NYT Pips Hints & Answers for August 15, 2026

Aug 15, 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, by Ian Livengood, is a confidence-builder — today's NYT Pips easy has one tight sum block that immediately forces a double, and the neighboring four-cell equals region turns that single deduction into a cascade. You will not hit forks; the board is compact and every remaining domino falls out in order.

Medium, also by Livengood, is a one-bottleneck puzzle. The lower less-than strip is the key: read its total carefully, orient the crossing dominoes, and the sum bridge plus the equals pair unlock the rest. It should feel moderate and satisfying.

Hard, by Rodolfo Kurchan, is the real climb. Sparse space, two tiny sum traps, two four-cell equals chains, and a top greater-than block that demands near-maximum pips. The bottlenecks are many; expect it to feel significantly harder than the other two.

💡 Progressive Hints

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

💡 Look for the sum that has to use a double
Start with the sum-type region. It is unusually tight, and one of the available doubles covers two of its three cells cleanly.
💡 Anchor the top-left block
The sum region at [0,0], [0,1], and [1,0] must be balanced by a single outside cell. That forces the left-column domino, and then the adjacent four-cell equals region inherits its value.
💡 Full easy solve
Place [4,4] at [0,0]/[1,0] and [0,6] at [0,1]/[1,1]. The equals block [0,2], [0,3], [1,1], [1,2] is all 6, so place [6,6] at [0,2]/[0,3]. Then [6,5] goes at [1,2]/[2,2] to satisfy the lone sum-5 cell, and [3,3] finishes [1,3]/[2,3].
💡 Hunt for the low-total region
Look for a multi-cell less-than region near the bottom. The dominoes that cross into it must all put their smaller pip inside that strip.
💡 Solve the bottom bottleneck first
The less-than region at [3,1]–[3,3] is the hinge. Orient the vertical dominoes there, then use the sum region above at [1,1]/[2,1] to place the [3,6] domino and bridge upward.
💡 Full medium solve
Place [1,3] at [3,1]/[2,1] with [3,1]=1, [2,1]=3; [2,0] at [4,2]/[3,2] with [4,2]=2, [3,2]=0; and [1,6] at [3,3]/[2,3] with [3,3]=1, [2,3]=6. The sum-6 region forces [1,1]=3, so place [3,6] at [1,0]/[1,1]. The equals pair at [1,2]/[1,3] forces [4,2] at [0,2]/[1,2] and [6,2] at [0,3]/[1,3]. Finally [4,6] goes at [0,0]/[0,1].
💡 Look for tiny sums and equals chains
Start with the sum-to-zero traps and four-cell equals regions. These tight totals do a lot of the placing before any single-cell greater region becomes useful.
💡 Anchor the middle and left chain
Begin with the sum-0 region at [2,4], [2,5], [3,4], then lock the four-cell equals chain at [1,1], [1,2], [2,1], [3,1]. These force several small dominos early.
💡 Attack the top greater block
The top row greater-than region crossing [0,4], [0,5], [1,4] needs near-maximum pips. That locks the top-right corners and the cell at [1,4].
💡 Work the right-center and lower zero strip
Next build the right-middle four-cell equals block, then clear the lower sum-to-zero pair. Finish with the sum-8 pair near column 5 and the equal pair in row 4.
💡 Full hard solve
Place [0,0] at [2,4]/[2,5] and [2,0] at [4,4]/[3,4] with [3,4]=0, [4,4]=2. Left equals chain: [1,1] at [1,1]/[2,1], [2,1] at [0,2]/[1,2] with [0,2]=2, and [0,1] at [4,1]/[3,1] with [4,1]=0, [3,1]=1. Top: [6,5] at [0,4]/[1,4] gives 6 at [0,4] and 5 at [1,4]; [6,4] at [0,5]/[0,6] gives 6 at [0,5] and 4 at [0,6]; [0,5] at [1,0]/[0,0] gives 0 at [1,0] and 5 at [0,0]. Right-middle: [4,4] at [6,2]/[7,2], [4,3] at [7,3]/[8,3] with [7,3]=4, [8,3]=3, and [4,5] at [8,2]/[9,2] with [8,2]=4, [9,2]=5. Lower zeros: [3,0] at [8,4]/[8,5] with [8,5]=0, [8,4]=3; [0,4] at [9,5]/[9,4] with [9,5]=0, [9,4]=4. Finish with [6,2] at [6,5]/[7,5] for the sum-8 pair, and [6,6] at [4,5]/[4,6].

