NYT Pips Hints & Answers for April 14, 2026

Apr 14, 2026

🚨 SPOILER WARNING

This page contains the final **answer** and the complete **solution** to today's NYT Pips puzzle. If you haven't attempted the puzzle yet and want to try solving it yourself first, now's your chance!

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Want hints instead? Scroll down for progressive clues that won't spoil the fun.

🎲 Today's Puzzle Overview

Ian Livengood's easy grid for April 14th is driven by a single greater-than constraint that anchors the entire right side. The cell at (2,5) must exceed 5 — the only valid pip is 6 — and only one domino in today's set can place a 6 there. That placement forces the equals pair directly above it, which in turn identifies the domino covering the empty cell. On the left, a three-cell equals region running down column 0 is seeded by the [2|2] double, and the only orientation of the final domino that satisfies the less-than-2 ceiling above closes the puzzle.

Rodolfo Kurchan's medium puzzle for April 14th is anchored by three doubles — [3|3], [1|1], and [0|0] — but the critical entry point is neither a double nor an equals region: it is the single greater-than-2 cell at (0,4). Only one tile in today's set can satisfy that constraint, and it deposits a 2 into the three-cell equals region below, pulling a chain of 2s through the center of the board. That cascade resolves the adjacent equals pair and the less-than-2 cell in one motion. Two remaining equals regions close from the bottom: the [3|3] double fills one outright, and the [4|1] domino closes the other.

Rodolfo Kurchan's hard puzzle for April 14th is dominated by four multi-cell equals regions, two sum=4 constraints, two sum=3 constraints, and a single less-than-4 cell. The board's most powerful entry point is the four-cell equals region at the center-left — four cells that must all carry the same pip. Only the [6|6] double can seed it, and from that single placement, six more dominoes cascade into position: two extend the 6 through the equals region, one deposits a 2 that forces a second four-cell equals of 2s, and two more complete a third four-cell equals of 0s, closing both sum=3 constraints on the bottom row. The remaining dominoes resolve from the sum=4 constraints and the lone less-than-4 cell in the upper left.

💡 Progressive Hints

Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!

