NYT Pips Hints & Answers for June 22, 2026

Jun 22, 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 opens the day with a compact easy puzzle that finds its spine in a pair of equals constraints. Rather than relying on sums, Livengood uses these twin equalities to force a logically tight chain: the two regions demand specific low numbers that lock down the available tiles. The design feels clean and instructive, a textbook demonstration of how equality constraints can anchor an entire grid without heavy arithmetic.

The medium puzzle, also from Livengood, shifts the focus to small sums and large-valued restrictions. A four-cell sum-4 region is the centerpiece—mathematically, it can only be filled with 1's, which in turn dictates the placement of the double-one domino. Combined with a greater-than-6 region that screams for the double-six, the architecture guides you to assemble the corners first before resolving the interior. It's a satisfying escalation that rewards solvers who read the constraints as a blueprint.

Rodolfo Kurchan's hard puzzle introduces a rare unequal constraint, signaling a different kind of challenge. This five-cell region, where every value must be distinct, weaves through the grid and forces a careful dance with the adjacent equals and sum regions. Kurchan layers equalities of three and four cells, a sum-0 singleton, and sum-12 and sum-10 pairs, creating a cascade of interdependencies. The result is a puzzle that feels like a precision instrument; each deduction reveals another locked-in piece. Today's NYT Pips hard is a masterclass in constraint interplay.

💡 Progressive Hints

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

💡 Hint 1: Follow the equals
Seek out the equals constraints—they demand two or more cells share the same pip. These are your anchor points today.
💡 Hint 2: Top-left and bottom-row equalities
The equals region in the top-left pairs a cell in row 0 with one directly below in row 1. The only way to satisfy that with the available low numbers is to use a domino whose 0-pip end covers one of those cells. Similarly, the equals region on the bottom row (row 2, spanning two adjacent cells) will require a 3 from the domino inventory.
💡 Hint 3: Complete answer
Place domino [0,1] horizontally with the 0 on [0,1] and 1 on [0,0]; place domino [0,3] vertically with the 0 on [1,1] and 3 on [2,1]. The equals region at [2,1]-[2,2] forces [2,2]=3, so the [5,3] domino goes there with 3 on [2,2] and 5 on [2,3]. The remaining equals region [2,3]-[2,4] forces both to 5, so the [6,5] domino covers [1,4]-[2,4] with 6 on [1,4] and 5 on [2,4]. Finally, the less-4 cell [0,5] gets 2 from the [4,2] domino placed horizontally with 4 on [0,4] and 2 on [0,5].
💡 Hint 1: A sum that speaks volumes
The four-cell sum-4 region at the bottom of the grid is the key. With all cells in this region required to sum to 4, the only possible values are all 1's. That immediately locks down which domino must cover two of those cells.
💡 Hint 2: A greater constraint calls for a specific double
Look to the greater-than-6 region in the top-center (cells [1,2] and [2,2]). The only pips greater than 6 are... well, only 6 itself. Thus the domino that occupies this region must be the double-six, and it can only sit there vertically across those two rows.
💡 Hint 3: Complete answer
Place the double-six ([6,6]) at [1,2] and [2,2]. The sum-4 region (cells [4,1]–[4,4]) must all be 1's, so place the double-one ([1,1]) horizontally across [4,2] and [4,3]. The remaining two cells in that region, [4,1] and [4,4], also need 1's; use [1,0] for [4,0]–[4,1] (1 on [4,1], 0 on [4,0]) and [1,2] for [4,4]–[3,4] (1 on [4,4], 2 on [3,4]). The equals region at [2,4]-[3,4] then forces [2,4]=2, so place [2,3] with 2 on [2,4] and 3 on [1,4]. To satisfy the sum-9 region ([0,4],[0,5],[1,4]=3), place [3,3] horizontally on [0,4]–[0,5]. Finally, the equals region at [3,0]-[4,0] forces both to 0; place [0,2] vertically from [3,0]=0 to [2,0]=2. The empty cells [1,0] and [1,1] take the last domino [1,5] with 1 on [1,0] and 5 on [1,1].
💡 Hint 1: Start with the odd constraint
The unequal region spanning [0,2],[1,0],[1,1],[1,2],[2,1] prohibits any repeated pip values. That means all five cells must be different. Combine that with the surrounding equal and sum constraints to force specific numbers into these spots.
💡 Hint 2: Zero and the right edge
A single-cell sum-0 at [2,7] tells you that cell must be 0. The only domino with a 0 is [6,0]; it must cover [2,7] with its 0 end, placing the 6 on [3,7] to satisfy the sum-12 region at [3,7]-[4,7] which then requires [4,7]=6.
💡 Hint 3: A triple of 4s
The equals region of three cells ([4,4],[4,5],[4,6]) demands a triple of the same number. The double-4 domino [4,4] can cover two of them, while the third must come from another domino that can supply a 4—and the adjacent sum-12 pair ([5,3],[5,4]) wants a double-6, leaving the [4,6] domino to give the third 4 to [4,6] with its 6 on [4,7] already used.
💡 Hint 4: A cascade of 3s and 1s
The equals region at [3,3]–[3,4]–[3,5]–[4,3] (four cells) forces all to be the same number, and the available 3's point to the double-3 [3,3] and the [3,5] domino. The unequal region and adjacent sums push that value to be 3. Meanwhile, the four-cell equals cluster around [3,1] will settle on 1s, supplied by the double-1 and the [3,1] domino.
💡 Hint 5: Complete answer
Place [4,4] horizontally at [4,4]-[4,5]; [4,6] at [4,6]-[4,7] (4 on [4,6], 6 on [4,7]); [6,0] vertically from [2,7]=0 to [3,7]=6; [3,3] at [3,3]-[3,4]; [3,5] at [3,5]-[3,6] (3 on [3,5], 5 on [3,6]); [6,6] at [5,3]-[5,4]; [5,5] at [5,5]-[5,6]; [4,0] vertically [0,2]=4 to [1,2]=0; [3,2] horizontally [1,0]=3 to [1,1]=2; [1,1] at [3,1]-[3,2]; [3,1] at [4,3]=3 and [4,2]=1; [6,1] vertically [2,1]=6 to [2,2]=1; [2,2] at [6,4]-[6,5]; [2,4] horizontally [7,4]=2 to [7,3]=4. All constraints satisfied.

