NYT Pips Hints & Answers for May 16, 2026

May 16, 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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🎲 Today's Puzzle Overview

Ian Livengood's easy grid opens on two independent footholds that quickly resolve the whole board. A triple-equals region spanning [0,2], [1,2], and [1,3] forces the [1,1] domino, distributing 1's across the top-right corner. Immediately below, a greater‑10 region on two cells in row 2 demands a sum exceeding 10; the only way to satisfy it uses the [6,2] and [5,1] dominoes in a cascade that also feeds the less‑3 cell and the bottom‑left equals triplet. The deduction graph is shallow but tightly interlocked, with each placement unlocking the next.

Rodolfo Kurchan's medium puzzle packs a row of sum constraints along row 0 that act like a combination lock. The leftmost sum‑6 pair, sum‑2 pair, sum‑10 pair, and an equals pair are all forced by the available pip values, but interestingly none of them is satisfied by a single domino. Instead, solo sum‑5 cells at [3,2] and [3,3] crack the puzzle first—only the [5,5] domino covers both—while a sum‑2 singleton at [2,0] pulls in the [1,2] domino. From there the top row unfolds as a precise domino‑sharing chain, with each region borrowing values from a neighboring domino placed in a crossing orientation. The result is a compact 4×8 solving graph with no unused slack.

Rodolfo Kurchan's hard grid is a 10×5 battlefield centered on a sprawling equals region that forces six separate cells to zero. Many individual cells carry fixed sum or inequality constraints—sum‑1 at [0,0], sum‑2 at [1,1], sum‑3 at [2,2], sum‑4 at [3,3], and so on—each locking a pip value early. The zero block at [5,0]–[5,3], [6,1], and [7,1] then becomes a distribution hub: dominoes carrying a 0 must be placed to hit each of those cells while simultaneously satisfying the adjacent sum‑4 and unequal regions. Solving this NYT Pips hard is an exercise in resource mapping, as you allocate zero‑bearing dominoes to exactly the right squares and then fill the non‑zero distinct digits of the unequal set. The final steps reconcile a sum‑9 vertical pair and an equals‑6 duo, completing a dense, satisfying deduction graph.

💡 Progressive Hints

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

💡 Hint 1: Spot the locks
Look for a region where every cell must share the same pip value, and another region that demands an unusually large total. These two constraints will drive your opening moves.
💡 Hint 2: Pinpoint the zones
The equals region sits in the top right, covering three cells including [0,2] and [1,3]. Directly below it, a two‑cell greater‑10 region occupies row 2, columns 2 and 3. The triple‑equals forces a specific double‑pip domino, while the greater‑10 needs a 6 plus something large enough to push the sum over 10.
💡 Hint 3: Full solve
Place domino [1,1] vertically at [0,2] and [1,2] to satisfy the equals region, making [1,3] also 1. Then set [6,2] vertically at [2,2]/[3,2] to put a 6 in the greater‑10 cell, followed by [5,1] horizontally at [2,3]/[1,3] for a 6+5=11 sum. Next, use [3,2] vertically at [2,1]/[1,1] to place 2 (less than 3) at [1,1] and 3 at [2,1]. Finally, [3,3] goes horizontally at [2,0]/[3,0], completing the bottom equals trio with all 3s.
