NYT Pips Hints & Answers for May 13, 2026

May 13, 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!

Click here to play today's official NYT Pips game first.

Want hints instead? Scroll down for progressive clues that won't spoil the fun.

🎲 Today's Puzzle Overview

Today’s NYT Pips easy puzzle, constructed by Ian Livengood, gives you a straightforward entry point with a massive sum-25 region. Right away, you’ll realize that five cells can only reach 25 by each holding a 5, which instantly locks the double-5 domino in place. From there, a less-than constraint and an equals pair fall into line around it, making the tiling unfold rapidly—a gentle but satisfying warm-up.

Livengood’s medium grid steps up the interplay between opposites. A row splits into a less-4 quartet and a single greater-4 cell, while a sum-13 trio below pulls in the high numbers. You’ll find yourself juggling the three 6-valued dominos early, using equals regions as anchors to distribute the 4s and 6s. It’s a delightful dance of constraint balancing that rewards each deduction with a clear next step.

Rodolfo Kurchan’s hard puzzle flips the script: every cell is its own sum constraint, effectively handing you all the pip values. The challenge shifts from “what number goes here?” to “which domino connects these known values?” It becomes a pure domino-tiling logic exercise, where you’ll start by hunting the five 0-cells and their forced pairings. The zero-dominos chain together like a puzzle skeleton, after which the rest of the board clicks into place with elegant precision.

💡 Progressive Hints

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

💡 Look at the Sum
Identify which region forces all its cells to share the same high value. There's only one pip that can appear five times to reach that total.
💡 Lock the Five-Fold Path
The sum-25 region in the lower right can only be satisfied by five 5s. The double-5 domino is your key to filling two of those cells at once.
💡 Full Solution
Place the [5,5] domino horizontally covering [2,2] and [2,3]. Then place [0,5] horizontally at [1,4]-[1,3] (0 in the less-3 cell, 5 in the sum region). Next, [5,2] at [1,1]-[1,0] gives a 5 and a 2, matching the equals region with [2,0] via the [2,3] domino placed across [2,0]-[2,1] (2 and 3). Finally, [5,6] vertically at [1,2]-[0,2] supplies the empty top cell with a 6.
💡 Separate High and Low
A row splits into a strict less-than region and a single greater-than cell, while below it a sum-13 region pulls in the big numbers.
💡 Find the 6's Home
The sum-13 in row 2 needs a 6, a 5, and a 2. The two equals regions lock in another 6 and a pair of 4s. Focus on which dominos can supply those values.
💡 Full Solution
Place [6,1] vertically at [2,0]-[1,0] to start the sum-13 and less-4. Use [6,5] horizontally at [2,3]-[2,2] (6 for the equals, 5 for the sum-13). [4,2] goes at [3,1]-[2,1] (4 for the equals, 2 completes sum-13). [6,4] at [3,3]-[3,2] finishes the equals regions. Then [1,5] at [1,3]-[1,4] (1 in less-4, 5 in greater-4). Place [0,1] horizontally at [1,2]-[0,2] (0 for less-4, 1 for less-2). Finally [1,2] at [1,1]-[0,1] completes the less-4 and the empty cell.
💡 A Grid of Known Values
Every cell has a fixed sum target, so all pip values are pre-determined. The puzzle is about pairing them correctly with the given domino set.
💡 Zero In on the Zeros
Five cells must hold a 0, and five dominos start with 0. Look for a 0-cell with only one adjacent number matching an available 0-domino.
💡 The First Forced Pairings
Cell [1,1] is 0 and only [1,2] (value 1) can pair using [0,1]. Then [2,5] (0) must pair with [2,4] (2) via [0,2]. With those used, [4,3] (0) has to link to [4,2] (5) using [0,5].
💡 Finish the 0-Dominos
That leaves [0,3] (0) to pair with [0,2] (4) via [0,4], and [1,5] (0) to pair with [0,5] (3) via [0,3]. All 0-cells are now paired.
💡 Complete the Tiling
With zeros resolved, place the remaining dominos: [1,2] at [4,4]-[3,4], [1,3] at [2,0]-[1,0], [1,4] at [4,0]-[3,0], [1,5] at [2,3]-[1,3], [2,3] at [4,1]-[3,1], [2,4] at [0,0]-[0,1], [2,5] at [3,3]-[3,2], [3,4] at [2,2]-[2,1], [3,5] at [0,4]-[1,4], and [4,5] at [3,5]-[4,5].

🎨 Pips Solver

May 13, 2026

Click a domino to place it on the board. You can also click the board, and the correct domino will appear.

