NYT Pips Hints & Answers for May 29, 2026

May 29, 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

In today's NYT Pips easy by Ian Livengood, you're greeted by a broad sum-20 region that immediately dictates the entire bottom row—only one combination of four numbers fits, turning that row into a solid wall of 5s. With that anchor locked, the less-than-2 column to the left forces two zeros, and the equals pair in the top row quickly claims the sixes. The remaining dominos slot in almost automatically, making the solve feel like snapping together a puzzle with only one valid configuration.

Rodolfo Kurchan's medium puzzle plays a craftier game. At first, the empty cells look permissive, but the equals pairs form a backbone: a double-zero domino must serve the top-center pair outright, and a second equals region later demands matching sixes from two different dominos. Small sum constraints then distribute 1, 2, and 3 with precision, using every available pip. The grid becomes a tidy chain of deductions where each placement pulls the next.

Kurchan's hard puzzle on the same day is an intricate weave. A massive vertical equals column of five cells forces an equal value throughout, which you'll resolve with a double-1 domino sandwiched between horizontal partners. Nearby, a pair of sum-10 regions demands 5-5 and 5-5 arrangements, while linked less-than and sum-2 clusters push zeros into place. Solving this hard is like untangling a knot—each new discovery, such as the column of 1s or the 3s equals group, sets off a cascade that eventually snaps every domino into place.

💡 Progressive Hints

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

💡 Find the Heavy Lifter
Look for a sum constraint with a large target—it will force a very specific set of pip values, limiting your options drastically.
💡 The Bottom Row Demands 5s
The four‑cell region spanning row 2 must sum to 20. Given the dominos you have, the only viable combination is to put a 5 in every one of those cells.
💡 Full Logical Lock
Place the double‑5 (5|5) domino across the two inner cells of row 2, then give the outer cells the 5 from 1|5 and 5|0. The less‑than‑2 column in rows 0‑1 forces zeros—set the 0 from 0|6 at [0,0] and the 0 from 5|0 at [1,0]. The equals pair at top‑center then claims the 6s: 0|6’s 6 to [0,1] and 2|6’s 6 to [0,2]. The greater‑than‑2 region wraps up with the leftover 2 and 1.
💡 Equals Region Tactic
Spot an equals constraint that touches only two cells—it will require a domino with two identical pips, because no other cell shares the region’s requirement.
💡 Top‑Center Lockdown
The equals region covers cells [0,1] and [0,2]. Among the available dominos, only the double‑0 can make both cells equal on its own. That placement anchors the top of the grid.
💡 Deduction Chain to Completion
After placing the 0|0, the equals pair at [1,2] and [2,2] must be filled; both get 6 from the 6|1 and 0|6 dominos—6|1 provides 6 to [1,2] and 0|6 provides 6 to [2,2]. The sum‑3 region (1,3)/(1,4) then takes 1 from 6|1 and 2 from 2|0. The sum‑4 region (1,5)/(1,6) uses 1 from 0|1 and 3 from 2|3. The remaining singles: 0|1’s 0 goes to [0,5]; 5|0’s 5 to [1,0] and its 0 to the equals region [1,1]/[2,1]; 2|0’s 0 to [2,4]; and 2|3’s 2 to the greater‑than‑0 cell [2,6].
💡 Constraint Radar
Survey for sum constraints with a high fixed target and equals regions that span many cells—they will lock in values that can only come from specific domino combinations.
💡 Sum‑10 Pairs & the Equals Column
The sum‑10 region in rows 1‑2, column 0 forces both cells to be 5s, demanding the double‑5 domino. Meanwhile, the five‑cell equals column (starting at [0,1] down to [4,1]) will require a central double‑1 domino because it’s the only way to fill the column’s length with an equal number.
💡 Zero Clusters
A less‑6 single at [0,0] and a less‑2 pair on [0,3]/[0,4] create a zero‑heavy top row. The 0|1 domino places 0 at [0,0] and 1 at [0,1], while the 0|2 takes [0,3] and [0,2] to satisfy sum‑2 at [0,2]. This sets the equals column’s top value to 1.
💡 Cascading Equals and Sums
With the column’s 1s confirmed and the bottom rows receiving the double‑1 (1|1) at [2,1]/[3,1], the 4|1 domino covers [4,1] and [5,1] to keep the equals region intact. The other sum‑10 region at rows 1‑2, column 4 then forces 5 from the 5|0 domino at [1,4] and the 3|5’s 5 at [2,4].
💡 Final Placement Order
Lock the equals region of 3s: use 3|3 at [1,3]/[2,3] and the 3 from 3|5 at [3,4] and 3|0 at [3,3] (with 0 to [4,3]). The sum‑6 region takes the 6 from 2|6 at [4,4] and 0 from 3|0 at [4,3]. The bottom equals clusters finish: 4|2 places 4 at [6,1] and 2 at [6,2], 2|2 at [6,3]/[6,4] for the equals‑2 group, and 4|0’s 4 at [6,0] with 0 at [5,0] for the sum‑2 region. Complete by placing the 3|2 at [3,0]/[4,0] and the 1|3 at [1,1]/[1,2].

