NYT Pips Hints & Answers for June 9, 2026

Jun 9, 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.

🎲 Today's Puzzle Overview

The easy grid gives you two gifts up front—single-cell sum-4 regions at [2,2] and the pair [3,2]/[4,2] force immediate domino placements. You'll lock in a column of identical pips, then use an equals region to the left to mirror a double. The top-right sum-11 region and its neighboring equals pair create a short domino chain that falls neatly into place without guesswork. Ian Livengood's construction turns a handful of constraints into a brisk, satisfying solve.

Stepping into the medium, you're greeted by two tiny sum regions in the top row—a sum-5 at [0,0] and a sum-3 at [0,1]—each spilling out a single pip. That immediately defines the equals pairs in row 1, creating a ripple effect that cascades down. From there, an empty cell in column 0 and a sum-9 region at the bottom left force a specific 0/6 split, while a greater-than-9 pair in the bottom right demands maximum pips, pulling a 6/5 domino into place. Livengood's layout feels like a set of dominoes tipping over in sequence, a satisfying midweek challenge.

Rodolfo Kurchan's hard puzzle throws you into a thicket of constraints, starting with a three-cell sum-12 in the top-left corner that must be carved out of the available dominoes. A 'greater than 10' pair on the right, a less-than-4 pair beneath, and a rare four-cell unequal region in the center demand careful pip allocation. The solving path weaves through a single sum-4 cell at [2,0], a locked equals column on the far right, and a sum-10 pair at the bottom. This NYT Pips hard is a delightful exercise in constraint juggling that rewards systematic elimination.

💡 Progressive Hints

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

💡 Start with the loners
Keep an eye out for regions that contain just one cell and a sum target — those cells' pip values are instantly known.
💡 Pin down column 2
The single cell at [2,2] has a sum-4 target, so it must be 4. Then the two-cell sum-4 region directly below it forces both of those cells to be 2s. That pattern nearly fills the column.
💡 Complete the chain
Place domino [2,4] vertically in column 2 with the 4 at [2,2] and 2 at [3,2]. Then domino [2,2] goes horizontally across [4,2] and [5,2] with 2s. Domino [6,6] fills the equals pair [4,0]/[5,0]. For the sum-11 region at [1,4]/[2,4], use domino [6,2] (6 at [1,4], 2 at [0,4]) and domino [5,4] (5 at [2,4], 4 at [3,4]); the equals pair [3,4]/[4,4] then takes domino [3,4] with 4 at [4,4] and 3 at [5,4].
💡 Top-row triggers
Single-cell sum regions in the top row hand you two pip values without any pairing — they're your first footholds.
💡 Build the top-left corner
Cell [0,0] is a sum-5 alone, so it's 5; cell [0,1] is a sum-3 alone, so it's 3. The two adjacent equals regions in row 1 then force the cells directly below them to be 1s, creating a matched set.
💡 Full placement guide
Domino [5,1] goes at [0,0]=5 and [1,0]=1; domino [3,1] at [0,1]=3 and [1,1]=1, satisfying the equals pair [1,0]/[1,1]. The other equals pair [1,2]/[1,3] gets domino [1,1] horizontally. In column 0, domino [0,6] sits vertically with 0 at [2,0] and 6 at [3,0]; that [3,0]=6 teams with [3,1]=3 from domino [3,6] (3 at [3,1], 6 at [3,2]) to hit the sum-9 region. The greater-9 pair at [3,2]/[3,3] takes domino [6,5] (6 at [3,3], 5 at [2,3]), and the remaining equals region in row 2 fills with domino [5,5] at [2,1]=5, [2,2]=5.
💡 The big sum
Search for the region with the largest sum target — it involves three cells and will dictate how the early dominoes must be allocated.
💡 Corner constraints
The top-left sum-12 region across [0,0], [0,1], and [1,0] forces a combination of 3, 4, and 5. The only domino that can cover two of these cells with those pips is [5,3], and the remaining 4 must sit alone at [1,0] from a domino that also serves the single sum-4 cell at [2,0].
💡 Unequal and greater-than
Once the top-left is resolved, the greater-than-10 pair on the right ([0,3]/[0,4]) forces a 6 at [0,3] and a 5 at [0,4]; that pulls in the [0,6] and [5,5] dominoes. Then the less-than-4 pair below it forces 1 and 2, handled by [1,2]. The unequal region in the center must hold four different values, so watch for the pip cross-check.
💡 Center and right column
The unequal 2x2 block uses dominoes [6,1] and [0,4] to provide a 1, 6, 4, and 0. The rightmost equals column forces three 3s from domino [3,3], and the sum-10 at the bottom uses [4,6] with 4 and 6. The remaining greater-10 pair [3,7]/[3,8] snaps up the [6,6] double.
💡 Solution summary
Place [5,3] at [0,1]=5 & [0,0]=3; [4,4] at [1,0]=4 & [2,0]=4; [0,6] at [0,2]=0 & [0,3]=6; [5,5] at [0,4]=5 & [0,5]=5; [1,2] at [1,5]=1 & [2,5]=2. The unequal block: [6,1] at [4,4]=6 & [4,3]=1; [0,4] at [5,3]=4 & [5,4]=0. Right side: [3,3] at [5,7]=3 & [6,7]=3; [4,6] at [7,6]=4 & [7,7]=6; [0,3] at [4,8]=0 & [4,7]=3; [6,6] at [3,7]=6 & [3,8]=6.

