NYT Pips Hints & Answers for April 9, 2026

Apr 9, 2026

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🎲 Today's Puzzle Overview

Ian Livengood's easy grid for April 9th is anchored at the bottom by the tightest possible constraint: a three-cell region that must total exactly 1. With three cells and a sum of 1, two cells must be zero and one must be a single pip. The [0|0] double handles the two zeros, and the only remaining tile with a 1-pip face locks the final cell. Everything else on the board resolves upward from there: an equals pair in the center takes the other double, and the top row settles through a two-cell equals check and a loose less-than ceiling.

Ian Livengood's medium puzzle is controlled by three doubles — [0|0], [1|1], and [5|5] — and each one owns a distinct constraint. The [5|5] double is the most constrained: a sum=10 region demands exactly 5+5, and there is no other way to reach that total with the available tiles. The [0|0] double follows from a sum=4 cell that pins its neighbor to zero, seeding a three-cell equals region of zeros. The [1|1] double fills the top-right corner, closing a sum=5 pair that was otherwise ambiguous. Two greater-than cells at opposite corners of the board each narrow to a single possible domino, and the remaining sums fall in order.

Rodolfo Kurchan's hard puzzle for April 9th is saturated with sum=3 constraints — six of the fifteen regions on the board require a total of three. The board's only less-than constraint (less than 3, bottom row) is the sharpest entry point: it limits its cell to pip 0 or 1, and only one domino in today's set can satisfy that while pairing with the adjacent sum=3 region. From that placement, a chain of ones cascades through a three-cell sum=3 region mid-board. Meanwhile, the single-cell sum=3 at the top-right corner names its domino outright and sets off a cascade through three equals regions that runs from the top-right all the way down the right side of the board. Two doubles — [3|3] and [4|4] — each close a two-cell equals region, with the [4|4] double additionally satisfying four separate greater-than-3 constraints in the bottom-left corner.