🎨 Pips Solver

May 29, 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 29, 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 29, 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‑20 Anchor
The four‑cell row at the bottom (row 2) must sum to 20. With pips 0‑6, the only way to reach 20 with four cells is 5+5+5+5. So every cell in that row must be a 5. This immediately forces all three dominos that contain a 5 to contribute to this row.
2
Step 2: Placing the Double‑5
The double‑5 domino (5|5) can cover two of the row’s 5s in one move. Place it horizontally across [2,1] and [2,2] to satisfy two of the four required 5s. The remaining 5s will come from the 1|5 and 5|0 dominos.
3
Step 3: Zeros for the Less‑2 Column
The less‑than‑2 region contains [0,0] and [1,0]; both must be 0 or 1, but available dominos force 0s. The 5|0 domino places its 5 at [2,0] and its 0 at [1,0]. The 0|6 domino places its 0 at [0,0] (and its 6 will go to [0,1] later).
4
Step 4: Sixes for the Equals and Finish
The equals region at [0,1] and [0,2] must hold identical values; the only remaining match is 6. Place the 6 from 0|6 at [0,1] and the 6 from 2|6 at [0,2]. Then the 2|6’s 2 fills [0,3], the 1|5’s 1 fills [1,3], and the 1|5’s 5 goes to the last open row‑2 cell [2,3]. All constraints are satisfied.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Double‑0 Equals
The equals region at [0,1] and [0,2] demands the same pip in both cells. The only domino with two identical numbers is the 0|0. Place it vertically (or horizontally) covering those two cells, locking them as zeros.
2
Step 2: Matching Sixes in the Side‑by‑Side Equals
The equals region [1,2] and [2,2] now must be filled. With no more double dominos, both cells will get 6 from different dominos: the 6|1 domino puts its 6 at [1,2] and the 0|6 domino puts its 6 at [2,2].
3
Step 3: Filling the Sum‑3 Region
The sum‑3 region covers [1,3] and [1,4]. Already the 6|1 domino places its 1 at [1,3] (since [1,2] got 6). Thus [1,4] must be a 2 to reach 3. Use the 2|0 domino with its 2 at [1,4] and its 0 at [2,4].
4
Step 4: Sum‑4 Region Allocation
The sum‑4 region at [1,5] and [1,6] must total 4. The 0|1 domino places its 1 at [1,5] (and its 0 at [0,5], an empty single-cell region). Then [1,6] needs a 3, which comes from the 2|3 domino placed with 3 at [1,6].
5
Step 5: Remaining Singles and Fills
Finish the grid: the 5|0 domino puts its 5 at [1,0] (empty) and its 0 at [1,1] to meet the equals region [1,1]/[2,1] (which gets its matching 0 from the 0|6 domino at [2,1]). The 2|3’s 2 goes to the greater‑0 cell [2,6]. All dominos are placed.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Sum‑10 in Column 0
The region [1,0] and [2,0] must sum to 10. With pips up to 6, the only combination is 5+5. The double‑5 domino (5|5) is the only way to place two 5s adjacent. Orient it vertically from [1,0] to [2,0].
2
Step 2: The Big Equals Column of 1s
Region [0,1] to [4,1] must all hold the same value. A vertical domino can cover two adjacent cells, but five cells need horizontal links. Place the 1|1 double domino vertically at [2,1]–[3,1] as the core. Then [0,1] gets its 1 from the 0|1 domino (covering [0,0]/[0,1]), [1,1] gets 1 from the 1|3 domino (covering [1,1]/[1,2]), and [4,1] gets 1 from the 4|1 domino (covering [4,1]/[5,1]).
3
Step 3: Zero‑Heavy Top Row
A less‑6 single at [0,0] and a less‑2 pair at [0,3]/[0,4] force zeros. Place the 0 from 0|1 at [0,0] (giving 1 to [0,1] as above). The sum‑2 region at [0,2] alone forces 2 there, so the 0|2 domino places its 2 at [0,2] and its 0 at [0,3]. Then [0,4] gets 0 from the 5|0 domino, which will also place its 5 at [1,4] later.
4
Step 4: Second Sum‑10 and the 3s Equals
The region [1,4]/[2,4] must sum to 10, so both become 5. Use the 5|0 domino’s 5 at [1,4] (0 already at [0,4]) and the 3|5 domino’s 5 at [2,4]. The equals region spanning [1,2],[1,3],[2,3],[3,3],[3,4] now forces 3s. Place the 3|3 domino vertically at [1,3]–[2,3], the 3 from 1|3 at [1,2] (1 already at [1,1]), the 3 from 3|5 at [3,4], and the 3 from 3|0 at [3,3].
5
Step 5: Lower Sums and Equals Wrap‑Up
The sum‑6 region [4,3]/[4,4] takes the 6 from 2|6 at [4,4] and the 0 from 3|0’s 0 at [4,3] (its 3 used at [3,3]). The sum‑3 single [3,0] takes 3 from the 3|2 domino, which gives its 2 to [4,0]. The sum‑2 region [4,0]/[5,0] then gets 0 from 4|0 at [5,0] and 4 from the same 4|0 at [6,0]. The equals‑4 group [5,1]/[6,0]/[6,1] gets 4 from 4|0 already at [6,0], 4 from 4|1 at [5,1] (1 used above), and 4 from 4|2 at [6,1] (2 goes to [6,2]). Finally, the equals‑2 group [5,4]/[6,2]/[6,3]/[6,4] gets 2 from 2|6 at [5,4] (6 at [4,4]), 2 from 4|2 at [6,2], and the 2|2 domino at [6,3]/[6,4].

💡 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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