NYT Pips Hints & Answers for May 18, 2026

May 18, 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 NYT Pips puzzle is a compact 3x3 grid built around a sum-6 triplet that elegantly forces a uniform row of identical pips. With just four dominoes, every placement is dictated by the interplay of sum and less constraints, making it a clean, instructive solve.

Rodolfo Kurchan's medium grid expands to 4x6, showcasing his signature use of equals and unequal regions. A three-cell equals region in the upper right locks down a uniform set, while an unequal triplet and a sum-8 pair ripple outward, resolving the dominoes with a satisfying logical flow.

Kurchan's hard puzzle is a brilliant lattice of thirteen regions, including a three-cell equals column that cascades zeros and a sum-14 triplet that forces a precise 5-5-4 distribution. The grid interweaves greater-than, less-than, and sum constraints, demanding solvers trace delicate chains of deduction across its 10-row expanse.

💡 Progressive Hints

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

💡 Hint 1: Start with the tightest sum
Consider the sum region that spans a whole row; its target severely limits the pip combination. The top row must sum to 6 across three cells. With the available domino pips, think about the only way to reach that total.
💡 Hint 2: Focus on the triple 2s
Once you realize the top row needs all 2s, you'll place the double-2 domino horizontally and the 1-2 domino with its 2 in the remaining top cell, leaving the 1 below.
💡 Hint 3: Complete the grid with sums
Place the [2,2] domino at [0,0]-[0,1]; the [1,2] domino with 2 at [0,2] and 1 at [1,2]; the [2,4] domino spanning [1,0] (2) and [2,0] (4); and the [5,5] domino filling [2,1] and [2,2] with 5s.
💡 Hint 1: Find the sum region that dictates exact pips
Look for a sum region whose target can only be met by one specific pip combination given the available dominoes. This will anchor the first placements.
💡 Hint 2: The sum-8 pair forces two 4s
The sum-8 region at the right edge ([2,5] and [3,5]) must use 4+4. This dictates the two dominoes carrying 4s—[4,3] and [4,0]—which then interact with the adjacent equals region.
💡 Hint 3: The full domino map
Place [4,3] with 4 on [2,5] and 3 on [1,5]; [4,0] with 4 on [3,5] and 0 on [3,4]; [0,3] with 0 on [2,4] and 3 on [1,4] (creating the equals pair). Then [5,3] gives 3 to [0,4] and 5 to [0,5]; [5,5] fills [2,0]-[2,1]; [6,6] fills [3,0]-[3,1]; [3,2] finishes [3,2]-[3,3] with 3 and 2.
💡 Hint 1: Scan for single-cell sum regions
Identify isolated cells with sum constraints—they give you instant pip values and will dictate the first domino placements.
💡 Hint 2: Lock in the sum-3 cells
The sum-3 cells at [4,0], [4,2], and [6,2] force 3s, directing the [5,3], [6,3], and [2,3] dominoes into place with their partner pips.
💡 Hint 3: Sum-11 and greater regions unfold
With [4,0]=3, [3,0] gets 5; the sum-11 region above demands [2,0]=6, which pulls in the [4,6] domino. The greater region on [2,2]-[3,2] then forces maximum pips there.
💡 Hint 4: The bottom-right complex
The sum-14 vertical triple [6,4]-[8,4] must split as 5+5+4; the equals column [6,6]-[8,6] is all zeros. These two anchors resolve the bottom-right corner and dictate the double-zero and single-zero placements.
💡 Hint 5: Complete solution for hard
Dom 0 [5,3] at [3,0]-[4,0] (5,3); Dom 5 [6,3] at [3,2]-[4,2] (6,3); Dom 11 [2,3] at [6,1]-[6,2] (2,3); Dom 10 [4,6] at [1,0]-[2,0] (4,6); Dom 1 [0,0] at [6,6]-[7,6]; Dom 6 [5,5] at [6,4]-[7,4]; Dom 8 [4,4] at [8,4]-[8,5]; Dom 9 [1,0] at [9,6]-[8,6] (1,0); Dom 7 [0,6] at [8,2]-[8,1] (0,6); Dom 4 [5,2] at [0,7]-[0,6] (5,2); Dom 13 [2,1] at [1,7]-[2,7] (2,1); Dom 3 [3,3] at [3,7]-[4,7] (3,3); Dom 12 [4,0] at [2,4]-[2,5] (4,0); Dom 2 [6,2] at [2,2]-[1,2] (6,2).

