NYT Pips Hints & Answers for June 4, 2026

Jun 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 is a tight 5x3 layout where only ten cells are active, shaped by five dominoes. The solving graph starts at the single-cell sum-2 region at [0,2], which acts as an immutable root, forcing a specific domino and cascading upward into a sum-8 pair. The remainder of the deduction flows through a sum-15 triple along the left edge, which interacts with an equals pair to lock the final domino orientations. The entire chain is linear, with no branching—every constraint funnels into a single forced path.

Rodolfo Kurchan's medium puzzle spreads across a sparse 4x8 strip, using isolated cells and small clusters. The deduction graph splits after an initial single-cell sum-5 anchor at [1,7] solves the lower-right corner. From there, the sum-15 vertical triple demands a 6+6 pairing, which in turn forces the remaining domino values through a series of equals and sum regions. The equals chain along the top row (cells [0,2]–[0,4]) and the sum-5 pair at [2,7]–[3,7] act as synchronizing points that resolve the grid's last ambiguities. This puzzle rewards solvers who trace connectivity, as many cells only touch one neighbor.

Kurchan's hard puzzle is a dense 9x9 field peppered with single-cell sum-1 and sum-0 regions—six such orphans in total. These instantaneously force six domino placements, scattering 1s and 0s across the board and seeding a rhythmic propagation. The greater-than-10 pair on the top row narrows to a single combination, while layered equals regions along the right and bottom edges form a nested cascade: solving row 4's equals block pushes values into the column-1 equals chain, which then feeds the bottom-right interconnected equals region. The result is a puzzle where every step is deterministic, yet the spatial layout makes the deduction feel like a domino avalanche. Today's NYT Pips hard exemplifies how a profusion of rigid constraints can create an elegant, unstoppable solve.

💡 Progressive Hints

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

💡 Look for lone wolves
Scan for regions that consist of a single cell with a fixed sum. That tiny constraint immediately sets the value of one cell, and the pip is small enough to appear only on a few dominoes.
💡 Trace the top row
The isolated cell at [0,2] demands a 2. Once that 2 claims a domino, the adjacent sum-8 pair [0,0]–[0,1] resolves quickly. Then the sum-15 cluster down the left side gets its first number, forcing a specific domino with a 5.
💡 Fill it in
