NYT Pips Hints & Answers for April 28, 2026

Apr 28, 2026

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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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🎲 Today's Puzzle Overview

The NYT Pips easy grid by Ian Livengood opens on two independent deduction footholds. A single-cell sum-5 region at [3,3] demands exactly five pips, forcing the [5,6] domino to cover that cell and [3,4]. Simultaneously, a three-cell equals region in the left column forces three identical values, which only the [1,1] domino can satisfy, pinning ones into [1,0], [2,0], and [3,0]. From these anchors, the remaining equals pairs and the sum-5 at [0,1] cascade logically, resolving the grid with minimal branching.

Livengood's medium puzzle scales the complexity with a less-4 region and a sum-1 singleton. The sum-1 cell at [2,0] acts as the primary key; it must be 1, which assigns the [5,1] domino across [1,0] and [2,0], immediately satisfying the equals pair in column 0. That 5 then forces the adjacent equals pair [1,1] to also be 5, activating the [5,2] domino into the top row. The less-4 region (cells [0,3] and [0,4]) then accepts only 0–3 pips, narrowing the remaining placements to a clean, deterministic chain.

Rodolfo Kurchan's hard grid is a dense lattice of 25 single-cell sum targets, turning the puzzle into a tightly constrained deduction graph. The many sum-0 cells (at [0,5], [3,0], [4,2], [4,3]) must be zero, immediately fixing four dominoes that contain a 0-pip, including the [0,5] and [0,4] and [0,2]/[0,3] combinations. These initial placements propagate constraints through the sum-1 and sum-3 cells in the top two rows, binding the low-value dominoes into place. What remains is a precise unlocking of the sum-9, sum-7, and sum-5 pairs in the middle rows, resulting in a pure step-by-step solve that rewards methodical chain-tracing.