🎨 Pips Solver

Aug 15, 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 15, 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 15, 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: Lock the top-left sum
The sum region at [0,0], [0,1], [1,0] must reach its exact total. Placing the [4,4] domino in [0,0] and [1,0] gives eight from those two cells, which forces [0,1] to be 0. That means the [0,6] domino must provide the 0 at [0,1] and a 6 at [1,1].
2
Step 2: Cascade through the equals block
With [1,1] set to 6, the adjacent equals region [0,2], [0,3], [1,1], [1,2] must all show the same pip count. Therefore [0,2], [0,3], and [1,2] are also 6. The top of that block is completed by placing [6,6] horizontally at [0,2]/[0,3].
3
Step 3: Satisfy the single-cell sum
Because [1,2] is 6 and the cell directly below at [2,2] is a lone sum region, the only way to hit that single-cell total is with the [6,5] domino placed vertically.
4
Step 4: Finish the last equals pair
The remaining uncovered cells are [1,3] and [2,3], which form a two-cell equals region. Place [3,3] there, and the board is complete.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Start with the bottom less-than strip
The region covering [3,1], [3,2], and [3,3] has a very small total, so no crossing domino can put its high pip into that row. The [1,3] domino must therefore place a 1 at [3,1] and the 3 above at [2,1].
2
Step 2: Keep the strip small
With [3,1]=1, the middle cell [3,2] cannot be a 2 or the total would blow past the limit. The [2,0] domino is forced to put its 0 at [3,2], leaving a 2 at [4,2].
3
Step 3: Close out the strip
The right cell [3,3] has to stay small as well, so the [1,6] domino goes in with its 1 at [3,3] and 6 at [2,3]. The lower less-than region is now satisfied.
4
Step 4: Bridge upward through the sum
The sum region at [1,1]/[2,1] sees the 3 already at [2,1], so [1,1] must be 3. That places the [3,6] domino with 6 at [1,0]. The equals pair [1,2]/[1,3] then forces each of their lower dominoes to show a 2 in row 1.
5
Step 5: Finish the top row
With [0,2]=4 and [1,2]=2 from [4,2], and [0,3]=6 and [1,3]=2 from [6,2], the greater-than region at [0,1]/[0,2] requires [0,1]=6. Place [4,6] at [0,0]/[0,1], completing the grid.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Zero the middle sum
The sum-to-zero region [2,4], [2,5], [3,4] is merciless: every cell in it must be 0. Place the [0,0] domino across [2,4]/[2,5]. The [2,0] domino then must put its 0 at [3,4], which leaves a 2 at [4,4] and satisfies the less-than region there.
2
Step 2: Equalize the left chain
The four-cell equals region at [1,1], [1,2], [2,1], [3,1] is locked by the double [1,1] at [1,1]/[2,1]. To keep all four equal, the [2,1] domino goes at [0,2]/[1,2] with its 1 at [1,2] and 2 at [0,2]; the [0,1] domino goes at [4,1]/[3,1] with its 1 at [3,1] and 0 at [4,1]. The greater-than single at [0,2] is also handled.
3
Step 3: Max out the top greater block
The region [0,4], [0,5], [1,4] has a very high greater-than target, so it needs near-maximum pips. Place [6,5] at [0,4]/[1,4] with 6 at [0,4] and 5 at [1,4]. Then [6,4] goes at [0,5]/[0,6] with 6 at [0,5] and 4 at [0,6]. The [0,5] domino also locks [1,0]=0 and [0,0]=5.
4
Step 4: Build the right-center equals block
The four-cell equals region [6,2], [7,2], [7,3], [8,2] all share one value. The [4,4] domino occupies [6,2]/[7,2]. Then [4,3] must put its 4 at [7,3] and 3 at [8,3]; [4,5] puts 4 at [8,2] and 5 at [9,2].
5
Step 5: Clear the lower zero strip
The sum-to-zero region [8,5]/[9,5] forces two more 0s. Place [3,0] at [8,4]/[8,5] with 0 at [8,5] and 3 at [8,4], and [0,4] at [9,5]/[9,4] with 0 at [9,5] and 4 at [9,4].
6
Step 6: Finish with the themed pairs
The sum-8 region [6,5]/[7,5] is satisfied by [6,2] with 6 at [6,5] and 2 at [7,5]. The final equals pair [4,5]/[4,6] takes [6,6], and all open single cells now sit within their constraints.

💡 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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