NYT Pips Hints & Answers for June 20, 2026

Jun 20, 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 with a triple-equals region on [0,2], [0,3], [1,3] that demands three identical pips, forcing the [2,2] double domino into the top corners. That placement then pulls the [2,5] tile to complete the trio, with a cascade through a second equals pair and a sum-6 constraint.

Livengood's medium puzzle sharpens the equals logic with a triple-equals cluster at top-left [0,0], [0,1], [1,1] that consumes the [3,3] double and recruits [3,5] for the third cell, which simultaneously feeds a greater-3 region. The equals ring on [1,2]-[2,2]-[2,3] locks down the 1-values, while a sum-3 single-cell anchors the final 3.

Rodolfo Kurchan's hard grid in today's NYT Pips presents two sum regions that no single domino can satisfy—a sum-1 at [1,0]-[2,0] and a sum-11 at [2,1]-[2,2]. This forces a cross-domino opening: the [3,1] and [0,6] tiles split the sum-1, and [0,6]'s 6 meets a 5 from [4,5] for sum-11. The deduction graph then ripples through interlocking sum, greater, and equals constraints across the 10x8 board.

💡 Progressive Hints

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

💡 Hint 1: Spot the Triple
Look for a region where all cells must share the same value—especially one spanning three cells in an L-shape.
💡 Hint 2: Anchor the Double
The equals region at [0,2], [0,3], [1,3] forces the piece with two identical pips into the top corners. Which domino has matching faces?
💡 Hint 3: Complete the Cascade
Place the double-2 domino on [0,2]-[0,3]. Then [1,3] must also be 2, so the [2,5] domino goes horizontally there with 2 on [1,3] and 5 on [1,4]. The sum-6 region on [1,4]-[1,5] gets a 1 at [1,5] from the [1,1] domino, and the equals pair [1,0]-[1,1] become 0s using [0,5] and [0,3].
💡 Hint 1: Identify the Clusters
The medium grid features two triple-equals clusters; locate the one at the top-left where three cells must be identical.
💡 Hint 2: Lock the First Dominoes
The [3,3] domino fits neatly across the equals region [0,1]-[1,1], leaving [0,0] to be filled by a domino that also carries a 5 for the adjacent greater-3 spot.
💡 Hint 3: Unfold the Full Layout
Place the double-3 on [0,1]-[1,1] vertically. Then the [3,5] domino goes with 3 on [0,0] and 5 on [1,0]. The equals ring [1,2],[2,2],[2,3] all become 1s: use [6,1] to give 6 at [1,3] and 1 at [1,2], [4,1] to place 4 at [2,1] and 1 at [2,2], and [1,0] to place 1 at [2,3] and 0 at [3,3]. The sum-3 lone cell [0,3] forces [2,3] with 3 at [0,3] and 2 at [0,2].
💡 Hint 1: Phantom Sums
Scan for sum regions that cannot be covered by a single domino. The sum-1 and sum-11 constraints stand out, because no piece carries both needed pips.
💡 Hint 2: Split the Pairs
The sum-1 cells [1,0]-[2,0] must be 0 and 1. Since no [0,1] domino exists, [1,0] must receive its 1 from a domino extending upward, while [2,0] gets its 0 from a tile going right. The adjacent sum-11 pair [2,1]-[2,2] demands 5 and 6—also split across dominos.
💡 Hint 3: The Opening Salvo
The equals region [0,0]-[0,1] forces both cells to share the same pip, and only two 3s remain available. The [3,1] domino can place a 3 at [0,0] and a 1 at [1,0]. Then the [0,6] domino fits perfectly with its 0 on [2,0] and its 6 on [2,1]—solving both sum-1 and providing the 6 for sum-11.
💡 Hint 4: Building the Spine
With [2,1]=6, the sum-11 partner [2,2] must be 5. The [4,5] domino delivers 5 to [2,2] and 4 to [3,2], which later combines with 1 at [4,2] for the sum-5 region. Then the equals need for [0,1] forces the [3,2] domino to place a 3 there and a 2 at [0,2] (satisfying less-3). The sum-7 on [2,4]-[2,5] is solved by [5,2] (2 at [2,4], 5 at [2,5]).
💡 Hint 5: Solution — The Cross-Domino Opening
Place [3,1] vertically on [0,0]=3, [1,0]=1. Place [0,6] horizontally on [2,0]=0, [2,1]=6. Place [4,5] vertically on [3,2]=4, [2,2]=5. Place [3,2] horizontally on [0,1]=3, [0,2]=2. Place [5,2] horizontally on [2,5]=5, [2,4]=2. Start the right-column equals: [3,3] on [6,7]-[7,7] (3,3) and [3,6] on [8,7]-[9,7] (3,6). Then [4,0] to [1,7]=4, [2,7]=0; [2,1] to [4,7]=2, [3,7]=1 (sum-1 on [2,7]-[3,7]). [6,6] on [7,5]-[8,5] for greater-10. [1,4] places 1 at [4,2] and 4 at [5,2] (sum-5). [2,6] covers [5,0]=2, [5,1]=6. Finally, [4,6] on [9,2]=4, [9,3]=6 for greater-9.

