NYT Pips Hints & Answers for May 10, 2026

May 10, 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 puzzle is a masterclass in minimal anchoring. By placing a solitary less-than-1 region at the edge, he locks the board instantly, then uses equals pairs to propagate values downward. The architecture is so tight that each domino is forced in sequence, leaving no room for ambiguity—a clean, elegant solve that rewards methodical deduction.

For the medium, Livengood expands the canvas with sprawling equals regions that act like magnetic poles, pulling the solver toward the only viable number. The top-left 2x2 block of identical cells sets off a chain reaction, where a single double-six domino decides the fate of half the grid. It’s a structural gambit that showcases how a well-placed repeated-value constraint can compress a puzzle’s solving space dramatically.

Rodolfo Kurchan’s hard puzzle is a dense lattice of interdependent constraints. The standout feature is a three-cell sum-zero region that instantly seeds a column of zeros, but the real artistry is in how that zero interacts with a web of sum and equals regions across the grid. Kurchan weaves these elements so tightly that every domino feels earned; it’s a puzzle that demands you trace the logic thread by thread. Today’s NYT Pips hard exemplifies his talent for designing richly layered, consequential grids.

💡 Progressive Hints

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

💡 Hint 1
Focus on the single-cell region with a strict 'less than' constraint; it forces the smallest possible pip.
💡 Hint 2
The cell at top-right [0,3] must be 0 because of the less-than-1 rule. This forces you to place a domino containing a 0 there, linking it to [1,3].
💡 Hint 3 (Full Solve)
Place the [6,0] domino at [0,3]=0, [1,3]=6. The equals region at [1,3]/[2,3] forces [2,3]=6, so [6,2] domino goes [2,3]=6, [3,3]=2. Then equals [3,3]/[4,3] forces [4,3]=2; the less-2 [4,4] must be 1, so [1,2] domino goes [4,4]=1, [4,3]=2. The equals region [1,1]/[2,1] must be 4: place [4,0] at [1,1]=4, [1,2]=0; then [3,4] at [3,1]=3, [2,1]=4. Finally sum-8 [3,1]/[4,1] needs 5: [5,3] domino at [4,1]=5, [4,0]=3.
💡 Hint 1
Look for large 'equals' regions that cover multiple cells; they demand a number that appears on several dominoes.
💡 Hint 2
The top-left 2×2 equals block (cells [0,1],[0,2],[1,1],[1,2]) all must share the same value. The domino [6,6] fits perfectly, so place it horizontally at the top row, locking the entire block to 6.
💡 Hint 3 (Full Solve)
With top equals block set to 6, place [6,6] at [0,1]=6, [0,2]=6. The domino covering [1,1] must carry a 6: [6,1] goes [1,1]=6, [1,0]=1 (greater-0). [1,2] also must be 6, so [5,6] goes [1,2]=6, [1,3]=5 (greater-3). For the middle equals block, all four cells need 6: [6,0] domino at [2,1]=6, [2,0]=0; [3,6] domino at [2,3]=3, [2,2]=6; [2,6] domino at [3,0]=2, [3,1]=6 (sum-2 satisfied); [4,6] domino at [3,3]=4, [3,2]=6. The final equals region [3,3],[4,2],[4,3] must be 4; use [4,4] at [4,2]=4, [4,3]=4.
💡 Hint 1
A sum-zero region spanning three cells is the key; it forces all involved cells to 0 immediately.
💡 Hint 2
The sum-zero region is at column 6, rows 1–3. You’ll need the double-zero domino [0,0] to cover two of these cells, and another zero-domino to cover the third.
💡 Hint 3
Place [0,0] at [2,6]=0, [3,6]=0. The remaining cell [1,6] must be 0, so the [0,1] domino (0 and 1) goes there: [1,6]=0, [1,7]=1 (since [1,7] is less-than-2). Now focus on column 0: [2,0] is less-than-2 → must be 1; the sum-2 pair [3,0] and [4,0] must be 1+1; [5,0] is >2 → at least 3.
💡 Hint 4
Column 0 resolves with [1,1] domino (two 1s) covering [2,0]=1, [3,0]=1; then [1,3] domino (1 and 3) covers [4,0]=1, [5,0]=3. Row 1’s sum-10 region is resolved by a 6+4 combo, using [5,6] at [0,0]=5, [1,0]=6 and [4,4] at [1,1]=4, [2,1]=4.
💡 Hint 5 (Full Solve)
From zeros: [0,0] at [2,6]/[3,6]=0; [0,1] at [1,6]=0, [1,7]=1. Column 0: [1,1] at [2,0]=1, [3,0]=1; [1,3] at [4,0]=1, [5,0]=3. Row 1: [5,6] gives [0,0]=5, [1,0]=6; [4,4] gives [1,1]=4, [2,1]=4. Equals region [2,2],[2,3],[2,4] all 1: [1,5] at [2,2]=1, [3,2]=5; [1,2] at [2,4]=1, [3,4]=2; [1,4] at [2,3]=1, [1,3]=4 (greater-3 satisfied). Equals column [3,4],[4,4],[5,4] all 2: [2,2] at [4,4]=2, [5,4]=2. Column 8: [6,6] at [4,8]=6, [5,8]=6; [3,3] at [1,8]=3, [2,8]=3 (sum-6). Sum-9 at [3,7],[3,8]: [4,5] at [3,7]=4, [3,8]=5. Remaining: [0,4] at [0,4]=0, [1,4]=4; [2,4] at [4,6]=2, [5,6]=4.