💡 Progressive Hints

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

💡 Hint 1: Identify the tightest constraint
Focus on regions that contain only a single cell. A sum target on a lone cell forces that cell to take an exact pip value. Look for one such cell that demands a specific high pip.
💡 Hint 2: Single-cell sum and equals chain
The single-cell sum-5 at the bottom row, third column (row 3, col 3) needs a 5. That will force a domino with a 5. Then check the three-cell equals region in the leftmost column—it needs three identical numbers, which severely limits your domino choices.
💡 Hint 3: Full solution path
Place the [5,6] domino so that 5 sits in the sum-5 cell at [3,3] and 6 fills [3,4]. The left equals region (cells [1,0], [2,0], [3,0]) must all be 1, so use the [1,1] domino on [1,0] and [2,0], then the [3,1] domino gives 1 to [3,0] and 3 to the empty [3,1]. The right equals pair [2,4]/[3,4] forces 6 at [2,4], so place the [6,2] domino with 6 at [2,4] and 2 at [1,4]. Finally, the sum-5 at [0,1] takes the [4,5] domino (5 at [0,1], 4 at [0,0]) and the [0,2] domino fills [0,3]=0, [0,4]=2 to match the equals pair.
💡 Hint 1: Find the tightest numeric lock
Search for a single-cell sum region. A sum-1 cell requires exactly one pip, so the domino connected to that cell must carry a 1. This anchor will dictate the placement of a high-value partner.
💡 Hint 2: The sum-1 anchor and adjacent equals
The sum-1 cell sits at the start of row 3 (row 2, col 0). It must be 1, so the domino covering it also goes above it. That forces the cell above (row 1, col 0) to be the domino's other pip. Because row 1, col 0 is part of an equals pair with its right neighbor, that neighbor gets the same value, which must be a 5 from the available dominos.
💡 Hint 3: Complete solving chain
Place [5,1] vertically with 1 at [2,0] and 5 at [1,0]; then [5,2] with 5 at [1,1] and 2 at [0,1]. The top equals pair [0,1]/[0,2] forces [0,2] to 2, so [0,2] domino places 2 at [0,2] and 0 at [0,3]. The less-4 region (<4) requires [0,4] to be low, so [0,4] domino places 0 at [0,4] and 4 at [1,4] (satisfying the vertical equals pair). The row-2 equals triple ([2,1],[2,2],[2,3]) gets 2s from [2,2] domino and the [2,4] domino’s 2 at [2,3] (with 4 at [2,4]). Finally, the rightmost equals pair takes [3,3] with 3s at [1,5] and [2,5].
💡 Hint 1: Scan for extreme sum targets
This grid is built from single-cell sum regions. The most restrictive are sum-0 and sum-1 cells, which can only accept pips of 0 and 1 respectively. Identify all such cells—they will force the placement of every domino containing a 0 or a 1.
💡 Hint 2: Zero in on the sum-0 cells
The sum-0 cells are at the top right [0,5], far-left middle [3,0], and two in the bottom row: [4,2] and [4,3]. Each must be 0. The dominoes available with a 0 are [0,1], [0,2], [0,3], [0,4], [0,5]. Start by placing the [0,4] domino so that its 0 fills [0,5] and its 4 fills [0,4]—this satisfies the sum-4 next to it.
💡 Hint 3: Lock down the bottom zeros
Next, the sum-0 at [3,0] forces the [0,5] domino to cover it vertically, placing 0 at [3,0] and 5 at [4,0] (satisfying sum-5). Then the two adjacent sum-0 cells [4,2] and [4,3] take [0,2] and [0,3] in a chain: [0,2] with 0 at [4,3], 2 at [3,3]; [0,3] with 0 at [4,2], 3 at [3,2].
💡 Hint 4: Use the sum-1 locks
Now the top rows have several sum-1 cells. The [0,1] domino covers [2,0] (empty) and [1,0] with 1 at [1,0] (sum-1). The top-left sum-1 [0,0] takes the [1,3] domino: 1 at [0,0], 3 at [0,1] (sum-3). Next, the [1,5] domino fills sum-1 at [1,2] with 1 and places 5 at [0,2] (sum-5); the [1,4] domino fills sum-1 at [1,3] with 1 and 4 at [0,3] (sum-4). The [1,2] domino then covers sum-1 at [3,5] with 1 and 2 at [3,4] (sum-2).
💡 Hint 5: Final resolution
Complete the interlocking sums: sum-3 at [1,4] takes [3,5] (3 at [1,4], 5 at [2,4]); sum-3 at [1,5] takes [2,3] (3 at [1,5], 2 at [2,5]); sum-9 pair [2,1]/[2,2] uses [3,4] (3 at [1,1], 4 at [2,1]) and [2,5] (5 at [2,2]); sum-7 pair [2,3]/[2,4] gets 2 from [2,5] at [2,3] and 5 from [3,5] at [2,4]; sum-4 at [3,1] and sum-5 at [4,1] take [4,5] (4 at [3,1], 5 at [4,1]); finally [2,4] domino places 2 at [4,4] (sum-2) and 4 at [4,5] (sum-4).

