NYT Pips Hints & Answers for July 14, 2026

Jul 14, 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

In today's NYT Pips easy, Ian Livengood constructs a compact puzzle around a triple-equals region that demands a single-value domino. That demand funnels the solver toward the [4,4] domino, which then interlocks with a sum-4 region requiring a zero entry point and a sum-2 region that pulls in the [0,0] and [2,6] dominoes. The entire grid becomes a tidy mechanical proof — each piece clicks into place with no ambiguity, showcasing Livengood's skill at packing logical necessity into a small space.

Rodolfo Kurchan's medium puzzle pivots on a deceptively simple single-cell sum-3 in the lower left. That lone cell forces a [3,2] domino, which ripples outward through a sum-4 pair, a sum-11 pair, and an interlocking set of greater- and less-than constraints. Kurchan turns a modest grid into a chain reaction, where every subsequent deduction follows from the one before, giving the puzzle a breezy but deeply logical rhythm.

For the hard, Kurchan orchestrates three separate equals regions — a column of three sixes on the right, a cluster of three fours in the middle, and a pair of threes at bottom right — threaded through sum-9 and sum-8 top-row constraints. The [6,6] domino naturally anchors the rightmost column, while a [0,6] at top-left sets off a calculation that resolves the entire top edge. The puzzle exemplifies Kurchan's ability to weave multiple constraint types into a cohesive, interlocking narrative that rewards methodical deduction.

💡 Progressive Hints

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

💡 Hint 1: Equal footing
Concentrate on the region that forces all its cells to be identical. That requirement severely restricts which domino can fit, and it’s the natural place to start.
💡 Hint 2: The identity trio
The equals region occupies (0,3), (1,2), and (1,3). Only a domino with two matching numbers — think 4s or 2s — can cover two of those cells. The adjacent sum-4 region will then help decide which one.
💡 Hint 3: Solving step-by-step
Place the [4,4] domino horizontally at (1,2)-(1,3). Place [4,0] horizontally at (0,3)-(0,4) to finish the equals and plant a 0 into the sum-4 region. The sum-4 at (0,4),(0,5),(1,5) now needs 2+2, so place [2,2] vertically at (0,5)-(1,5). Finally, the sum-2 region at (2,1),(2,2),(3,1) gets [0,0] horizontally at (2,1)-(2,2), then the last cell (3,1) must be 2; place [2,6] horizontally at (3,1)-(3,0) with the 6 at (3,0) to satisfy greater-4.
💡 Hint 1: Solo sum cell
