NYT Pips Hints & Answers for August 28, 2026

Aug 28, 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.

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🎲 Today's Puzzle Overview

Easy: Ian Livengood's grid opens on two independent footholds — a four-cell less-than block that behaves as a rigid zero spine and a pair of singleton sum cells that force adjacent domino pips. From those anchors the deduction graph is almost linear: each forced value narrows the remaining domino pool until the tiling locks. Medium: Rodolfo Kurchan's grid is a two-track solve. A vertical three-cell equals chain is pinned by a greater-than singleton, while a separate two-cell equals chain and a singleton sum cell feed a high-sum bridge and a low sum closure. The domino placements are less about search and more about reading which shared edge carries the forced repeated pip. Hard: Rodolfo Kurchan's hard puzzle is the most constraint-dense; four anchors — a zero-sum pair, singleton sum cells, a greater-than pair, and a bottom-row less-than band — create a left-to-right cascade. This NYT Pips hard rewards solvers who chain single-cell deductions through domino adjacency before attacking the bottom row.

💡 Progressive Hints

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

💡 A rigid zero spine
Look for a less-than region whose upper bound forces every cell in it to the same pip. That region is the puzzle's first domino magnet.
💡 Singletons name their pips
The four-cell region includes [0,1], [1,1], [1,2], and [2,1]. The singleton sum cells at [0,2] and [1,0] each pin exact values, pulling specific 0-carrying and 2-carrying dominoes to their edges.
💡 Lock the full tiling
Place [2,0] across [0,2]=2 and [0,1]=0, [3,2] across [1,0]=2 and [2,0]=3, [0,0] at [1,1]/[2,1], [0,5] at [1,2]=0/[1,3]=5, and [1,1] at [2,2]/[3,2].
💡 Follow the forced singletons
Start with the single-cell greater/less/sum constraints. One cell has a lower bound that only the highest pip can satisfy, and a sum singleton nearby names its exact pip.
💡 Pin the equals ladders
The greater-than singleton at [1,4] must be 6, and it drags the adjacent triple-equals column [0,3]/[1,3]/[2,3] to 4. The sum-3 singleton at [0,1] does similar work for the two-cell equals chain at [0,2]/[1,2].
💡 Complete the Kurchan chain
Place [6,4] at [1,4]=6/[1,3]=4; [0,4] at [0,4]=0/[0,3]=4; [5,4] at [3,3]=5/[2,3]=4; [3,3] at [0,1]=3/[0,2]=3; [3,5] at [1,2]=3/[2,2]=5; [6,1] at [2,1]=6/[2,0]=1; [1,3] at [3,1]=3/[3,2]=1.
💡 Find the small anchors
The grid is anchored by small sum and zero constraints. Look for a two-cell sum-zero region and a few single-cell sum cells; they create tight toeholds that ripple across the top row.
💡 Zero corner and sum singletons
The sum-0 pair at [0,0]/[1,0] takes the double-zero domino. Then [0,1]'s singleton sum-1 and the sum-10 pair [0,2]/[1,2] force a 4/1 split, while the singleton sum-1 at [0,5] pins the 6/1 domino beside the greater-than-10 region.
💡 Top-left and top-middle cascades
After [0,1]=1 and [0,2]=4, the sum-10 pair makes [1,2]=6; since [1,1] is a singleton sum-5, the [5,6] domino runs across [1,1]/[1,2]. On the middle top, [0,5]=1 drags [0,4]=6, so [0,3]=5 satisfies the greater-than-10 pair; the singleton sum-2 at [1,3] then takes [5,2] across [0,3]/[1,3].
💡 Resolve the right side
Move to the right side. The sum-0 singleton [1,4] forces a zero, and the sum-11 pair [1,5]/[1,6] resolves as 6/5: place [0,6] at [1,4]/[1,5], then [3,5] at [0,6]/[1,6] to satisfy the top sum-7 pair with [0,7]=4. The [4,0] domino then covers [0,7]/[1,7], which feeds the sum-2 pair [1,7]/[2,7].
💡 Dense final answer
Place [0,0] at [0,0]/[1,0]; [4,1] at [0,1]=1/[0,2]=4; [5,6] at [1,1]=5/[1,2]=6; [6,1] at [0,4]=6/[0,5]=1; [5,2] at [0,3]=5/[1,3]=2; [0,6] at [1,4]=0/[1,5]=6; [3,5] at [0,6]=3/[1,6]=5; [4,0] at [0,7]=4/[1,7]=0; [0,2] at [2,7]=2/[3,7]=0; [3,1] at [2,0]=3/[3,0]=1; [1,2] at [3,1]=1/[3,2]=2; [5,0] at [3,3]=5/[3,4]=0; [1,0] at [3,5]=1/[3,6]=0.

