NYT Pips Hints & Answers for May 28, 2026

May 28, 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 for today's NYT Pips deploys a set of greater‑sum constraints, where a region's total must exceed a target rather than requiring every cell to be individually greater. The deduction graph opens with two isolated single‑cell greater‑sum regions — [0,4] (target >3) and [1,2] (target >2) — each forcing a high pip from the available pool. Their placements in turn activate an equals pair at [2,2]‑[2,3] via a shared 1, which cascades into the remaining equal‑cell regions and the final greater‑sum region at [3,4]‑[3,5].

Rodolfo Kurchan's medium puzzle weaves two unequals regions across rows 1–4, creating a web of forced distinct values. The chain begins with a sum‑4 singleton at [2,1] that locks a 4 on a double‑4 domino, while the equals pair at [3,1] and [4,1] latches onto the number 2, splitting it across two different dominoes. A sum‑5 singleton at [2,5] then fixes a double‑5, after which the unequals constraints slot the remaining dominoes into precise positions based on value exclusion.

For the hard, Kurchan constructs a grid around two massive equals clusters — a 2×2 block in the northwest and a 5‑cell equals bar on the eastern edge — with a long unequals column slicing through the center. The solve is anchored by the equal‑pip dominoes [3,3] and [3,6], which seed the top‑left block and the adjacent sum‑6 cell. From there, a greater‑1 cell forces a 2 that splits a [3,2] domino, a sum‑6 cell forces a 6 that feeds into a sum‑6 pair, and the right‑side equals row demands a synchronized cascade of [0,6], [6,6], [4,6], and [6,1] before bottom equals and a sum‑3 close the grid.

💡 Progressive Hints

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

💡 Hint 1
Look for regions whose sum must exceed a target number (greater‑sum constraints). Single‑cell greater‑sum spots will instantly dictate specific pip values.
💡 Hint 2
The isolated greater‑3 cell at [0,4] demands a domino with a pip greater than 3; only the [4,0] and [4,4] fit. Meanwhile, the greater‑2 cell at [1,2] forces another high pip.
💡 Hint 3 (final answer)
Place the [4,0] domino horizontally at [0,4]‑[0,5] (4 in [0,4], 0 in [0,5]). Then place [3,1] vertically at [1,2]‑[2,2] (3 in [1,2], 1 in [2,2]), which activates the equals pair at [2,2]‑[2,3]: both must be 1, so [2,1] goes at [1,3]‑[2,3] (2,1). The greater‑4 region [0,0]‑[0,1] takes [3,3] (both 3). The equals region [3,0]‑[3,1] gets [2,2] (both 2). Finally, the greater‑6 region [3,4]‑[3,5] uses [4,4] (both 4).
💡 Hint 1
Kurchan's medium puzzle features two overlapping unequals regions that force distinct values across several rows. Pay attention to the single‑cell sum regions – they lock in exact pips immediately.
💡 Hint 2
The sum‑4 cell at [2,1] must be a 4, so the domino covering it must supply a 4. Simultaneously, the equals pair at [3,1] and [4,1] demands a shared value, which can only come from the [2,5] and [2,2] dominoes.
💡 Hint 3 (final answer)
Place [4,4] horizontally at [2,1]‑[2,2] to satisfy the sum‑4 cell with a 4, putting 4 in both. The equals pair needs [3,1] and [4,1] to match: put [2,5] vertically at [3,0]‑[3,1] (5 in [3,0], 2 in [3,1]). For [4,1], place [2,2] horizontally at [4,1]‑[4,2] (both 2). The sum‑5 cell at [2,5] takes [5,5] vertically at [1,5]‑[2,5] (both 5). The unequals regions are completed with [4,3] at [0,4]‑[1,4] (4,3), [5,6] at [1,3]‑[2,3] (5,6), and [3,3] at [3,2]‑[3,3] (3,3).
💡 Hint 1
Kurchan's hard puzzle is anchored by a large equals block in the top‑left and a sprawling equals row on the right, with a long unequals column weaving between them. Single sum and greater/less cells on the edges provide the initial toeholds.
💡 Hint 2
The top‑left equals block (cells [0,0],[0,1],[1,0],[2,0]) must all hold the same pip. Available equal‑pip dominoes are [3,3], [0,0], [1,1], and [6,6]. The adjacent sum‑6 cell at [0,2] will constrain the choice.
💡 Hint 3
Placing [3,3] vertically at [0,0]‑[1,0] sets two of the equals cells to 3. Then [0,1] must also be 3; the only way is to use the [3,6] domino horizontally at [0,1]‑[0,2] with 3 and 6, which also satisfies the sum‑6 at [0,2]. Now [2,0] still needs 3, and the greater‑1 cell at [2,1] requires a pip >1.
💡 Hint 4
Place [3,2] at [2,0]‑[2,1] horizontally, giving 3 to [2,0] and 2 to [2,1]. The sum‑6 at [2,2] then forces a 6; use [6,5] vertically at [2,2]‑[3,2] (6,5). This starts the sum‑6 pair at [3,2],[4,2] where [3,2]=5, so [4,2] must be 1, provided by [1,0] at [4,2]‑[5,2] (1,0). Meanwhile, the right‑side equals row starts with [0,6] at [0,6]‑[1,6] (0,6) and [6,6] at [1,7]‑[2,7] (6,6).
💡 Hint 5 (final answer)
Complete the grid: [4,1] at [0,4]‑[1,4] (1,4) satisfies less‑3 and unequals. [4,6] at [0,8]‑[1,8] (4,6) fulfills greater‑3 and adds to the right equals. [6,1] at [3,7]‑[4,7] (6,1) extends the equals row. [0,0] at [5,0]‑[5,1] (0,0) starts bottom equals; [1,0] already covered; [1,1] at [4,8]‑[5,8] (1,1) finishes bottom equals. [2,6] at [4,4]‑[5,4] (2,6) completes unequals column, and [1,3] at [4,6]‑[5,6] (1,3) ends with sum‑3 at [5,6]=3.

