NYT Pips Hints & Answers for September 4, 2026

Sep 4, 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 easy grid opens on two independent footholds: a one-cell empty region at the top edge forces a vertical domino, and the attached sum region propagates downward into a multi-cell equality block. The remaining sum and greater regions then collapse in cascade.

Rodolfo Kurchan's medium puzzle clusters three equality regions around a large central sum. The top-left equality block is self-locking because its topmost cell has only one neighbor; that determines a shared value and feeds the central sum, while a two-cell equality and a single greater/less pair resolve the right and bottom flanks.

This NYT Pips hard from Kurchan expands the same constraint-graph approach into a sparse field. Two separate cascades dominate: a top-right chain built from a greater-than singleton and paired sum regions, and a bottom-right chain triggered by a single-cell sum region, which locks a five-cell equality strip and then radiates through the lower rows. The deduction graph is almost entirely forced once these anchors are placed.

💡 Progressive Hints

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

💡 Start at the top edge
Look for an isolated empty cell and a short vertical sum region nearby. Edge cells with only one possible neighbor create the first foothold.
💡 The top-right column
The empty cell at [0,3] can only use [0,4], which sits in the sum region with [1,4]. That forces the value in [1,4] and then the vertical domino below it.
💡 Full easy answer
Domino 3/1 goes [0,3]=1, [0,4]=3. Sum-7 forces [1,4]=4, paired with [2,4]=2 via 2/4. The equals block then makes [3,3]=[3,4]=2 with 2/2. Greater-4 puts [2,2]=6, so 6/4 gives [2,1]=4; sum-4 makes [1,1]=0, and 0/0 covers [1,0]=0, [1,1]=0.
💡 Find the equality clusters
Three equality regions and a large central sum do most of the work. Focus first on equality regions whose top cell has only one neighbor.
💡 Top-left lock
The top equality block at [0,1]/[1,1]/[1,2] starts with [0,1] forced into [1,1]. That shared value then forces [1,2] and feeds down into the central sum at [2,2].
💡 Full medium answer
Use 5/5 on [0,1]/[1,1], making [1,2]=5; then 5/1 on [1,2]/[2,2] gives [2,2]=1. The 3/3 domino covers [1,3]/[2,3]. Greater-3 at [2,5] takes 0/4 with [2,4]=0, [2,5]=4; equality forces [3,4]=0, and 1/0 covers [3,3]=1/[3,4]=0. The large sum-5 forces 1/1 on [2,0]/[2,1] and leaves [3,2]=1; less-4 at [3,1] uses 3/1 with [3,1]=3/[3,2]=1.
💡 Spot the singleton triggers
Look for single-cell greater-than and sum regions. The upper-right greater-than cell and the lower-right zero-sum cells are the tightest triggers, with equality regions waiting behind them.
💡 Anchor the bottom-right corner
The sum-0 singleton at [4,8] has only one neighbor, [3,8]. That locks a vertical domino and then forces the adjacent [3,7] cell, which feeds the five-cell equality strip [3,6]/[4,6]/[5,6]/[6,6]/[7,6].
💡 Top-right sum chain
The greater-0 cell at [0,5] must pair with [1,5]. The two paired sum-11 regions then force the top-right 5/5 horizontal and the 0/6 vertical; watch how one high value propagates across [2,5] to [2,6].
💡 Bottom-left sum-4 funnel
Once the equals block fixes [6,4], the lone sum-4 cell at [7,4] has only [7,3] open. That forces the 2/4 domino, then the sum-4 region below makes [8,3]=2, and the [8,1]/[8,2] sum-11 plus [8,0] sum-3 complete the lower left.
💡 Full hard answer
Place 0/4 at [4,8]=0/[3,8]=4; 6/3 at [3,7]=6/[3,6]=3. Use 3/3 at [4,6]/[5,6]; 5/0 at [8,8]=5/[8,7]=0; 2/3 at [8,6]=2/[7,6]=3; 1/3 at [6,5]=1/[6,6]=3; 1/1 at [6,3]/[6,4]=1. Top-right: 1/6 at [0,5]=1/[1,5]=6; 5/5 at [2,5]/[2,6]=5; 0/6 at [1,7]=0/[1,6]=6. Bottom-left: 2/4 at [7,3]=2/[7,4]=4; 6/2 at [8,2]=6/[8,3]=2; 5/3 at [8,0]=3/[8,1]=5. Center-left: 2/0 at [4,5]=0/[5,5]=2; 2/5 at [5,4]=2/[4,4]=5; 5/6 at [3,4]=5/[3,5]=6.

