NYT Pips Hints & Answers for September 9, 2026

Sep 9, 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!

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.

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🎲 Today's Puzzle Overview

Ian Livengood's easy grid hands you a compact solve in just a few moves. You'll likely notice the tight sum-4 L-shape first, because four cells with such a small total leaves almost no wiggle room; once that resolves, the single-cell greater-than clues and the mirror-image equals pair click into place like a short domino cascade.

Rodolfo Kurchan's medium puzzle feels like a row of dominoes with three vertical equals pairs at the bottom. From the solver's side, the tiny zero-sum singleton is a very generous anchor — it sends you upward into a sum-6 region and sideways into the lower columns. As you work, you're not so much searching for constraints as you are being handed the next locked pip value.

Kurchan's hard grid is a wide, spiky field of single-cell targets and long equals/vertical-sum regions. The solve opens in two far corners, then the middle gets squeezed: a top-row equals run, a sum-2 cluster, and bottom vertical sums all become one interlocking chain. This NYT Pips hard rewards solvers who trust the singleton constraints as doorways rather than dead ends.

💡 Progressive Hints

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

💡 Look low and tight
Scan for a multi-cell sum region with a very small target — that's where the smallest pip values in the available domino set have to cooperate.
💡 The L-shaped sum cauldron
The four-cell region at [0,2], [1,2], [1,3], and [2,2] is the key. Its tiny total means the vertical pair [1,2]-[1,3] likely takes the only domino with two equal minimum pips, while [2,2] and [2,1] pair to keep the total exact.
💡 Easy answer lock
Place [1,1] on [1,2]-[1,3]. Put [1,0] on [2,2]-[2,1] with 1 on [2,2] and 0 on [2,1]. Use [0,6] on [2,0]-[1,0] with 0 on [2,0] and 6 on [1,0]. Then [3,2] goes [0,0]=3, [0,1]=2, and [1,5] goes [0,2]=1, [0,3]=5.
💡 Find the forced singleton
Look for a one-cell region whose target is an extreme value — the kind that tells you exactly which pip must sit there before you place any dominoes.
💡 The zero-sum spark
The lone cell at [2,4] has a sum target of 0. Whatever domino supplies that cell will drag a specific partner into the adjacent sum-6 region at [2,3], which then unlocks the cell at [1,3].
💡 Medium full chain
Place [0,3] with 0 at [2,4] and 3 at [2,3]; place [2,3] with 3 at [1,3] and 2 at [0,3]. Then [3,1] gives 1 at [2,1] and 3 at [3,1]; [5,6] gives 5 at [2,2] and 6 at [1,2]. Finish with [5,5] on [3,0]-[4,0], [5,3] with 3 at [4,1] and 5 at [4,2], and [5,4] with 5 at [3,2] and 4 at [3,3].
💡 Anchor on singletons
Start by hunting for one-cell sum targets and long equals runs. In this hard grid, the tightest doors are single cells that force both their own pip value and the value of a neighbor in a larger region.
💡 Bottom-right sum-13 chain
The singleton sum-1 at [8,5] is the opener. Its only neighbor [8,4] belongs to the vertical sum-13 column with [6,4] and [7,4], so placing [8,5]=1 forces [8,4]=5 and leaves the other two cells as a matched pair.
💡 Run the top-right equals
The sum-3 singleton at [0,8] forces 3, and its neighbor [0,7] is part of the four-cell equals run [0,4]-[0,7]. That makes the whole run lock to the same small value, and the domino reaching down to [1,6] sets up a vertical equals pair.
💡 Top-left squeeze
The sum-11 pair [0,1]-[0,2] and the sum-2 cluster below it form one interlock. Use the empty [0,0] and the zero-heavy tile in the cluster to fix [0,1], [0,2], [1,2], [2,1], and [2,2]; that leaves [3,2] with the only value that makes the cluster sum work.
💡 Hard complete answer
Place [1,5] on [8,5]-[8,4] as 1-5; [4,4] on [6,4]-[7,4] as 4-4. Use [3,1] at [0,8]-[0,7] as 3-1, [1,1] at [0,4]-[0,5], and [0,1] at [1,6]-[0,6] as 0-1. Top-left: [6,2] gives 6 at [0,1], 2 at [0,0]; [5,0] gives 5 at [0,2], 0 at [1,2]; [0,0] covers [2,1]-[2,2]. Put [4,2] as 2 at [3,2], 4 at [4,2]; [0,6] as 0 at [2,6], 6 at [3,6]; [5,6] as 5 at [4,6], 6 at [5,6]. Finish with [2,2] on [4,4]-[5,4], [3,3] on [1,0]-[2,0], and [3,4] with 3 at [3,0], 4 at [4,0].

