NYT Pips Hints & Answers for July 5, 2026

Jul 5, 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 puzzle, by Ian Livengood, feels like a gentle introduction. You start with tiny sum targets that force a domino's pip value instantly: a sum-2 cell can only be 2. These single-cell constraints anchor the entire grid, and from there you just pair dominos to satisfy the remaining sums and one equals constraint. It's a satisfyingly straightforward solve.

The medium, also by Livengood, raises the stakes with a row of mixed constraints across the top. You'll encounter a sum-0 anchor, a greater-than-10 pair that demands high pips, and equals regions that lock in patterns quickly. The bottom row's equals cascades in columns 1–6 elegantly force dominos into place once the top row is settled.

Rodolfo Kurchan's hard puzzle is a dense thicket of interconnected limits. You're greeted by a triple-equals region on the left edge, a less-2 region that restricts three cells to 0 or 1, and a greater-than-10 pair demanding 6s. Threading the [0,0] domino through the less-2 cage and then dealing with the 11-sum pair makes you feel like a detective. It's a masterclass in deduction, where each placement unlocks the next with minimal guesswork.

💡 Progressive Hints

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

💡 The Singleton Clue
A region consisting of just one cell with a sum constraint is an immediate giveaway — it forces that cell to match the target number exactly.
💡 The 2-Cell Anchor
The top row's column 2 has a sum-2 single cell. That means it must hold a 2, so the domino covering it must include a 2 pip.
💡 Unlocking the Grid
Place the [2,6] domino horizontally with the 2 at (0,2). This fills the equals region below with a 6, so [5,6] goes vertically with 6 at (1,3). The sum-8 region now resolves to [3,1] vertically with 3 at (2,2) and 1 at (2,1). Finally, the sum-9 top-left pair and single sum-2 cell at (1,0) are satisfied by [2,5] vertical (5 at (0,0), 2 at (1,0)) and [1,4] vertical (4 at (0,1), 1 at (1,1)).
💡 Instant Singles
Single-cell constraints again provide instant values — look for sum-0 and sum-1 cells that can’t be anything else.
💡 Zero and One Seeds
The top left corner has a sum-0 cell and two columns over a sum-1 cell. These two forces will lock down the dominos covering them, including a [0,5] and a [1,6].
💡 Top Row Dominoes
Place [0,5] so 0 sits at (0,0) and 5 at (0,1). Place [1,6] horizontally with 1 at (0,2) and 6 at (0,3). The greater-10 neighbor demands another 6, so stack [5,6] vertically with 6 at (0,4). Then the sum-10 pair at columns 5-6 uses [2,5] and [5,4] to give 5+5=10, and the sum-4 region resolves to [2,0] giving 4+0. The bottom row’s equals groups then fill with [1,3], [3,5], and [2,0].
💡 Spot the Limits
The puzzle presents a triple-equals region and a less-2 region — these sharply limit the possible pip values.
💡 Triple Lock
The top-left triple equals across (0,0)-(0,1)-(1,0) must hold the same number. Combined with the adjacent sum-8 region, the value is forced to 2.
💡 Zero Zone
The less-2 region at row 4 columns 1–3 can only contain 0s. Since you have a [0,0] domino, it will occupy two of these cells, and the third must also be 0 via another domino.
💡 Column of Three
Once the less-2 region is all 0s, the column 0 equals region (rows 2–4) inherits the 3 from the domino that placed 0 at (4,1), forcing that entire column to 3. That lets you place the [2,3] and [3,1] dominoes.
💡 Full Deduction Chain
Full solution: [2,2] at (0,0)-(1,0) sets the triple equals to 2; [5,2] places 2 at (0,1) and 5 at (0,2); [3,4] places 3 at (0,3) and 4 at (1,3). The [0,0] domino fills (4,2)-(4,3) with zeros, and [0,3] places 0 at (4,1) and 3 at (4,0). The equals column then gets [2,3] with 3 at (2,0),2 at (2,1), and [3,1] with 3 at (3,0),1 at (3,1). Finally, [5,6] gives 6+5 for sum-11 at (5,0)-(5,1), [6,4] places 6 at (5,6) and 4 at (5,5), [2,6] places 5 and 3 for sum-8 at (2,5)-(2,6), [1,6] places 1 and 6 for sum-1 and sum-10, [6,6] gives 6,6 for greater-10, and [5,3] for sum-6 and less-3 regions.

