NYT Pips Hints & Answers for July 8, 2026

Jul 8, 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 equals regions — one three-cell block that forces uniformity and a two-cell block that quickly pins down a low-value anchor. A sum-11 region across the top then channels the remaining domino choices, leaving only one way to satisfy adjacent less-than constraints. The solving graph is pleasantly linear, each resolved region unlocking the next placement.

Livengood's medium offering builds on a four-cell sum-12 cluster that acts as the central hub. An equals column down the left side and a tight sum-3 trio along row 1 form the initial constraints that ripple outward. Greater-than single-cell regions on the right edge force specific pips into the sum-6 and sum-12 groups, making the midgame a careful cascade of domino placement without ambiguity.

Rodolfo Kurchan's hard puzzle presents a sprawling grid with a triple-equals region in the top row and a sum-1 two-cell region near the left edge that immediately forces a {0,1} split. A sum-10 pair on the far right and an equals-6 pair in the bottom right corner create a second foothold. The deduction architecture here is non-linear: solvers must toggle between the left-side sum-4, sum-1, and sum-4 chains and the right-side sum-10 and equals regions, gradually filling the center's empty and sum-3 cells. This NYT Pips hard requires tracking multiple constraints simultaneously.

💡 Progressive Hints

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

💡 Hint 1: Spot the force
Look for equals regions that force cells to share the same pip value, and a sum region that dictates the total across two cells.
💡 Hint 2: Zero in on the block
The three-cell equals region on the left-middle (rows 2 and 3) narrows possible dominoes with uniform pips. Check the less-than regions for upper bounds.
💡 Hint 3: Full chain
Place [2,2] vertically in [2,1]-[3,1] to start the equals chain; then [0,2] horizontally at [3,2]-[3,3] completes the three-cell equals and sets the two-cell equals to 0. The sum-11 forces [6,3] at [0,1]-[0,2] (6+3, with [0,2] less than 4), leaving [4,5] at [1,0]-[1,1] and [0,1] at [2,3]-[1,3].
💡 Hint 1: Look for grouped sums
Focus on the sum-3 trio in a single row and an equals column that constrains two cells to identical values.
💡 Hint 2: The row of ones
The sum-3 region along row 1 columns 1–3 must use three ones. The [1,1] domino can supply two of them; the third 1 must come from a [4,1] domino linking to the left equals column.
💡 Hint 3: Complete solve
Place [1,1] across [1,2]-[1,3]; [4,1] down [1,0]-[1,1] (4 top, 1 bottom). Then [4,0] down [0,4]-[1,4]; [3,3] down [2,3]-[3,3]; [6,5] across [2,4]-[3,4]; [3,5] across [3,1]-[3,2]; [3,4] across [2,0]-[2,1].
💡 Hint 1: Identify bottlenecks
Look for equals regions (three cells alike) and a sum-1 region that forces a 0 and 1 split.
💡 Hint 2: Top-row zeros
The top-row equals trio spans [0,1]-[0,2]-[0,3]; it must be all zeros because of the available domino pool. The sum-4 region in the top left corner ([0,0]-[1,0]) then pushes a 2+2 combo.
💡 Hint 3: Left chain unfolds
With the top left determined, the sum-1 at [2,0]-[3,0] becomes 1 and 0, domino [1,2] fits exactly. On the right, the equals pair [2,8]-[3,8] is all sixes from [6,6].
💡 Hint 4: Right-side tens
The far right sum-10 [4,8]-[5,8] gets a 6 from the [6,6] and a 4 from [0,4]; then the sum-10 [3,6]-[4,6] takes [1,5] (placing 5 at [3,6], 1 at [3,7]) and [5,0] (placing 5 at [4,6]).
💡 Hint 5: Full reveal
Solution: [0,2] across [0,0]-[0,1] (2-0); [0,0] across [0,2]-[0,3]; [1,2] down [2,0]-[1,0]; [0,4] across [5,7]-[5,8]; [6,6] down [3,8]-[4,8]; [1,5] down [3,7]-[3,6]; [0,6] down [1,8]-[2,8]; [3,1] down [5,1]-[5,0]; [4,2] down [4,4]-[3,4]; [0,3] down [3,0]-[4,0]; [2,6] across [5,2]-[5,3]; [1,0] across [2,4]-[1,4].