Place domino [4,2] covering [0,1]=4 and [0,2]=2. Then [4,5] goes [0,0]=4 and [1,0]=5. The sum-15 region then requires [2,0]=5 and [2,1]=5: use [6,5] for [3,0]=6 / [2,0]=5 and [5,1] for [2,1]=5 / [2,2]=1. Finally, [6,2] covers [4,0]=6 and [4,1]=2 to satisfy the equals pair and the last sum-2 cell.
💡 Single-cell beacons
Several regions contain only one cell and a fixed sum. One such sum-5 cell sits at the bottom of the grid; another sum-5 lurks further right. Their forced values activate dominoes that point toward the top row.
💡 Leverage the sum-15 triple
The vertical region [0,0]–[2,0] sums to 15. The only way to reach that total is with a double-6 domino covering two of these cells, which then forces the third to be 3. This 3 must come from the [0,3] domino, placing a 0 next to it.
💡 Complete the weave
Place domino [5,0] at [1,7]=5 and [0,7]=0 to satisfy the sum-5 and the less-than-3 region. Use [6,6] for [0,0]=6 and [1,0]=6, then [0,3] for [3,0]=0 and [2,0]=3. The equals zone [0,2]–[0,4] takes [1,1] (1s) and [1,0] (1 and 0). For the right side, [5,2] goes [3,5]=5 / [3,4]=2; [4,2] goes [2,5]=4 / [2,4]=2; [2,3] goes [2,7]=2 / [3,7]=3. The remaining [0,4] fills [0,6]=0 and [0,5]=4.
💡 Single-cell sweep
The grid is dotted with six sum-1 and two sum-0 single-cell regions. Each of these orphans forces an exact digit—1 or 0—immediately dictating which domino must cover it. Begin by marking all of these forced values.
💡 Top-row pressure
The sum-0 cells at [0,2] and [0,6] claim dominoes with 0, dragging along 4 and 6 into key positions. The adjacent sum-1 cells [0,0] and [0,1] are a dead giveaway for the [1,1] domino. This top row soon reveals a greater-than-10 pair that can only be {5,6} and a sum-10 pair that requires {4,6}.
💡 Midfield sum-10 triples
The sum-10 region spanning [1,6], [2,6], [3,6] interacts with a fixed 6 at [1,6] (from the earlier [0,6] domino) and a sum-1 at [2,7] that donates a 2 to [2,6]. That leaves [3,6] needing a 2, which pulls in the [6,2] domino and places a 6 at [4,6].
💡 Equals avalanche
Once [4,6] is 6, the entire three-cell equals region on row 4 becomes 6s, forcing the [6,6] domino at [4,7]–[4,8]. Meanwhile, the sum-1 cell at [5,1] feeds a 1, giving [6,1] a 4 that then propagates down the column-1 equals chain, placing the [4,4] domino. The sum-1 at [5,4] injects a 1 and 3, locking the large interconnected equals zone in the lower-right.
💡 Final placements
Place domino [1,1] at [0,0]–[0,1] (1/1). [4,0] at [0,2]=0 / [1,2]=4. [5,6] at [0,3]=5 / [0,4]=6. [0,6] at [0,6]=0 / [1,6]=6. [4,6] at [0,8]=4 / [0,7]=6. [2,1] at [2,6]=2 / [2,7]=1. [6,1] at [2,2]=6 / [3,2]=1. [1,4] at [5,1]=1 / [6,1]=4. [1,3] at [5,4]=1 / [6,4]=3. [6,2] at [4,6]=6 / [3,6]=2. [6,6] at [4,7]=6 / [4,8]=6. [4,4] at [7,1]=4 / [8,1]=4. [6,3] at [6,2]=6 / [7,2]=3. [3,3] at [7,3]=3 / [7,4]=3. [5,5] at [8,3]=5 / [8,4]=5.

