NYT Pips Hints & Answers for June 1, 2026

Jun 1, 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

Today’s NYT Pips easy, constructed by Ian Livengood, greets you with an immediate gift: a sum‑3 region of three cells that can only be three ones. The grid is small and the constraints—two equals, a less‑5, and open cells—unfold in a tidy domino chain. It’s a quick, satisfying solve where each placement telegraphs the next.

The medium puzzle, also from Livengood, introduces a large unequal region that demands six different numbers. This forces you to manage a tight set of available pips across two rows. Sum‑2, sum‑10, and sum‑12 regions provide rigid anchors, and you’ll find yourself eliminating candidates until the unequal cells slot into a perfect sequence. The solve has a nice rhythm of small clusters locking before the bottom falls into place.

Rodolfo Kurchan’s hard puzzle is a dense tangle of constraints: sum‑0 corners, a sum‑10 pair, a four‑cell equals block, and a sprawling five‑cell equals web near the bottom. As you tease out the first zeros, the puzzle cascades—a 5 here, a 4 there, and suddenly the unequal region forces a specific set of values. The equals quintet clinches the mid‑grid, and the bottom row resolves with a trio of zeros and a lone 4. It’s a mind‑bending but elegant lock‑and‑key design.

💡 Progressive Hints

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

💡 Think Small
A tiny sum is your entry point. Look for a region where the total is so low that only the smallest pips can possibly add up to it.
💡 Triple Ones
The three‑cell sum‑3 region in the upper left (cells [0,1], [0,2], [1,1]) demands all 1s. The [1,1] domino has to cover two of them—place it there, and then find a way to supply the third 1.
💡 Full Unlock
Place the [1,1] domino across [0,1] and [0,2]. Then the third 1 for [1,1] must come from the [1,2] domino, so place it vertically with 1 at [1,1] and 2 at [2,1]. This forces the equals pair [2,1]/[3,1] to both be 2, which pulls in [6,2] (2 at [3,1], 6 at [3,0]). The equals pair [1,3]/[2,3] now needs a match: [5,3] goes vertically (3 at [2,3], 5 at [2,2]), then [3,4] completes it (3 at [1,3], 4 at [0,3] — satisfying less‑5).
💡 All Different
A large 'unequal' region demands six distinct numbers. This heavy restriction will ripple through the entire grid. Start by finding a small sum that can’t escape a specific domino.
💡 A Pair of Ones
The sum‑2 region at [1,0]/[2,0] can only be satisfied by the [1,1] domino. Place it vertically. Now the unequal region can’t use 1, forcing its six cells to be chosen from the other six possible pips.
💡 Chain of Sums
With [1,1] set, the sum‑10 region [3,0]/[3,1] demands [5,5] (both 5). The sum‑12 region [3,3]/[4,3] needs a 6 at [4,3] and a 6 at [3,3], so place [6,6] vertically at [2,3]/[3,3] and later [1,6] at [4,2]/[4,3] (1,6). Fill the unequal set: [3,3] places 3,3 at [0,0]/[0,1] giving [0,1]=3; [1,5] covers [0,2]/[0,3] with 1 and 5; [4,0] covers [1,2]/[1,3] with 4 and 0. Finally, [1,0] supplies 1 at [4,1] and 0 at [4,0] for the last sum‑2.
💡 Zero Sum
The very top‑left region demands a sum of zero across two cells. Since pips can’t be negative, both cells must be 0. This will anchor your entire start.
💡 Double Fives Below
With zeros in [0,0] and [0,1], the two cells directly below them ([1,0] and [1,1]) must sum to 10. You’ll need to split two 5s here, using a vertical domino that also delivers a zero above.
💡 Square of Fours
The 2×2 equals block on the right ([0,3],[0,4],[1,3],[1,4]) can only be all 4s. The [4,4] domino handles the top row, and a clever use of [5,4] feeds a 4 to the bottom left of the square while also supplying a 5 to a sum‑10 pair.
💡 The Quintet Value
The large 5‑cell equals region near the bottom (cells [4,4],[5,1],[5,2],[5,3],[5,4]) needs the same number in every one. The only pip that appears often enough among the dominos to cover all five is 2. Start lining up the [2,2] domino and others to satisfy this equality.
💡 Full Solve
Top‑left sum‑0: place [5,0] vertically at [0,0]/[1,0] (0,5) and [2,0] horizontally at [0,2]/[0,1] (2,0). Sum‑10: [3,5] at [1,2]/[1,1] (5 on [1,1], 3 on [1,2]). Equals block: [4,4] at [0,3]/[0,4]; [5,4] at [2,4]/[1,4] (4 on [1,4], 5 on [2,4]); [4,0] at [1,3]/[2,3] (4,0) completing the block and sum‑0 cell. Unequal: [1,4] at [2,0]/[2,1] (1,4); [1,6] at [2,2]/[3,2] (1,6). Sum‑3 & sum‑2: [1,3] at [4,0]/[3,0] (3,1); [1,2] at [5,0]/[5,1] (1,2). Equals quintet all 2s: [2,2] at [5,2]/[5,3]; [2,5] at [4,4]/[3,4] (2,5); [4,2] at [6,4]/[5,4] (4,2). Bottom row: [0,3] at [6,1]/[6,0] (0,3); [0,0] at [6,2]/[6,3] (0,0); last cell [6,4]=4 from [4,2].

