NYT Pips Hints & Answers for July 4, 2026

Jul 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

Easy: Livengood's 3x5 grid launches with two isolated single-cell sum constraints—a sum‑3 at [1,2] and a sum‑2 at [2,0]—each directly instantiating a domino's pip value and orientation. The sum‑3 seed forces a vertical [3,2] domino that feeds an equals region above, while the sum‑2 key unlocks a three‑cell sum‑3 region through a shared intersection, closing the deduction loop with no ambiguity.

Medium: In today's NYT Pips medium, the same constructor weaves a network of intersecting equals and inequality gates. A three‑cell equals block in the top‑left forces a uniform digit that is resolved by the single less‑2 cell at [2,0] and the sum‑4 singleton at [0,4]; the latter pulls a [5,4] domino that also satisfies an equals pair. This creates a domino cascade where each placement opens the next, from the >4 cell to the multi‑cell sum‑10 anchor at the right.

Hard: Rodolfo Kurchan's hard grid is built on a tightly bound top‑row rhythm: three sum‑4 single cells interleaved with sum‑7 pairs. Because each 4 must be placed by a domino that also covers a partner cell in a sum‑7 pair, the whole row resolves sequentially—the [6,4], [1,4], and [2,4] dominos falling into place. Below, a vertical four‑cell equals column acts as a secondary backbone, channeling 3s from [5,3], [1,3], and [3,3] dominos, while scattered greater‑than and sum gates lock the remaining few moves.

💡 Progressive Hints

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

💡 Find the Lone Cells
Scan for single-cell sum regions—they force a specific pip value directly, creating an immediate anchor. Today's easy has two such cells, one demanding a 3 and one demanding a 2.
💡 Lock the Vertical
The cell at row 1, column 2 is the sum-3 singleton; it must be 3. This forces a domino with a 3 to cover it vertically, dropping a 2 into the equals region above.
💡 Full Solve
Place domino [3,2] with the 3 on [1,2] and 2 on [0,2]. Then [2,1] domino on [2,0] (2) and [1,0] (1) to satisfy the sum-2 cell and the three-cell sum-3 region. Next, [4,2] domino on [0,4] (4) and [0,3] (2) to complete the equals pair and set up the sum-4 pair. Finish with [0,3] domino on [1,4] (0) and [2,4] (3).
💡 Watch the Overlaps
Look for regions that share cells with multiple constraints—especially equals blocks adjacent to less-than or sum singles. These intersections force narrow choices for domino placement.
💡 The Top-Left Equals and the Less-2
The three-cell equals at [0,0]-[0,1]-[1,0] must all be the same number, but the cell [2,0] below is a less-2 region. The only number that works is 2 for the equals, pushing [2,0] to 0 via the [2,0] domino.
💡 Complete Medium Walkthrough
Start with [5,4] domino covering [0,4] (4) and [0,3] (5) to meet the sum-4 singleton and the equals pair with [1,3] (gets 5 from [0,5]). Next, [2,0] domino on [2,0] (0) and [1,0] (2), then [2,2] on [0,0]-[0,1] (both 2). [2,6] on [0,6] (6) and [1,6] (2); [2,1] on [2,6] (2) and [3,6] (1); [0,1] on [3,8] (0) and [3,7] (1). [3,3] fills [3,0]-[3,1] (3); [4,6] finishes with [3,3] (4) and [3,4] (6) to sum 10 alongside [2,3]=0.
💡 Top-Row Rhythm
Identify the pattern on the top row: alternating sum-4 single cells and sum-7 pairs. Each 4 must be placed by a domino that also covers a cell in a neighboring sum-7 pair, so solving the row is a chain reaction.
💡 The First 4
The sum-4 cell at [0,2] demands a 4; the only domino that can place a 4 there while also covering an adjacent cell is [6,4], which must sit horizontally to put 6 on [0,3] and start the sum-7 pair [0,3]-[0,4].
💡 Propagate the 7s
With 6 at [0,3], [0,4] must be 1 to sum to 7, so [1,4] domino places its 1 there and its 4 at [0,5]. Then the next 4 at [0,8] pulls in [2,4] to give 2 at [0,7] for the 5+2 sum-7 pair, with 5 coming from [5,3] at [0,6].
💡 Equality Underneath
Below the top row, the equals column at col 6 (rows 1–4) forces a single digit. The dominos [5,3], [1,3], and [3,3] all carry a 3, slotting into [1,6], [2,6], and [3,6]/[4,6] respectively. This satisfies the equals constraint and lets you place surrounding sum-2, sum-5, and greater-4 cells.
💡 Full Hard Solve
Domino [6,4] on [0,3] (6) & [0,2] (4); [1,6] on [0,0] (1) & [0,1] (6); [1,4] on [0,4] (1) & [0,5] (4); [5,3] on [0,6] (5) & [1,6] (3); [2,4] on [0,7] (2) & [0,8] (4). Then [5,2] on [2,3] (5) & [1,3] (2); [1,2] on [1,0] (1) & [2,0] (2); [5,0] on [3,0] (5) & [4,0] (0); [0,0] on [3,3] (0) & [4,3] (0); [6,0] on [2,8] (6) & [3,8] (0); [5,6] on [2,5] (5) & [1,5] (6); [6,2] on [4,5] (6) & [3,5] (2); [1,3] on [2,7] (1) & [2,6] (3); [3,3] on [3,6] (3) & [4,6] (3); [4,4] on [4,7] (4) & [4,8] (4).