💡 Hint 1: Single‑cell summons
Several cells carry their own sum or less/greater rule. Hunt for those that demand a specific pip value—especially cells that must equal a fixed number because of sum constraints on a single square.
💡 Hint 2: Double‑5 anchor
Two adjacent cells at [3,2] and [3,3] each have a sum‑5 rule, meaning both must be exactly 5. Only one domino can cover them both at once. Meanwhile, the isolated sum‑2 at [2,0] forces that cell to be 2, and the only way to supply it is with a domino carrying a 2 that also reaches the empty neighbor below.
💡 Hint 3: Full cascade
Place [5,5] horizontally at [3,2]/[3,3]. Put [1,2] vertically with 2 at [2,0] and 1 at [3,0]. The sum‑2 region [0,2]/[0,3] must be 1+1; use [4,1] horizontally at [0,1]/[0,2] (4/1) and part of [6,1] later. The sum‑6 pair [0,0]/[0,1] gets 2 from [2,6] vertically at [0,0]/[1,0] (6 at [1,0] satisfies its sum‑6). For sum‑10 at [0,4]/[0,5], place [6,1] horizontally at [0,4]/[0,3] (6 at [0,4]) and [3,4] horizontally at [0,6]/[0,5] (4 at [0,5], 3 at [0,6]). The equals pair [0,6]/[0,7] becomes 3 and 3; put [5,3] vertically at [1,7]/[0,7] (3 at [0,7], 5 at [1,7]). Finish with [4,6] horizontally [2,3]/[2,2] (6>5,4<5) and [6,5] vertically [2,7]/[3,7] (6 at [2,7] completes sum‑11, 5 at [3,7] satisfies sum‑5).
💡 Hint 1: Lock down the singles
Many cells are governed by a sum or less/greater constraint all by themselves. Identify which cells get a hard‑coded pip number before thinking about domino placement—they will anchor the entire grid.
💡 Hint 2: Zero block and early values
A huge equals region covering [5,0]–[5,3], [6,1], and [7,1] forces all those cells to 0. Nearby, [1,1] must be 2 (sum‑2), [2,2] must be 3 (sum‑3), [4,4] and [5,4] each must be 5. These fixed values will guide your first domino placements.
💡 Hint 3: First forced moves
To satisfy [1,1]=2, use the [2,3] domino vertically over [1,1]/[2,1] (2 and 3). For [2,2]=3, place [4,3] vertically at [3,2]/[2,2] (4 at [3,2] for later). The two sum‑5 cells are adjacent; cover them together with the [5,5] domino horizontally at [4,4]/[5,4].
💡 Hint 4: Populate the zero web
Now feed the equals‑zero region. Place [0,0] horizontally at [5,1]/[6,1] to claim two zeros. Put [4,0] vertically at [6,3]/[5,3]—it gives 4 to sum‑4 cell [6,3] and 0 to [5,3]. Place [0,3] vertically at [5,0]/[4,0] to drop 0 at [5,0] and 3 at [4,0] (for the unequal set). Then use [2,0] vertically at [8,1]/[7,1] to put 2 at sum‑2 cell [8,1] and 0 at [7,1].
💡 Hint 5: Complete grid solution
Continue by installing [2,5] horizontally at [3,1]/[4,1] (2 and 5) and [6,0] vertically at [4,2]/[5,2] (6 and 0), finishing the unequal set {2,4,3,5,6} with all distinct values. The sum‑9 region [1,0]/[2,0] is handled by [4,1] vertically [1,0]/[0,0] (4 and 1, giving 1 at sum‑1 [0,0]) and [0,5] vertically [3,0]/[2,0] (5 at [2,0], 0 at [3,0] under less‑2). For the equals‑6 at [7,0]/[8,0], place [6,2] vertically [7,0]/[6,0] (6 and 2, with 2 satisfying less‑3 at [6,0]) and [1,6] vertically [9,0]/[8,0] (1 at sum‑1 [9,0] and 6 at [8,0]). Finally, [3,6] goes horizontally [7,2]/[6,2] (3 at sum‑3 [7,2], 6 at empty [6,2]), and [2,4] horizontally [4,3]/[3,3] (2 at less‑3 [4,3], 4 at sum‑4 [3,3]).