🎨 Pips Solver

Jun 1, 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 1, 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 1, 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: Three Ones Are a Must
The sum‑3 region covering [0,1], [0,2], and [1,1] can only be satisfied by three 1s, since 0+0+3 or 0+1+2 would break the small target. You immediately know that two of these cells must be filled by the [1,1] domino, the only piece with two ones.
2
Step 2: Completing the Trio
Place the [1,1] domino on [0,1] and [0,2]. The remaining cell [1,1] still needs a 1, so the [1,2] domino must be placed vertically with the 1 at [1,1] and the 2 at [2,1]. This activates the equals region [2,1]/[3,1], which now forces [3,1] to also be a 2.
3
Step 3: Equal Pairs Stack
The 2 on [3,1] can only come from the [6,2] domino, so place it horizontally with 2 at [3,1] and 6 at [3,0] (an empty cell). Now look at the other equals region [1,3]/[2,3]. The available dominos with a matching pair are [5,3] and [3,4]. Since [2,3] is vertically above an empty cell, a 3 there allows the [5,3] to give 5 to [2,2]. Place [5,3] vertically (3 at [2,3], 5 at [2,2]). Then [1,3] must get a 3 from the [3,4] domino, which also places a 4 at [0,3].
4
Step 4: Final Check
The less‑5 cell [0,3] happily takes the 4, and all dominos are placed. The grid resolves cleanly with the empty cells absorbing the high pips 5 and 6.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Unequal Foundation
A large unequal region of six cells forces all its pips to be distinct. The small sum‑2 region at [1,0]/[2,0] can’t escape the [1,1] domino, so place it vertically, giving 1,1. Now the unequal region cannot use the number 1, leaving only the values {0,2,3,4,5,6} for its six cells.
2
Step 2: Sums of Ten and Twelve
The region at [3,0]/[3,1] sums to 10, so the only domino that can fit is [5,5] (both 5). Place it there. Next, the sum‑12 region at [3,3]/[4,3] requires two 6s. The [6,6] domino placed vertically at [2,3]/[3,3] supplies one 6, and the [1,6] domino later will cover [4,2]/[4,3] (1,6) to complete the 12.
3
Step 3: Populating the Unequal Set
The empty cell [0,0] and unequal cell [0,1] can be filled by the [3,3] domino, placing 3 at both. This gives [0,1]=3 for the unequal group. The top‑right unequal cells [0,2] and [0,3] then get [1,5] (1 and 5). The final unequal cells [1,2] and [1,3] are completed by [4,0] (4 and 0). The unequal region now holds {3,1,5,4,0,6} — all distinct.
4
Step 4: Bottom‑Left Sum‑2
The remaining sum‑2 region [4,1]/[4,2] needs 1+1, but the [1,1] domino is used. The [1,0] domino steps in: place it with 1 at [4,1] and 0 at [4,0] (an empty cell).
5
Step 5: Validation
Every domino finds a home, the unequal region remains all‑different, and all sum targets are met. The puzzle’s interlocking regions leave no loose ends.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Zeros at the Top
The sum‑0 region [0,0]/[0,1] forces both cells to be 0. Since no single domino with two zeros is used here, you must split the duty: [5,0] vertically at [0,0]/[1,0] puts 0 on [0,0] and 5 on [1,0]; [2,0] horizontally at [0,2]/[0,1] puts 2 on [0,2] and 0 on [0,1]. This also starts the unequal region with [0,2]=2.
2
Step 2: Stacking Fives
The sum‑10 region [1,0]/[1,1] already has a 5 from [1,0]; the only way to reach 10 is a second 5 at [1,1]. The [3,5] domino fits perfectly at [1,2]/[1,1] (5 on [1,1], 3 on [1,2]), giving the unequal region its 3.
3
Step 3: The Four‑Square Block
The equals region [0,3],[0,4],[1,3],[1,4] must be all the same. The [4,4] domino placed on [0,3]/[0,4] sets the value to 4. To fill [1,4] without a 4‑containing domino, [5,4] goes to [2,4]/[1,4] (4 on [1,4], 5 on [2,4]). The last cell [1,3] gets its 4 from the [4,0] domino (0 on [2,3], 4 on [1,3]), which also satisfies the sum‑0 cell [2,3].
4
Step 4: Completing the Unequal Chain
The sum‑1 cell [2,0] demands a 1. The [1,4] domino at [2,0]/[2,1] gives 1 on [2,0] and 4 on [2,1] (unequal). Now the unequal region needs a final distinct value at [2,2]; [1,6] placed at [2,2]/[3,2] gives 1 on [2,2] (distinct) and 6 on [3,2], fulfilling the sum‑6 cell [3,2].
5
Step 5: Lower‑Left Sums and the Quintet
The sum‑3 cell [3,0] gets 3 from [1,3] at [4,0]/[3,0] (3 on [3,0], 1 on [4,0]). The sum‑2 region [4,0]/[5,0] then gets the needed 1 from [4,0] and a 1 at [5,0] via [1,2] (1 on [5,0], 2 on [5,1]). The 5‑cell equals region now reveals its value as 2: [5,1]=2, so all cells must be 2. The [2,2] domino covers [5,2]/[5,3] (2,2); [2,5] goes to [4,4]/[3,4] giving 2 on [4,4] and 5 on [3,4] (solving sum‑10 pair [2,4]/[3,4] since [2,4] already 5). The last equals cell [5,4] gets 2 from [4,2] domino.
6
Step 6: Bottom Row Wrap‑Up
The sum‑3 cell [6,0] takes 3 from the [0,3] domino at [6,1]/[6,0] (0 on [6,1], 3 on [6,0]). The sum‑0 area [6,1],[6,2],[6,3] is satisfied: [6,1]=0, and the [0,0] domino supplies 0,0 to [6,2]/[6,3]. Finally, sum‑4 cell [6,4] gets 4 from the same [4,2] domino (4 on [6,4], 2 on [5,4]). All constraints are met.

💡 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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