💡 Greater-than-5 at (2,5) names one domino outright
The only pip value strictly greater than 5 is 6. Cell (2,5) must show a 6. Scan today's easy set: only one domino has a 6-pip face. Place it so that 6 lands at (2,5), and its other face sits directly above — that cell belongs to the equals pair at (0,5) and (1,5), so you can immediately read off what (1,5) must show.
💡 Equals cascades right to left: (0,5) locks a second domino, (1,2)–(1,3) take the double
With (1,5) fixed from the first placement, the equals region at (0,5) and (1,5) forces (0,5) to match. Only one remaining domino can place that value at (0,5) while parking its other face in the empty cell at (0,4). On the other side of the board, the equals pair at (1,2) and (1,3) is waiting for the [6|6] double — both faces identical, constraint satisfied instantly.
💡 Full solution
Greater>5 at (2,5)=6: [3|6] covers (1,5) and (2,5) — 3 at (1,5), 6 at (2,5) ✓. Equals (0,5),(1,5): (1,5)=3, so (0,5)=3. [3|5] covers (0,5) and (0,4): 3 at (0,5) ✓, 5 at (0,4) — empty ✓. Equals (1,2),(1,3): [6|6] double placed horizontally — 6 at (1,2) and 6 at (1,3) ✓. Equals (0,0),(1,0),(2,0): all three must match. [2|2] double covers (1,0) and (2,0): 2 at (1,0) and 2 at (2,0) ✓. So (0,0)=2. [2|1] covers (0,0) and (0,1): 2 at (0,0) ✓, 1 at (0,1) — less<2 ✓. All six constraints satisfied. Puzzle complete.
💡 Greater-than-2 at (0,4) is the sharpest entry point
Cell (0,4) must be strictly greater than 2 — so pip 3, 4, 5, or 6. Scan today's medium domino set and find the only tile whose higher face can satisfy that constraint without conflicting with the grid. Placing it fixes a value in the three-cell equals region below, and a chain of 2s immediately follows.
💡 A cascade of 2s runs through the center of the board
Once the tile from the first step deposits a 2 into the three-cell equals region at (1,3),(1,4),(2,3), all three cells must equal 2. Two of the remaining dominoes each have a 2-pip face: place them to fill (1,3) and (2,3). Each one leaves its other face in an adjacent equals or empty cell, and the value 1 flows automatically into the three-cell equals region above.
💡 Full solution
Greater>2 at (0,4)=4: [2|4] covers (1,4) and (0,4): 2 at (1,4), 4 at (0,4) ✓. Equals (1,3),(1,4),(2,3): (1,4)=2, all must be 2. [1|2] covers (1,2) and (1,3): 2 at (1,3) ✓, 1 at (1,2). [0|2] covers (3,3) and (2,3): 2 at (2,3) ✓, 0 at (3,3) — empty ✓. Equals (0,2),(1,1),(1,2): (1,2)=1, all must be 1. [1|1] double covers (1,0) and (1,1): 1 at (1,0) and 1 at (1,1) ✓. [0|1] covers (0,3) and (0,2): 0 at (0,3) — less<2 ✓ — 1 at (0,2) ✓. Equals (1,0),(2,0): (1,0)=1, so (2,0)=1. [4|1] covers (3,0) and (2,0): 4 at (3,0) — empty ✓ — 1 at (2,0) ✓. [3|3] double covers (3,1) and (3,2): 3 at (3,1) and 3 at (3,2) — equals ✓. All eight constraints satisfied. Puzzle complete.
💡 The four-cell equals region is the longest chain — start with [6|6]
Four cells at (4,2),(4,3),(5,2),(6,2) must all carry the same pip. With four cells needing a single value, the only way to lock two of them with one tile is a double. The [6|6] double is the only one in today's set that can anchor this region. From that placement, two more dominoes extend the 6 through (4,3) and (6,2), and the face left at (7,2) is 2 — which immediately seeds the next equals chain.
💡 2 at (7,2) forces a four-cell equals of 2s, which cascades to a four-cell equals of 0s
The equals region at (7,1),(7,2),(7,3),(8,1) now requires all four cells to be 2. The [2|2] double fills two of them; a second domino covers (7,3) and deposits a 0 at (8,3). That 0 seeds the final four-cell equals at (8,2),(8,3),(9,2),(9,3), which all become 0 via the [0|0] double, closing both sum=3 constraints on the bottom row.
💡 Full solution
[6|6] covers (4,2) and (5,2): 6 at both — equals ✓. [5|6] covers (3,3) and (4,3): 5 at (3,3) — empty ✓ — 6 at (4,3) ✓. [2|6] covers (7,2) and (6,2): 2 at (7,2), 6 at (6,2) ✓. Equals (7,1),(7,2),(7,3),(8,1): all must be 2. [2|2] covers (7,1) and (8,1): 2 at both ✓. [0|2] covers (7,3) and (8,3): 2 at (7,3) ✓, 0 at (8,3). Equals (8,2),(8,3),(9,2),(9,3): all must be 0. [0|0] covers (8,2) and (9,2): 0 at both ✓. (8,3)=0 ✓. (9,3)=0. [0|3] covers (9,3) and (9,4): 0 at (9,3) ✓, 3 at (9,4) — sum=3 ✓. Sum=3 at (9,0): [4|3] covers (9,1) and (9,0): 4 at (9,1) — empty ✓ — 3 at (9,0) ✓. [3|3] covers (3,1) and (4,1): 3 at both. [2|3] covers (2,2) and (3,2): 3 at (3,2) ✓ — equals (3,1)=3 ✓ — 2 at (2,2). Sum (1,2)+(2,2)=4: (2,2)=2, so (1,2)=2. [1|2] covers (0,2) and (1,2): 2 at (1,2) ✓, 1 at (0,2) — empty ✓. Sum=4 at (1,4): [4|6] covers (1,4) and (1,3): 4 at (1,4) ✓, 6 at (1,3) — empty ✓. Less<4 at (1,0): [1|6] covers (1,0) and (1,1): 1 at (1,0) — 1<4 ✓ — 6 at (1,1) — empty ✓. All fourteen constraints satisfied. Puzzle complete.