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Aug 28, 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 August 28, 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 August 28, 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: Zero-spine lock
The less-than-1 region at [0,1], [1,1], [1,2], and [2,1] admits only 0. This four-cell block is the puzzle's backbone: every adjacent non-zero clue must borrow a zero from it.
2
Step 2: Top sum-2 resolves a 2/0 domino
The singleton sum-2 at [0,2] must be 2. Its only sensible zero-spine neighbor is [0,1], so domino [2,0] is forced across [0,2]=2 and [0,1]=0.
3
Step 3: Lower sum-2 forces a 3/2 placement
The singleton sum-2 at [1,0] pins that cell to 2. With [2,0] already accounted for, the remaining [3,2] domino covers [1,0]=2 and [2,0]=3 — the empty region takes the extra pip.
4
Step 4: Close the spine and final pair
The greater-than-2 singleton [1,3] needs 5, so [0,5] spans [1,2]=0 and [1,3]=5. The last zero-spine pair [1,1]/[2,1] is [0,0], and the sum-2 pair [2,2]/[3,2] takes [1,1] as 1+1.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Greater-than singleton pins the equals column
The single-cell greater-than-5 region at [1,4] must be 6. Since its adjacent neighbor [1,3] sits in the three-cell equals chain, the [6,4] domino is forced across [1,4]=6 and [1,3]=4, setting the entire column's repeated value.
2
Step 2: Complete the three-cell equals chain
With [1,3]=4, [0,3] and [2,3] must also be 4. Domino [0,4] covers [0,4]=0/[0,3]=4, satisfying the less-than-5 singleton. Domino [5,4] covers [3,3]=5/[2,3]=4, satisfying the less-than-6 singleton at [3,3].
3
Step 3: Left equals pair from the sum singleton
The singleton sum-3 at [0,1] forces 3 there. It pairs with [0,2] via domino [3,3], so [0,2]=3; by the two-cell equals rule, [1,2] must also be 3.
4
Step 4: Sum-11 bridge resolves
The sum-11 region [2,1]/[2,2] needs 6 and 5. Domino [3,5] puts 3 at [1,2] and 5 at [2,2]; domino [6,1] puts 6 at [2,1] and 1 at [2,0], which fits the less-than-3 singleton.
5
Step 5: Bottom sum-4 closure
The final region [3,1]/[3,2] must sum to 4. The unused [1,3] domino is the only fit, placed as [3,1]=3 and [3,2]=1.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Top-left zero anchor
The sum-0 region [0,0]/[1,0] forces both cells to 0. Domino [0,0] is immediate across [0,0] and [1,0], creating the first fixed corner.
2
Step 2: Sum-1 and sum-10 force the 4/1 and 5/6 dominoes
The singleton sum-1 at [0,1] must be 1. It pairs with [0,2] via [4,1], so [0,2]=4. The sum-10 pair [0,2]/[1,2] then forces [1,2]=6. With [1,1]'s singleton sum-5 fixed to 5, [5,6] spans [1,1]=5 and [1,2]=6.
3
Step 3: Greater-than-10 pair and sum-1 at [0,5]
The singleton sum-1 at [0,5] forces 1 there, so [6,1] covers [0,4]=6/[0,5]=1. The greater-than-10 pair [0,3]/[0,4] then forces [0,3]=5; the singleton sum-2 at [1,3] takes [5,2] across [0,3]=5/[1,3]=2.
4
Step 4: Top-right sum-11 and sum-7 cascade
The sum-0 singleton [1,4] forces 0, and the sum-11 pair [1,5]/[1,6] must be 6+5. Place [0,6] at [1,4]=0/[1,5]=6, then [3,5] at [0,6]=3/[1,6]=5. That makes the sum-7 pair [0,6]/[0,7] force [0,7]=4, so [4,0] covers [0,7]=4/[1,7]=0.
5
Step 5: Bottom-left equals and sum-7 pair
The singleton sum-3 at [2,0] forces 3. Domino [3,1] covers [2,0]=3/[3,0]=1, so the equals pair [3,0]/[3,1] forces [3,1]=1. The sum-7 pair [3,2]/[3,3] then takes [1,2] at [3,1]=1/[3,2]=2, leaving [3,3]=5 for the final band.
6
Step 6: Close the less-than-2 band
With [5,0] spanning [3,3]=5/[3,4]=0, the less-than-2 band [3,4]-[3,7] locks to 0/1 values. Place [1,0] at [3,5]=1/[3,6]=0; the sum-2 pair [1,7]/[2,7] and the top-right [1,7]=0 force [0,2] across [2,7]=2/[3,7]=0.

💡 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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