NYT Pips Hints & Answers for June 8, 2026

Jun 8, 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

Today's NYT Pips easy by Ian Livengood opens on two independent deductions: a less‑1 singleton that immediately crystallizes as zero and a sum‑10 pair that forces a double‑five split across two dominoes. The constraint graph is small but neatly interlocking; solving flow proceeds from that zero to a sum‑5 region, then to a sum‑8 column, each step shrinking the possibility space until the remaining empty cells fill without friction.

Rodolfo Kurchan's medium puzzle weaves together equals and greater constraints in a larger, 14‑cell layout. The deduction engine starts with a single‑cell sum‑5 that locks a 5, which in turn forces a column of equal zeros, propagating through a sum‑7 region and a sum‑11 trio. A greater‑8 pair and a greater‑0 singleton tie off the right side, completing a chain where each constraint passes the baton to the next.

Kurchan's hard puzzle is the most demanding, built around a five‑cell equals block along the top row, flanked by a sum‑0 and a sum‑4 singleton. That block’s uniform value is deduced by the interaction of those extreme anchors, triggering a cascade through greater/less, unequal, and multiple small‑sum targets across four rows. The solving architecture is dense: mid‑grid equals force a zero quad, while sum‑6 and sum‑10 cells demand specific high pips, ultimately resolving a 30‑cell field with no ambiguity.

💡 Progressive Hints

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

💡 Hint 1
Look for a region where the permitted values are extremely restricted — a less‑than constraint with a tiny target.
💡 Hint 2
The less‑1 cell at [3,2] can only be 0. That forces the domino that supplies it, and the sum‑5 region below dictates the partner pip. Meanwhile, a sum‑10 pair in row 1 demands two fives.
💡 Hint 3
Place domino [2,0] (2‑0) horizontally at [3,3]‑[3,2] with 2 at [3,3] and 0 at [3,2]. Sum‑10 fills: domino [6,5] (6‑5) vertical [0,2]‑[1,2] gives 5 at [1,2]; domino [5,3] (5‑3) vertical [1,3]‑[2,3] gives 5 at [1,3] and 3 at [2,3]. Sum‑8 forces 2+6: domino [3,2] (3‑2) vertical [0,0]‑[1,0] gives 2 at [1,0]; domino [6,1] (6‑1) horizontal [2,0]‑[2,1] gives 6 at [2,0].
💡 Hint 1
Identify the single‑cell sum region that locks a value immediately, and observe the equals regions that force identical pips across separate cells.
💡 Hint 2
The sum‑5 cell at [3,1] must be 5. The equals pair in column 0 ([2,0] and [3,0]) must be identical. Since [3,1] is 5, its neighbor [3,0] cannot be 5 — it must be 0, forcing the whole column‑0 equals to 0. The sum‑7 pair [1,1]+[2,1] then resolves with a 1.
💡 Hint 3
Placements: Domino [0,5] (0‑5) horizontal [3,0]‑[3,1] (0,5). Domino [0,6] (0‑6) horizontal [2,0]‑[2,1] (0,6) — satisfies equals and gives 6 at [2,1]. Domino [2,1] (2‑1) horizontal [1,2]‑[1,1] (2 at [1,2], 1 at [1,1]). Sum‑11: Domino [4,4] (4‑4) horizontal [0,3]‑[0,4] (4,4); Domino [3,0] (3‑0) vertical [1,3]‑[2,3] (3,0). Equals in row‑2: [2,2] and [2,3] both 0, so Domino [0,4] (0‑4) horizontal [2,2]‑[3,2] (0,4). Greater‑8 region gets 5 from Domino [1,5] (1‑5) horizontal [3,4]‑[3,3] (1,5).
💡 Hint 1
Focus on the enormous equals region that spans five cells, and the tiny sum‑0 and sum‑4 singletons that bracket it.
💡 Hint 2
The sum‑0 cell at [0,5] forces a 0 there, which forces its partner cell [0,4] to be a 1 (from the only available domino with a 0 and a pip that fits the equals group). That 1 then must be the uniform value of the entire equals block.
💡 Hint 3
With the equals group fixed at 1, the sum‑4 cell [0,0] requires a 4, so the domino covering [0,0] and one of the equals cells must be a 1‑4. This pins down the beginnings of the top row. The greater‑4 cell at [1,0] then must be 5 or 6, while the less‑4 cell at [1,1] is ≤3.
💡 Hint 4
The equals region in the middle left (cells [2,0],[2,1],[2,2],[3,0]) all become 0, fed by dominoes with a 0 pip. The sum‑2 at [4,0] and sum‑3 at [4,1] fix two corners. The sum‑6 at [3,4] demands a 6, which will pair with a 0.
💡 Hint 5
Full placement: Row‑0: [1,4] (1‑4) covers [0,1]=1,[0,0]=4; [1,1] (1‑1) covers [0,2]=1,[0,3]=1; [1,0] (1‑0) covers [0,4]=1,[0,5]=0; [6,2] (6‑2) covers [1,6]=6,[0,6]=2. Row‑1: [5,0] (5‑0) covers [1,0]=5,[2,0]=0; [3,0] (3‑0) covers [1,1]=3,[2,1]=0; [4,0] (4‑0) covers [1,2]=4,[2,2]=0; [3,1] (3‑1) covers [1,4]=3,[1,3]=1; [3,3] (3‑3) covers [1,5]=3,[2,5]=3. Lower section: [2,0] (2‑0) covers [4,0]=2,[3,0]=0; [5,3] (5‑3) covers [4,2]=5,[4,1]=3; [5,5] (5‑5) covers [4,3]=5,[4,4]=5; [6,0] (6‑0) covers [3,4]=6,[3,5]=0; [4,4] (4‑4) covers [4,5]=4,[4,6]=4; [2,2] (2‑2) covers [2,6]=2,[3,6]=2.

