NYT Pips Hints & Answers for August 1, 2026

Aug 1, 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

Today's NYT Pips easy grid greets you with a tight sum-2 region that practically whispers the answer—two cells must add to 2, and the available pip numbers leave only one possibility. From there, an equals region in the same row hands you a matching pair, and the rest of the dominos cascade into place with gentle nudges. Ian Livengood's design feels like a warm-up, giving you early wins that build confidence.

The medium puzzle, also by Livengood, ratchets up the tension with an unequal trio that demands three distinct values, while a less-than-2 cell and a sum-10 region pull you in opposite directions. You'll find yourself juggling a handful of doubles and mixed dominos, and the solve becomes a satisfying dance of elimination as each forced placement opens the next.

Rodolfo Kurchan's hard puzzle is a sprawling masterpiece of equals regions—two massive five-cell blocks, one destined for 0s, the other for 3s, plus two single-cell sum anchors that hand you starting numbers outright. Untangling the web of equalities while respecting scattered sum and equals constraints makes this NYT Pips hard a deeply layered challenge that rewards systematic thinking.

💡 Progressive Hints

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

💡 Hint 1: Spot the tightest constraint
Scan the board for the region that demands the smallest possible total. That's your in—it will severely limit which numbers can go there.
💡 Hint 2: Zero in on the sum-2 duo
The sum-2 region covers cells [2,1] and [2,2]. With the available pip values, the only way to reach a total of 2 is with 0 and 2. No domino holds two 1s, so look for a domino that delivers one of these numbers while an adjacent region supplies the other.
💡 Hint 3: Full solution chain
Place the [0,3] domino with 0 at [2,2] and 3 at [2,3]. That fulfills the sum-2's 0, forcing [2,1] to be 2; slide the [5,2] domino vertically with 5 at [1,1] and 2 at [2,1]. The equals region at [2,3]-[2,4] needs another 3, so the [3,4] domino goes to [2,4]=3 and [1,4]=4. Up top, the equals pair [0,1]-[0,2] must be identical; place [0,6] horizontally as [0,0]=0, [0,1]=6, then [6,5] fills [0,2]=6 and [0,3]=5 (satisfying greater-4).
💡 Hint 1: Find the most restrictive single cell
Look for a cell that has a strict upper limit. That tiny number will clip your options and set off a chain reaction.
💡 Hint 2: The less-than-2 cell and a bottom equals pair
Cell [2,0] must be < 2, so it's 0 or 1. Meanwhile, the equals region at [2,3]-[3,3] needs two matching numbers—only a double domino can fill it. Check which doubles you have.
💡 Hint 3: Full solution path
Start with the [0,0] domino vertically at [2,3]=0, [3,3]=0 (bottom equals). The less-2 cell [2,0] must be 0, so the [5,0] domino goes [2,0]=0, [2,1]=5. The sum-10 region [1,1]-[2,1] then demands 5 at [1,1]; place [5,3] with 5 at [1,1] and 3 at [1,0]. Now the left equals [0,0]-[1,0] forces [0,0]=3, so [3,3] goes [0,0]=3, [0,1]=3, completing sum-6. Use [3,4] for [0,2]=3, [0,3]=4. For the unequal trio, set [1,2]=1, [1,3]=2 via [1,2] domino, then [2,2]=6, [3,2]=6 from [6,6], satisfying greater-4.
💡 Hint 1: Single-cell sums give away numbers
Some regions contain just one cell but demand a specific total. Those cells are forced to a single pip value—find them first.
💡 Hint 2: Two anchors in opposite corners
The sum-3 cell at [0,2] must be 3, and the sum-4 cell at [5,0] must be 4. These hard numbers will radiate through the large equals blocks.
💡 Hint 3: Place the top domino and start the 0-equals block
With [0,2]=3, the only domino that can supply a 3 there is [3,5]; lay it horizontally to also put 5 at [0,3], partially satisfying the sum-9 pair. Now look at the lower five-cell equals region—every cell in that clump will end up holding the same value. That value will be 0.
💡 Hint 4: Build the 3-equals block and sum-10
Use [0,0] to place two zeros in the lower block at [4,3]-[4,4]. The sum-4 cell at [5,0] needs a 4, so place [3,4] with 3 at [5,1] and 4 at [5,0]; that kicks off the left five-cell equals block (all 3s). Place [3,3] at [4,0]=3,[4,1]=3 and [6,3] at [2,0]=6,[3,0]=3. The sum-10 region [2,0]-[2,1] then demands [2,1]=4, forcing the [1,4] domino.
💡 Hint 5: Full solve set
Place [3,5] h: [0,2]=3,[0,3]=5. [0,0] v: [4,3]=0,[4,4]=0. [3,4] h: [5,0]=4,[5,1]=3. [3,3] h: [4,0]=3,[4,1]=3. [6,3] v: [2,0]=6,[3,0]=3. [1,4] v: [1,1]=1,[2,1]=4. [2,1] h: [1,3]=2,[1,2]=1 (equals-2 region). [4,2] h: [0,4]=4,[1,4]=2 (sum-9). [6,2] h: [2,5]=6,[1,5]=2 (sum-10). [4,4] v: [2,6]=4,[3,6]=4 (sum-9). [1,0] h: [3,5]=1,[3,4]=0 (completing 0-block). [3,0] h: [5,2]=3,[5,3]=0. [5,5] h: [5,4]=5,[5,5]=5 (sum-10). [0,2] h: [2,4]=0,[2,3]=2. [1,5] v: [4,5]=1,[4,6]=5 (sum-9).