🎨 Pips Solver

May 10, 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 10, 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 10, 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 solitary less-than anchor
The region at [0,3] has a less-than-1 constraint, so the only possible pip is 0. The only domino with a 0 is [6,0]. Place it vertically with 0 at [0,3] and 6 at [1,3].
2
Step 2: Equals cascade downward
The equals region tying [1,3] and [2,3] forces [2,3]=6. The domino [6,2] is the only one left with a 6, so place it with 6 at [2,3] and 2 at [3,3].
3
Step 3: Right-side chain to the bottom corner
The equals region [3,3] and [4,3] now requires [4,3]=2. The less-than-2 region at [4,4] must be 1. Domino [1,2] fits perfectly, placing 1 at [4,4] and 2 at [4,3].
4
Step 4: Left-side pairs and the sum-8 finish
The equals region [1,1] and [2,1] needs a common value — 4 is the only candidate among remaining dominoes. Place [4,0] at [1,1]=4 and [1,2]=0. Then [3,4] goes at [3,1]=3 and [2,1]=4, fulfilling the equals. Finally, the sum-8 region at [3,1] and [4,1] totals 3+5; domino [5,3] supplies 5 at [4,1] and 3 at [4,0].

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: The top-left equals magnet
The large equals region covering [0,1],[0,2],[1,1],[1,2] must be filled with the same number. The domino [6,6] is the only double; place it horizontally across [0,1] and [0,2], setting the entire block’s value to 6.
2
Step 2: Forcing the 6 into row 1
Now [1,1] and [1,2] are forced to 6. Cell [1,0] is greater-than-0, so the domino covering [1,1] must deliver a 6 and a small value. Place [6,1] with 6 at [1,1] and 1 at [1,0]. Then cover [1,2] with the 6 from domino [5,6], placing its 5 at [1,3] (which requires >3).
3
Step 3: The middle equals block demands sixes
The second large equals region at [2,1],[2,2],[3,1],[3,2] also requires a uniform number, and 6 is the only value that appears on enough remaining dominoes. Place [6,0] with 6 at [2,1] and 0 at [2,0] (empty cell).
4
Step 4: Completing the middle grid
To fill [2,2] with 6, use domino [3,6] placing 6 at [2,2] and 3 at [2,3] (greater-1 satisfied). For [3,1] and [3,2], the domino [2,6] provides 6 at [3,1] and 2 at [3,0] — perfect, because [3,0] has a sum-2 constraint that demands exactly 2. Then domino [4,6] finishes the block by placing 6 at [3,2] and 4 at [3,3].
5
Step 5: The final equals set and the double-four
The equals region [3,3],[4,2],[4,3] must match the 4 at [3,3]. The only remaining domino is [4,4]; place it to cover both [4,2] and [4,3] with 4s, wrapping up the puzzle.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: The zero triumvirate
The sum-0 region at [1,6],[2,6],[3,6] forces all three cells to 0. The double-zero domino [0,0] covers two of them — place it on [2,6] and [3,6] both as 0. The third cell [1,6] must also be 0, so the other zero-carrying domino, [0,1], goes there: [1,6]=0, [1,7]=1 (since [1,7] is less-than-2).
2
Step 2: Anchoring column 0 with law of small numbers
The less-than-2 region at [2,0] requires 1. The sum-2 pair [3,0] and [4,0] must be 1+1. The >2 region at [5,0] needs at least 3. Place the double-1 domino [1,1] at [2,0]=1 and [3,0]=1. Then the [1,3] domino (1,3) covers [4,0]=1 and [5,0]=3, satisfying all column 0 constraints.
3
Step 3: Sum-10 and the 4-4 connector
Row 1’s sum-10 region at [1,0] and [1,1] must total 10. The [5,6] domino handles [0,0] (sum-5 cell) with 5 and [1,0] with 6. To reach 10, [1,1] needs 4. The double-4 domino [4,4] fits perfectly, giving 4 at [1,1] and also at [2,1] (which is >2, so 4 works).
4
Step 4: The triple-equals chain across row 2
The equals region [2,2],[2,3],[2,4] must all share a value. The domino [1,5] places 1 at [2,2] and 5 at [3,2] (>4 constraint met). Now the region forces [2,3]=1 and [2,4]=1. Use [1,2] to give 1 at [2,4] and 2 at [3,4]. The [1,4] domino then covers [2,3]=1 and [1,3]=4 (greater-3 satisfied).
5
Step 5: Vertical equals and the double-two
The equals column [3,4],[4,4],[5,4] must all be 2, matching the 2 just placed at [3,4]. The domino [2,2] carries two 2s, so place it at [4,4]=2 and [5,4]=2, finishing that constraint.
6
Step 6: Closing the bookends: column 8, sum-9, and the final gaps
The equals pair [4,8],[5,8] needs identical values; the last high domino is [6,6], so place it to give 6s there. The sum-6 region [1,8],[2,8] now needs 3+3, so [3,3] domino goes at [1,8]=3, [2,8]=3. The sum-9 at [3,7]-[3,8] uses [4,5] (4+5). Remaining: [0,4]<2 demands 0, so [0,4] domino with 0 at [0,4] and 4 at [1,4] (greater-2). Finally, [4,6] (<3) must be 2, so [2,4] domino places 2 at [4,6] and 4 at [5,6] (>3).

💡 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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