NYT Pips Hints & Answers for August 22, 2026

Aug 22, 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

Ian Livengood's NYT Pips easy is built on deliberate domino scarcity: one double, one low-cap singleton, and a handful of equal-pip regions that interlock. The result is a compact chain where spotting the only tile that can satisfy an equal pair immediately sends a signal through neighboring less-than and greater-than regions.

Rodolfo Kurchan's medium shifts the design toward mirrored bands. An all-equal trio along the top row sets a high anchor, while a low-cap trio just below it quietly forces a cascade of zeroes and ones. Kurchan's placement of the double is particularly elegant — it is the key that turns an ambiguous row into a rigid frame.

By the hard puzzle, Kurchan opens the board into a lattice of small sum constraints, including a central column with a near-minimal total and a neighboring two-cell anchor with the smallest possible nonzero sum. The design keeps solvers zooming between distant corners, but each cluster ultimately dovetails into a single right-side ladder and top-band resolution.

💡 Progressive Hints

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

💡 Spot the all-equal anchor
Scan for a region that forces two adjacent cells to match; in this early grid that constraint is extremely restrictive because only one double exists.
💡 Let the double lock the top row
Look at the top-right pair at [0,3] and [0,4]. They demand the double, which tells you exactly which domino starts there. Then the less-than singleton at [0,0] must lean on [0,1].
💡 Complete the cascade
Place [2,2] at [0,3]-[0,4]; then [2,6] covers [0,0]-[0,1] with 2 in [0,0] and 6 in [0,1]; [0,6] goes vertically at [2,1]-[1,1] with 0 below and 6 above; [0,5] spans [2,2]-[2,3] with 0 and 5; and [1,3] closes [3,4]-[2,4] with 1 and 3.
💡 Follow the equals trio
In today's medium, the most restrictive feature is a three-cell region that all must share the same value. Only one tile can begin that line; find it and you've found the puzzle's hinge.
💡 Work the top two bands
The three-cell equals at [0,2]-[0,4] needs the [4,4] double across [0,2]-[0,3], plus another 4 at [0,4]. The greater-than singleton at [0,5] then forces a high-low vertical pair.
💡 Complete the mirrored rows
Put [4,4] at [0,2]-[0,3]; [0,4] at [1,4]-[0,4] with 0 below and 4 above; [6,0] at [0,5]-[1,5] with 6 above and 0 below. On row 1, use [2,1] at [0,1]-[1,1] for 2 above and 1 below, [1,4] at [1,0]-[0,0] for 1 below and 4 above, and [1,0] at [1,2]-[1,3] for 1 and 0. Finish [3,4] at [2,2]-[2,3] with 3 and 4.
💡 Locate the sum-2 column
Start with the tightest multi-cell sum on the board — a vertical three-cell region that must total only a hair above zero. That tiny budget dictates where the lone [1,1] double belongs.
💡 Anchor the central column
The region at [3,3], [4,3], and [5,3] can only split as 1,1,0. Put the [1,1] double at [3,3]-[4,3], then a zero-carrying domino must supply [5,3]; its other end lands in the empty [5,2].
💡 Exploit the sum-1 hinge
The pair at [5,5]-[5,6] is even more rigid: only 0 and 1 add to its target. The 1 must come from the [1,5] domino along [5,6]-[5,7], which leaves the 0 for a vertical [6,0] running up into [4,5].
💡 Climb the right-side sum ladder
With [5,7] set high and [4,5] set to its maximum, the sum-8 and sum-10 regions force [4,7] and [3,5]. That ripples into the sum-5 and sum-12 pairs along rows 2 and 3, so the upper right corner cascades.
💡 Finish the top and left clusters
Place [1,1] at [3,3]-[4,3]; [3,0] at [5,2]-[5,3] for 3 and 0; [1,5] at [5,6]-[5,7] for 1 and 5; [6,0] at [4,5]-[5,5] for 6 and 0. Then [3,4] at [2,5]-[3,5] for 3 and 4, [2,6] at [2,6]-[2,7] for 2 and 6, and [3,6] at [4,7]-[3,7] for 3 and 6. On top, [4,6] at [2,3]-[1,3] for 4 and 6, [1,6] at [0,2]-[0,3] for 1 and 6, [5,4] at [0,0]-[0,1] for 5 and 4, [0,5] at [2,0]-[1,0] for 0 and 5, [2,4] at [3,0]-[4,0] for 2 and 4, and [4,4] at [5,0]-[5,1] for 4 and 4. Finish [0,4] at [6,7]-[7,7] for 0 and 4, and [2,2] at [6,2]-[7,2] for 2 and 2.

