NYT Pips Hints & Answers for May 5, 2026

May 5, 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

The easy grid gives you an immediate gift: a wide equals region spanning the entire top row, so right away you know those four cells share the same number. Ian Livengood’s design then funnels you into a tidy chain of single-cell constraints below—greater than, less than, and a sum—that lock each vertical domino into place as you scan the available tiles. It’s a smooth opening, perfectly calibrated to get you into the Pips mindset.

Rodolfo Kurchan’s medium puzzle turns up the heat with a solitary sum-3 cell that forces a specific domino, and a three-cell sum-2 region that demands two 1s and a 0. As you navigate the irregular board, each small region feels like a little puzzle box: the sum-9 pair coaxes out a 4-and-5 split, while a complementary sum-10 pair nearby resolves with a double-5. The interplay is delightful, with each placement opening the next, and the board contracts satisfyingly until the final tile drops.

The hard puzzle, also from Kurchan, is a masterclass in cascading constraints. The centerpiece is a triple-cell sum-12 region that can only be satisfied by a double-5 and a 2, a revelation that echoes outward. From there, a less-than-4 region claims a 1-2 domino, a vertical sum-8 column orchestrates a 3-5-0 chain, and disparate sum-7 regions tie the top and bottom corners together. You’ll feel like a detective connecting dots across the grid, with inequality and unequal regions adding a final layer of deduction. This NYT Pips hard is a deeply rewarding solve.

💡 Progressive Hints

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

💡 Hint 1
Look for a large region that forces multiple cells to be identical—that uniformity will dictate your first moves.
💡 Hint 2
The four-cell equals region on the top row narrows the possibilities dramatically. When you check the available dominoes, only one pip value can appear in all four cells. Then, the single-cell constraints below (greater than 4, less than 4) will pick out which domino goes where.
💡 Hint 3
Place the [0,4] domino vertically at [0,0]-[1,0] with 0 on top and 4 below. The [6,0] domino goes to [0,1]-[1,1] with 0 above and 6 below (satisfying the >4). The [2,0] fills [0,2]-[1,2] with 0 and 2 (<4). The [0,5] lands at [0,3]-[1,3] with 0 and 5 (the 5 forms part of the sum-10). Finally, the remaining [1,5] domino completes the grid at [2,2]=1 and [2,3]=5, hitting the sum-10 exactly at 5+5.
💡 Hint 1
Search for a tiny region that dictates an exact number—a single-cell sum constraint leaves no wiggle room.
💡 Hint 2
The sum-3 cell at [1,6] must be a 3, so the domino covering it is the [3,5] tile, placing 5 in [0,6]. Nearby, the three-cell sum-2 region at [1,4]-[1,5]-[2,4] can only be two 1s and a 0, which forces the double-1 domino into [1,4]-[1,5].
💡 Hint 3
Place [3,5] at [1,6]=3 and [0,6]=5. Put [1,1] at [1,4]=1,[1,5]=1. That leaves [2,4]=0, which pairs with [3,4]=6 via the [0,6] domino. The sum-3 at [5,0]-[6,0] is met with [5,2] giving 2 at [5,0] and [6,1] giving 1 at [6,0]. The sum-10 pair [5,1]-[5,2] gets 5+5 from the double-5 placed vertically at [4,2]-[5,2], while [4,2]’s 5 completes the sum-9 region with a 4 at [3,2] from the [4,3] domino (which also puts 3 at [3,3]).
💡 Hint 1
Zero in on a sum region with three cells—the total forces a repeated digit on one domino.
💡 Hint 2
The sum-12 region at [5,4], [5,5], [5,6] can only be 5+5+2. That means the double-5 domino [5,5] must sit in [5,4]-[5,5], and [5,6] will need a 2 from another tile.
💡 Hint 3
With the double-5 placed, the less-than-4 region at [2,3]-[2,4] eagerly accepts the [1,2] domino (1 and 2). Then the 2 in [5,6] emerges from the [2,3] domino, which also puts a 3 in [6,6], kicking off a vertical sum-8 chain.
💡 Hint 4
The sum-8 column at [6,6]-[7,6]-[8,6] already has a 3 from that [2,3] domino. It needs 5 more across two cells; the [3,5] tile provides 5 at [7,6] and 3 at [7,5], forcing [8,6] to be 0 via the [0,2] domino. Meanwhile, the bottom-right sum-7 region at [9,5]-[9,6] resolves to 5+2 using the [5,1] and [0,2] tiles.
💡 Hint 5
Complete solution: Place [5,5] at [5,4]-[5,5] (both 5). Put [1,2] at [2,3]=1, [2,4]=2. Place [3,3] at [6,1]-[6,2] (both 3). Use [5,1] at [9,5]=5, [9,4]=1. Set [0,2] at [8,6]=0, [9,6]=2. Place [3,5] at [7,5]=3, [7,6]=5. [2,4] goes to [2,8]=2, [3,8]=4. [3,6] at [1,1]=3, [2,1]=6. [4,5] at [0,1]=4, [0,0]=5. [2,3] at [5,6]=2, [6,6]=3. [4,1] at [0,8]=4, [1,8]=1. [0,5] at [8,2]=0, [8,1]=5. [4,3] at [0,6]=4, [1,6]=3. [5,2] at [2,7]=5, [2,6]=2. [1,6] at [3,1]=1, [4,1]=6.