🎨 Pips Solver

Sep 4, 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 4, 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 4, 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: Top edge empty forces the first domino
The empty cell at [0,3] has only one available neighbor, [0,4], so the top row must use the 3/1 domino with [0,3]=1 and [0,4]=3.
2
Step 2: Sum region sends the chain downward
With [0,4]=3 fixed, the sum-7 region [0,4]/[1,4] forces [1,4]=4. The only neighbor left for [1,4] below is [2,4], so the 2/4 domino runs vertically with [2,4]=2 and [1,4]=4.
3
Step 3: Equality block fills the right side
The equality region [2,4], [3,3], [3,4] now requires all cells to be 2. Therefore [3,3] and [3,4] must be the 2/2 domino.
4
Step 4: Greater cell and final sum close the grid
The greater-4 cell at [2,2] must be 6, so it pairs with [2,1] via the 6/4 domino, putting [2,1]=4. That satisfies the sum-4 region [1,1]/[2,1] by forcing [1,1]=0; the 0/0 domino covers [1,0] and [1,1].

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Top equality block locks vertically
In the top equality region [0,1]/[1,1]/[1,2], the top cell [0,1] has only one neighbor, [1,1], so the 5/5 domino must cover that vertical pair. The equality rule then forces [1,2] to be 5 as well.
2
Step 2: Feeding the central sum
With [1,2]=5, its only open neighbor is [2,2], so the 5/1 domino runs vertically, putting [2,2]=1. That places the first known value inside the large sum-5 region.
3
Step 3: Two-cell equality on the right
The two-cell equality [1,3]/[2,3] has no other cells, so it must be the 3/3 domino, placing both cells as 3.
4
Step 4: Right-flank greater pair
The single greater-3 cell [2,5] has only one neighbor, [2,4]. The 0/4 domino covers them with [2,5]=4 and [2,4]=0. That makes the equality region [2,4]/[3,4] force [3,4]=0; [3,4]'s only remaining neighbor is [3,3], so the 1/0 domino places [3,3]=1 and [3,4]=0.
5
Step 5: Bottom-left and final sum
The sum-5 region now has [2,2]=1 and [3,3]=1, leaving [2,0]/[2,1]/[3,2] to total 3. The edge cell [2,0] has only neighbor [2,1], so they take the 1/1 domino, making [3,2]=1 by the sum. The less-4 singleton [3,1] then has only [3,2] left, so the 3/1 domino gives [3,1]=3 and [3,2]=1.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Bottom-right zero locks the five-cell equal strip
The single sum-0 cell at [4,8] has only one neighbor, [3,8], so it must take the 0/4 domino with [4,8]=0 and [3,8]=4. The adjacent sum-10 region [3,7]/[3,8] then forces [3,7]=6; since [3,7] can only pair left with [3,6], the 6/3 domino places [3,6]=3. That settles the [3,6] cell of the five-cell equals region, so [4,6], [5,6], [6,6], and [7,6] all become 3. The adjacent [4,6]/[5,6] pair is the 3/3 domino.
2
Step 2: The lower-right sum-5 and equality block
The single sum-5 cell [8,8] has only neighbor [8,7], so the 5/0 domino gives [8,8]=5 and [8,7]=0. The sum-2 region [8,6]/[8,7] then forces [8,6]=2, and [8,6] must pair upward with [7,6], using the 2/3 domino. That leaves [6,6]=3, whose only remaining neighbor is [6,5], so the 1/3 domino puts [6,5]=1 and [6,6]=3. The three-cell equals region [6,3]/[6,4]/[6,5] is now fixed at 1, with [6,3]/[6,4] taking the 1/1 domino.
3
Step 3: Top-right greater/sum cascade
The greater-0 singleton [0,5] can only pair with [1,5]. Because [1,5]/[2,5] sum to 11, that vertical domino must be the 1/6, placing [0,5]=1 and [1,5]=6. Then [2,5]=5; it pairs with [2,6] via the 5/5 domino. The second sum-11 region [1,6]/[2,6] now yields [1,6]=6, and [1,6] pairs with the empty cell [1,7] through the 0/6 domino.
4
Step 4: Bottom-left sum-4 chain
The single sum-4 cell [7,4] has only one open neighbor left, [7,3], since [6,4] is already used by the equality block, so the 2/4 domino places [7,4]=4 and [7,3]=2. The sum-4 region [7,3]/[8,3] then forces [8,3]=2. [8,3] pairs with [8,2] using the 6/2 domino, so [8,2]=6; the sum-11 region [8,1]/[8,2] makes [8,1]=5. Finally [8,1] pairs with [8,0] via the 5/3 domino, giving [8,0]=3 and satisfying its single-cell sum-3.
5
Step 5: Central left zero-sum closes the loop
The single sum-0 cell [4,5] must be 0. It cannot pair upward with [3,5] because that would need the already-used 0/6, and cannot pair left with [4,4] because the sum-10 region would become impossible; so its only viable partner is [5,5]. The 2/0 domino places [4,5]=0 and [5,5]=2. The sum-4 region [5,4]/[5,5] then forces [5,4]=2, and [5,4] pairs upward with [4,4] via the 2/5 domino, giving [4,4]=5. The sum-10 region [3,4]/[4,4] makes [3,4]=5; finally the single sum-6 cell [3,5] pairs left with [3,4] through the 5/6 domino, placing [3,5]=6 and [3,4]=5.

💡 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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