🎨 Pips Solver

Sep 9, 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 Sept 9, 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 Sept 9, 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: Sum-4 squeeze
Four cells in the L-shaped region at [0,2], [1,2], [1,3], and [2,2] must sum to just 4. The tightest fit is to put the [1,1] domino on [1,2]-[1,3], locking two of them as 1s.
2
Step 2: Complete the sum-4 tail
The remaining sum-4 cells [0,2] and [2,2] must total 2, so each should be 1. Domino [1,0] can place its 1 on [2,2] and its 0 on [2,1], which starts the equals pair at [2,1].
3
Step 3: Equals and sum-9 column
With [2,1]=0, the equals region [2,0]-[2,1] makes [2,0]=0. The only remaining neighbor for [2,0] is [1,0], so place [0,6] with 0 at [2,0] and 6 at [1,0]. Together with [0,0]'s later 3, the sum-9 target is satisfied.
4
Step 4: Close the top row
Finally, [1,5] covers [0,2]-[0,3] with 1 at [0,2] and 5 at [0,3], satisfying the greater-4 cell. Then [3,2] covers [0,0]-[0,1] with 3 at [0,0] and 2 at [0,1], making the sum-9 and greater-1 constraints both work.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Zero-sum ignition
The singleton sum-0 at [2,4] can only ever be 0. The only domino that can cover that cell usefully is [0,3], so put its 0 on [2,4] and its 3 on [2,3]. This immediately gives half of the adjacent sum-6 region.
2
Step 2: Finish the right-side sum 6
With [2,3]=3, the sum-6 pair [1,3] and [2,3] forces [1,3] to also be 3. Place domino [2,3] on [0,3]-[1,3], putting 3 at [1,3] and 2 at [0,3]. That 2 satisfies the greater-1 singleton.
3
Step 3: Cross the middle sum 6
Now attack [2,1]-[2,2], also sum 6. Domino [3,1] can place 3 at [3,1] and 1 at [2,1], setting up the equals pair below. With [2,1]=1, [2,2] must be 5; put [5,6] there, giving 6 to the empty cell [1,2].
4
Step 4: Equal columns begin
The first vertical equals pair [3,0]-[4,0] needs matching pips. Domino [5,5] covers both with 5s. Then [3,1] from earlier is 3, so [4,1] must match; domino [5,3] supplies 3 at [4,1] and 5 at [4,2].
5
Step 5: Close the lower equal chain
The final equals pair [3,2]-[4,2] gets [4,2]=5 from the previous step, so [3,2] must be 5. Domino [5,4] covers [3,2]-[3,3] with 5 at [3,2] and 4 at [3,3], which is safely below the less-6 cap.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Bottom-right sum-1 lever
Start at [8,5], the singleton sum-1. It has to be 1. Because [8,4] is its only partner and sits in the sum-13 column, place [1,5] with 1 at [8,5] and 5 at [8,4]. That leaves [6,4]+[7,4]=8, so [4,4] covers them with 4 and 4.
2
Step 2: Top-right equals ladder
The singleton sum-3 at [0,8] is forced to 3. Its neighbor [0,7] belongs to the four-cell equals run [0,4]-[0,7], so [0,7] must be 1 and the whole run becomes 1. Use [3,1] on [0,8]-[0,7] as 3-1, then [1,1] on [0,4]-[0,5], and [0,1] on [1,6]-[0,6] with 0 at [1,6], 1 at [0,6].
3
Step 3: Top sum-11 and sum-2 cluster
The sum-11 pair [0,1]-[0,2] can be closed with [6,2] and [5,0]: [6,2] puts 6 at [0,1], 2 at [0,0]; [5,0] puts 5 at [0,2], 0 at [1,2]. Now the sum-2 region [1,2],[2,1],[2,2],[3,2] already has 0 at [1,2]. Use [0,0] on [2,1]-[2,2] for two more 0s, leaving [3,2] to be 2.
4
Step 4: Greater-3 bridge at [4,2]
With [3,2]=2, the [4,2] domino fits perfectly: place 2 at [3,2] and 4 at [4,2]. That satisfies the greater-3 singleton at [4,2] while completing the sum-2 region.
5
Step 5: Left equals stack
The three-cell equals column [1,0],[2,0],[3,0] needs a common value. Domino [3,3] covers [1,0]-[2,0] with 3s, and [3,4] covers [3,0]-[4,0] with 3 at [3,0] and 4 at [4,0]. That also clears the greater-3 singleton at [4,0].
6
Step 6: Right-side sums and final equals
From Step 2, [1,6]=0, so the vertical equals pair [1,6],[2,6] makes [2,6]=0. Place [0,6] on [2,6]-[3,6] with 0 at [2,6], 6 at [3,6]; then the sum-11 pair [3,6],[4,6] forces [4,6]=5. Use [5,6] with 5 at [4,6], 6 at [5,6] to satisfy the greater-3 at [5,6]. Finally, [2,2] covers [4,4]-[5,4] with both 2s for the last equals pair.

💡 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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