🎨 Pips Solver

Jun 8, 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 8, 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 8, 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 less‑1 handcuff
Cell [3,2] has a less‑1 constraint, so its pip must be 0. The only domino containing a 0 is [2,0], so it must occupy [3,2] with the 0, placing its 2 in the adjacent cell [3,3]. This also immediately satisfies the sum‑5 region (cells [2,3] and [3,3]) because [3,3]=2, meaning [2,3] must become 3.
2
Step 2: Sum‑10 in the top right
Region [1,2] and [1,3] must sum to 10. With pips ≤6, the only possibility is 5+5. Two different dominoes supply the fives: [6,5] goes vertically [0,2]‑[1,2] (6 on top, 5 at [1,2]), and [5,3] goes vertically [1,3]‑[2,3] (5 at [1,3], 3 at [2,3]). Now [2,3]=3 works with [3,3]=2 to make the sum‑5.
3
Step 3: Sum‑8 in the left column
Region [1,0] and [2,0] must total 8. The remaining unused dominoes are [3,2] and [6,1]. The combination 2+6 gives exactly 8. So [3,2] is placed vertically [0,0]‑[1,0] with pips 3 and 2, putting the 2 at [1,0]. [6,1] goes horizontally [2,0]‑[2,1] with 6 at [2,0] and 1 at [2,1]. All constraints met.
4
Step 4: Final sweep
The empty cells [0,0] and [0,2] receive 3 and 6 respectively from their placed dominoes, and [2,1] gets 1, completing the grid without further conflict.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: The sum‑5 anchor
Cell [3,1] is a single‑cell sum‑5 region, so it must contain exactly 5. The only dominoes with a 5 are [0,5] and [1,5]. This cell will be pivotal.
2
Step 2: Equals zeros in the left column
The equals region covering [2,0] and [3,0] forces both cells to share a value. Since [3,1] is 5, its neighbor [3,0] cannot be 5, so it is forced to 0, making both [2,0] and [3,0] 0. To place these zeros, domino [0,5] (0‑5) is placed horizontally [3,0]‑[3,1] (0,5). The remaining zero for [2,0] must come from the only other domino containing a 0, [0,6]. It goes horizontally [2,0]‑[2,1], placing 0 at [2,0] and 6 at [2,1].
3
Step 3: Sum‑7 drives a 1
Region [1,1] and [2,1] sum to 7. With [2,1]=6, [1,1] must be 1. The domino [2,1] (2‑1) is perfect; placed horizontally [1,2]‑[1,1] gives 2 at [1,2] and 1 at [1,1].
4
Step 4: Sum‑11 and row‑2 equals
The three‑cell sum‑11 region ([0,3],[0,4],[1,3]) must total 11. The remaining dominoes are [4,4], [3,0], [1,5], [0,4]. To reach 11, 4+4+3 is the only viable combination. Place [4,4] horizontally [0,3]‑[0,4] (4,4) and [3,0] vertically [1,3]‑[2,3] (3,0). Now the equals region [2,2] and [2,3] must match; [2,3] is 0, so [2,2] becomes 0. That forces domino [0,4] (0‑4) horizontally [2,2]‑[3,2] with 0 at [2,2] and 4 at [3,2].
5
Step 5: Greater‑8 and remaining
The greater‑8 region [3,2] and [3,3] must be satisfied. [3,2] is already 4; [3,3] needs a high pip. Domino [1,5] (1‑5) fits horizontally [3,4]‑[3,3] giving 5 at [3,3] and 1 at [3,4], where [3,4] satisfies greater‑0. All cells covered.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: The sum‑0 and equals cascade
