NYT Pips Hints & Answers for June 16, 2026

Jun 16, 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 4×3 grid where an equals region linking three cells acts as the linchpin. This triple‑equals constraint forces a single pip value, which then dovetails with two sum‑9 regions to lock each of the five dominos into a unique spot, creating a clean, linear deduction.

Livengood's medium puzzle expands to a 4×4 layout and interweaves column‑wise equals pairs with a solo sum‑5 cell and a greater‑than region. The solving chain ignites at the isolated sum‑5 cell, which mandates the only domino containing a 5, then feeds the equals columns and propagates through a sum‑4 region—the puzzle rewards tracking pip parity across rows and columns.

Rodolfo Kurchan's hard puzzle stretches across eight columns, dominated by a massive sum‑22 region in the lower‑left. This block forces a double‑6 domino and a subsequent 6 from another piece, while a network of sum‑2, less‑2, and equals regions weaves a dense web of constraints. The deduction graph is tightly coupled; each placed domino unlocks one or two of the remaining restrictive niches.

💡 Progressive Hints

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

💡 Focus on the tightest constraint
Look for a region that demands all its cells share the same pip value—it’s the only way to break into the grid.
💡 Pinpoint the equals cluster
The three‑cell equals region sits at [1,1], [1,2], and [2,2]. Whatever value they share, it must come from a domino that can reach [2,2].
💡 Full solve chain
Place the [3,2] domino vertically with 2 at [2,2] and 3 at [3,2], forcing the equals value to 2. The sum‑9 region on [3,1]-[3,2] then demands 6 at [3,1], so [4,6] goes horizontally on row 3 with 4 at [3,0], 6 at [3,1]. The other sum‑9 at [2,0]-[3,0] needs 5 at [2,0], so [5,0] goes at [2,0]-[1,0] with 5 and 0. Finally, the equals cells [1,1] and [1,2] must be 2—set [2,1] at [1,1]-[0,1] (2,1) and [2,6] at [1,2]-[0,2] (2,6) to complete.
💡 Seek a solo constraint
One region contains only a single cell with a fixed sum target—this is your first forced value.
💡 The sum‑5 cell
The cell at [1,3] must be exactly 5. The only domino with a 5 is [5,1], so it must cover [1,3] and its neighbor [1,2], giving [1,2]=1.
💡 Full solve chain
With [1,2]=1, the equals column at [0,2]-[1,2] forces [0,2]=1. Place [0,1] horizontally with 1 at [0,2] and 0 at [0,3]. Col‑0 equals: [0,0] and [1,0] are equal; the [1,3] domino (1,3) sits at [0,0]-[0,1], giving [0,0]=1, then [1,0]=1 via [1,4] at [1,0]-[2,0] (1,4). The greater‑3 at [2,0] accepts 4. Col‑1 equals: [0,1]=3 forces [1,1]=3; place [3,0] at [1,1]-[2,1] (3,0). The sum‑4 region [2,1],[3,1],[3,2] now has 0 at [2,1], so [3,1]+[3,2]=4. Place [6,1] at [3,0]-[3,1] (6,1), forcing [3,1]=1, then [3,2]=3. Finish with [3,2] at [3,2]-[3,3] (3,2) and [2,2] at [2,3]-[2,4] (2,2) to satisfy the equals‑2 cluster.
💡 Start big
The puzzle’s heaviest constraint is a sum region demanding a very high total—look for the dominoes that can supply the largest pips.
💡 Sum‑22 region
The region covering [5,0], [5,1], [5,2], and [6,0] must sum to 22. With only 0–6 pips, almost every cell here must be a 6.
💡 Place the double‑6
The only way to get two 6s into [5,0] and [5,1] is the [6,6] domino. Place it horizontally so [5,0]=6, [5,1]=6.
💡 The third six
The sum‑22 region still needs a 6 on [6,0] (and a 4 on [5,2]) to reach 22. The [6,0] domino must cover [6,0] and [7,0], giving [6,0]=6, [7,0]=0, and the less‑2 at [7,0] is satisfied.
💡 Full solve chain
After placing [6,6] and [6,0], [5,2] must be 4 from [4,2] placed at [5,2]-[6,2] (the sum‑2 at [6,2] then forces the 2). The equals region [3,4]-[4,4] demands 4s, so [4,0] goes at [3,4]-[2,4] (4,0). The sum‑2 at [3,2] forces [2,0] (2,0) at [3,2]-[2,2], while the large equals region on row 0 and parts of rows 1‑2 forces 0s from [0,3] (0,3) at [0,1]-[0,0] and [0,0] (0,0) at [0,2]-[1,2]; then [0,1] covers [0,3]-[0,4] (0,1) to satisfy less‑2 at [0,4]. The remaining sum‑2 cells lock [2,6] at [3,8]-[2,8] (2,6), [2,5] at [4,8]-[5,8] (2,5), [2,1] at [6,5]-[5,5] (2,1), and [5,3] at [5,7]-[6,7] (5,3). The equals‑1 group forces [5,4]-[5,6] via [4,1] at [4,4]-[5,4] (4,1) and the double‑2 finishes at [7,1]-[7,2].