🎨 Pips Solver

May 18, 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 18, 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 18, 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 sum-6 row
The top row is a sum-6 region spanning three cells. With available pips, the only way to reach 6 with three numbers from the domino set is three 2s. This forces the double-2 domino to cover two of those cells.
2
Step 2: Placing the 1-2 domino
The third cell of the top row, [0,2], must also be a 2. The only domino with a 2 that isn't already used is [1,2]. Place it so its 2‑pip sits at [0,2] and its 1‑pip occupies the cell below, [1,2] (an empty region).
3
Step 3: The sum-9 anchor
The region at [2,0] and [2,1] must sum to 9. The remaining dominoes are [2,4] and [5,5]; the only way to sum to 9 is 4+5. The [2,4] domino gives 4 to [2,0] and 2 to [1,0] (satisfying the less‑5 constraint), while [5,5] places a 5 at [2,1] and the remaining cell [2,2].
4
Step 4: Final checks
All constraints are now satisfied: the sum-6 row holds 2,2,2; the less‑5 cell holds 2; the sum-9 region sums to 9; and the empty cells are filled as shown.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Sum-8 forces two 4s
The sum-8 region at [2,5] and [3,5] can only be achieved with two 4s (4+4). This dictates that the two dominoes bearing 4—[4,3] and [4,0]—must place their 4‑pips in those cells.
2
Step 2: Equals region spreads zeros
With [4,0] providing 4 to [3,5], its 0‑partner goes to [3,4]. The adjacent equals region [2,4] and [3,4] demands both cells match, forcing [2,4] to also be 0. The only remaining 0 is on [0,3], so place it with 0 at [2,4] and its 3 at [1,4].
3
Step 3: Triple-equals L locks in threes
The [4,3] domino's 3 sits at [1,5]. The L-shaped equals region ([0,4], [1,4], [1,5]) now has two 3s, so [0,4] must be 3. The [5,3] domino can supply this 3 to [0,4] while its 5 goes to the empty cell [0,5].
4
Step 4: Unequal region and empty cells
The unequal region ([2,1], [3,1], [3,2]) must have all different pips. The empty cell [2,0] is filled by the [5,5] domino, giving 5 to [2,0] and [2,1]. Then [6,6] fills [3,0] and [3,1] with 6s, leaving the unequal set as 5,6,3—all distinct.
5
Step 5: Final less-5 cell
The cell [3,3] has a less‑5 constraint. The only unused domino is [3,2]; place it horizontally with the 3 at [3,2] and the 2 at [3,3], satisfying all regions.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Single-cell sum-3 anchors
The cells [4,0], [4,2], and [6,2] are each isolated sum‑3 regions, meaning they must hold a 3. This instantly places domino [5,3] with 3 at [4,0] and 5 at [3,0]; domino [6,3] with 3 at [4,2] and 6 at [3,2]; and domino [2,3] with 3 at [6,2] and 2 at [6,1].
2
Step 2: Sum-11 and greater regions
With [3,0]=5, the sum‑11 region [2,0]‑[3,0] needs 6 at [2,0]. The [4,6] domino supplies that 6 while placing 4 at [1,0] (satisfying greater‑2). The adjacent greater region at [2,2]‑[3,2] now has [3,2]=6, so [2,2] must also be as high as possible, yielding 6 from the [6,2] domino, which sends its 2 to [1,2].
3
Step 3: Sum-14 vertical triple
The column [6,4],[7,4],[8,4] sums to 14. With the remaining high pips, the only feasible split is 5+5+4. Place [5,5] horizontally covering [6,4]‑[7,4] with 5s, and [4,4] horizontally from [8,4] to [8,5] with 4 at [8,4].
4
Step 4: Equals zero column
The equals region [6,6],[7,6],[8,6] must be all identical. The [0,0] domino can cover [6,6]‑[7,6] with zeros. The third zero comes from the [1,0] domino, which places 0 at [8,6] and its 1 at [9,6] (empty).
5
Step 5: Sum-7 pairs and triplet
The sum‑7 region [0,7]‑[1,7] needs two numbers adding to 7. Use [5,2] horizontally to put 5 at [0,7] and 2 at [0,6] (greater‑1 satisfied). The needed 2 at [1,7] arrives via [2,1] domino placed vertically, with 2 at [1,7] and 1 at [2,7]. The second sum‑7 region [2,7],[3,7],[4,7] now has 1 at [2,7], so the remaining sum of 6 is split as 3+3, provided by [3,3] placed horizontally at [3,7]‑[4,7].
6
Step 6: Remaining less/greater cells
The [4,0] and [0,6] dominoes fill the final gaps: [2,4] (greater‑2) gets 4, and [2,5] (less‑2) gets 0 from [4,0] placed horizontally. [8,2] (less‑2) gets 0, and [8,1] (greater‑4) gets 6 from [0,6] placed horizontally. All cells are now locked.

💡 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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