🚨 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!
Click here to play today's official NYT Pips game first.
Want hints instead? Scroll down for progressive clues that won't spoil the fun.
🎲 Today's Puzzle Overview
Easy in today's NYT Pips is a confidence-builder — Livengood hands you two single-cell sums and a one-cell greater-than gate, then funnels everything through a large unequal region. It should feel quick, with no real forks once the singles are down.
Medium, from Rodolfo Kurchan, is a one-bottleneck puzzle. The key is a vertical sum-6 pair in the middle column; solvers should expect to stall briefly there. Once that hinge splits correctly, the adjacent less-than and equals regions collapse in a very satisfying chain.
Hard, also from Kurchan, is the day's heavy build. The upper half is a web of equals constraints, highlighted by a five-cell equals block, and the bottom row is a sequence of sum locks. The main difficulty is psychological: matching the top equals web while keeping the bottom sums coherent.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Scan for the singles
Start by scanning for single-cell regions. One asks for a specific sum, another for a sum one above, and a third demands a value above a threshold. These are your free gifts.
💡 Top row and left-edge singles
The single-cell sum regions are at [1,0] and [0,2]; the greater-than gate is at [1,3]. Solve those cells first, then let their domino halves reach into the neighboring unequal region.
💡 Full easy chain
Domino [3,6] lies horizontally at [1,0]-[1,1] with 3 on the left and 6 on the right; [1,4] spans [0,1]-[0,2] with 1 then 4; [2,6] spans [1,2]-[1,3] with 2 then 6. The remaining dominoes are [5,3] at [2,1]-[2,2] (5 left, 3 right) and [5,5] at [2,3]-[2,4].
💡 Find the lone gate and the vertical sum
Find the lone greater-than cell first. Then look for a two-cell sum region arranged vertically; its sum restriction is the puzzle's central squeeze. Focus on how the available pip values can split.
💡 The middle-column hinge
The key cells are [0,1] for the greater-than gate and the stacked pair [0,2]/[1,2] for the sum. The sum's two cells must take the same pip count, so it forces two different dominoes to contribute that matching value. The less-than pair to the right then fixes the vertical chain.
💡 Full medium chain
Complete medium: [5,5] spans [2,4]-[3,4] vertically, both 5; [2,2] spans [2,2]-[2,3] horizontally; [3,1] spans [0,2]-[0,3] with 3 then 1; [2,6] spans [1,1]-[0,1] vertically with 2 below 6; [1,5] spans [0,4]-[1,4] with 1 over 5; [2,4] spans [2,1]-[2,0] with 2 and 4; [5,3] spans [1,2]-[1,3] with 3 and 5.
💡 Equals first
Before placing anything, identify the clusters where cells must match. Today's hard has a tall left-edge equals stack, a top-left pair, and a sprawling five-cell equals region. Let those repeated-value constraints frame your solve.
💡 Top equals blocks force early placements
The vertical stack at [0,0], [1,0], [2,0] must all match, and the pair at [0,1]/[0,2] must match. Only high-value doubles and adjacent mixed dominoes can satisfy these tight blocks. Also check the single-cell sum at [1,2] for an early lock.
💡 The five-cell cascade
Once the top-left pairs are set, follow the five-cell equals region covering [0,3], [1,3], [1,4], [2,3], [2,4]. It is the puzzle's core, and it shares a boundary with the sum-7 pair [0,4]/[0,5]. That adjacency narrows which pip value all five can hold.
💡 Bottom-row sums and right-side stairs
After the upper equals cascade, use the sum-6 pair at [3,0]/[4,0], then the equal triple at [3,3]/[3,4]/[3,5], and finish with the bottom-row sum locks at [5,0]/[5,1], [5,2]/[5,3], and [5,4]/[5,5]. The zero sum is a giveaway.
💡 Full hard chain
Place [5,5] vertically at [0,0]-[1,0]; [5,4] vertically at [2,0]-[3,0] with 5 over 4; [4,4] horizontally at [0,1]-[0,2]. The five 3-equals: [3,6] at [0,3]-[0,4] (3,6), [3,3] at [1,4]-[2,4], [3,5] at [1,3]-[1,2] (3,5), [3,1] at [2,3]-[2,2] (3,1). Then [5,1] at [1,5]-[0,5] (5,1), [5,6] at [2,5]-[3,5] (5,6); [6,6] at [3,4]-[4,4], [6,1] at [3,3]-[3,2] (6,1). Finish: [2,0] at [4,0]-[5,0] (2,0), [3,0] at [5,2]-[5,1] (3,0), [2,1] at [5,4]-[5,3] (2,1), [2,4] at [4,5]-[5,5] (2,4).
🎨 Pips Solver
Aug 27, 2026
Click a domino to place it on the board. You can also click the board, and the correct domino will appear.