💡 Progressive Hints

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

💡 Sum=1 across three cells — two zeros and a one
The three-cell region at the bottom must total exactly 1. With three cells and only one pip to distribute, two cells must be zero and one must show a single pip. Today's easy set contains exactly one double — and it's the [0|0], which can supply both zeros. Find the only tile with a 1-pip face to complete the region, and the entire bottom section is locked.
💡 One double left, one equals region in the middle
With the bottom settled, the remaining double — [5|5] — belongs in the two-cell equals region in the center of the grid. Both cells must match, and the double guarantees it. The top row closes through the other equals pair and a less-than ceiling that rules out only one possible value.
💡 Full solution
Sum=1 at (3,4)+(4,4)+(5,4)=1: the [0|0] double covers (3,4) and (4,4) horizontally — 0+0=0. Cell (5,4) must supply the remaining 1. The [1|5] domino orients with 1 at (5,4) and 5 at (5,3) (empty ✓). Equals at (2,2),(3,2): the [5|5] double placed vertically — 5 at (2,2) and 5 at (3,2) ✓. Equals at (0,0),(1,0): both cells must match. The [2|3] domino covers (0,0) and (0,1): 2 at (0,0), 3 at (0,1). Less-than-4 at (0,1)=3 ✓. The [2|4] domino covers (1,0) and (2,0): 2 at (1,0) — equals (0,0)=2 ✓ — and 4 at (2,0) (empty ✓). All six constraints satisfied. Puzzle complete.
💡 Sum=10 is only reachable one way — start there
Three doubles sit in today's medium set. The sum=10 region at the right side of row 2 names which double goes there immediately: only 5+5 reaches 10. Place [5|5] horizontally in those two cells, and its presence seeds the equals pair directly above it.
💡 Sum=4 triggers the three-cell equals region of zeros
Once the [5|5] double is placed, look at the single-cell sum=4 at the far right — it names an exact pip value and identifies the domino. That domino deposits a 0 into the adjacent cell, which seeds the three-cell equals region below it: all three cells must match, and they all match at zero. The [0|0] double fills two of those three cells, and the third is already set.
💡 Full solution
Sum=10 at (2,3)+(2,4)=10: the [5|5] double placed horizontally — 5 at (2,3) and 5 at (2,4) ✓. Sum=4 at (3,4)=4: the [4|0] domino covers (3,4) and (3,3): 4 at (3,4) ✓, 0 at (3,3). Three-cell equals (3,2),(3,3),(4,2): (3,3)=0, so all three must be 0. The [0|0] double covers (3,2) and (4,2): 0 at (3,2) ✓, 0 at (4,2) ✓. Greater-than-4 at (1,0) and (4,1): both cells need pip 5 or 6. [5|6] covers (1,1) and (1,0): 5 at (1,1), 6 at (1,0) — 6>4 ✓. [6|2] covers (4,1) and (3,1): 6 at (4,1) — 6>4 ✓ — and 2 at (3,1). Equals (1,1),(1,2): (1,1)=5, so (1,2)=5. [5|4] covers (1,2) and (0,2): 5 at (1,2) ✓, 4 at (0,2). Sum=5 at (0,2)+(0,3): (0,2)=4, so (0,3)=1. [1|1] double covers (0,3) and (1,3): 1 at (0,3) ✓, 1 at (1,3) (empty ✓). Sum=3 at (2,0)=3: the [3|2] domino covers (2,0) and (2,1): 3 at (2,0) ✓, 2 at (2,1). Equals (2,1),(3,1): (3,1)=2 ✓. All ten constraints satisfied. Puzzle complete.
💡 The less-than-3 cell at the bottom is the sharpest constraint on the board
Six of the fifteen regions require a sum of exactly 3 — but none is as restrictive as the single less-than-3 cell in the bottom row. A pip value of 0, 1, or 2 is required there. Scan the domino set: only one tile can place a sub-3 face in that position while satisfying the adjacent sum=3 region. That placement cascades immediately into a three-cell sum=3 region higher up.
💡 Sum=3 in the top-right corner names its domino outright
The single-cell sum=3 at (0,3) forces an exact pip count. Only one domino in today's set can deliver a 3 there and still have its other face place usefully into the equals region directly below. From that placement, a chain of equals constraints flows: two cells lock to 6, which forces a third, which seeds another sum=3 further down the right side.
💡 Full solution
Less-than-3 at (6,2): must be 0 or 1. [3|2] places 2 at (6,2) — 2<3 ✓ — and 3 at (6,3). Sum=3 at (6,3)+(6,4): (6,3)=3, so (6,4)=0. [0|1] covers (6,4) and (5,4): 0 at (6,4) ✓, 1 at (5,4). Sum=3 at (4,3)+(4,4)+(5,4)=3: (5,4)=1. [3|1] covers (3,4) and (4,4): 3 at (3,4), 1 at (4,4). So (4,3) must be 1. [1|5] covers (4,3) and (4,2): 1 at (4,3) ✓, 5 at (4,2). Greater-than-3 at (4,2)=5 ✓. Sum=3 at (0,3)=3: [3|6] covers (0,3) and (0,4): 3 at (0,3) ✓, 6 at (0,4). Equals (0,4),(1,4): (0,4)=6, so (1,4)=6. [6|0] covers (1,4) and (2,4): 6 at (1,4) ✓, 0 at (2,4). Sum=3 at (2,4)+(3,4): (2,4)=0, (3,4)=3 — 0+3=3 ✓. [3|3] double covers (2,2) and (2,3) horizontally — equals region, both 3 ✓. Sum=3 at (3,0)+(4,0)=3: [5|0] covers (2,0) and (3,0): 5 at (2,0), 0 at (3,0). [3|4] covers (4,0) and (5,0): 3 at (4,0) ✓ — 0+3=3 ✓ — and 4 at (5,0). Greater-than-3 at (5,0)=4 ✓. [4|4] double covers (6,0) and (6,1): 4 at (6,0) — greater-than-3 ✓ — and 4 at (6,1) — greater-than-3 ✓. Equals (1,0),(2,0): (2,0)=5, so (1,0)=5. [5|2] covers (1,0) and (0,0): 5 at (1,0) ✓, 2 at (0,0). Equals (0,0),(0,1): (0,0)=2, so (0,1)=2. [2|6] covers (0,1) and (0,2): 2 at (0,1) ✓, 6 at (0,2). Greater-than-3 at (0,2)=6 ✓. All fifteen constraints satisfied. Puzzle complete.