Look for a region containing just one cell but with a sum constraint. That cell’s value is forced to a precise number, and the domino containing that number becomes your anchor.
💡 Hint 2: The 3 in the corner
The single-cell sum-3 sits at (3,0). It demands a 3, so the [3,2] domino must be placed there. That sets off a domino effect: the neighboring sum-4 pair at (2,1)-(3,1) will lock into place immediately.
💡 Hint 3: Full solution cascade
Place [3,2] horizontally at (3,0)-(3,1) — 3 at (3,0), 2 at (3,1). For sum-4, (2,1) becomes 2; use [2,6] horizontally at (2,1)-(2,2) with 6 at (2,2). Sum-11 at (2,2)-(3,2) then forces 5 at (3,2); place [6,5] horizontally at (3,3)-(3,2) with 6 at (3,3). Now [6,0] vertical at (1,3)-(2,3) (0 at (2,3) for less-2), and [4,5] horizontal at (0,3)-(0,2) (4 at (0,3), 5 at (0,2)). Place [4,2] vertical at (1,0)-(2,0): 4 at (1,0) for greater-3, 2 at (2,0) for less-3. Finish with [0,5] horizontal at (0,1)-(1,1): 0 at (0,1) empty, 5 at (1,1) greater-4.
💡 Hint 1: Triple-equals sighting
Three separate equals regions exist. Start by identifying the vertical column where three cells must all be the same number. That will dictate which domino can supply two of them, and the third will follow from a top-row domino.
💡 Hint 2: Column of sixes
The equals region at (0,6), (1,6), (2,6) practically begs for the [6,6] domino. Place it vertically at (1,6)-(2,6). The top cell (0,6) still needs a 6, which will come from a domino that also contributes to the top-row sum-8.
💡 Hint 3: Top-row arithmetic
The sum-9 region at (0,0)-(0,2) needs total 9. The [0,6] domino fits nicely: place it horizontally at (0,2)-(0,1) with 0 at (0,2) and 6 at (0,1). Now (0,0) must be 3 to reach 9; that 3 will be supplied by [2,3]. The sum-8 pair at (0,4)-(0,5) will use [5,6] giving 6 at (0,4).
💡 Hint 4: Filling the gaps
Complete the top row: place [2,3] vertical at (1,0)-(0,0) with 2 at (1,0) (<3) and 3 at (0,0). Place [5,6] vertical at (1,4)-(0,4) with 5 at (1,4) (>3) and 6 at (0,4). The remaining top cell (0,5) needs 2, and (0,6) needs 6 — place [2,6] horizontal at (0,5)-(0,6). Now the left sum-8 column (2,0),(3,0),(4,0) must sum to 8. Use [1,6] horizontal at (2,0)-(2,1): 1 at (2,0), 6 at (2,1) (>3). Then [3,4] vertical at (3,0)-(4,0): 3 at (3,0), 4 at (4,0). That column now yields 1+3+4=8.
💡 Hint 5: Full solution
Complete the grid: place [4,4] horizontally at (2,4)-(2,5) for the middle equals region. Place [4,5] horizontally at (3,4)-(4,4): 4 at (3,4) to complete that equals trio, and 5 at (4,4) for greater-2. Finally, place [3,3] vertically at (3,6)-(4,6) to satisfy the bottom-right equals of threes. All dominoes placed: [6,6], [0,6], [2,3], [5,6], [2,6], [1,6], [3,4], [4,4], [4,5], [3,3].