✅ 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
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
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: Cash in the single-cell sums
The lone sum-3 cell at [1,0] must be 3, and the lone sum-4 cell at [0,2] must be 4. For today's intended path, the 3-bearing [3,6] domino locks across [1,0]-[1,1] with 6 on the right, while the [1,4] domino settles across [0,1]-[0,2] with 1 on the left.
2
Step 2: Open the greater-than gate
The one-cell greater-than-5 region at [1,3] must be 6. Place the [2,6] domino across [1,2]-[1,3], putting 2 at [1,2] and 6 at [1,3] to feed the big unequal region without repeating a value.
3
Step 3: Let the unequal region sort itself
The unequal region now holds [0,1]=1, [1,1]=6, and [1,2]=2. Its remaining cells [2,1] and [2,2] must be different from those and from each other, so the [5,3] tile fits across [2,1]-[2,2] with 5 then 3.
4
Step 4: Close the equal pair
The last unused domino [5,5] completes the equal region at [2,3]-[2,4], placed horizontally with both cells showing 5.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Solve the lone greater-than cell
The single-cell region at [0,1] demands a value above 4, so it must be 6. The only 6-bearing domino is [2,6], so it locks vertically with [0,1]=6 and [1,1]=2.
2
Step 2: Extend the equal pair to the empty cell
Because [1,1] is 2, the equals pair at [1,1]/[2,1] forces [2,1]=2. Place the [2,4] domino across [2,1]-[2,0] with 2 at [2,1] and 4 at [2,0], filling the empty single-cell region.
3
Step 3: Crack the vertical sum-6 hinge
The two-cell sum region at [0,2]/[1,2] must total 6. Both cells need a 3 to make that sum work with the remaining tiles: [3,1] spans [0,2]-[0,3] (3 then 1), and [5,3] spans [1,2]-[1,3] (3 then 5).
4
Step 4: Settle the less-than and equals runs
With [1,3]=5, the equals pair at [1,3]/[1,4] forces [1,4]=5. That locks [1,5] vertically at [0,4]-[1,4] with 1 over 5, which also satisfies the less-than-4 region at [0,3]/[0,4] since [0,4]=1.
5
Step 5: Finish with the two doubles
Remaining dominoes [2,2] and [5,5] slot into the two open equals corners. Put [2,2] horizontally at [2,2]-[2,3] for its equal pair, and [5,5] vertically at [2,4]-[3,4] to complete the bottom-right corner.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Crack the left-edge equals stack
The vertical equals region at [0,0],[1,0],[2,0] must be three identical pips. Use [5,5] vertically across [0,0]-[1,0], then the [5,4] domino vertically across [2,0]-[3,0] with 5 at [2,0] and 4 at [3,0]. The top-left pair [0,1]/[0,2] also must match, so [4,4] goes horizontally with 4s.
2
Step 2: Unlock the five-cell equals web
The region [0,3],[1,3],[1,4],[2,3],[2,4] is one big equals block, and it must sit next to the sum-7 pair [0,4]/[0,5] and the sum-5 single [1,2]. That forces the common pip to be 3. Place [3,6] across [0,3]-[0,4] (3,6) to help satisfy sum 7, then [3,3] vertically at [1,4]-[2,4], [3,5] horizontally at [1,3]-[1,2] (3,5), and [3,1] at [2,3]-[2,2] (3,1).
3
Step 3: Complete the right-edge equals chain
The sum-7 pair [0,4]/[0,5] now has [0,4]=6, so [0,5] must be 1. Place [5,1] vertically at [1,5]-[0,5] with 5 below and 1 on top. The equal pair [1,5]/[2,5] then forces [2,5]=5, so [5,6] settles vertically at [2,5]-[3,5] with 5 then 6.
4
Step 4: Build the center-right 6-equals triple
The equals triple [3,3],[3,4],[3,5] must all be 6. With [3,5]=6, [6,6] locks vertically at [3,4]-[4,4] (both 6) and [6,1] locks horizontally at [3,3]-[3,2] (6,1). The equals pair [2,2]/[3,2] then confirms [2,2]=1, already sitting on the [3,1] domino.
5
Step 5: Work the lower-left sums
The sum-6 region [3,0]/[4,0] has [3,0]=4, so [4,0]=2 and [2,0] must cover [4,0]-[5,0] with 0 at [5,0]. The sum-0 pair [5,0]/[5,1] then makes [5,1]=0, so [3,0] spans [5,2]-[5,1] with 3 and 0. The sum-4 pair [5,2]/[5,3] then forces [5,3]=1, placing [2,1] across [5,4]-[5,3] with 2 and 1.
6
Step 6: Finish the lower-right sums
The sum-8 pair [4,4]/[4,5] has [4,4]=6, so [4,5]=2 and [2,4] goes vertically at [4,5]-[5,5] with 4 at [5,5]. The bottom-right sum-6 pair [5,4]/[5,5] is then fulfilled by [5,4]=2 and [5,5]=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.
💬 Community Discussion
Leave your comment