NYT Pips Hints & Answers for June 19, 2026

Jun 19, 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

Today’s NYT Pips easy from Ian Livengood is a confidence-builder. Two “less 1” corners immediately hand you zeros, and a pair of equals regions lock a triple 5 and a double 3. There’s no branching, no heavy arithmetic—just read the pip list, follow the zero, and you’ll fill the grid in minutes.

The medium puzzle, also by Livengood, has one tight bottleneck: a sum‑12 region in the upper center that forces two 6s. Spot that and the rest unravels cleanly—a less‑2 cell, an equals doubleton, and a greater‑4 cell all fall into place. Expect a satisfying “aha” moment, then rapid completion.

Rodolfo Kurchan’s hard is a dense tangle of equals and sum chains. You’ll propagate values through a triple of 1s, a sum‑0 cell that forces a 0/4 pair, and a remote sum‑9 on the right edge. The solve is methodical and demands patience, but the cascading logic is clean once you find the anchors.

💡 Progressive Hints

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

💡 Hint 1: The forced zeros
Look at the two less‑1 regions—single cells that must hold a pip less than 1. That’s an immediate giveaway.
💡 Hint 2: Lock in the equals
With zeros at [0,4] and [3,0], the adjacent equals region across [1,3], [1,4], [2,4] has to be all 5s because the only domino with a 0 is [5,0]. The other equals pair at [1,2] and [2,2] will become 3s.
💡 Hint 3: Full solve
Domino [5,0] covers [1,4]=5 and [0,4]=0. [5,3] goes to [1,3]=5 and [1,2]=3. [5,1] fills [2,4]=5 and [2,3]=1. [3,6] completes [2,2]=3 and [2,1]=6. Finally [2,0] places [3,1]=2 and [3,0]=0.
💡 Hint 1: Find the giveaway sum
The puzzle has a sum‑12 region—with the available dominoes, there’s only one way to reach that total. Focus there first.
💡 Hint 2: The 6‑6 bottleneck
Cells [1,2] and [1,3] must both be 6. That forces [1,3] to be 6 from the [6,4] domino, placing [0,3]=4 vertically, and [1,2] to be 6 from [1,6] with [0,2]=1 horizontally—satisfying the less‑2 at [0,2].
💡 Hint 3: Full solve
Play [6,4] at [1,3]/[0,3]; [1,6] at [0,2]/[1,2]; [0,3] at [4,4]=0 and [4,3]=3 (pairing with [3,3]=3 from [3,6] at [3,3]/[3,2]); [2,6] at [5,1]=2 and [5,0]=6; [5,2] at [4,2]=5 and [4,1]=2; [4,1] at [5,4]=4 and [5,5]=1; [5,4] at [5,2]=5 and [5,3]=4.
💡 Hint 1: Look for the forced single‑digit cells
A sum‑0 region and several equals chains will hand you your first values. Check cells that have no choice.
💡 Hint 2: Zero sets off a chain
[1,0] must be 0. The only domino with a 0 is [4,0], giving [1,1]=4 and forcing the equals pair [1,1] and [1,2] to be 4. That will then set [2,2] to 1.
💡 Hint 3: The triple‑1s and domino [1,3]
The equals region [2,1],[2,2],[2,3] is all 1s. With [2,2]=1, domino [1,3] fills [2,1]=1, [3,1]=3. Then [1,2] fills [2,3]=1, [2,4]=2, triggering the equals‑2 region at [2,4],[2,5],[3,5].
💡 Hint 4: Propagate out to the edges
The equals [2,6],[2,7] become 1s, placing [1,5] at [2,7]=1, [2,8]=5 (sum‑5) and [1,6] at [2,6]=1, [3,6]=6. Equals [3,6],[4,6] gets 6 from [4,6] at [4,6]=6, [5,6]=4. The sum‑9 at [0,7]/[1,7] resolves to 3+6 via [6,3].
💡 Hint 5: Complete solution
[4,0] at [1,1]=4,[1,0]=0; [4,1] at [1,2]=4,[2,2]=1; [1,3] at [2,1]=1,[3,1]=3; [1,2] at [2,3]=1,[2,4]=2; [2,2] at [2,5]=2,[3,5]=2; [1,5] at [2,7]=1,[2,8]=5; [1,6] at [2,6]=1,[3,6]=6; [4,6] at [5,6]=4,[4,6]=6; [6,3] at [1,7]=6,[0,7]=3; [5,5] at [0,1]=5,[0,2]=5; [5,4] at [5,2]=5,[4,2]=4; [2,6] at [6,1]=2,[6,2]=6; [2,3] at [3,2]=2,[3,3]=3; [5,0] at [6,6]=5,[6,5]=0.