🎨 Pips Solver

Jun 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 June 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 June 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: The lone sum-2 cell [0,2]
The region at [0,2] is a single cell with sum target 2. Only a 2 can occupy it. The only domino containing a 2 is [4,2], so that domino must cover [0,2] and an adjacent cell. The only neighbor is [0,1] (to the left). Therefore [0,1]=4, [0,2]=2.
2
Step 2: Sum-8 pair forces [0,0]
The region [0,0]–[0,1] sums to 8. With [0,1]=4, we get [0,0]=4. [0,0] is adjacent only to [1,0] below. The remaining domino with a 4 is [4,5], so it must cover [0,0]=4 and [1,0]=5.
3
Step 3: Sum-15 triple along the left
The region [1,0], [2,0], [2,1] must sum to 15. We already have [1,0]=5, so the remaining two cells need to sum to 10. They are not adjacent to each other horizontally, because the only domino covering them would sum to something not matching 10. Therefore each must pair vertically with cells below: [2,0] with [3,0], and [2,1] with [2,2]. The available dominoes include [6,5] and [5,1]. The only way to reach sum 10 from two different dominoes is with a 5 in each position: [2,0]=5 and [2,1]=5. Thus [6,5] places 5 at [2,0] and 6 at [3,0]; [5,1] places 5 at [2,1] and 1 at [2,2].
4
Step 4: Equals pair and final sum-2
The equals region [3,0]–[4,0] forces both cells to be the same. [3,0]=6, so [4,0]=6. The only remaining domino with a 6 is [6,2], which also carries a 2. It covers [4,0]=6 and [4,1]=2. [4,1] is a single-cell sum-2 region, which matches perfectly.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Single sum-5 cell [1,7]
The cell [1,7] is a single-cell sum-5 region, so its value must be 5. The only dominoes containing a 5 are [5,0] and [5,2]. [1,7] has only one neighbor: [0,7] above. The [5,0] domino fits, giving [1,7]=5 and [0,7]=0. The [0,7] cell belongs to a less-than-3 region with [0,6], which will later constrain [0,6].
2
Step 2: Sum-15 triple demands double-6
The vertical region [0,0], [1,0], [2,0] sums to 15. The highest possible values are 6,6,3 (since 6+6+3=15). Two 6s must come from the [6,6] domino, covering [0,0] and [1,0] (the only two vertically adjacent cells). So [0,0]=6, [1,0]=6, leaving [2,0]=3. The [0,3] domino supplies the 3 at [2,0] and 0 at its neighbor [3,0].
3
Step 3: Top-row equals chain
Region [0,2]–[0,4] is an equals trio. The [1,1] domino, which holds two 1s, can cover [0,3] and [0,4] (the only adjacent pair in that row). Thus all three cells must be 1, forcing [0,2]=1. The [1,0] domino then covers [0,2]=1 and [0,1]=0.
4
Step 4: Right-side sum-5 and equals propagation
Single-cell sum-5 at [3,5] forces 5 there. The [5,2] domino covers [3,5]=5 and [3,4]=2. The equals pair [2,4]–[3,4] then forces [2,4]=2, which is covered by the [4,2] domino with its 2, putting 4 at [2,5]. The sum-5 region [2,7]–[3,7] takes the [2,3] domino (2+3=5).
5
Step 5: Cleanup
The less-than-3 region [0,6]–[0,7] already has [0,7]=0; any subsequent placement for [0,6] must be <3. The remaining domino [0,4] covers [0,6]=0 and [0,5]=4, satisfying the constraints. All cells are filled.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Anchor the single-cell sums
Six regions each contain one cell with sum 1: [0,0], [0,1], [2,7], [3,2], [5,1], [5,4]. Each must be exactly 1. Two regions contain one cell with sum 0: [0,2] and [0,6]. These force immediate domino attachments. [0,0] and [0,1] are adjacent, so the [1,1] domino covers both. [0,2] takes [4,0]’s 0, giving [1,2]=4. [0,6] takes [0,6]’s 0, giving [1,6]=6. [2,7] takes [2,1]’s 1, giving [2,6]=2. [3,2] takes [6,1]’s 1, giving [2,2]=6. [5,1] takes [1,4]’s 1, giving [6,1]=4. [5,4] takes [1,3]’s 1, giving [6,4]=3.
2
Step 2: Top-row high-sum pairs
The greater-than-10 region [0,3]–[0,4] can only be 5+6 or 6+5. The [5,6] domino fits, setting [0,3]=5, [0,4]=6. The sum-10 pair [0,7]–[0,8] must sum to 10; the [4,6] domino provides 4+6, so [0,7]=6, [0,8]=4.
3
Step 3: The triple sum-10 in column 6
Region [1,6],[2,6],[3,6] sums to 10. We have [1,6]=6 from Step 1, so [2,6]+[3,6]=4. [2,6] is already 2 (from [2,1] domino), so [3,6]=2 forced. The [6,2] domino covers [4,6]=6 and [3,6]=2.
4
Step 4: Row-4 equals cascade
Region [4,6],[4,7],[4,8] is equal. [4,6]=6 from Step 3, so all are 6. The [6,6] domino covers [4,7]=6 and [4,8]=6.
5
Step 5: Column-1 equals and the bottom-right interconnected equals
Equals region [6,1],[7,1],[8,1]: [6,1]=4 from Step 1, so [7,1]=4 and [8,1]=4. The [4,4] domino covers those. The large equals region [6,4],[7,2],[7,3],[7,4] is all equal; [6,4]=3 from Step 1, so all are 3. [7,2] gets 3 from the [6,3] domino (with [6,2]=6). [7,3] and [7,4] get 3 from the [3,3] domino. The equals region [8,3],[8,4] then takes the [5,5] domino (5 and 5).
6
Step 6: Final check
All constraints are satisfied with the given placements. The empty cell at [6,2] ends up with 6, which is unrestricted. The cascade from single sums through high-value targets and interconnected equals chains produces a fully determined grid.

💡 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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