NYT Pips Hints & Answers for August 3, 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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🎲 Today's Puzzle Overview

Today's NYT Pips easy puzzle, by Ian Livengood, is a confidence‑builder. A triple‑equals region dominates the board — three cells must match, so the double‑4 domino is the inevitable anchor. A pair of sum regions (11 and 10) with a less‑2 restriction force every placement cleanly and without forks. Spot the double‑4 lock and the rest opens in a minute.

Rodolfo Kurchan’s medium grid hides a crafty bottleneck. The top‑right region demands a sum greater than 8 with only two cells — the domino menu whittles the candidates down to a single viable pair. Meanwhile, a sum‑5 region in the opposite corner constrains the small values. Once you crack the top‑row tandem, the equals and sum‑8 cascades ripple outward smoothly.

Kurchan’s hard puzzle is a dense web of micro‑regions. Sum targets of 0, 1, and 3 populate the lower‑left and center, forcing exact low numbers. A key equals region ties three cells to the same value, and a sum‑10 pair near the right edge interlocks with a sum‑6 across the top. The real test is managing the high‑value dominoes without breaking the many tiny constraints — a satisfying positional grind that will stretch your deducing stamina.

💡 Progressive Hints

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

💡 Spot the shape‑shifter

Look for a region where every cell must show the same pip number. That kind of region can only be satisfied by one special type of domino.

💡 Triple threat territory

The triple‑equals region lives at column 2, rows 0 through 2. It needs three identical values. The only domino in the menu that can supply two identical pips is the [4,4] double.

💡 Full lock

Place [4,4] vertically covering [0,2] and [1,2] as 4s. The [6,4] domino must run horizontally from [2,2]=4 to [2,3]=6 to complete the equals sequence. The sum‑11 region [1,0]‑[2,0] gets 6 at [2,0] from the [5,6] domino (placed vertically, 6 at [2,0], 5 at [3,0]) and a 5 at [1,0] from the [5,2] domino (vertical: 5 at [1,0], 2 at [1,1] — the 2 satisfies less‑3). Finally, sum‑10 region [3,0]‑[3,1] uses [5,0] horizontally: 5 at [3,1], 0 at [3,2] (less than 2).

💡 Threshold guardian

Pay attention to the greater‑than‑8 constraint — it really acts like a sum threshold. Two cells must together beat 8, which dramatically narrows the domino choices.

💡 Top‑row tango

Cells [0,4] and [0,5] in the top right must jointly exceed 8. Among the dominoes, only [5,5] (sum 10) and [3,6] (sum 9) are candidates. But the top‑left sum‑5 region [0,0]‑[0,1] will force the [4,1] domino, and the equal‑3 region in row 3 needs two 3s — that steals the [3,6] domino. So [5,5] must take the greater‑8 spot.

💡 Full cascade

Place [4,1] horizontally at [0,0]=1, [0,1]=4 (sum 5). Next, [5,5] horizontally at [0,4]=5, [0,5]=5 (sum >8). For the equals pair [3,2]‑[3,3], use [3,2] vertically: [3,2]=3, [2,2]=2; and [3,6] vertically: [3,3]=3, [2,3]=6 — that also satisfies the sum‑8 region (2+6=8). On the bottom row, place [5,3] horizontally at [6,2]=5, [6,3]=3. Then [0,2] vertically links empty [5,1]=0 to [6,1]=2, and [0,5] vertically links empty [5,4]=0 to [6,4]=5.

💡 Zero in

Scan for single‑cell sum regions — these demand an exact pip, no arithmetic needed. Find the cells that must be 0.

💡 The sum‑1 squeeze

Cells [2,4] and [5,4] are both sum‑0 regions, so each must be 0. The domino [0,1] can put 0 at [2,4] and 1 at [3,4]; the domino [0,6] puts 0 at [5,4] and 6 at [6,4]. Meanwhile, the three‑cell sum‑1 region at [0,1],[1,1],[2,1] forces a combination of two 0s and one 1. You already used [0,1] and [0,6]; the remaining zero‑bearing dominoes are [0,2] and [0,5].

💡 Three of a kind

The equals region at [1,3],[1,4],[2,3] needs three identical pips. Check the remaining dominoes: [3,3] is a double‑3, perfect for two of the cells. The third 3 will come from the [2,3] domino (the pair, not the cell), placed so its 3 lands at [1,4] and its 2 at [0,4].

