NYT Pips Hints & Answers for July 12, 2026

Jul 12, 2026

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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 footholds: a single-cell sum-3 region at [1,3] that locks the pip value to 3, forcing the [3,2] domino into that column, and a greater-4 constraint at [0,1] that couples with a sum-9 two-cell region. The deduction graph then branches: the [6,6] domino is forced into the lower row after the sum-9 requires a 6, leaving the [3,1] domino to complete the pair, while the second sum-3 region with two cells mops up the remaining low pips.

Rodolfo Kurchan's medium puzzle leans on an equals region (column 0 rows 2–3) that compels a twin pair, propagating through a sum-7 triplet and a single-cell sum-2 lock. The [2,0] domino anchors the single-cell 2, and the [4,1] domino feeds the equals chain; the rest of the grid then resolves through a series of sum constraints that funnel the double-zero and [5,3] dominoes into precise vertical placements.

Kurchan's hard is an all-sum lattice where every cell's value is dictated by its region's target—most are single-cell sums, so the grid becomes a fixed-value mosaic. Zero-sum cells at [0,2], [3,0], [4,1], [4,2], and [7,3] immediately lock five dominoes with 0-pips; then the 1-valued cells pull in [1,5], [1,3], [1,2], and [1,4] in a cascade. The central sum-18 quad consumes the remaining high dominoes [3,4], [2,5], [3,6], and [4,5], making this NYT Pips hard a striking example of an all-sum design.

💡 Progressive Hints

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

💡 Hint 1: Identify the Anchor
Scan for sum constraints on a single cell—these directly set the pip value, giving you a fixed starting point. Look at row 1 for a region whose target equals its only cell's value.
💡 Hint 2: Work the Ripple
The sum-3 region at [1,3] must be exactly 3. That forces the [3,2] domino to cover [1,3] and its adjacent empty cell [1,2], placing 2 there. Next, the greater-4 cell [0,1] demands a pip >4; the only domino that can supply a 5 or 6 there is [3,5], so it goes horizontally at [0,0]-[0,1] with 5 in [0,1].
💡 Hint 3: Complete the Chain
The sum-9 region [3,2]-[3,3] needs a total of 9. The [6,6] domino fits perfectly at [3,1]-[3,2] (giving [3,2]=6), so [3,3] must be 3—provided by the [3,1] domino at [3,3]-[3,4]. Finally, the sum-3 region [2,4]-[3,4] gets the [0,2] domino vertically at [1,4]-[2,4] with 0 and 2.
💡 Hint 1: Single-Cell Lock
Focus on sum regions containing only one cell; they nail the pip value. In row 2, a sum-2 region will force a specific domino.
💡 Hint 2: Equals and Neighbors
The sum-2 region at [2,3] must be 2. The only way to cover that cell with a domino containing a 2 and respect its neighbors is to place the [2,0] domino horizontally at [2,3]-[2,4] (2 in [2,3], 0 in [2,4]). Then the equals region in column 0 rows 2–3 compels [2,0] and [3,0] to match; a [4,1] domino at [2,0]-[2,1] (4,1) starts that equality chain.
💡 Hint 3: Full Solve
Continue with equals: place [0,4] vertically at [4,0]-[3,0] (4 in [3,0], 0 in [4,0]). Now the sum-7 region (cells [2,1]=1, [2,2], [3,1]) needs 6 more; place [2,5] vertically at [1,2]-[2,2] (5 in [2,2], 2 in [1,2]), giving sum-7 1+5=6, so [3,1] must be 1—use [0,1] vertically at [4,1]-[3,1] (0,1). The sum-2 region [1,2] & [1,3] now has [1,2]=2, so [1,3]=0 from [0,0] domino at [0,3]-[1,3]. Finish with sum-8 region using [5,3] at [0,4]-[1,4].
💡 Hint 1: The All-Sum Board
Every cell's value is dictated by single-cell sum targets. Start by listing all cells with zero-sum: [0,2], [3,0], [4,1], [4,2], [7,3]—they must contain 0.
💡 Hint 2: Zero Partners
Each zero needs a domino with a 0 and a matching neighbor. [0,2]=0 pairs with adjacent [0,3]=2, forcing the [0,2] domino. Similarly, [3,0]=0 with [4,0]=3 forces [0,3] domino, and [4,1]=0 with [5,1]=1 forces [0,1] domino.
💡 Hint 3: One-Value Cascade
After placing the zero dominoes, focus on cells with target sum 1. [2,1]=1, [4,3]=1, [5,3]=1, [6,1]=1. [2,1] pairs with [2,0]=5 via [1,5] domino; [4,3] will later pair with [3,3]=4 via [1,4]; [5,3] pairs with [6,3]=3 via [1,3]; [6,1] pairs with [6,2]=2 via [1,2].
💡 Hint 4: Two- and Four-Values
Now 2-valued cells: [1,0]=2 pairs with [0,0]=3 via [2,3]; [1,3]=2 pairs with [2,3] (value 5) via [2,5]; [7,0]=2 pairs with [7,1]=4 via [2,4]. The 4s: [1,1]=4 with [0,1]=5 via [4,5]; [1,2]=4 with [2,2]=3 via [3,4]; [7,1] already paired. Then [3,3]=4 pairs with [4,3]=1 via [1,4].
💡 Hint 5: Complete Solution
Finish with remaining 3s and 5s: [5,0]=3 & [6,0]=5 via [3,5]; [3,1]=3 & [3,2]=6 via [3,6]. The central sum-18 region automatically checks out with [2,2]=3, [2,3]=5, [3,2]=6, [3,3]=4. All domino placements listed: [0,1] at [4,1]-[5,1]; [0,2] at [0,2]-[0,3]; [0,3] at [3,0]-[4,0]; [0,4] at [7,3]-[7,2]; [0,5] at [4,2]-[5,2]; [1,2] at [6,1]-[6,2]; [1,3] at [5,3]-[6,3]; [1,4] at [4,3]-[3,3]; [1,5] at [2,1]-[2,0]; [2,3] at [1,0]-[0,0]; [2,4] at [7,0]-[7,1]; [2,5] at [1,3]-[2,3]; [3,4] at [2,2]-[1,2]; [3,5] at [5,0]-[6,0]; [4,5] at [1,1]-[0,1]; [3,6] at [3,1]-[3,2].

