NYT Pips Hints & Answers for June 12, 2026

Jun 12, 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 puzzle, Ian Livengood constructs a compact deduction graph. The central 2×2 equals block forces a precise pip value—only the number 1 appears on enough domino faces to fill four cells—immediately coupling with the adjacent less‑2 restriction. That anchor cascades into a vertical [0,1] bridging domino, a horizontal [1,1] placement, and a clean finish through the sum‑3 and greater‑6 regions.

Livengood’s medium puzzle escalates to an eight‑domino interlock. A solitary sum‑4 cell at [1,0] demands the only domino carrying a 4, which simultaneously feeds a greater‑9 pair with a 6. The resulting 5‑1 domino triggers a top‑row triple‑equals block that must collapse to all zeros, steering the sum‑6 and empty regions into place and ultimately locking the bottom‑row equals with two fives.

Rodolfo Kurchan’s hard puzzle splinters tiny sum‑2 targets across the grid, weaving a dense fabric of constraints. A top‑left 2‑0‑0 sum tangles with a sum‑5 cell, forcing a cross‑zone 2‑5 domino, while an independent row‑4 chain of greater‑2 and three sum‑2 segments anchors the middle. The bottom‑right four‑cell equals block demands all fives, commanding exact multi‑pip placements that resolve greater‑4, sum‑10, and final equals threads. The solving chain requires careful pip‑availability bookkeeping across otherwise disconnected regions.

💡 Progressive Hints

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

💡 Spot the forcing constraint
Focus on the central equals region—it forces every cell within it to share the same pip value. That demand narrows which domino faces can possibly appear there.
💡 Connect the less-2 cell
The less‑2 region at [0,2] touches the equals cell at [1,2]. The domino linking them must place a value that obeys both: the equals region already forces 1, so the less‑2 cell must take the 0 from the [0,1] domino, planted vertically.
💡 Full layout
Place domino [0,1] vertically at [0,2]‑[1,2] (0 on top, 1 below). Place [1,1] horizontally at [2,1]‑[2,2] (both 1). Then [1,2] horizontally at [1,1]‑[1,0] (1 at [1,1], 2 at [1,0]). Finish with [3,3] vertically at [1,3]‑[2,3] (both 3) and [0,6] vertically at [1,4]‑[2,4] (0 above, 6 below).
💡 Find the tight sum
Search for a cell with a sum constraint that, by itself, can only be satisfied by a single pip. Then watch how that value activates a neighbouring greater‑than region.
💡 Start at the sum-4
Cell [1,0] must sum to 4 alone, so it must be exactly 4. Only the [4,6] domino can deliver that 4, which places its 6 on [1,1] and triggers the greater‑9 pair. That forces a 5 onto [2,1] from the [1,5] domino, whose 1 then dictates the equals region at [1,2]‑[2,2].
💡 Complete solution
Place [4,6] at [1,0]‑[1,1] (4,6). [1,5] at [2,1]‑[2,2] (5,1). [1,0] at [1,2]‑[1,3] (1,0). [0,0] at [0,1]‑[0,2] (0,0). [2,0] at [0,4]‑[0,3] (2 at [0,4], 0 at [0,3]). [3,4] at [0,6]‑[0,5] (3,4). Finally [5,6] at [2,3]‑[2,4] (6,5) and [5,3] at [2,5]‑[2,6] (5,3).
💡 Look for the sum-2 seeds
Several sum‑2 regions are scattered across the board, each clamping down the smallest pips. The interplay between the top‑left sum‑2 and a lone sum‑5 cell will determine the first placements.
💡 Unlock the top-left corner
The sum‑2 region at row 0 columns 0‑2 must be 2,0,0. The sum‑5 at [1,0] must be exactly 5. These two constraints can only be satisfied together by placing the [2,5] domino across [0,0]‑[1,0] (2 above, 5 below) and the [0,0] domino on [0,1]‑[0,2].
💡 Conquer the next two rows
With the top locked, the sum‑8 at [2,0]‑[2,1] demands 4+4 from the [4,4] domino. Directly below, the greater‑9 region at [3,0]‑[4,0] requires a sum >9; only [6,5] works, placed vertically with 6 at [3,0] and 5 at [4,0].
💡 Unfold the row-4 sum-2 chain
The greater‑2 cell [4,4] must be >2, and the next three cells form a sum‑2 trio. Place [3,0] at [4,4]‑[4,5] (3,0), then [4,2] at [5,6]‑[4,6] (4,2) to give 2 at [4,6], and [0,2] at [4,7]‑[4,8] (0,2) to finish the set, satisfying the free sum‑2 cell at [4,8] with its 2.
💡 Full grid assembly
Top: [2,5] at [0,0]‑[1,0] (2|5); [0,0] at [0,1]‑[0,2] (0|0); [4,4] at [2,0]‑[2,1] (4|4); [6,5] at [3,0]‑[4,0] (6|5). Row‑4: [3,0] at [4,4]‑[4,5] (3|0); [4,2] at [5,6]‑[4,6] (4|2); [0,2] at [4,7]‑[4,8] (0|2). Bottom‑right: [5,5] at [9,2]‑[9,3] (5|5); [5,3] at [9,4]‑[8,4] (5|3); [5,0] at [9,1]‑[8,1] (5|0). Then [6,2] at [6,1]‑[7,1] (6|2) meets greater‑4 and sum‑2; [3,6] at [7,4]‑[6,4] (3|6) for equals plus greater‑4; [6,1] at [6,6]‑[7,6] (6|1) and [1,2] at [8,6]‑[9,6] (1|2) handle sum‑10, equals, and the final sum‑2.