💡 High numbers lock in

After placing the low numbers, fill the sum‑9 region [2,0]‑[3,0]: [2,0]=5 (from [2,5] domino), so [3,0]=4 (via [4,4] double). The sum‑12 region [2,2]‑[3,2] then forces [2,2]=6 and [3,2]=6. The single‑cell sum‑5 at [3,3] gets a 5 from the [5,6] domino, which also supplies the 6 at [3,2].

💡 Full blueprint

Exact placements: [0,1] vert [2,4]=0, [3,4]=1. [0,6] vert [5,4]=0, [6,4]=6. [1,1] horiz [4,3]=1, [4,4]=1 (sum‑3). [0,2] vert [0,1]=0, [0,0]=2. [0,5] vert [1,1]=0, [1,2]=5. [1,6] vert [2,1]=1, [2,2]=6. [2,5] vert [1,0]=2, [2,0]=5. [4,4] vert [3,0]=4, [4,0]=4. [1,3] vert [6,0]=1, [5,0]=3. [3,3] vert [1,3]=3, [2,3]=3. [2,3] horiz [0,4]=2, [1,4]=3. [4,5] horiz [0,3]=4, [0,2]=5. [5,6] horiz [3,3]=5, [3,2]=6. [6,6] horiz [5,1]=6, [6,1]=6. [1,4] horiz [6,2]=1, [6,3]=4.

🎨 Pips Solver

Aug 3, 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

Pips hint for August 3, 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 3, 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

Step 1: The equals lock

The equals region spans [0,2], [1,2], [2,2] — all three cells must show the same pip. Among the available dominoes, only [4,4] can supply two identical values, so it must cover two of those cells.

Step 2: Double to the rescue

Place the [4,4] domino vertically, setting [0,2]=4 and [1,2]=4. Now [2,2] must also be 4. The only unused domino carrying a 4 is [6,4]. Place it horizontally: 4 at [2,2], 6 at [2,3] — the 6 also satisfies the greater‑3 constraint on [2,3].

Step 3: Sum‑11 and sum‑5

The region [1,0]‑[2,0] must sum to 11. Place the [5,6] domino vertically: 6 at [2,0], 5 at [3,0]. To reach 11, [1,0] needs a 5. The [5,2] domino fits vertically: 5 at [1,0], 2 at [1,1] — the 2 respects the less‑3 rule.

Step 4: Sum‑10 clean‑up

The last region [3,0]‑[3,1] needs a total of 10. With [3,0] already 5, [3,1] must be 5. Use the [5,0] domino horizontally: 5 at [3,1], 0 at [3,2] (less than 2, perfect). All dominoes placed.

🔧 Step-by-Step Answer Walkthrough For Medium Level

Step 1: Top‑left sum‑5

The region [0,0]‑[0,1] must sum to exactly 5. Scanning the domino list, only [4,1] (4+1=5) can hit that small total. So place [4,1] horizontally: [0,0]=1, [0,1]=4.

Step 2: Top‑right threshold

The [0,4]‑[0,5] region demands a sum greater than 8. Pips max at 6, so the sum must be at least 9. Candidates: [5,5] (10) and [3,6] (9). Later an equals region needs two 3s, which will require the [3,6] domino. Thus [5,5] must go here: place it horizontally at [0,4]=5, [0,5]=5.

Step 3: Equals and sum‑8 tandem

Now tackle the equals pair [3,2]‑[3,3] and the sum‑8 region [2,2]‑[2,3]. Use the [3,2] domino vertically: 3 at [3,2], 2 at [2,2]. Then use the [3,6] domino vertically: 3 at [3,3], 6 at [2,3]. This satisfies equals (3=3) and sum‑8 (2+6=8) in one shot.

Step 4: Bottom‑row anchors

The bottom row has two sum regions: [6,1]‑[6,2] must sum to 7, and [6,3]‑[6,4] must sum to 8. Place the [5,3] domino horizontally over [6,2]=5, [6,3]=3. Then [6,1] needs a 2 (to total 7 with the 5), and [6,4] needs a 5 (to total 8 with the 3).