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Jul 12, 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 12, 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 12, 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: Lock the Single-Cell Sum-3
The region at [1,3] is a sum-3 with only that cell, so its pip must be exactly 3. The domino [3,2] contains a 3 and can cover [1,3] and the adjacent empty cell [1,2]; place it horizontally, putting 3 in [1,3] and 2 in [1,2].
2
Step 2: Satisfy the Greater-4
Cell [0,1] requires a pip >4 (5 or 6). The only remaining domino with a 5 or 6 is [3,5]; thus it must cover [0,1] and the empty [0,0]. Place it horizontally with 5 at [0,1] and 3 at [0,0].
3
Step 3: Resolve the Sum-9 Region
The region [3,2]-[3,3] must sum to 9. The [6,6] domino fits at [3,1]-[3,2], giving [3,2]=6, so [3,3] must be 3. The [3,1] domino covers [3,3] and [3,4], placing 3 in [3,3] and 1 in [3,4].
4
Step 4: Finish the Second Sum-3
The region [2,4]-[3,4] sums to 3; with [3,4]=1, we need [2,4]=2. The last domino [0,2] goes vertically: 0 at [1,4] and 2 at [2,4].

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Single-Cell Sum-2
Region [2,3] is a sum-2 single cell, so it must be 2. The [2,0] domino (2,0) can cover [2,3] and the empty [2,4]; place it horizontally with 2 at [2,3] and 0 at [2,4].
2
Step 2: Establish the Equals Twin
The equals region [2,0] and [3,0] forces identical values. Place [4,1] horizontally at [2,0]-[2,1] with 4 in [2,0] and 1 in [2,1]. Then [3,0] must be 4, so use the [0,4] domino vertically at [4,0]-[3,0] (0 at [4,0], 4 at [3,0]).
3
Step 3: Build the Sum-7 Triplet
The sum-7 region comprises [2,1]=1, [2,2] (empty), and [3,1] (empty). Place [2,5] vertically at [1,2]-[2,2] (2 at [1,2], 5 at [2,2]), so now sum-7 has 1+5=6, needing 1 at [3,1]. Use [0,1] vertically at [4,1]-[3,1] (0,1) to place 1 at [3,1].
4
Step 4: Handle the Sum-2 Region
The two-cell sum-2 region at [1,2] and [1,3] now has [1,2]=2, so [1,3]=0. Place the [0,0] domino vertically at [0,3]-[1,3] (0,0).
5
Step 5: Final Sum-8
The region [0,4]-[1,4] must sum to 8. The remaining [5,3] domino goes vertically: 5 at [0,4], 3 at [1,4].

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Map the Zero Cells
Many single-cell sum regions: [0,2]=0, [3,0]=0, [4,1]=0, [4,2]=0, [7,3]=0. Their adjacent cells have targets that match available 0-pip dominoes. Place [0,2] at [0,2]-[0,3] (0,2); [0,3] at [3,0]-[4,0] (0,3); [0,1] at [4,1]-[5,1] (0,1); [0,5] at [4,2]-[5,2] (0,5); [0,4] at [7,3]-[7,2] (0,4).
2
Step 2: Pair the Ones
Cells with target 1: [2,1]=1 f [2,0]=5 → place [1,5] at [2,1]-[2,0] (1,5). [5,3]=1 & [6,3]=3 → place [1,3] at [5,3]-[6,3] (1,3). [6,1]=1 & [6,2]=2 → place [1,2] at [6,1]-[6,2] (1,2). [4,3]=1 will later pair with [3,3]=4 via [1,4].
3
Step 3: Place the Twos and Fours
2s: [1,0]=2 with [0,0]=3 → [2,3] at [1,0]-[0,0] (2,3). [1,3]=2 with [2,3] (free, value 5) → [2,5] at [1,3]-[2,3] (2,5). [7,0]=2 with [7,1]=4 → [2,4] at [7,0]-[7,1] (2,4). Fours: [1,1]=4 with [0,1]=5 → [4,5] at [1,1]-[0,1] (4,5). [1,2]=4 with [2,2] (free, 3 later) → [3,4] at [2,2]-[1,2] (3,4). Then [3,3]=4 with [4,3]=1 → [1,4] at [4,3]-[3,3] (1,4).
4
Step 4: Complete the Threes and Fives
Place [3,5] at [5,0]-[6,0] (3,5) for targets 3 and 5. Place [3,6] at [3,1]-[3,2] (3,6) for 3 and 6; this supplies the 6 needed for the central sum-18.
5
Step 5: The Central Sum-18 Quad
Now check the four-cell region [2,2],[2,3],[3,2],[3,3]. [2,2]=3 (from [3,4]), [2,3]=5 (from [2,5]), [3,2]=6 (from [3,6]), [3,3]=4 (from [1,4]) — they sum to 18, confirming all placements.
6
Step 6: Verify Complete Coverage
All 16 dominoes are placed and every cell satisfies its sum target. The grid is 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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