🎨 Pips Solver

May 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 May 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 May 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: Feed the singleton greater‑3
Locate the single‑cell greater‑3 at [0,4] — it must have a pip >3, so the only dominoes with a value greater than 3 are [4,4] and [4,0]. Since [0,5] is empty and can absorb a 0, place the [4,0] domino horizontally at [0,4]‑[0,5] with 4 in [0,4] and 0 in [0,5].
2
Step 2: Satisfy the greater‑2 cell
The greater‑2 cell at [1,2] needs a pip >2. Domino [3,1] can supply the 3, so place it vertically at [1,2]‑[2,2], giving 3 at [1,2] and 1 at [2,2].
3
Step 3: Activate the equals pair
Now the equals region at [2,2]‑[2,3] forces [2,3] to equal [2,2]'s 1. The only remaining domino with a 1 is [2,1]; place it vertically at [1,3]‑[2,3] with 2 above and 1 below, satisfying the equality.
4
Step 4: Close the grid with greater‑sums
The greater‑4 region at [0,0]‑[0,1] requires sum >4. The [3,3] domino yields 3+3=6, so place it horizontally there. The equals region [3,0]‑[3,1] takes [2,2] vertically (both 2). Finally, the greater‑6 region [3,4]‑[3,5] uses [4,4] horizontally for two 4s (sum 8).

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Lock the equals pair
The equals pair at [3,1] and [4,1] must share the same value. Only the [2,5] and [2,2] dominoes contain a 2, so both cells will be 2. Place [2,5] vertically at [3,0]‑[3,1], putting 5 in [3,0] and 2 in [3,1].
2
Step 2: Fulfill the sum‑4 singleton
The sum‑4 cell at [2,1] must hold exactly 4. The domino [4,4] can cover [2,1] and its right neighbor [2,2] with 4 in both, so place it horizontally there.
3
Step 3: Fix the sum‑5 cell
The sum‑5 cell at [2,5] forces a 5. Use the [5,5] domino vertically at [1,5]‑[2,5], giving 5 in both cells.
4
Step 4: Complete the equals pair
Now [4,1] still needs a 2 to match [3,1]. Place [2,2] horizontally at [4,1]‑[4,2] (both 2), completing the equals constraint and introducing a 2 into the large unequals region.
5
Step 5: Resolve the unequals regions
Fill the unequals regions: place [4,3] at [0,4]‑[1,4] (4 and 3), [5,6] at [1,3]‑[2,3] (5 and 6), and [3,3] at [3,2]‑[3,3] (two 3s). All region constraints now check out.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Anchor the top‑left equals block
The top‑left equals block of four cells ([0,0], [0,1], [1,0], [2,0]) must all be equal. The equal‑pip dominoes [3,3] and [0,0] are the only candidates that won't conflict with the adjacent sum‑6 cell [0,2]. Place [3,3] vertically at [0,0]‑[1,0], setting both to 3.
2
Step 2: Feed equals and sum‑6 together
Cell [0,1] must also be 3. The [3,6] domino can provide the 3 while placing 6 in [0,2] to satisfy its sum‑6. Place it horizontally at [0,1]‑[0,2] (3 and 6).
3
Step 3: Fill the last equals cell and the greater‑1
The remaining equals cell [2,0] needs a 3, and the greater‑1 cell [2,1] must be >1. The [3,2] domino gives both: place it horizontally at [2,0]‑[2,1] (3 in [2,0], 2 in [2,1]).
4
Step 4: Build the sum‑6 chain
Cell [2,2] is a sum‑6 singleton, so it must be 6. Use [6,5] vertically at [2,2]‑[3,2] (6 top, 5 bottom). This also kicks off the sum‑6 pair at [3,2] and [4,2]: since [3,2]=5, [4,2] must be 1. Achieve that by placing [1,0] horizontally at [4,2]‑[5,2] (1 and 0).
5
Step 5: Erect the right‑side equals row
Build the right‑side equals row: [0,6] goes at [0,6]‑[1,6] (0 and 6) to put 0 in the less‑3 cell and 6 in the first equals cell. Then [6,6] at [1,7]‑[2,7] (both 6), [4,6] at [0,8]‑[1,8] (4 and 6) for the greater‑3 at [0,8]=4 and to extend equals, and [6,1] at [3,7]‑[4,7] (6 and 1) to fill the remaining equals cells.
6
Step 6: Close with bottom equals and remaining cells
Complete with the bottom equals and remaining constraints: [0,0] at [5,0]‑[5,1] (two 0s) creates the bottom‑left equals; [1,1] at [4,8]‑[5,8] (two 1s) finishes the bottom‑right equals; [2,6] at [4,4]‑[5,4] (2 and 6) fills the unequals column; and [1,3] at [4,6]‑[5,6] (1 and 3) yields 3 in the sum‑3 cell [5,6].

💡 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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