🎨 Pips Solver

Jun 19, 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 19, 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 19, 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: Zero in on the less‑1 cells
Cells [0,4] and [3,0] are less‑1 regions—they must be 0, no other pip fits.
2
Step 2: Place the zero‑carrying domino
The only domino with a 0 is [5,0]. To satisfy [0,4]=0, place it vertically with 5 at [1,4] and 0 at [0,4]. That also primes the adjacent equals region.
3
Step 3: Complete the triple‑5 equals region
The equals at [1,3],[1,4],[2,4] needs all 5s. [1,4] is already 5, so place [5,3] with 5 at [1,3] and 3 at [1,2]; then [5,1] with 5 at [2,4] and 1 at [2,3].
4
Step 4: Fill the 3s pair and final zero
The equals pair [1,2],[2,2] must be 3s. [1,2] is 3, so place [3,6] with 3 at [2,2] and 6 at [2,1]. Finally, [2,0] places 2 at [3,1] and 0 at [3,0].

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: The sum‑12 demands 6+6
The region [1,2] and [1,3] sums to 12. With a maximum pip of 6, both cells must be 6. That’s the critical bottleneck.
2
Step 2: Anchor [1,3] and [0,3]
[1,3]=6 can only come from domino [6,4] placed vertically. So [1,3]=6 and [0,3] gets 4, satisfying the empty region there.
3
Step 3: Resolve [1,2] and the less‑2
[1,2]=6 needs [1,6] placed horizontally, putting 1 at [0,2]—perfect for the less‑2 region. Now [0,2]=1 and [1,2]=6.
4
Step 4: The equals chain on the right
The equals pair [3,3] and [4,3] forces a 3. Domino [3,6] places 3 at [3,3] and 6 at [3,2] (empty). Then [0,3] places 0 at [4,4] (less‑1) and 3 at [4,3].
5
Step 5: Finish the bottom layer
Greater‑4 at [5,0] forces 6, so [2,6] gives [5,1]=2, [5,0]=6. Equals [4,1],[5,1] force [5,1]=2, so [5,2] gives [4,2]=5, [4,1]=2. Equals [4,2],[5,2] forces [5,2]=5, so [5,4] gives [5,2]=5, [5,3]=4. Finally [4,1] gives [5,4]=4, [5,5]=1 (less‑2).

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Sum‑0 marks the start
Cell [1,0] must be 0 (sum‑0). The only domino with a 0 is [4,0], so place it as [1,1]=4, [1,0]=0. The equals [1,1],[1,2] then forces [1,2]=4.
2
Step 2: A triple of 1s
Place [4,1] at [1,2]=4, [2,2]=1. That triggers the equals [2,1],[2,2],[2,3] all 1s. Then [1,3] gives [2,1]=1, [3,1]=3; [1,2] gives [2,3]=1, [2,4]=2. Now the equals [2,4],[2,5],[3,5] are all 2s: [2,2] goes [2,5]=2, [3,5]=2.
3
Step 3: Right‑side equals and sum‑5
The equals [2,6],[2,7] must be 1s. Domino [1,5] gives [2,7]=1, [2,8]=5 (satisfying sum‑5 at [2,8]). Domino [1,6] gives [2,6]=1, [3,6]=6, which forces [4,6]=6 via [4,6] at [5,6]=4, [4,6]=6.
4
Step 4: The top‑right sum‑9
Region [0,7],[1,7] sums to 9. The only combo is 3+6, so [6,3] places [1,7]=6, [0,7]=3.
5
Step 5: Greater‑9 forces two 5s
The greater‑9 at [0,1],[0,2] requires pips >4, so both are 5. Domino [5,5] places [0,1]=5, [0,2]=5.
6
Step 6: The final corner
Greater‑9 at [5,2],[6,2] needs 5+6. [5,4] gives [5,2]=5, [4,2]=4 (completing the sum‑9 at [3,1],[3,2],[4,2] with [3,2]=2 from [2,3] at [3,2]=2, [3,3]=3, and [3,3] satisfies sum‑3). [2,6] gives [6,1]=2, [6,2]=6 (sum‑2 at [6,1]). Finally [5,0] gives [6,6]=5, [6,5]=0 (sum‑0 and sum‑9 with [5,6]=4).

💡 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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