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Jun 12, 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 12, 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 12, 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: Count pips for the central equals
The equals region covers cells [1,1], [1,2], [2,1], [2,2] and requires the same pip on all four. Tally the pips on the available dominoes: only the value 1 appears on enough faces (three dominoes supply a total of four 1’s) to populate the entire region, while no other number appears more than twice. Hence the region must settle on all ones.
2
Step 2: Bridge the less-2 cell
Cell [0,2] has a less‑2 restriction and is adjacent only to [1,2], which now must be a 1. The only domino that can cover both cells with a value under 2 on [0,2] and a 1 on [1,2] is [0,1]. Place it vertically: [0,2]=0, [1,2]=1.
3
Step 3: Fill the remaining equals corners
With [1,2] already a 1, the other three equals cells still need 1’s. The [1,1] domino provides two adjacent 1’s; it must occupy two of the remaining cells. The only way to complete the region is to lay [1,1] horizontally at [2,1]‑[2,2] and then place the final 1 using [1,2] horizontally at [1,1]‑[1,0], giving [1,1]=1 and [1,0]=2.
4
Step 4: Resolve the sum‑3 and greater‑6 regions
Cell [1,3] is a single‑cell sum‑3 target, so it must hold 3. Use the [3,3] domino (the only source of a 3) vertically at [1,3]‑[2,3] (both 3). The greater‑6 pair [2,3],[2,4] demands a sum >6; [2,3]=3 forces [2,4] to be as large as possible. The [0,6] domino places its 6 on [2,4] and its 0 on the empty cell [1,4], satisfying the sum 3+6=9>6.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Sum‑4 forces the 4–6 domino
The sum‑4 target at [1,0] is a single cell, so it must contain exactly 4. The only domino with a 4 is [4,6]. Place it horizontally covering [1,0]‑[1,1], putting 4 at [1,0] and 6 at [1,1].
2
Step 2: Greater‑9 drives a 5–1 placement
Region [1,1],[2,1] has a greater‑9 target, meaning their sum must exceed 9. With [1,1]=6, we need [2,1] >3. The domino [1,5] can supply a 5. Placing it horizontally at [2,1]‑[2,2] sets 5 on [2,1] (sum 11) and 1 on [2,2]. The equals region [1,2],[2,2] now forces [1,2] to also be 1, which is provided by the [1,0] domino placed horizontally at [1,2]‑[1,3] (1 at [1,2], 0 at [1,3]).
3
Step 3: Top‑row equals turns all zero
The equals region in row 0 columns 1‑3 must all hold the same value. The remaining pips allow only 0 to appear on three cells (the [0,0] domino gives two 0’s, and [2,0] has one 0). Place [0,0] horizontally at [0,1]‑[0,2]; then the [2,0] domino must give its 0 to [0,3], spanning [0,4]‑[0,3] with 2 at [0,4] and 0 at [0,3].
4
Step 4: Sum‑6 at top requires 4 and 3
The sum‑6 region at [0,4]‑[0,5] now has [0,4]=2. To reach 6, [0,5] must be 4. The [3,4] domino carries a 4 and a 3; place it horizontally at [0,6]‑[0,5] with 3 at [0,6] (empty) and 4 at [0,5].
5
Step 5: Final sum‑6 and equals lock bottom
The sum‑6 pair [1,3],[2,3] already has [1,3]=0, so [2,3] must be 6. That 6 comes from the [5,6] domino, placed horizontally at [2,3]‑[2,4] (6 at [2,3], 5 at [2,4]). The equals region [2,4],[2,5] now forces [2,5] to be 5, which is supplied by the [5,3] domino horizontally at [2,5]‑[2,6] (5 at [2,5], 3 at [2,6] empty).