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May 16, 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 16, 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 16, 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: Triple‑equals lock
The equals region at [0,2], [1,2], [1,3] requires all three cells to have identical pips. The only available domino with matching numbers is [1,1], so it must cover two of these cells. Place [1,1] vertically across [0,2] and [1,2], setting both to 1. By the constraint, [1,3] also becomes 1.
2
Step 2: Greater‑10 demand
The greater‑10 region at [2,2] and [2,3] must sum to more than 10. The cell [2,2] is adjacent to the empty [3,2]; placing the [6,2] domino vertically at [2,2]/[3,2] gives 6 to [2,2] and 2 to [3,2] (no constraint on [3,2]). Now [2,2] is 6, so [2,3] needs at least 5 to exceed 10.
3
Step 3: Bridging the constraints
The domino [5,1] can place a 5 at [2,3] and a 1 at [1,3]—which is already set to 1, so the pip fits. Place it horizontally over [2,3] and [1,3]. Now 6+5=11 satisfies greater‑10. The less‑3 cell at [1,1] must be under 3; the only remaining domino that can reach it is [3,2]. Place [3,2] vertically at [2,1]/[1,1] with 3 at [2,1] and 2 at [1,1] (2<3).
4
Step 4: Bottom‑left equals trio
The equals region [2,0], [2,1], [3,0] must all be equal. Since [2,1] is already 3 from the previous step, [2,0] and [3,0] must also be 3. Place the last domino [3,3] horizontally across [2,0] and [3,0], completing the puzzle with all constraints satisfied.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Twin sum‑5 cells
The single‑cell sum‑5 regions at [3,2] and [3,3] force those cells to be exactly 5. They sit side by side, and the only domino that can supply two 5's is [5,5]. Place [5,5] horizontally covering both cells, satisfying both constraints at once.
2
Step 2: Isolated sum‑2 singleton
The cell [2,0] has its own sum‑2 rule, so it must be 2. The empty cell [3,0] sits directly below it with no restriction. The domino [1,2] can place 2 at [2,0] and 1 at [3,0]; place it vertically to nail down these values.
3
Step 3: Top‑row sum‑2 and sum‑6
The sum‑2 region [0,2]/[0,3] needs 1+1. The sum‑6 region [0,0]/[0,1] needs values that add to 6. Place [4,1] horizontally at [0,1]/[0,2] (4 and 1), giving 1 to [0,2]. Then [0,0] must be 2 to reach sum‑6 with 4, and [1,0] is a single‑cell sum‑6 needing 6. Domino [2,6] fits vertically at [0,0]/[1,0]—2 at [0,0] and 6 at [1,0]—perfectly.
4
Step 4: Sum‑10 and equals pair
For sum‑10 at [0,4]/[0,5], place [6,1] horizontally at [0,4]/[0,3] to put 6 at [0,4] (and 1 at [0,3]—matching the needed 1). Then [0,5] needs 4; use [3,4] horizontally at [0,6]/[0,5] with 4 at [0,5] and 3 at [0,6]. The equals region [0,6]/[0,7] now gets 3 at [0,6], so [0,7] must also be 3. Place [5,3] vertically at [1,7]/[0,7] with 3 at [0,7] and 5 at [1,7].
5
Step 5: Final right‑side constraints
The sum‑11 region [1,7]/[2,7] needs 5+6; we have 5 at [1,7], so [2,7] must be 6. Place [6,5] vertically at [2,7]/[3,7] with 6 at [2,7] and 5 at [3,7] (satisfying the sum‑5 at [3,7]). The last unsettled region is greater‑5 at [2,2] and less‑5 at [2,3]. Domino [4,6] placed horizontally [2,3]/[2,2] gives 6 at [2,2] (>5) and 4 at [2,3] (<5), closing the puzzle.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Pin down forced singles
Numerous cells carry rigid single‑cell constraints. [0,0] sum‑1 → 1; [1,1] sum‑2 → 2; [2,2] sum‑3 → 3; [3,3] sum‑4 → 4; [4,4] sum‑5 → 5; [5,4] sum‑5 → 5; [6,3] sum‑4 → 4; [7,2] sum‑3 → 3; [8,1] sum‑2 → 2; [9,0] sum‑1 → 1; [3,0] less‑2 → 0 or 1; [6,0] less‑3 → 0,1,2; [4,3] less‑3 → 0,1,2. The giant equals region [5,0],[5,1],[5,2],[5,3],[6,1],[7,1] forces all these cells to 0. These fixed values form the skeleton of the solution.
2
Step 2: First three domino placements
Begin with [1,1]=2. Place domino [2,3] vertically at [1,1]/[2,1], yielding 2 at [1,1] and 3 at the empty cell [2,1]. Next, [2,2]=3. The domino [4,3] placed vertically at [3,2]/[2,2] supplies 4 to [3,2] (which will join the unequal group) and 3 to [2,2]. The two sum‑5 cells [4,4]/[5,4] are adjacent; cover them with [5,5] horizontally to give both a 5.
3
Step 3: Filling the zero block
The equals‑0 region must get zeros from dominoes that also deliver other needed pips. Place [0,0] horizontally at [5,1]/[6,1], setting two zeros. Place [4,0] vertically at [6,3]/[5,3]—this puts the required 4 at sum‑4 cell [6,3] and 0 at [5,3]. Place [0,3] vertically at [5,0]/[4,0] to give 0 to [5,0] and 3 to [4,0] (3 is needed for the unequal pentad). Finally, use [2,0] vertically at [8,1]/[7,1] to satisfy [8,1]=2 (sum‑2) while placing 0 at [7,1].
4
Step 4: Complete the unequal region
The five‑cell unequal region [3,1],[3,2],[4,0],[4,1],[4,2] must contain all distinct values. We already have 4 at [3,2] and 3 at [4,0]. Place [2,5] horizontally at [3,1]/[4,1] to put 2 at [3,1] and 5 at [4,1]. Then place [6,0] vertically at [4,2]/[5,2] to put 6 at [4,2] and 0 at [5,2] (contributing another zero to the equals chain). The set becomes {2,4,3,5,6}—all distinct, satisfying the constraint.
5
Step 5: Sum‑9 and top‑left corner
The region [1,0]/[2,0] must sum to 9. [0,0] is already fixed as 1; it must be covered by a domino with an adjacent cell. Place [4,1] vertically at [1,0]/[0,0]—4 at [1,0] and 1 at [0,0] (satisfying sum‑1). Then [2,0] must be 5 to reach 9. Use [0,5] vertically at [3,0]/[2,0] to set 5 at [2,0] and 0 at [3,0] (0 < 2 fulfills the less‑2 rule).
6
Step 6: Right‑side equals and remainders
The equals region [7,0]/[8,0] must both be 6. Place [6,2] vertically at [7,0]/[6,0]—6 at [7,0] and 2 at [6,0] (2 < 3, satisfying less‑3). Then place [1,6] vertically at [9,0]/[8,0]—1 at sum‑1 cell [9,0] and 6 at [8,0], completing the equals pair. The sum‑3 cell [7,2] is covered by [3,6] horizontally at [7,2]/[6,2]—3 at [7,2] and 6 at empty [6,2]. The last loose cell is less‑3 at [4,3]; use [2,4] horizontally at [4,3]/[3,3] with 2 at [4,3] (<3) and 4 at sum‑4 cell [3,3]. All regions now satisfied.

💡 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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