🎨 Pips Solver

Jul 5, 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 July 5, 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 5, 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: Seed the Singleton
The sum-2 region at (0,2) is a single cell, so it must contain exactly 2. The only domino with a 2 and no conflicting constraints is [2,6], so place it horizontally with 2 at (0,2) and 6 at (0,3).
2
Step 2: Propagate the Six
The equals region at (0,3)-(1,3) now forces (1,3) to also be 6. That requires the [5,6] domino placed vertically with 6 at (1,3) and 5 at (2,3).
3
Step 3: Complete the Bottom Sum
The sum-8 region at (2,2)-(2,3) now has 5 at (2,3), so (2,2) must be 3. Place [3,1] vertically with 3 at (2,2) and 1 at (2,1).
4
Step 4: Tie Up the Left
The sum-2 region at (1,1)-(2,1) now has 1 at (2,1), so (1,1)=1; the sum-9 region at (0,0)-(0,1) needs 9 total. With a single sum-2 cell at (1,0) forcing 2, the [2,5] domino fits with 5 at (0,0) and 2 at (1,0). Then [1,4] fills the remaining cells with 1 at (1,1) and 4 at (0,1).

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Instant Zero and One
The top row holds two single-cell sum regions: (0,0) sum-0 ⇒ 0, (0,2) sum-1 ⇒ 1. Place [0,5] vertically with 0 at (0,0) and 5 at (0,1). Place [1,6] horizontally with 1 at (0,2) and 6 at (0,3).
2
Step 2: Greater Than Ten Push
The greater-10 region at (0,3)-(0,4) demands both cells >10, so (0,4) must be 6. The [5,6] domino fits vertically with 6 at (0,4) and 5 at (1,4).
3
Step 3: Sum to Ten
The sum-10 pair at (0,5)-(0,6) needs a total of 10. With (0,5) already receiving 5 from the [2,5] domino (placed next), (0,6) must be 5. Use [5,4] horizontally with 5 at (0,6) and 4 at (0,7). Then place [2,5] so 5 goes to (0,5) and 2 to (1,5).
4
Step 4: Four on the Right
The sum-4 region at (0,7)-(1,7) now has 4 at (0,7), so (1,7) must be 0. Place [2,0] vertically with 0 at (1,7) and 2 at (1,6).
5
Step 5: Equals Cascade
The row 1 equals regions lock into place: (1,5)-(1,6) equals so both 2; (1,3)-(1,4) equals so both 5; (1,1)-(1,2) equals so both 3. Place [3,5] vertically with 3 at (1,2) and 5 at (1,3) (completing the 5-equals). Finally, [1,3] covers (1,0)-(1,1) with 1 at (1,0) and 3 at (1,1).

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Triple Equals Lock
The region (0,0)-(0,1)-(1,0) must have three equal pips. With the adjacent sum-8 region at (0,2)-(0,3), the only workable value is 2. Place [2,2] vertically at (0,0)-(1,0) to set two 2s.
2
Step 2: Third Two and Sum Eight
(0,1) must also be 2, so place [5,2] with 2 at (0,1) and 5 at (0,2). Now the sum-8 region has 5 at (0,2), so (0,3) must be 3. Place [3,4] vertically with 3 at (0,3) and 4 at (1,3).
3
Step 3: Less Than Two Block
The less-2 region (4,1)-(4,2)-(4,3) requires each cell <2, so 0 or 1. The [0,0] domino fits perfectly by placing it horizontally at (4,2)-(4,3) (both 0). Then (4,1) must be 0; use [0,3] vertically with 0 at (4,1) and 3 at (4,0).
4
Step 4: Column of Threes
The equals region in column 0 for rows 2–4 now has (4,0)=3, so (3,0) and (2,0) must also be 3. Place [3,1] horizontally with 3 at (3,0) and 1 at (3,1) to satisfy the adjacent sum-1 at (3,1). Place [2,3] horizontally with 3 at (2,0) and 2 at (2,1).
5
Step 5: Bottom Row Sums
The sum-11 at (5,0)-(5,1) needs two pips summing to 11: the only option is 6+5, so place [5,6] with 6 at (5,0) and 5 at (5,1). The sum-6 at (5,6) forces a 6 there, so place [6,4] vertically with 6 at (5,6) and 4 at (5,5).
6
Step 6: Fill the Remaining
The sum-8 at (2,5)-(2,6) takes [2,6] with 5 at (2,6) and 3 at (2,5). The sum-1 at (3,5) requires [1,6] with 1 at (3,5) and 6 at (4,5) to help the sum-10 at (4,5)-(5,5) (4+6=10). The greater-10 at (1,1)-(1,2) gets [6,6] with both 6s. The less-3 at (5,2) and sum-6 at (5,3) are satisfied by [5,3] horizontally with 2 at (5,2) and 6 at (5,3).

💡 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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