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Top‑left sum‑2 and sum‑5 interact
The sum‑2 region [0,0],[0,1],[0,2] must total 2 with three cells; the only viable split with 0‑6 pips is 2+0+0. The sum‑5 single cell at [1,0] must be 5. To satisfy both, the [2,5] domino must span [0,0] (2) and [1,0] (5), while the [0,0] domino covers [0,1]‑[0,2] (0,0). This fixes the top‑left corner.
2
Step 2: Sum‑8 takes the 4‑4 domino
Region [2,0],[2,1] has a sum‑8 target. The only domino that sums to 8 is [4,4]. Place it horizontally across [2,0]‑[2,1], putting 4 in both cells.
3
Step 3: Greater‑9 forces 6‑5
The greater‑9 region at [3,0],[4,0] needs a sum >9. With the remaining high‑pip dominoes, only [6,5] (6+5=11) qualifies. Place it vertically at [3,0]‑[4,0], assigning 6 to [3,0] and 5 to [4,0].
4
Step 4: Row‑4’s sum‑2 chain unfolds
The greater‑2 cell [4,4] must be >2; the next three cells [4,5],[4,6],[4,7] form a sum‑2 region, and [4,8] is a solo sum‑2. Start with [3,0] (3,0) at [4,4]‑[4,5] (3,0). The sum‑2 trio now has [4,5]=0, so the remaining two must be 2 and 0. Place [4,2] at [5,6]‑[4,6] (4 at [5,6], 2 at [4,6]), then [0,2] at [4,7]‑[4,8] (0 at [4,7], 2 at [4,8]) to finish the set, perfectly satisfying both sum‑2 targets.
5
Step 5: Bottom‑right equals demands all fives
The four‑cell equals region [9,1]‑[9,4] must all be the same value; the only remaining pip with four copies is 5 (coming from [5,5], [5,3], and [5,0]). Place [5,5] horizontally at [9,2]‑[9,3] (two 5’s). Then place [5,3] at [9,4]‑[8,4] (5 at [9,4], 3 at [8,4]), making the equals pair [7,4],[8,4] both 3. Finally, [5,0] at [9,1]‑[8,1] puts 5 at [9,1] and 0 at [8,1].
6
Step 6: Wrap up greater‑4, sum‑2, sum‑10, and final equals
Now [6,1] has greater‑4 and must be 6; the [6,2] domino at [6,1]‑[7,1] supplies that 6 and puts 2 on [7,1], completing the sum‑2 region [7,1],[8,1] (2+0). The greater‑4 at [6,4] needs 6, so place [3,6] vertically at [7,4]‑[6,4] (3 at [7,4], 6 at [6,4]). Sum‑10 at [5,6],[6,6] already has [5,6]=4, so [6,6] must be 6; use [6,1] at [6,6]‑[7,6] (6 at [6,6], 1 at [7,6]) to satisfy the sum‑10 and give the equals pair [7,6],[8,6] both 1. Finally, [1,2] at [8,6]‑[9,6] places 1 at [8,6] and 2 at [9,6] to meet the sum‑2 target.

💡 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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