🎨 Pips Solver

Apr 14, 2026

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Final Answer & Complete Solution For Hard Level

The key to solving today's hard puzzle was identifying the placement for the critical dominoes highlighted in the starting grid. Once those were in place, the rest of the puzzle could be solved logically. See the final grid below to compare your solution.

Starting Position & Key First Steps

Pips hint for April 14, 2026 – hard level puzzle grid with critical first placements and strategy

This image shows the initial puzzle grid for the hard level, with a few critical first placements highlighted.

Final Answer: The Solved Grid for Hard Mode

NYT Pips April 14, 2026 hard puzzle full solution grid showing final answer with hints

Compare this final grid with your own solution to see the correct placement of all dominoes.

🔧 Step-by-Step Answer Walkthrough For Easy Level

1
Step 1: Greater>5 at (2,5) — only pip 6 qualifies, and only [3|6] delivers it
Cell (2,5) must be strictly greater than 5. The maximum pip is 6, so (2,5)=6. Scan today's domino set for a tile with a 6-pip face: [3|6] is the only candidate. Place it with 6 at (2,5) — greater>5 ✓ — and 3 at (1,5).
2
Step 2: Equals at (0,5),(1,5) locks [3|5] and the empty cell
(1,5)=3 from Step 1. The equals region at (0,5) and (1,5) requires (0,5)=3. Only [3|5] has a 3-pip face and can cover (0,5) and the adjacent (0,4). Place it with 3 at (0,5) ✓ and 5 at (0,4) — empty constraint ✓.
3
Step 3: [6|6] double fills the two-cell equals at (1,2),(1,3)
The equals region at (1,2) and (1,3) requires both cells to match. The [6|6] double placed horizontally delivers 6 at (1,2) and 6 at (1,3) ✓.
4
Step 4: [2|2] double and [2|1] close the three-cell equals and the less-than
The equals region at (0,0),(1,0),(2,0) requires all three cells to match. The [2|2] double covers (1,0) and (2,0): 2 at both ✓. So (0,0) must also be 2. The only remaining domino is [2|1]: place it with 2 at (0,0) ✓ and 1 at (0,1). Less<2 at (0,1)=1 ✓. All six constraints satisfied. Puzzle complete.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Greater>2 at (0,4) identifies [2|4]
Cell (0,4) must be strictly greater than 2 — so pip 3, 4, 5, or 6. Scan today's medium dominoes: [2|4] is the only tile whose higher face can satisfy the constraint while fitting the grid. Orient it with 4 at (0,4) — greater>2 ✓ — and 2 at (1,4).
2
Step 2: (1,4)=2 seeds the three-cell equals region at pip 2
The equals region at (1,3),(1,4),(2,3) requires all three cells to match. (1,4)=2 from Step 1, so all three must equal 2. Two dominoes in today's set have a 2-pip face: [1|2] and [0|2].
3
Step 3: [1|2] and [0|2] fill the equals region and the empty cell
[1|2] covers (1,2) and (1,3): 2 at (1,3) ✓, 1 at (1,2). [0|2] covers (3,3) and (2,3): 2 at (2,3) ✓, 0 at (3,3) — empty constraint ✓.
4
Step 4: Equals at (0,2),(1,1),(1,2) cascades to [1|1] double and [0|1]
