NYT Pips Hints & Answers for May 4, 2026

May 4, 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 opens on two independent footholds: a four-cell sum-4 region that forces the double-1 immediately, and a three-cell sum-17 chute that locks in a 6-6-5 triple. These anchors resolve all dominoes without ambiguity.

Livengood's medium puzzle deploys a greater-than-10 gate on the left column, demanding 6-6 from two dedicated 6-carrying dominos, which then feeds a sum-2 anchor and a chain of equals regions. The result is a clean cascade that sweeps across the grid.

Rodolfo Kurchan's hard puzzle is a constraint-dense lattice. The standout feature is a triple-equals column that requires three identical values, satisfied by a double-6 and a 6-4 domino. From there, a series of sum-4 and sum-5 singletons interact with equals and greater-than regions to propagate values outward. This NYT Pips hard rewards careful pip accounting and sharp deduction.

💡 Progressive Hints

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

💡 Spot the low sum
Look for a region whose sum is so small that it can only be achieved by placing the double-? domino. This will immediately narrow the possibilities for adjacent cells.
💡 Anchor on the four-cell sum-4
The sum-4 region at [1,2], [2,1], [2,2], [3,2] demands four pip values that add to 4. The only way to meet this without breaking the domino bank is to plant the [1,1] domino inside it, then force the other two cells to 1s as well.
💡 Complete the chain
Place [1,1] vertically at [1,2]-[2,2]. The sum-17 region then forces [4,6] horizontally at [2,3]-[2,4] (6 in sum, 4 for >3), [3,6] vertically at [0,3]-[1,3] (6 in sum), and [5,1] vertically at [3,3]-[3,2] (5 in sum, 1 in sum-4). Finish with [2,1] vertical at [2,0]-[2,1].
💡 Chase the high sum
A greater-than region restricts its cells to the highest pip values. Identify the pair of cells that must both be 6, and see which dominoes can deliver them.
💡 Left-column lock
The greater-than-10 region at [1,0] and [2,0] forces 6 and 6. The [6,0] domino must go to [1,0]-[0,0] (giving 0 above) and [6,3] to [2,0]-[2,1]. This then sets the top-left sum-2 and the equals pair in motion.
💡 Solve the cascade
With [0,0]=0, [0,1] must be 2 from [2,2] vertical at [0,1]-[1,1]. The equals pushes [2,1] horizontally at [1,2]-[1,3] (2,1). The equals triple forces [1,1] vertical at [1,4]-[2,4]. Finally [4,3] at [0,4]-[0,5] and [3,0] at [1,5]-[2,5] complete the grid.
💡 Crack the triple-equals
Start by locating a column of three cells locked into an equals constraint. The only way to satisfy three identical high pips is with the double-6 and a 6-4 domino.
💡 Sum-4 singles ignite row 1
The isolated sum-4 cell at [2,1] demands a 4. Placing the [4,1] domino there forces a 1 into [1,1], which then turns the adjacent equals region on row 1 into a trio of 1s. This propagates a 1 into [1,0] via the [1,3] domino.
💡 Mid-grid sum-5 meets double-1
The sum-4 singleton at [3,4] forces [3,4]=4 via the [3,4] domino, dropping a 3 into [4,4]. The sum-5 region on [4,4],[5,4],[5,5] then needs 2 more — solved perfectly by the double-1 domino.
💡 Equal zeros and closing sums
The sum-5 at [3,5] pulls in the [5,0] domino (5,0), which forces an equals pair at [3,6]=[4,6]=0 via the [4,0] domino, also satisfying sum-4 at [5,6]. Meanwhile, the [5,4] domino gives 5 to [6,6] and 4 to [7,6].
💡 Final snap
Place [6,6] vertically at [5,0]-[6,0], [6,4] horizontally at [7,0]-[7,1]; [4,1] at [2,1]-[1,1]; [1,3] at [1,0]-[0,0]; [3,4] at [4,4]-[3,4]; [1,1] at [5,4]-[5,5]; [5,0] at [3,5]-[3,6]; [4,0] at [5,6]-[4,6]; [5,4] at [6,6]-[7,6]; [1,5] at [1,2]-[0,2]; [5,3] at [0,4]-[1,4]; [3,6] at [1,5]-[0,5]; [3,3] at [5,2]-[6,2]; [2,4] at [4,0]-[4,1].