Cell [0,5] is a sum‑0 region, so it must be 0. It is adjacent to [0,4], which belongs to the huge equals region (also [0,1],[0,2],[0,3],[1,3]). The only domino that can supply a 0 to [0,5] and a matching pip to [0,4] is [1,0] (1‑0). Placing it horizontally [0,4]‑[0,5] forces [0,4]=1 and [0,5]=0. Consequently, every cell in the equals region must be 1.
2
Step 2: Filling the equals block
With the equals value locked at 1, the sum‑4 cell [0,0] requires a 4. The domino [1,4] (1‑4) goes vertically [0,1]‑[0,0], putting 1 at [0,1] and 4 at [0,0]. The remaining equals cells [0,2] and [0,3] get 1s via domino [1,1] (1‑1) placed horizontally [0,2]‑[0,3]. Finally, [1,3] is part of the equals region and must be 1; it will receive that from a later domino.
3
Step 3: Top‑row right and Row‑1 greater/less
The empty cell [0,6] and greater‑2 cell [1,6] must be filled. The only remaining domino with a 6 is [6,2] (6‑2). Place it vertically [1,6]‑[0,6] giving 6 at [1,6] (satisfying >2) and 2 at [0,6]. Now Row‑1 has greater‑4 at [1,0] and less‑4 at [1,1]. The greater‑4 cell needs a pip >4, so 5 or 6. With 6 used, it must be 5. The domino [5,0] (5‑0) goes vertically [1,0]‑[2,0], placing 5 at [1,0] and 0 at [2,0]. For less‑4 at [1,1], it must be ≤3; the remaining pips allow 3 from [3,0] (3‑0), placed vertically [1,1]‑[2,1] giving 3 and 0, which also contributes to the equals region below.
4
Step 4: Middle equals zero block
The equals region [2,0],[2,1],[2,2],[3,0] is already getting forced to 0: [2,0] is 0 from step 3, [2,1] becomes 0 from [3,0], [2,2] needs a 0, and [3,0] will be 0. The sum‑4 cell [1,2] needs a 4, so domino [4,0] (4‑0) goes vertically [1,2]‑[2,2] with 4 at [1,2] and 0 at [2,2]. Now all equals cells are 0.
5
Step 5: Sum‑6 and unequal patch
Cell [3,4] is a sum‑6 singleton, so it must be 6. The domino [6,0] (6‑0) fits vertically [3,4]‑[3,5] with 6 at [3,4] and 0 at [3,5], where 0 is allowed in the unequal region. The sum‑6 region [1,4]‑[1,5] must total 6. With [1,3]=1 already set, domino [3,1] (3‑1) is placed horizontally [1,3]‑[1,4] giving 1 at [1,3] and 3 at [1,4]. Then [1,5] must be 3, so domino [3,3] (3‑3) goes vertically [1,5]‑[2,5] placing 3 at both.
6
Step 6: Lower right and final sums
Sum‑2 at [4,0] needs 2; domino [2,0] (2‑0) goes horizontally [4,0]‑[3,0] with 2 at [4,0] and 0 at [3,0]. Sum‑3 at [4,1] needs 3; domino [5,3] (5‑3) goes horizontally [4,2]‑[4,1] with 5 at [4,2] and 3 at [4,1]. Sum‑10 region [4,2]+[4,3] needs 10; with [4,2]=5, [4,3] must be 5, so place [5,5] (5‑5) horizontally [4,3]‑[4,4]. The unequal region includes [4,4]=5 and [4,5]; domino [4,4] (4‑4) goes horizontally [4,5]‑[4,6] with 4 at both, satisfying sum‑4 at [4,6] and the unequal constraint. Sum‑2 at [3,6] gets 2 from [2,2] (2‑2) placed vertically [2,6]‑[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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