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Apr 9, 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 9, 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 9, 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: Sum=1 forces the [0|0] double and the [1|5] domino
The region at (3,4), (4,4), and (5,4) must total 1. Two cells must be zero and one must be 1. The [0|0] double is the only tile that can supply two zeros in adjacent cells — place it horizontally at (3,4) and (4,4). The remaining 1 pip must come from cell (5,4). Only [1|5] has a 1-pip face; orient it with 1 at (5,4) and 5 at (5,3). Cell (5,3) is an empty constraint ✓.
2
Step 2: [5|5] double fills the center equals region
The equals region at (2,2) and (3,2) requires both cells to match. The [5|5] double placed vertically satisfies this immediately: 5 at (2,2) and 5 at (3,2) ✓.
3
Step 3: The equals pair at the top and the less-than ceiling
The equals region at (0,0) and (1,0) requires both cells to share one pip value. The [2|3] domino covers (0,0) and (0,1) horizontally: 2 at (0,0), 3 at (0,1). Less-than-4 at (0,1)=3 — just under the limit ✓.
4
Step 4: The last domino confirms the equals pair and fills the empty cell
Cell (1,0) must equal (0,0)=2. The [2|4] domino covers (1,0) and (2,0) vertically: 2 at (1,0) — equals check ✓ — and 4 at (2,0). Cell (2,0) is an empty constraint ✓. All six constraints across the board are satisfied. Puzzle complete.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Sum=10 identifies the [5|5] double immediately
The region at (2,3) and (2,4) must total 10. No single domino can produce 10 from two distinct pip faces when the maximum is 6+5=11, but the only tile that hits 10 cleanly with equal faces is [5|5]. Place it horizontally: 5 at (2,3) and 5 at (2,4) ✓.
2
Step 2: Sum=4 at (3,4) and the zero cascade
Cell (3,4) must equal exactly 4. The [4|0] domino covers (3,4) and (3,3): 4 at (3,4) ✓, 0 at (3,3). That zero seeds the three-cell equals region at (3,2),(3,3),(4,2): all three must equal 0. The [0|0] double covers (3,2) and (4,2): 0 at (3,2) ✓, 0 at (4,2) ✓.
3
Step 3: Two greater-than-4 cells each lock a domino
Cells (1,0) and (4,1) both require pip values above 4 — so 5 or 6. [5|6] covers (1,1) and (1,0) horizontally: 5 at (1,1), 6 at (1,0) — 6>4 ✓. [6|2] covers (4,1) and (3,1): 6 at (4,1) — 6>4 ✓ — and 2 at (3,1).
4
Step 4: Equals pair at (1,1),(1,2) feeds sum=5 at the top
Cell (1,1)=5 is now fixed. The equals region at (1,1) and (1,2) forces (1,2)=5. [5|4] covers (1,2) and (0,2): 5 at (1,2) ✓, 4 at (0,2). Sum=5 at (0,2)+(0,3): (0,2)=4, so (0,3)=1. The [1|1] double covers (0,3) and (1,3): 1 at (0,3) ✓, 1 at (1,3) — empty constraint ✓.
5
Step 5: Sum=3 and the center equals pair close the puzzle
Sum=3 at (2,0)=3: the [3|2] domino covers (2,0) and (2,1): 3 at (2,0) ✓, 2 at (2,1). Equals at (2,1),(3,1): (3,1)=2 from Step 3 — and (2,1)=2 ✓. All ten constraints satisfied. Puzzle complete.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Less-than-3 at (6,2) — the board's tightest cell
Cell (6,2) must be strictly less than 3: so pip 0, 1, or 2. Dominoes with faces in that range: [0,1], [3,2], [3,1], [3,2]. Only [3|2] can place 2 at (6,2) while putting a usable face in the adjacent position. Place [3|2] with 2 at (6,2) — 2<3 ✓ — and 3 at (6,3).
2
Step 2: Sum=3 at (6,3)+(6,4) and the one-cascade begins
Cell (6,3)=3. Sum=3 requires (6,4)=0. [0|1] covers (6,4) and (5,4): 0 at (6,4) ✓, 1 at (5,4). Sum=3 at (4,3)+(4,4)+(5,4)=3: (5,4)=1. [3|1] covers (3,4) and (4,4): 3 at (3,4), 1 at (4,4). So (4,3)=3−1−1=1. [1|5] covers (4,3) and (4,2): 1 at (4,3) ✓, 5 at (4,2). Greater-than-3 at (4,2)=5 ✓.
3
Step 3: Sum=3 at (0,3) names its domino and cascades down the right side
Cell (0,3) must equal exactly 3. [3|6] covers (0,3) and (0,4): 3 at (0,3) ✓, 6 at (0,4). Equals (0,4),(1,4): (0,4)=6, so (1,4)=6. [6|0] covers (1,4) and (2,4): 6 at (1,4) ✓, 0 at (2,4). Sum=3 at (2,4)+(3,4): (2,4)=0, (3,4)=3 — 0+3=3 ✓ (already placed in Step 2).
4
Step 4: [3|3] double fills the center equals region
The equals region at (2,2) and (2,3) requires both cells to match. The [3|3] double placed horizontally delivers 3 at (2,2) and 3 at (2,3) ✓. No other arrangement satisfies the equals constraint with the remaining tiles.
5
Step 5: Sum=3 at (3,0)+(4,0) and the left-column greater-than chain
The region at (3,0) and (4,0) must total 3. [5|0] covers (2,0) and (3,0): 5 at (2,0), 0 at (3,0). [3|4] covers (4,0) and (5,0): 3 at (4,0) — 0+3=3 ✓ — and 4 at (5,0). Greater-than-3 at (5,0)=4 ✓.
6
Step 6: [4|4] double handles four greater-than-3 constraints at once
Cells (6,0) and (6,1) both require pip values strictly above 3. The [4|4] double placed horizontally satisfies both simultaneously: 4 at (6,0) — 4>3 ✓ — and 4 at (6,1) — 4>3 ✓.
7
Step 7: Equals chain at (1,0),(2,0) and (0,0),(0,1) closes the top-left
Equals (1,0),(2,0): (2,0)=5, so (1,0) must also be 5. [5|2] covers (1,0) and (0,0): 5 at (1,0) ✓, 2 at (0,0). Equals (0,0),(0,1): (0,0)=2, so (0,1)=2. [2|6] covers (0,1) and (0,2): 2 at (0,1) ✓, 6 at (0,2). Greater-than-3 at (0,2)=6 ✓. All fifteen 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.

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