, 3 at row 4, column 5
  11. 4–0 — 4 at row 2, column 5, 0 at row 1, column 5
  12. 1–1 — 1 at row 5, column 4, 1 at row 6, column 4
  13. 3–0 — 3 at row 1, column 3, 0 at row 1, column 4
  14. 4–4 — 4 at row 7, column 5, 4 at row 7, column 6

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