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Aug 22, 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 22, 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 22, 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: Find the lone double
The only all-equal clue that can force an immediate tile is the top-right pair [0,3]-[0,4]. Since the domino list has exactly one double, [2,2], it must lie horizontally there with both cells as 2.
2
Step 2: Use the less-than singleton
The single cell [0,0] must stay under 3. Its only available neighbor is [0,1], so the [2,6] domino spans [0,0]-[0,1], placing the low 2 in [0,0] and the high 6 in [0,1].
3
Step 3: Satisfy the vertical equals
With [0,1] now 6, the equals region shared with [1,1] forces [1,1] to be 6. The [0,6] domino therefore drops vertically over [2,1]-[1,1], giving [2,1] the 0.
4
Step 4: Cascade through the low row
The equals pair [2,1]-[2,2] forces [2,2] to 0, so [0,5] spans [2,2]-[2,3] with 5 in [2,3]. The greater-6 pair [2,3]-[2,4] then takes the 3 in [2,4], and the last domino [1,3] fills [3,4]-[2,4] with 1 below and 3 above.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Lock the top equals trio
The trio at [0,2]-[0,4] must all match, and only the [4,4] double can provide two adjacent equal cells. Place it horizontally on [0,2]-[0,3].
2
Step 2: Complete the top row anchor
The remaining cell [0,4] must also be 4. The [0,4] domino supplies it vertically from [1,4], giving [1,4] a 0 and satisfying part of the less-than-2 band.
3
Step 3: Resolve the far-right singleton
The greater-than-3 cell at [0,5] can only take the 6 from the [6,0] domino. That tile drops to [1,5] with 0, completing the less-than-2 region at [1,5].
4
Step 4: Build the row-1 equality chain
The equals trio [1,0]-[1,2] needs all 1s. Three separate dominoes cooperate: [2,1] covers [0,1]-[1,1], [1,4] covers [1,0]-[0,0], and [1,0] covers [1,2]-[1,3], simultaneously satisfying [0,1]'s less-than-4 cap and [1,3]'s less-than-2 cap.
5
Step 5: Finish the lower pair
The final two singletons [2,2] and [2,3] are pinned by their own constraints: [2,2] must stay under 4, while [2,3] must exceed 3. The [3,4] domino spans them with 3 in [2,2] and 4 in [2,3].

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Anchor the tiny three-cell sum
The central column region [3,3], [4,3], and [5,3] has a sum of only 2, so its only possible arrangement is 1,1,0. Since [1,1] is the lone double available, it must sit vertically at [3,3]-[4,3], leaving [5,3] to be 0.
2
Step 2: Complete the central sum with a zero
With [5,3] forced to 0, the [3,0] domino must cover [5,2]-[5,3], placing 3 in the empty cell [5,2]. This locks the lower center even before the surrounding sums are touched.
3
Step 3: Crack the sum-1 and sum-10 bridge
The two-cell region [5,5]-[5,6] must sum to 1, so it can only be 0 and 1. The 1 has to come from [1,5] laid along [5,6]-[5,7], giving [5,7] a 5. Then [5,5] is 0, so [6,0] runs vertically [4,5]-[5,5] with 6 above, setting up the sum-10 partner [3,5]-[4,5].
4
Step 4: Climb the right-side ladder
From [5,7] and [4,5], the adjacent sums resolve outward. The sum-8 region [4,7]-[5,7] forces [4,7] to 3, which puts the 6 of [3,6] in [3,7]. The sum-10 pair then makes [3,5] 4 with [3,4] spanning [2,5]-[3,5]; sum-5 at [2,5]-[2,6] makes [2,6] 2, and sum-12 at [2,7]-[3,7] makes [2,7] 6, so [2,6] covers [2,6]-[2,7].
5
Step 5: Unlock the top band
The high three-cell sum [0,3]-[2,3] now needs two 6s and a 4. Since [2,3] takes 4 from the [4,6] domino along [2,3]-[1,3], [1,3] gets 6. Then [1,6] covers [0,2]-[0,3] to give [0,3] its 6 and [0,2] a 1. The adjacent sum-5 pair [0,1]-[0,2] then forces [0,1] to 4, so [5,4] spans [0,0]-[0,1] with 5 in [0,0].
6
Step 6: Close the left and bottom clusters
The equals region [0,0]-[1,0] forces [1,0] to 5, so [0,5] drops [2,0]-[1,0] with 0 in [2,0]. Sum-2 at [2,0]-[3,0] then makes [3,0] 2, so [2,4] covers [3,0]-[4,0] with 4 at [4,0]. The equals trio [4,0], [5,0], and [5,1] is completed by [4,4] on [5,0]-[5,1]. Finally, [0,4] goes [6,7]-[7,7] for sum-4, and [2,2] covers [6,2]-[7,2] for the bottom equals.

💡 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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