Step 5: Empty bridges

Cells [5,1] and [5,4] are empty (no constraint). Use the [0,2] domino vertically: 0 at [5,1], 2 at [6,1] — giving [6,1] its 2. Similarly, use [0,5] vertically: 0 at [5,4], 5 at [6,4] — completing the board.

🔧 Step-by-Step Answer Walkthrough For Hard Level

Step 1: Zero‑in on forced zeros

Single‑cell sum‑0 regions at [2,4] and [5,4] must be 0. Place [0,1] vertically: 0 at [2,4], 1 at [3,4]. Place [0,6] vertically: 0 at [5,4], 6 at [6,4].

Step 2: Sum‑3 cluster

The region [3,4],[4,3],[4,4] must sum to 3. We already have 1 at [3,4], so [4,3] and [4,4] must sum to 2. The only way with available numbers is two 1s. Use the [1,1] double horizontally: [4,3]=1, [4,4]=1.

Step 3: Sum‑1 trio

The three cells [0,1],[1,1],[2,1] must sum to exactly 1 — that means two 0s and one 1. We have zero‑bearing [0,2] and [0,5] left. Place [0,2] vertically: 0 at [0,1], 2 at [0,0]. Place [0,5] vertically: 0 at [1,1], 5 at [1,2]. Now [2,1] needs the 1 — use [1,6] vertically: 1 at [2,1], 6 at [2,2].

Step 4: Sum‑4 and equals chain

With [0,0]=2, the sum‑4 pair [0,0]‑[1,0] forces [1,0]=2. The [2,5] domino handles it vertically: 2 at [1,0], 5 at [2,0]. Next, the equals trio [1,3],[1,4],[2,3] needs three identical pips. Place the [3,3] double vertically: 3 at [1,3], 3 at [2,3]. The third 3 comes from the [2,3] domino placed horizontally: 2 at [0,4], 3 at [1,4] (also setting [0,4]=2 for sum‑6).

Step 5: Middle numbers fall in

The sum‑6 region [0,3]‑[0,4] needs [0,3]=4 (since [0,4]=2). Place [4,5] horizontally: 4 at [0,3], 5 at [0,2] — that 5 pairs with the earlier 5 at [1,2] to satisfy the sum‑10 region [0,2]‑[1,2] (5+5=10). The single‑cell sum‑5 at [3,3] demands a 5, and sum‑12 at [2,2]‑[3,2] needs 6+6 since [2,2] is already 6. The [5,6] domino fits horizontally: 5 at [3,3], 6 at [3,2].

Step 6: Lower‑left and final edges

Sum‑9 region [2,0]‑[3,0]: [2,0] is 5, so [3,0] needs 4. Place [4,4] vertically: 4 at [3,0], 4 at [4,0]. Then sum‑7 pair [4,0]‑[5,0] gives 4+?=7, so [5,0]=3. Use [1,3] vertically: 1 at [6,0] (satisfying sum‑1 cell [6,0]), 3 at [5,0]. The greater‑10 region [5,1]‑[6,1] must exceed 10 — two 6s. Place [6,6] horizontally: 6 at [5,1], 6 at [6,1]. Finally, sum‑1 cell [6,2] needs a 1 and sum‑10 pair [6,3]‑[6,4] needs 4+6 (since [6,4] is already 6). Use [1,4] horizontally: 1 at [6,2], 4 at [6,3]. Puzzle solved.

💡 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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📖 More to Read

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NYT Pips Hints & Answers for August 3, 2026

🚨 SPOILER WARNING

🎲 Today's Puzzle Overview

💡 Progressive Hints

🎨 Pips Solver

✅ Final Answer & Complete Solution For Easy Level

Starting Position & Key First Steps

Final Answer: The Solved Grid for Easy Mode

✅ Final Answer & Complete Solution For Medium Level

Starting Position & Key First Steps

Final Answer: The Solved Grid for Medium Mode

✅ Final Answer & Complete Solution For Hard Level

Starting Position & Key First Steps

Final Answer: The Solved Grid for Hard Mode

🔧 Step-by-Step Answer Walkthrough For Easy Level

🔧 Step-by-Step Answer Walkthrough For Medium Level

🔧 Step-by-Step Answer Walkthrough For Hard Level

💡 Pro Tips for Similar Puzzles

🎓 Keep Learning & Improve

🧠 Strategy Guide

🎮 How to Play

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