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Jun 20, 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 20, 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 20, 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: The Triple-Equals Trigger
The region [0,2], [0,3], [1,3] forces equality. Only the [2,2] domino has identical pips, so it must cover two of them. The only adjacent pair inside that region is [0,2]-[0,3] horizontally—place the double-2 there.
2
Step 2: Forcing the Third 2
With [0,2]=2 and [0,3]=2, the third cell [1,3] must also be 2. The only unused domino containing a 2 is [2,5]; place it horizontally with 2 on [1,3] and 5 on [1,4].
3
Step 3: Sum-6 Consequences
The sum-6 region [1,4]-[1,5] now has [1,4]=5, so [1,5] must be 1. The unused [1,1] domino provides the 1; place it vertically on [1,5]=1 and [2,5]=1.
4
Step 4: The Equals Pair and Less-5 Cleanup
The equals pair [1,0]-[1,1] both need the same value. Place [0,3] horizontally on [1,1]=0 and [1,2]=3 (satisfying less-5). Then [0,5] must cover [1,0]=0 and [2,0]=5, completing the grid.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Top-Left Triple-Equals
The cluster [0,0], [0,1], [1,1] demands three identical pips. The double [3,3] fits the adjacent pair [0,1]-[1,1] vertically—place it, giving both cells 3.
2
Step 2: Feeding the Greater-3
[0,0] must also be 3; the greater-3 cell [1,0] needs >3. The [3,5] domino holds a 3 and a 5—place it vertically with 3 on [0,0] and 5 on [1,0].
3
Step 3: The Sum-3 Lone Cell
The sum-3 single-cell region at [0,3] forces a 3 there. Only the [2,3] domino can supply this 3; place it horizontally on [0,2]=2 and [0,3]=3.
4
Step 4: The Equals Ring
The equals group [1,2], [2,2], [2,3] must all be 1. Place [6,1] horizontally on [1,2]=1 and [1,3]=6 (greater-3). Then place [4,1] horizontally on [2,1]=4 (greater-2) and [2,2]=1.
5
Step 5: Final Equals Pair and Gap Fill
The equals pair [3,2]-[3,3] requires identical values. Place [1,0] vertically on [2,3]=1 and [3,3]=0—completing the 1s and starting the 0s. Then [0,6] must fill the remaining cells: place it horizontally on [3,1]=6 (greater-2) and [3,2]=0.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Sum-1 Forces a Split
The sum-1 region [1,0]-[2,0] must be 0+1, but no [0,1] domino exists. [1,0]'s only upward neighbor is [0,0] (part of the equals pair [0,0]-[0,1] that will both be 3). The [3,1] domino fits: place it vertically on [0,0]=3 and [1,0]=1.
2
Step 2: Sum-1's 0 Links to Sum-11
With [1,0]=1, [2,0] must be 0. The adjacent [2,1] is in the sum-11 pair [2,1]-[2,2] that needs 5+6. The [0,6] domino can supply the 0 and the 6: place it horizontally on [2,0]=0 and [2,1]=6.
3
Step 3: Completing Sum-11
Sum-11 now demands [2,2]=5. The [4,5] domino holds 4 and 5; place it vertically on [2,2]=5 and [3,2]=4. The sum-5 region [3,2]-[4,2] will later use that 4 with a 1.
4
Step 4: The Top Equals and Less-3
The equals pair [0,0]-[0,1] already has [0,0]=3, so [0,1] must be 3. The [3,2] domino can give 3 to [0,1] and 2 to the less-3 cell [0,2]: place it horizontally on [0,1]=3 and [0,2]=2.
5
Step 5: Sum-7 and the Right-Column Equals
The sum-7 pair [2,4]-[2,5] fits the [5,2] domino (5+2=7). Place it horizontally on [2,5]=5 and [2,4]=2. Next, the equals column [6,7],[7,7],[8,7] demands 3s; place [3,3] on [6,7]-[7,7] and [3,6] on [8,7]-[9,7] (with 3 and 6).
6
Step 6: Cleanup of Sum-1, Sum-5, and Greater Blocks
The sum-1 at [2,7]-[3,7] needs 0+1: place [4,0] on [1,7]=4,[2,7]=0 and [2,1] on [4,7]=2,[3,7]=1. For sum-5 [4,2], place [1,4] on [4,2]=1,[5,2]=4. For greater-10 [7,5]-[8,5], place [6,6] on [7,5]-[8,5] (6s). The less-3 [5,0] and sum-10 [5,1]-[5,2] use [2,6] on [5,0]=2,[5,1]=6. Finally, the greater-9 [9,2]-[9,3] is solved by [4,6] on [9,2]=4,[9,3]=6.

💡 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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