🎨 Pips Solver

Jul 4, 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 July 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 July 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: Force the Sum‑3 Singleton
The single-cell region at [1,2] with target sum 3 demands a pip of exactly 3. The only domino carrying a 3 is [3,2]. It must cover [1,2] and an adjacent cell; the cell directly above at [0,2] is the only legal neighbor, so place the domino vertically with 3 on [1,2] and 2 on [0,2].
2
Step 2: Force the Sum‑2 Singleton
The single-cell region at [2,0] with target sum 2 demands a pip of 2. The remaining domino [2,1] carries a 2 and a 1. It must cover [2,0] and an adjacent cell; the only reachable neighbor is [1,0], which is part of a three-cell sum‑3 region. Place the domino with 2 on [2,0] and 1 on [1,0].
3
Step 3: Satisfy the Three‑Cell Sum‑3
The region comprising [0,0],[0,1],[1,0] now has [1,0]=1, needing two more cells to sum to 2. The only available values are two 1s. The remaining domino with two identical digits is [1,1]. Place it horizontally on [0,0] and [0,1] with 1s.
4
Step 4: Complete the Equals and Sum‑4 Pair
The equals region [0,2],[0,3] already has [0,2]=2, forcing [0,3]=2. The remaining [4,2] domino fits perfectly: place it horizontally with 2 on [0,3] and 4 on [0,4], matching the sum‑4 pair with [1,4] yet to be filled. Finally, place domino [0,3] vertically with 0 on [1,4] (summing to 4 with [0,4]) and 3 on [2,4] (empty).

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Single Sum‑4 and Equals Pair
The [0,4] sum‑4 cell forces a 4. The [5,4] domino can place 4 there and 5 on the adjacent cell [0,3]. Since [0,3] is in an equals region with [1,3], [1,3] must also be 5. This pulls the [0,5] domino vertically: 5 on [1,3] and 0 on [2,3].
2
Step 2: Three‑Cell Equals and Less‑2
The equals region [0,0],[0,1],[1,0] forces a uniform value. The less‑2 cell at [2,0] must be 0, so the [2,0] domino places 0 on [2,0] and the 2 on [1,0]. The three‑cell equals thus becomes all 2s, and domino [2,2] covers [0,0] and [0,1] with 2s.
3
Step 3: Greater‑4 and Sum‑4 Pair
The [0,6] greater‑4 cell requires a pip >4 (5 or 6). The [2,6] domino provides a 6, so place it horizontally with 6 on [0,6] and 2 on [1,6]. The sum‑4 pair [1,6]-[2,6] now needs 2 at [2,6]; the [2,1] domino fits with 2 on [2,6] and 1 on [3,6].
4
Step 4: Bottom Row Equals and Less‑1
The equals region [3,0],[3,1] requires identical digits; domino [3,3] places 3s on both cells. The less‑1 cell at [3,8] must be 0, so [0,1] domino puts 0 on [3,8] and 1 on [3,7], completing the equals region [3,6],[3,7] (both 1).
5
Step 5: Sum‑10 Anchor
The region [2,3],[3,3],[3,4] needs sum 10. We already have [2,3]=0 from Step 1. Domino [4,6] places 4 on [3,3] and 6 on [3,4], giving 0+4+6=10, finishing the grid.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Unlock the Top‑Row Singles
Three single sum‑4 cells at [0,2], [0,5], and [0,8] each require a 4. Starting with [0,2], the only reachable adjacent cell is [0,3]; domino [6,4] places 4 on [0,2] and 6 on [0,3]. This triggers the sum‑7 pair [0,3]-[0,4].
2
Step 2: Continue the Row Chain
With 6 at [0,3], the sum‑7 pair demands 1 at [0,4]. Domino [1,4] provides 1 there and its 4 at [0,5], fulfilling both that sum‑7 pair and the second sum‑4 single. The third sum‑4 single at [0,8] uses domino [2,4] to place 4 on [0,8] and 2 on [0,7]; the adjacent [0,6] must then be 5 to sum to 7, so domino [5,3] places 5 on [0,6] and 3 on [1,6].
3
Step 3: Anchor the Leftmost Pair
The sum‑7 pair [0,0]-[0,1] remains. The only available domino that sums to 7 and fits is [1,6], placing 1 on [0,0] and 6 on [0,1]. Now the entire top row is set.
4
Step 4: Build the Equals Column
The vertical equals region at column 6 rows 1–4 forces all cells to be identical. [1,6] already holds 3 from Step 2. To continue, domino [1,3] places 3 on [2,6] and 1 on [2,7]; domino [3,3] places 3 on [3,6] and [4,6]. The equals column is locked as all 3s.
5
Step 5: Resolve Scattered Singles
The sum‑2 at [2,0] forces 2, so domino [1,2] places 2 on [2,0] and 1 on [1,0] (>0). The sum‑2 at [1,3] gets 2 from [5,2] which also gives 5 to [2,3] (sum‑5). The greater‑4 at [1,5] gets 6 from [5,6], placing 5 at [2,5] for the sum‑7 pair with [3,5]; [6,2] gives 2 to [3,5] and 6 to [4,5] (>4). Sum‑5 at [3,0] places [5,0] with 5 on [3,0] and 0 on [4,0] (sum‑0). Sum‑0 at [3,3] uses [0,0] dominos with 0 on both [3,3] and [4,3].
6
Step 6: Finish the Remaining Sum‑7 and Sum‑4
The sum‑7 pair [2,7]-[2,8] gets 1 from [2,7] (already placed) and 6 from [6,0] at [2,8]; [3,8] gets 0 (empty). Finally, the sum‑4 cells [4,7] and [4,8] each need 4, fulfilled by [4,4] domino placing 4 on both. The grid is complete.

💡 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.

🎓 Keep Learning & Improve