(1,2)=1 from Step 3. The equals region at (0,2),(1,1),(1,2) requires all three to equal 1. The [1|1] double covers (1,0) and (1,1): 1 at (1,0) and 1 at (1,1) ✓. [0|1] covers (0,3) and (0,2): 1 at (0,2) ✓, 0 at (0,3) — less<2 ✓.
5
Step 5: Equals at (1,0),(2,0) and the [3|3] double close the puzzle
(1,0)=1 from Step 4. The equals region at (1,0) and (2,0) forces (2,0)=1. [4|1] covers (3,0) and (2,0): 1 at (2,0) ✓, 4 at (3,0) — empty ✓. The [3|3] double covers (3,1) and (3,2): 3 at both — equals ✓. All eight constraints satisfied. Puzzle complete.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: [6|6] double seeds the four-cell equals at (4,2),(4,3),(5,2),(6,2)
The four-cell equals region requires all four cells to carry the same pip. Only a double can guarantee two identical faces from a single tile. [6|6] is today's largest double and the only one that can anchor this region. Place it vertically at (4,2) and (5,2): 6 at both ✓.
2
Step 2: [5|6] and [2|6] extend the equals to (4,3) and (6,2), depositing 2 at (7,2)
The remaining cells in the equals region — (4,3) and (6,2) — must also be 6. [5|6] covers (3,3) and (4,3): 6 at (4,3) ✓, 5 at (3,3) — empty ✓. [2|6] covers (7,2) and (6,2): 6 at (6,2) ✓, 2 at (7,2).
3
Step 3: (7,2)=2 forces the four-cell equals at (7,1),(7,2),(7,3),(8,1) to pip 2
All four cells must equal 2. [2|2] double covers (7,1) and (8,1): 2 at both ✓. [0|2] covers (7,3) and (8,3): 2 at (7,3) ✓, 0 at (8,3).
4
Step 4: (8,3)=0 forces the four-cell equals at (8,2),(8,3),(9,2),(9,3) to pip 0
All four cells must equal 0. [0|0] double covers (8,2) and (9,2): 0 at both ✓. (8,3)=0 ✓. (9,3) must also be 0.
5
Step 5: (9,3)=0 closes sum=3 at (9,4); sum=3 at (9,0) places [4|3]
[0|3] covers (9,3) and (9,4): 0 at (9,3) ✓, 3 at (9,4) — sum=3 ✓. Sum=3 at (9,0): [4|3] covers (9,1) and (9,0): 3 at (9,0) ✓, 4 at (9,1) — empty ✓.
6
Step 6: [3|3] double and [2|3] anchor the three-cell equals at (3,1),(3,2),(4,1)
The region requires all three cells to match. [3|3] double covers (3,1) and (4,1): 3 at both ✓. [2|3] covers (2,2) and (3,2): 3 at (3,2) ✓, 2 at (2,2).
7
Step 7: Sum (1,2)+(2,2)=4 with (2,2)=2 forces (1,2)=2; [1|2] covers the top
(2,2)=2 from Step 6. The sum=4 region at (1,2) and (2,2) requires (1,2)=2. [1|2] covers (0,2) and (1,2): 2 at (1,2) ✓, 1 at (0,2) — empty ✓.
8
Step 8: Sum=4 at (1,4) and less<4 at (1,0) place the final two dominoes
Sum=4 at (1,4): [4|6] covers (1,4) and (1,3): 4 at (1,4) ✓, 6 at (1,3) — empty ✓. Less<4 at (1,0): [1|6] covers (1,0) and (1,1): 1 at (1,0) — 1<4 ✓ — 6 at (1,1) — empty ✓. All fourteen constraints satisfied. Puzzle complete.

💡 Pro Tips for Similar Puzzles

Start with Constraints
Always begin with the most constrained regions - sum regions with small numbers or tight spaces.
Use Equal Regions
Use "equal" regions as anchors - they eliminate many possibilities quickly.
Work Systematically
Let the rules guide your placement rather than guessing randomly.
Double-Check
Verify each region's rules are satisfied before moving to the next.

🎓 Keep Learning & Improve