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Apr 28, 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 28, 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 28, 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: Place the sum-5 anchor
The single-cell sum-5 region at [3,3] must contain exactly 5 pips. The only domino capable of placing a 5 there is [5,6]. Place it so that the 5 lands on [3,3] and the 6 automatically occupies the adjacent cell [3,4].
2
Step 2: Resolve the left equals triplet
The three-cell equals region [1,0], [2,0], [3,0] forces all three to hold the same pip value. The [1,1] domino can place two 1s at [1,0] and [2,0]. The third cell [3,0] still needs a 1, so the [3,1] domino must contribute its 1 there and place its 3 in the empty cell [3,1].
3
Step 3: Lock the right-side equals pair
The equals region [2,4] and [3,4] must match. Since [3,4] already holds a 6 from Step 1, [2,4] must also be 6. This forces the [6,2] domino, placing 6 at [2,4] and 2 at [1,4]—now the top-right equals pair ([0,4]=[1,4]) is satisfied as both will be 2.
4
Step 4: Finish the top row
The sum-5 region at [0,1] needs a 5, so the [4,5] domino fits with 5 at [0,1] and 4 at the empty cell [0,0]. The last unused domino [0,2] then completes the grid: 0 at [0,3] and 2 at [0,4], confirming the equals pair.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Anchor the sum-1 cell
The sum-1 region at [2,0] must be 1. The only domino with a 1 that can cover this cell is [5,1]. Place it vertically, with 1 on [2,0] and 5 on the cell above it, [1,0].
2
Step 2: Propagate the 5 into the top row
[1,0] is now 5 and is part of an equals pair with [1,1]. Therefore [1,1] must also be 5. The [5,2] domino fulfills this, placing 5 at [1,1] and 2 at [0,1].
3
Step 3: Equalize the top 2s
The equals region [0,1] and [0,2] now has 2 at [0,1], forcing [0,2] to 2. The [0,2] domino places 2 at [0,2] and 0 at [0,3].
4
Step 4: Satisfy the less-4 region and vertical equals
The less-4 region covers [0,3] and [0,4] and must be <4. With [0,3]=0, [0,4] must also be <4. The [0,4] domino places a 0 at [0,4] and a 4 at [1,4]. This 4 at [1,4] dictates that the vertical equals partner [2,4] must also be 4, which will come from the [2,4] domino (with its 2 at [2,3]).
5
Step 5: Finalize the middle-row equals triple
The equals triple [2,1],[2,2],[2,3] all take 2s. Use the [2,2] domino to fill [2,1] and [2,2] with 2s; the [2,4] domino (already partially placed) gives 2 at [2,3] and 4 at [2,4]. The rightmost equals pair [1,5],[2,5] receives the [3,3] domino, placing 3s in both cells.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Place the top-right zero
The sum-0 cell at [0,5] must be 0. The [0,4] domino is the only one that can place a 0 here while also satisfying the adjacent sum-4 at [0,4] with a 4. Place it horizontally: 0 at [0,5], 4 at [0,4].
2
Step 2: Zero in column 0
The sum-0 at [3,0] forces a 0. The [0,5] domino covers it vertically, placing 0 at [3,0] and 5 at [4,0], satisfying the sum-5 requirement there.
3
Step 3: Zeros in the bottom row
The two adjacent sum-0 cells [4,2] and [4,3] both require 0. Place [0,2] vertically: 0 at [4,3], 2 at [3,3] (sum-2). Place [0,3] vertically: 0 at [4,2], 3 at [3,2] (sum-3).
4
Step 4: Sum-1 cells on the left and top
The sum-1 at [1,0] needs a 1. Use [0,1] vertically: 0 at [2,0] (empty) and 1 at [1,0]. The sum-1 at [0,0] then forces [1,3] horizontally: 1 at [0,0], 3 at [0,1] (sum-3).
5
Step 5: Middle sum-1 cells and top completion
Sum-1 at [1,2] takes [1,5] horizontally: 1 at [1,2], 5 at [0,2] (sum-5). Sum-1 at [1,3] takes [1,4] horizontally: 1 at [1,3], 4 at [0,3] (sum-4). Sum-1 at [3,5] takes [1,2] horizontally: 1 at [3,5], 2 at [3,4] (sum-2).
6
Step 6: Resolve the remaining interlocking sums
Sum-3 at [1,4] forces [3,5]: 3 at [1,4], 5 at [2,4] (sum-7 part). Sum-3 at [1,5] forces [2,3]: 3 at [1,5], 2 at [2,5] (empty). The sum-9 pair [2,1]/[2,2] uses [3,4] (4 at [2,1]) and [2,5] (5 at [2,2]) with [3,4]'s 3 at [1,1] (sum-3). The sum-7 pair [2,3]/[2,4] gets 2 from [2,5] at [2,3] and 5 from [3,5] at [2,4]. Finally, [4,5] places 4 at [3,1] (sum-4) and 5 at [4,1] (sum-5), and [2,4] places 2 at [4,4] (sum-2) and 4 at [4,5] (sum-4).

💡 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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