🎨 Pips Solver

Jun 9, 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 9, 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 9, 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 isolated sum-4 cell
The region at [2,2] is a single cell with sum target 4, so it must contain a 4. The only way to place a 4 there is with a domino that includes a 4 pip and covers that cell. Domino [2,4] fits perfectly when placed vertically, with the 4 at [2,2] and the 2 at [3,2].
2
Step 2: The paired sum-4 below
With [3,2] now a 2, the adjacent two-cell sum-4 region at [3,2] and [4,2] requires [4,2] to be another 2. The only remaining domino with two 2s is [2,2], so it must occupy [4,2] and [5,2] horizontally, placing a 2 in each.
3
Step 3: Left-side equals
The equals region at [4,0] and [5,0] demands identical pips. With the double-2 just spent, the [6,6] domino is the only double left; placing it vertically at [4,0] and [5,0] satisfies the equals constraint and gives both cells a 6.
4
Step 4: Tying up the right
The sum-11 region covers [1,4] and [2,4]. The equals region to its left covers [3,4] and [4,4]. Domino [5,4] (5 and 4) can be placed at [2,4] (5) and [3,4] (4); this puts a 4 in [3,4], forcing [4,4] to be 4. The domino [3,4] (3 and 4) then goes at [4,4] (4) and [5,4] (3) to complete the equals pair. Finally, the sum-11 region gets its remaining 6 from domino [6,2] placed at [1,4] (6) and the empty cell [0,4] (2).

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Top-row sum cells
Cell [0,0] has a sum-5 target and stands alone, so it must be exactly 5. Similarly, cell [0,1] is a lone sum-3, forcing a 3. These two cells anchor the top-left corner.
2
Step 2: Creating the equals pairs in row 1
The equals region at [1,0] and [1,1] demands matching values. The only way to cover [0,0]=5 and [1,0] is with domino [5,1], placing 1 at [1,0]. Likewise, [0,1]=3 pairs with [1,1] using domino [3,1], putting another 1 at [1,1]. Now the second equals pair at [1,2] and [1,3] must both be 1, so the [1,1] domino (double 1) fits there horizontally.
3
Step 3: Building column 0 and the bottom-left sum-9
Cell [2,0] is empty, but [3,0] is part of a sum-9 region with [3,1]. Domino [0,6] can bridge [2,0] and [3,0] vertically, placing 0 at [2,0] and 6 at [3,0]. That 6 helps the sum-9: now [3,1] must be 3. Domino [3,6] covers [3,1] with 3 and [3,2] with 6.
4
Step 4: Greater-than-9 pair on the right
The region at [3,2] and [3,3] requires both cells >9, so each must be 6. With [3,2] already 6 from Step 3, [3,3] gets its 6 from domino [6,5] placed vertically at [3,3]=6 and [2,3]=5.
5
Step 5: Finishing the central equals region
Row 2 has an equals region at [2,1], [2,2], [2,3]. With [2,3]=5 from the previous step, the other two must also be 5. Domino [5,5] (double 5) is placed horizontally at [2,1]=5 and [2,2]=5, completing the puzzle.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Breaking down the sum-12 corner
The three-cell sum-12 region at [0,0], [0,1], [1,0] can only be made from 3, 4, and 5. Since no domino contains all three, the domino covering [0,0] and [0,1] must be [5,3] (placed with 3 and 5), leaving [1,0] to be 4. That 4 comes from domino [4,4], which also must cover the single sum-4 cell at [2,0] below it.
2
Step 2: Top-right greater-than and empty
The greater-than-10 pair at [0,3] and [0,4] requires two numbers that sum to more than 10; the only possibility with available dominoes is 6 and 5. Domino [0,6] covers the empty cell [0,2] with 0 and [0,3] with 6. Then domino [5,5] fills [0,4] and [0,5] with 5s, the latter satisfying the greater-than-4 single cell.
3
Step 3: Less-than-4 pair below
Directly below, the less-than-4 region at [1,5] and [2,5] needs two numbers both <4. The only remaining domino that fits is [1,2], placed vertically with 1 at [1,5] and 2 at [2,5].
4
Step 4: Tackling the unequal central block
The four-cell unequal region at [4,3]/[4,4]/[5,3]/[5,4] must contain four distinct values. Domino [6,1] provides 6 and 1; placing it horizontally at [4,4]=6 and [4,3]=1. Then domino [0,4] gives 4 and 0, placed horizontally at [5,3]=4 and [5,4]=0, satisfying the unequal constraint.
5
Step 5: The right-side column of equals
Column 7 has an equals region on [4,7], [5,7], [6,7], all requiring the same number. Domino [3,3] (double 3) fits perfectly when placed vertically at [5,7]=3 and [6,7]=3. This fills two of the three cells and forces the third, [4,7], to be 3 as well, which comes from domino [0,3] placed at [4,8]=0 and [4,7]=3 (since [4,8] is a less-than-4 cell, 0 works).
6
Step 6: Bottom-right sum-10 and greater-than-10
The sum-10 region at [7,6] and [7,7] can be achieved with 4 and 6, so domino [4,6] goes there with 4 at [7,6] and 6 at [7,7]. Finally, the greater-than-10 pair at [3,7] and [3,8] requires sixes; domino [6,6] (double 6) placed horizontally completes 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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