🎨 Pips Solver

Jul 14, 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 14, 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 14, 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 on the equals trio
The equals region spans (0,3), (1,2), (1,3). Since all three cells must have identical pips, a single domino can’t cover all three, but a domino with two matching pips must cover two of them. The only such dominoes are [4,4] and [2,2] and [0,0]. The [4,4] fits perfectly because an adjacent [4,0] domino can supply the third 4 while also serving the sum-4 region.
2
Step 2: Complete the equals and drop a zero
Place the [4,4] domino horizontally at (1,2)-(1,3). This leaves (0,3) needing a 4. The [4,0] domino can be placed horizontally at (0,3)-(0,4), putting a 4 at (0,3) and a 0 at (0,4). The equals region is now satisfied, and the sum-4 region at (0,4),(0,5),(1,5) gains a 0.
3
Step 3: Sum-4 demands a double 2
The sum-4 region requires a total of 4. With a 0 already at (0,4), the remaining two cells (0,5) and (1,5) must sum to 4. The only remaining domino that can provide that is [2,2]. Place it vertically at (0,5)-(1,5) to give 2+2. Now the sum-2 region in the lower left comes into focus.
4
Step 4: Sum-2 and the final greater-than
The sum-2 region includes (2,1), (2,2), (3,1). The only unused domino with low enough pips is [0,0]. Place it horizontally at (2,1)-(2,2) to get 0+0, leaving (3,1) needing a 2. Take the [2,6] domino and place it horizontally at (3,1)-(3,0): the 2 satisfies the sum-2, and the 6 goes to (3,0) where a greater-4 constraint demands a pip larger than 4. All constraints are resolved.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: The solo sum-3 ignition
The single-cell sum-3 region at (3,0) must hold exactly a 3. The [3,2] domino is the sole domino containing a 3, so place it horizontally at (3,0)-(3,1) with 3 at (3,0) and 2 at (3,1).
2
Step 2: Sum-4 propagates
The sum-4 pair at (2,1)-(3,1) already has a 2 at (3,1). Therefore (2,1) must be 2. That forces the [2,6] domino to be placed horizontally at (2,1)-(2,2), with 2 at (2,1) and 6 at (2,2).
3
Step 3: Sum-11 pulls a five
With 6 at (2,2), the sum-11 region (2,2)-(3,2) demands a 5 at (3,2). The [6,5] domino can go horizontally at (3,3)-(3,2), placing 6 at (3,3) (satisfying greater-4) and 5 at (3,2).
4
Step 4: Resolving the top and right
The less-2 constraint at (2,3) requires a 0 or 1; the [6,0] domino is the only one with a 0. Place it vertically at (1,3)-(2,3): 0 at (2,3), 6 at (1,3). The sum-10 at (0,3)-(1,3) now has 6 at (1,3), so (0,3) must be 4, which fits [4,5] placed horizontally at (0,3)-(0,2): 4 at (0,3), 5 at (0,2) (satisfying greater-4).
5
Step 5: Left side and empty cell
With remaining dominoes [4,2] and [0,5], address greater-3 at (1,0) and less-3 at (2,0). Place [4,2] vertically at (1,0)-(2,0): 4 at (1,0), 2 at (2,0). Finally, place [0,5] horizontally at (0,1)-(1,1): 0 at (0,1) (empty region) and 5 at (1,1) (greater-4). The grid is complete.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Equals on the right
The vertical equals region at (0,6),(1,6),(2,6) demands three identical pips. Among the dominoes, [6,6] is the only one with two identical high pips that can anchor this column. Place [6,6] vertically at (1,6)-(2,6), giving a 6 at (1,6) and (2,6). The top cell (0,6) will later receive a 6 from a top-row domino.
2
Step 2: Sum-9 at top left
The top row sum-9 region covers (0,0),(0,1),(0,2). The [0,6] domino fits the need for a 0 and a 6 to reach 9 when combined with another number. Place it horizontally at (0,2)-(0,1) with 0 at (0,2) and 6 at (0,1). Now (0,0) must be 3 to sum to 9. That 3 will come from the [2,3] domino.
3
Step 3: Completing the top row
Place [2,3] vertically at (1,0)-(0,0) — the 2 at (1,0) satisfies the less-3 constraint, and the 3 at (0,0) completes sum-9. Next, the sum-8 pair (0,4)-(0,5) needs total 8. The [5,6] domino can place 6 at (0,4) and 5 at (1,4) (satisfying greater-3). So place [5,6] vertically at (1,4)-(0,4). The remaining top cell (0,5) must be 2, and (0,6) must be 6. Place [2,6] horizontally at (0,5)-(0,6): 2 at (0,5), 6 at (0,6).
4
Step 4: The left sum-8 column
Now address the sum-8 region (2,0),(3,0),(4,0). The only unused dominoes that can contribute are [1,6], [3,4], [4,4], [4,5], [3,3]. The [1,6] can go horizontally at (2,0)-(2,1): 1 at (2,0), 6 at (2,1) for greater-3. That leaves (3,0) and (4,0) needing sum 7. The [3,4] domino placed vertically at (3,0)-(4,0) provides 3+4=7, completing the column sum 1+3+4=8.
5
Step 5: Middle equals fours
The equals region at (2,4),(2,5),(3,4) requires three equal values. The [4,4] domino can cover (2,4)-(2,5) horizontally for two 4s. Then the [4,5] domino placed horizontally at (3,4)-(4,4) puts a 4 at (3,4) to complete the equals trio and a 5 at (4,4) which satisfies the greater-2 constraint.
6
Step 6: Final equals threes
The last unused domino is [3,3], and the last equals region is at (3,6)-(4,6). Place [3,3] vertically at (3,6)-(4,6) to satisfy the equals-3 requirement. All constraints are now met.

💡 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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