🎨 Pips Solver

Jun 16, 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 June 16, 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 June 16, 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 region
The region at [1,1], [1,2], [2,2] forces those three cells to share the same pip. Examine which dominos can cover [2,2]—the only adjacent cells are [1,2] and [3,2], so the domino with a 2 must connect [2,2] to one of those neighbors.
2
Step 2: Resolve equals and sum‑9 at right
Place the [3,2] domino vertically at [3,2] and [2,2], giving [2,2]=2 and [3,2]=3. This forces the equals value to 2, and the sum‑9 region [3,1]-[3,2] now demands 6 at [3,1].
3
Step 3: Use the other sum‑9
With [3,1]=6, the [4,6] domino must go horizontally at [3,0]-[3,1] (4,6). The second sum‑9 region [2,0]-[3,0] then requires 5 at [2,0], so place [5,0] vertically at [2,0]-[1,0] (5,0); the less‑2 at [1,0] accepts the 0.
4
Step 4: Complete the equals cells
Now [1,1] and [1,2] must be 2 to match. Place [2,1] at [1,1]-[0,1] (2,1) and [2,6] at [1,2]-[0,2] (2,6), satisfying the less‑2 at [0,1] with 1 and the empty region at [0,2] with 6.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Lock the solo sum‑5 cell
The region containing only [1,3] has sum target 5, so that cell must be 5. The only domino with a 5 is [5,1]; place it horizontally from [1,3] to [1,2], setting [1,3]=5 and [1,2]=1.
2
Step 2: Propel column‑2 equals
The equals region at [0,2]-[1,2] forces [0,2]=1. Place [0,1] horizontally with 1 at [0,2] and 0 at [0,3] (empty region at [0,3] accepts the 0).
3
Step 3: Chain column‑0 equals and greater‑3
Column 0 equals: [0,0] and [1,0] must match. Place [1,3] at [0,0]-[0,1] (1,3), giving [0,0]=1. Then [1,0] must be 1; place [1,4] vertically at [1,0]-[2,0] (1,4). The greater‑3 at [2,0] accepts 4.
4
Step 4: Resolve column‑1 equals and sum‑4
Column 1 equals: [0,1]=3 forces [1,1]=3. Place [3,0] at [1,1]-[2,1] (3,0). The sum‑4 region [2,1],[3,1],[3,2] now has 0 at [2,1], so [3,1]+[3,2]=4. Place [6,1] horizontally at [3,0]-[3,1] (6,1) to satisfy greater‑4 at [3,0]=6 and set [3,1]=1, forcing [3,2]=3.
5
Step 5: Finish the equals‑2 cluster
Place [3,2] at [3,2]-[3,3] (3,2), giving the equals region [2,3],[2,4],[3,3] the value 2. Place [2,2] horizontally at [2,3]-[2,4] to complete the grid.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Tackle the sum‑22 giant
The region [5,0],[5,1],[5,2],[6,0] must sum to 22. The only domino with two high pips is [6,6]; place it at [5,0]-[5,1] (6,6) to supply both 6s.
2
Step 2: Insert the third 6 and a 4
The remaining sum to reach 22 is 10 across [5,2] and [6,0]. The [6,0] domino provides a 6, so place it at [6,0]-[7,0] (6,0), satisfying less‑2 at [7,0]=0. Then [5,2] must be 4—place [4,2] at [5,2]-[6,2] (4,2); the sum‑2 at [6,2] forces the 2.
3
Step 3: Unlock top‑left equals and empties
The large equals region across [0,1],[0,2],[0,3],[1,2],[2,2] must be 0 to avoid conflict with nearby less‑2 and sum‑2 regions. Place [0,3] at [0,1]-[0,0] (0,3) to set the chain—[0,0]=3 (empty), [0,1]=0. Then place [0,0] at [0,2]-[1,2] (0,0) to spread the 0s, covering all five equals cells.
4
Step 4: Manage the bottom equals and sum‑2s
The [0,1] domino places 0 at [0,3] and 1 at [0,4] (less‑2 satisfied). The equals pair [3,4]-[4,4] demands 4s; place [4,0] at [3,4]-[2,4] (4,0). The sum‑2 cells at [3,2], [3,8], [4,8], [6,5] each lock specific dominos: [2,0] at [3,2]-[2,2] (2,0), [2,6] at [3,8]-[2,8] (2,6), [2,5] at [4,8]-[5,8] (2,5), [2,1] at [6,5]-[5,5] (2,1).
5
Step 5: Settle the equals‑1 and greater‑2 groups
The equals‑1 region [5,4],[5,5],[5,6] requires three 1s. Place [4,1] at [4,4]-[5,4] (4,1) to set [5,4]=1; then [2,1] already gave [5,5]=1, and [5,1] from [5,6] (part of the greater‑2 at [4,6]) completes the triplet with [5,1] domino’s 1 on [5,6] (the greater‑2 at [4,6]=5 is satisfied).
6
Step 6: Clean up equals‑5, sum‑2, and bottom row
The equals pair [5,7]-[5,8] forces 5s—[5,3] at [5,7]-[6,7] (5,3) sets [5,7]=5, [6,7]=3 (empty). The remaining sum‑2 at [6,2] already holds 2 from [4,2]; sum‑2 at [4,8] was 2; less‑2 at [0,4] is 1. Finally, the equals‑2 at [7,1]-[7,2] locks the [2,2] domino there, finishing the puzzle.

💡 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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