✅ 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 13, 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 13, 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: Sum-25 forces all 5s
The region [1,1]-[1,2]-[1,3]-[2,2]-[2,3] with sum target 25 can only be achieved with five 5s (0-6 scale). The [5,5] domino must cover two of these cells. The logical placement is horizontally at [2,2]-[2,3] because it leaves room for other 5s.
2
Step 2: Less-3 demands a 0
The cell [1,4] has a 'less 3' constraint, so it can be 0, 1, or 2. Adjacent [1,3] is in the sum-25 region and must be 5. The [0,5] domino fits perfectly — place it horizontally at [1,4]-[1,3] giving [1,4]=0 and [1,3]=5.
3
Step 3: Equals region needs a match
Cells [1,0] and [2,0] must be equal. The remaining sum-25 cells need two more 5s. The [5,2] domino can supply a 5 and a 2. Place it at [1,1]-[1,0] (5 at [1,1], 2 at [1,0]). To match, [2,0] must be 2, so use the [2,3] domino placed horizontally at [2,0]-[2,1] (2 and 3, the 3 filling the empty cell [2,1]).
4
Step 4: Top cell and final 5
The empty region [0,2] needs a value, and the sum-25 region still needs a 5 at [1,2]. The [5,6] domino vertically at [1,2]-[0,2] supplies the 5 and gives 6 to [0,2], completing the grid.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Sum-13 needs a 6
The region [2,0],[2,1],[2,2] sums to 13. With available numbers, the only combination is 6+5+2. So a 6 and a 5 must occupy two of these cells. The [6,1] domino provides a 6 and a 1. Placing it vertically at [2,0]-[1,0] injects 6 into the sum region and 1 into the less-4 region in row 1.
2
Step 2: Capture the 6 and 5
The equals region [2,3]=[3,3] will also need a high number; the other 6 is perfect. Place [6,5] horizontally at [2,3]-[2,2] — this fills [2,3] with 6 and [2,2] with 5. Now the sum-13 has 6 (at [2,0]) and 5 (at [2,2]), so [2,1] must be 2.
3
Step 3: Lock the equals regions
The equals pairs [3,1]=[3,2] need identical values; the [4,2] and [6,4] dominos fit. Place [4,2] at [3,1]-[2,1] — [3,1]=4 and [2,1]=2 (completing sum-13). Then place [6,4] at [3,3]-[3,2], giving 6 to match [2,3] and 4 to match [3,1].
4
Step 4: Satisfy the less-4 and greater-4
Row 1's less-4 region still needs [1,1],[1,2],[1,3] under 4, and [1,4] must be >4. The [1,5] domino placed at [1,3]-[1,4] gives 1 and 5 — exactly one less and one greater. The [0,1] domino at [1,2]-[0,2] puts 0 at [1,2] and 1 at [0,2] (satisfying less-2).
5
Step 5: Finishing touches
The last domino [1,2] fills [1,1]-[0,1] with 1 (completing less-4) and 2 (the empty cell [0,1]). All constraints are now met.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Isolate the first 0-pair
Cell [1,1] must be 0. Its neighbors are [1,2]=1, [0,1]=4, [2,1]=4. The only 0-X domino including 1 is [0,1], so place it horizontally at [1,1]-[1,2].
2
Step 2: Anchor the bottom-right 0
Cell [2,5] must be 0. It neighbors [2,4]=2, [3,5]=4, [1,5]=0 (can't pair two 0s). The [0,2] domino is needed, placed at [2,5]-[2,4].
3
Step 3: Resolve the 0 at [4,3]
Cell [4,3] is 0, with neighbors [4,2]=5, [3,3]=2, [4,4]=1. [0,1] and [0,2] are used, so only [0,5] works with the adjacent 5. Place [0,5] vertically at [4,3]-[4,2].
4
Step 4: Pair the remaining 0s
The 0 at [0,3] neighbors [0,2]=4, [0,4]=3, [1,3]=5. With [0,5] used, [0,4] fits with 4: place [0,4] at [0,3]-[0,2]. Finally, [1,5]=0 pairs with [0,5]=3 using [0,3] placed at [1,5]-[0,5].
5
Step 5: Fill in the 1-2-3 dominoes
Now place the dominoes with 1s and 2s. [1,2] at [4,4]-[3,4] (1+2). [1,3] at [2,0]-[1,0] (1+3). [1,4] at [4,0]-[3,0] (1+4). [1,5] at [2,3]-[1,3] (1+5). [2,3] at [4,1]-[3,1] (2+3). [2,4] at [0,0]-[0,1] (2+4). [2,5] at [3,3]-[3,2] (2+5).
6
Step 6: The high-value finish
Complete with [3,4] at [2,2]-[2,1] (3+4), [3,5] at [0,4]-[1,4] (3+5), and [4,5] at [3,5]-[4,5] (4+5). All cells tiled, all sums 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.

🎓 Keep Learning & Improve

🧠 Strategy Guide

Master the core techniques to solve any Pips puzzle, no matter the difficulty.

🎮 How to Play

New to Pips? Learn all the rules and conditional regions in minutes.

📖 More to Read

May 12, 2026

NYT Pips Hints & Answers for May 12, 2026

View Answer →
May 11, 2026

NYT Pips Hints & Answers for May 11, 2026

View Answer →
May 10, 2026

NYT Pips Hints & Answers for May 10, 2026

View Answer →
May 9, 2026

NYT Pips Hints & Answers for May 9, 2026

View Answer →
May 8, 2026

NYT Pips Hints & Answers for May 8, 2026

View Answer →
May 7, 2026

NYT Pips Hints & Answers for May 7, 2026

View Answer →

💬 Community Discussion

Leave your comment