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Jul 8, 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 8, 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 8, 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: Anchor via equals regions
Identify the two equals regions: a three-cell block at [2,1], [3,1], [3,2] and a two-cell block at [2,3], [3,3]. The only domino with a matching pair that can feed a three-cell uniform region is the [2,2].
2
Step 2: Resolve the bottom
Place [2,2] vertically at [2,1]-[3,1], giving both cells a pip of 2. The third equals cell [3,2] still needs a 2; the [0,2] domino places its 2 there horizontally at [3,2] and its 0 at [3,3], satisfying the two‑cell equals as 0.
3
Step 3: Sum-11 and less-than-4
The top region [0,1]-[1,1] sums to 11. [0,2] has a less-than-4 constraint, so it must be 3. Place [6,3] horizontally at [0,1]-[0,2] with 6 at [0,1] and 3 at [0,2]. Now [1,1] must be 5 to reach the sum, forcing [4,5] across [1,0]-[1,1] (4 and 5).
4
Step 4: Final check
The last domino [0,1] fills the remaining cells [2,3] and [1,3] vertically. [2,3] is already 0 from the equals, so [1,3] becomes 1, respecting its less-than-6 cap. All regions satisfied.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Sum-3 trio
Row 1 cells [1,1]-[1,2]-[1,3] sum to 3, requiring three 1s. The [1,1] domino must cover two of them. Place it horizontally at [1,2]-[1,3].
2
Step 2: Equals column meets sum-3
The left equals region [1,0]-[2,0] demands identical values. To get the third 1 for the sum-3 trio, [1,1] must be 1, so the column top [1,0] needs a 4 from a [4,1] domino. Place [4,1] vertically with 4 at [1,0] and 1 at [1,1].
3
Step 3: Sum-12 hub begins
The sum-12 region spans [1,4],[2,3],[2,4],[3,3]. The greater-than-3 cell [0,4] forces a 4. Place [4,0] down [0,4]-[1,4] supplying 4 at top and 0 at [1,4] (part of sum-12).
4
Step 4: Complete sum-12
Now sum-12 needs 12 total. Place [3,3] vertically at [2,3]-[3,3] (3,3) and [6,5] horizontally at [2,4]-[3,4] (6,5). Sum becomes 0+3+6+3=12, and [3,4] satisfies greater-than-4.
5
Step 5: Tie up remaining regions
The sum-6 region [2,1]-[3,1] and greater-than-4 [3,2] are fixed by [3,5] across [3,1]-[3,2] (3+5=6, 5>4). Finally, [3,4] across [2,0]-[2,1] (4 and 3) completes the equals column and the grid.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Top-row triple equals
The equals region [0,1]-[0,2]-[0,3] must be three identical pips. The pool contains [0,0] and [0,2]; the only possible uniform value is 0. Place [0,0] across [0,2]-[0,3] (0,0) and [0,2] across [0,0]-[0,1] leaving 2 at [0,0] and 0 at [0,1]. The triple equals is now all zeros.
2
Step 2: Sum-4 top left
The sum-4 region [0,0]-[1,0] now has [0,0]=2, so [1,0] must be 2. The [1,2] domino provides the 2; place it vertically with 2 at [1,0] and 1 at [2,0] (feeding the sum-1 region).
3
Step 3: Left-side sum chains
The sum-1 [2,0]-[3,0] forces [3,0]=0. Place [0,3] down [3,0]-[4,0] (0,3). The sum-4 [4,0]-[5,0] now needs 1 for [5,0]; use [3,1] down [5,1]-[5,0] (3,1) giving 1 at [5,0].
4
Step 4: Right-side equals and tens
The equals [2,8]-[3,8] demands identical pips; the [6,6] placed vertically at [3,8]-[4,8] gives 6,6. Sum-10 [4,8]-[5,8] uses [0,4] across [5,7]-[5,8] (0,4) for 6+4=10. Sum-10 [3,6]-[4,6] gets [1,5] down [3,7]-[3,6] (1,5) and [5,0] placed at [4,6] (5) and [5,6] (0).
5
Step 5: Central region and empties
The sum-3 [2,4]-[3,4] is solved by [1,0] across [2,4]-[1,4] (1,0) and [4,2] down [4,4]-[3,4] (4,2), with [4,4]>0. Empty cells [1,4] and [1,8] fill with 0 from [1,0] and [0,6] down [1,8]-[2,8].
6
Step 6: Bottom row finish
The sum-5 [5,1]-[5,2] uses the 3 already at [5,1], so [5,2] must be 2. Place [2,6] across [5,2]-[5,3] (2,6) satisfying greater-than-4 at [5,3]. The equals [5,6]-[5,7] is zeroed by the 0 from [5,0] at [5,6] and [0,4] at [5,7]. All regions closed.

💡 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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