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Jun 22, 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 22, 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 22, 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: Identify the equalities
The top-left equals region ([0,1] and [1,1]) must share the same pip. The available dominoes with a 0 are [0,1] and [0,3]; since zero is the only low number that appears on two tiles, the pair must be zeros, forcing [0,1] to cover one of those cells with its 0 end.
2
Step 2: Place the 0-provider
Domino [0,1] placed with 0 on [0,1] and 1 on [0,0] satisfies the less-2 constraint at [0,0]=1. Then domino [0,3] must cover [1,1] with its 0 end, placing 3 on [2,1]. The equals region at [2,1]-[2,2] then forces [2,2]=3.
3
Step 3: Resolve the lower equals
With [2,2] forced to 3, the [5,3] domino is the only one with a 3 and a 5. Place it with 3 on [2,2] and 5 on [2,3]. The equals region [2,3]-[2,4] now requires [2,4]=5, so the [6,5] domino must cover [1,4]-[2,4] with 5 on [2,4] and 6 on [1,4].
4
Step 4: Finish with the remaining domino
The final domino [4,2] goes to [0,4] and [0,5]. Place 4 on [0,4] (empty region) and 2 on [0,5], satisfying the less-4 constraint. All placements are forced.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: The greater-6 lock
Cells [1,2] and [2,2] must each be >6, so they can only be 6. The double-six domino [6,6] fits exactly, placed vertically with 6 on both.
2
Step 2: Sum-4 forces all 1s
The region [4,1],[4,2],[4,3],[4,4] must sum to 4, so every cell is 1. The double-one domino [1,1] covers [4,2] and [4,3] horizontally. The other two cells need 1's from other dominoes.
3
Step 3: Complete the sum-4 and trigger an equals
Domino [1,0] provides a 1 on [4,1] (with 0 on [4,0]), and domino [1,2] provides 1 on [4,4] (with 2 on [3,4]). The equals region at [2,4]-[3,4] forces [2,4]=2 because [3,4] is 2. So [2,3] goes there: 2 on [2,4], 3 on [1,4].
4
Step 4: Sum-9 resolves the top-right
The sum-9 region includes [0,4],[0,5],[1,4]. We have [1,4]=3, so the remaining sum for those two cells is 6, and they must both be 3. Place [3,3] horizontally on [0,4]-[0,5] with 3 on both.
5
Step 5: Left column and final empties
Equals region [3,0]-[4,0] forces both to be the same; [0,2] provides 0 and 2. Place vertically with 0 on [3,0], 2 on [2,0] (empty). The last cells [1,0] and [1,1] take [1,5]: 1 on [1,0] and 5 on [1,1] (satisfying greater-3).

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Right-edge sum-0 and sum-12
Cell [2,7] must be 0, so [6,0] covers it with 0 and places 6 on [3,7]. The sum-12 pair [3,7]-[4,7] then requires [4,7]=6, supplied by the [4,6] domino's 6.
2
Step 2: The triple-4 strip
Equals region [4,4],[4,5],[4,6] must all be the same number. Domino [4,4] covers two of them horizontally with 4's; domino [4,6] gives the third 4 on [4,6], using its 6 on [4,7].
3
Step 3: Bottom-center sums
Sum-12 pair [5,3]-[5,4] takes the double-6 [6,6]; sum-10 pair [5,5]-[5,6] takes the double-5 [5,5]. These are forced by the target sums.
4
Step 4: Unequal region begins
Place [4,0] with 4 on [0,2] and 0 on [1,2]. This feeds distinct values into the unequal region. The equals cluster below (cells [2,2],[3,1],[3,2],[4,2]) needs a uniform value. The double-1 [1,1] fits at [3,1]-[3,2].
5
Step 5: Cascade of 1s and 3s
With [3,1]=1 and [3,2]=1, the equals region forces [2,2]=1 and [4,2]=1. Domino [6,1] provides [2,1]=6, [2,2]=1; [3,1] provides [4,3]=3, [4,2]=1. The four-cell equals [3,3]-[3,5],[4,3] now gets 3 from [4,3], so place [3,3] at [3,3]-[3,4]; [3,5] at [3,5]-[3,6] (3 and 5).
6
Step 6: Bottom equals and final offshoot
Equals region [6,4]-[6,5],[7,4] takes [2,2] at [6,4]-[6,5] and [2,4] at [7,4]-[7,3] (2 on [7,4], 4 on [7,3] satisfying less-6). The last domino [3,2] fills [1,0]-[1,1] with 3 and 2, completing the unequal region.

💡 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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