🎨 Pips Solver

May 4, 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 May 4, 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 May 4, 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: Lock the low sum
The sum-4 region of four cells can only be satisfied with the double-1 domino, because even a single 2 would require at least two 0s to reach 4. Thus, the [1,1] domino is forced into that region.
2
Step 2: Plant the double-1
Place the [1,1] domino vertically covering [1,2] and [2,2] to inject two 1s. The remaining sum-4 cells [2,1] and [3,2] must also be 1, which will later narrow the available dominoes.
3
Step 3: Tackle the sum-17
The sum-17 region spanning [1,3], [2,3], [3,3] must be 6+6+5 due to pip limits. The only dominoes with 6s are [4,6] and [3,6]; the 5 comes from [5,1]. Thus, [4,6] goes horizontal on [2,3]-[2,4] with 6 in column 3, and [3,6] vertical on [0,3]-[1,3] with 6 in column 3. The 5 from [5,1] lands at [3,3].
4
Step 4: Finish the weave
[5,1] vertical at [3,3]-[3,2] supplies the 5 and the final 1 for sum-4. [2,1] vertical at [2,0]-[2,1] gives 2 in the empty cell and 1 in sum-4. The greater-3 singleton at [2,4] naturally gets the 4 from [4,6].

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Force the sixes
The greater-than-10 region at [1,0] and [2,0] demands both cells be 6. The [6,0] and [6,3] dominoes are the only sources. So [6,0] must cover [1,0] and its neighbor [0,0]; [6,3] covers [2,0] and [2,1] (an empty cell).
2
Step 2: Unlock the sum-2
With [0,0]=0 from the [6,0] domino, the sum-2 region [0,0]-[0,1] forces [0,1]=2. The only way to get a 2 there is using the [2,2] domino, placing it vertically at [0,1]-[1,1] with 2 and 2.
3
Step 3: Equals pours right
The equals region [1,1]-[1,2] now demands both cells be 2 (since [1,1]=2). Thus [1,2]=2, forcing the [2,1] domino horizontally on [1,2]-[1,3] with 2 in column 2 and 1 in column 3.
4
Step 4: Triple ones drop
The equals triple [1,3],[1,4],[2,4] must now all be 1. The [1,1] domino (double-1) fits perfectly at [1,4]-[2,4] vertically, satisfying both cells.
5
Step 5: Cap the grid
Finally, the [4,3] domino covers [0,4]-[0,5] (4 for greater-3, 3 for sum-6), and [3,0] covers [1,5]-[2,5] (3 and 0) to complete sum-6.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Triple-equals column
The equal region at [5,0],[6,0],[7,0] forces all three cells to be 6. The [6,6] domino covers [5,0]-[6,0] vertically, and [6,4] domino covers [7,0]-[7,1] with 6 in column 0, 4 in column 1.
2
Step 2: Sum-4 anchor sets row 1
The sum-4 singleton at [2,1] requires a 4. The [4,1] domino places 4 there and 1 at [1,1]. The equals region [1,0],[1,1],[1,2] then forces all three to 1, so [1,0]=1, [1,2]=1. The [1,3] domino fills [1,0]-[0,0] (1 and 3).
3
Step 3: Mid-grid sum-4 feeds sum-5
The sum-4 singleton at [3,4] demands a 4. The [3,4] domino (3-4) covers [4,4]-[3,4] with 4 in [3,4] and 3 in [4,4], which is part of the sum-5 region [4,4],[5,4],[5,5].
4
Step 4: Double-1 locks the sum-5
With [4,4]=3, the remaining sum-5 cells [5,4] and [5,5] must sum to 2, so both are 1. The [1,1] domino (double-1) fits perfectly there.
5
Step 5: Equal zeros cascade
The sum-5 singleton at [3,5] forces a 5. The [5,0] domino placed at [3,5]-[3,6] gives 5 and 0. The equals region [3,6],[4,6] then forces [4,6]=0, satisfied by the [4,0] domino at [5,6]-[4,6] (4 and 0). The sum-4 singleton [5,6] gets 4 from that domino.
6
Step 6: Final tight fills
Remaining cells lock in: [1,5] domino gives 5 at [0,2] (>4) and 1 at [1,2] (equals). [5,3] domino gives 5 at [0,4] (sum-5) and 3 at [1,4]. [3,6] domino gives 6 at [0,5] (>5) and 3 at [1,5]. [3,3] fills [5,2]-[6,2]. [5,4] gives 5 at [6,6] (sum-5) and 4 at [7,6] (sum-4). [2,4] sets [4,0]=2 (<4) and [4,1]=4 (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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