🎨 Pips Solver

Aug 1, 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 1, 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 1, 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: The sum-2 region
Cells [2,1] and [2,2] must add up to exactly 2. With the available pip numbers (0,2,3,4,5,6), the only possible pair is 0 and 2. That's your starting lock.
2
Step 2: The equals region below
Cells [2,3] and [2,4] must hold the same value. The domino set contains no double, but you can assemble matching numbers from two different dominos. The only workable match is two 3s.
3
Step 3: Placing the first dominos
The [0,3] domino delivers a 0 and a 3. Place it with 0 at [2,2] (fulfilling half of the sum-2) and 3 at [2,3]. Now [2,1] must be 2, so the [5,2] domino goes vertically with 5 at [1,1] and 2 at [2,1]. For the equals region, the [3,4] domino provides a 3 at [2,4] and a 4 at [1,4].
4
Step 4: The top equals and greater constraints
The region [0,1]-[0,2] requires equal numbers. The only way to achieve this is with two 6s. Lay the [0,6] domino horizontally with 0 at [0,0] and 6 at [0,1]. Then place [6,5] to populate [0,2]=6 and [0,3]=5. The greater-4 restriction on [0,3] is satisfied by the 5.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: The less-than-2 cell
Cell [2,0] must be less than 2, so it's forced to be 0 or 1. Given the available dominoes, 0 is the natural choice because the [5,0] domino can supply it while also providing a 5 for an adjacent region.
2
Step 2: Fulfill the bottom equals region
Cells [2,3] and [3,3] must be equal. The only double domino that fits here is [0,0]; place it vertically with both cells as 0. That also means [2,0] cannot be 0? Wait, [2,0] can still be 0—there are multiple zeros available. The [0,0] domino uses two 0s, but the [5,0] still has a 0. So proceed.
3
Step 3: The sum-10 region and its partner
Cell [2,0] must be 0 (less-2), so use the [5,0] domino with 5 at [2,1] and 0 at [2,0]. The sum-10 region [1,1]-[2,1] now demands 5 at [1,1]. The [5,3] domino covers that with 5 at [1,1] and 3 at [1,0].
4
Step 4: The left equals and the sum-6
The region [0,0]-[1,0] requires equal values. Since [1,0] is already 3, [0,0] must be 3. The [3,3] domino places 3 at [0,0] and 3 at [0,1]. That gives the sum-6 region [0,1]-[0,2] a total of 3+?—so [0,2] must be 3 as well. Use the [3,4] domino to put 3 at [0,2] and 4 at [0,3] (empty region).
5
Step 5: The unequal trio and greater-4
The unequal region [1,2], [1,3], [2,2] needs three distinct numbers. Place the [1,2] domino with 1 at [1,2] and 2 at [1,3]. The remaining cell [2,2] must differ, so the [6,6] domino fills [2,2]=6 and [3,2]=6, which also satisfies the greater-4 constraint on [3,2].

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Single-cell sum anchors
The region at [0,2] demands sum 3 → value 3. The region at [5,0] demands sum 4 → value 4. These are unconditional and lock the first two numbers.
2
Step 2: Place the domino containing the 3
Only the [3,5] domino can deliver a 3 to [0,2]. Place it horizontally: [0,2]=3, [0,3]=5. Now the sum-9 pair [0,3]-[0,4] has 5; the partner must be 4—later.
3
Step 3: Initiate the lower 0-equals block
The massive equals region covering (2,4),(3,4),(4,3),(4,4),(5,3) will all hold the same digit. Start with the [0,0] domino at [4,3]=0, [4,4]=0. This forces the rest of the block to become 0 as we fill.
4
Step 4: The left 3-equals block and sum-10
The equals region (3,0),(4,0),(4,1),(5,1),(5,2) will all be 3. Place [3,4] at [5,0]=4, [5,1]=3 (giving the 3). Use [3,3] for [4,0]=3,[4,1]=3 and [6,3] for [2,0]=6,[3,0]=3. The sum-10 pair [2,0]-[2,1] now has 6, so [2,1] must be 4; place [1,4] with 1 at [1,1] and 4 at [2,1].
5
Step 5: Middle equals-2 and top sum-9 completion
The equals region (1,3),(1,4),(1,5),(2,3) will all be 2. Place [2,1] at [1,3]=2,[1,2]=1 (satisfying the 1-equals). Place [4,2] at [0,4]=4,[1,4]=2, completing the top sum-9. Use [6,2] for [2,5]=6,[1,5]=2 (sum-10 at [2,5]-[2,6] gets 6) and [4,4] for [2,6]=4,[3,6]=4 (sum-9 at right).
6
Step 6: Final completion of 0 and 3 blocks
Place [1,0] at [3,5]=1,[3,4]=0, adding to the 0-block. Place [3,0] at [5,2]=3,[5,3]=0, finishing both blocks. Use [5,5] at [5,4]=5,[5,5]=5 (sum-10). Place [0,2] at [2,4]=0,[2,3]=2 linking the 2-equals and 0-block. Finally, [1,5] at [4,5]=1,[4,6]=5 completes the sum-9 at [3,6]-[4,6].

💡 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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