🎨 Pips Solver

May 5, 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 5, 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 5, 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: Spot the equals region
The top row [0,0]-[0,3] all share the same value. Scanning the domino pool, the only pip that appears on one end of four different dominoes is 0 (dominoes [0,4], [6,0], [2,0], [0,5]). So the top row must be all 0s.
2
Step 2: Match single-cell constraints
The cell [1,1] must be >4, so its vertical domino must carry a 6 — that’s the [6,0] tile, placed at [0,1]=0, [1,1]=6. Similarly, [1,2] must be <4, so the [2,0] tile with a 2 goes there: [0,2]=0, [1,2]=2. The empty [1,0] can be any value, so the [0,4] tile fits naturally with 4: [0,0]=0, [1,0]=4. The remaining top cell [0,3] pairs with [1,3] using the [0,5] tile: [0,3]=0, [1,3]=5.
3
Step 3: Address the sum-10 region
The sum-10 region consists of [1,3] (already 5) and [2,3]. To reach 10, [2,3] must be exactly 5. The only domino left is [1,5], which provides a 5 and a 1. So place it as [2,2]=1, [2,3]=5.
4
Step 4: Verify
The equals region holds all 0s, all constraints are satisfied, and every domino is used.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: The sum-3 cell
The single-cell region at [1,6] must sum to 3, so that cell is exactly 3. The only domino containing a 3 is [3,5], so place it as [1,6]=3, [0,6]=5 (the empty region at [0,6] happily takes the 5).
2
Step 2: The three-cell sum-2
Region [1,4], [1,5], [2,4] sums to 2. With pips 0-6, the only possible combination is two 1s and a 0. The double-1 domino [1,1] is perfect for [1,4]-[1,5], both set to 1. This forces [2,4] to be 0, which pairs with [3,4] via the [0,6] domino, giving [3,4]=6.
3
Step 3: Bottom-left sum-3
The region [5,0]-[6,0] must total 3. The [5,2] tile (pips 5,2) can give a 2 at [5,0], and the [6,1] tile (pips 6,1) gives a 1 at [6,0]; 2+1=3. So place [5,2] horizontally with 5 at [5,1] and 2 at [5,0]; place [6,1] vertically as [7,0]=6, [6,0]=1.
4
Step 4: The sum-10 pair
[5,1] already has a 5 from the previous step, so the sum-10 region [5,1]-[5,2] needs a 5 at [5,2]. The double-5 domino [5,5] can supply that, placed vertically at [4,2]=5 and [5,2]=5. This also provides the 5 needed for the sum-9 region [3,2]-[4,2] which now has a 5 at [4,2] and needs a 4 at [3,2].
5
Step 5: Finish the sum-9
The remaining domino [4,3] goes to [3,2]=4, [3,3]=3, satisfying the sum-9 total of 4+5 and leaving the empty [3,3] with 3. All regions are met.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Anchor on the sum-12 triple
The region [5,4], [5,5], [5,6] must total 12. With available pips, the only feasible combination is 5+5+2. Therefore, the double-5 domino [5,5] fills [5,4]=5 and [5,5]=5. The third cell [5,6] will need a 2.
2
Step 2: The less-than-4 pair
Region [2,3]-[2,4] requires both cells to be less than 4, so they can only be 1 and 2 (or two identical pips less than 4, but the only available such domino is [1,2]). Place domino [1,2] at [2,3]=1, [2,4]=2.
3
Step 3: Cascade the 2
To give [5,6] its 2, domino [2,3] (pips 2 and 3) is placed at [5,6]=2, [6,6]=3. Now the vertical sum-8 region at [6,6], [7,6], [8,6] holds a 3 at [6,6], leaving 5 to be split between [7,6] and [8,6]. The [3,5] domino (3,5) goes to [7,5]=3, [7,6]=5, contributing 5. Then [8,6] must be 0, provided by the [0,2] domino (0,2) placed as [8,6]=0, [9,6]=2.
4
Step 4: Bottom-right sum-7
The region [9,5]-[9,6] sums to 7. With [9,6] already 2, [9,5] needs 5. The [5,1] domino (5,1) fits perfectly at [9,5]=5, [9,4]=1, satisfying the greater-than-0 constraint on [9,4].
5
Step 5: Resolve top-left sum-7 and surrounding singles
The sum-7 region [0,1]-[1,1] needs 4+3. The [4,5] domino (4,5) places 4 at [0,1] and 5 at [0,0] (>4 region satisfied). The [3,6] domino (3,6) gives 3 at [1,1] and 6 at [2,1] (>5 region). Single-cell sum-4 regions at [0,6] and [0,8] each need a 4. Use [4,3] at [0,6]=4, [1,6]=3, which with [1,6] and [2,6] sum-5 region gives [2,6]=2 from [5,2] domino placed at [2,7]=5, [2,6]=2 (sum-7 region [2,7]-[2,8] gets 5+2; [2,8]=2 from [2,4] domino placed at [2,8]=2, [3,8]=4, satisfying sum-4 at [3,8]). For [0,8]=4, use [4,1] domino at [0,8]=4, [1,8]=1 (less-than-3 region happy).
6
Step 6: Clean up remaining placements
The [3,3] domino goes to [6,1]=3, [6,2]=3 (both empty). The [0,5] domino satisfies the unequal region at [8,1]-[8,2] with 5 and 0 (order doesn’t matter). The [1,6] domino covers [3,1]=1, [4,1]=6, completing the sum-7 region there. All dominoes placed, all constraints honored.

💡 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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