🚨 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
Easy, by Ian Livengood, is a swift warm-up. A three-cell equals region and a lone sum-3 cell give you immediate handholds — lock those in and the rest unfolds without branching. If you're new to today's NYT Pips, this is a confidence-builder.
Medium, by Rodolfo Kurchan, has a clean board but one tight squeeze: the left column's equals block forces a single pip across three cells. Once that snaps into place, the sum-11 duo and the rare unequal 2×2 region (all four pips different) create a satisfying mid-puzzle puzzle.
Hard, also by Kurchan, is the dense one. Sum-1 islands and cascading sum-11 stripes leave very little wiggle room. You'll be threading double-6s through tight corridors and double-5s into stacked vertical sums. Expect to pause often — it's a rewarding grind.
💡 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 whose constraint is so tiny it can only be satisfied one way — that'll give you a critical value and a domino orientation.
💡 Hint 2
The sum-3 cell at [0,2] is a single-cell region; to hit that sum, the pip must be exactly 3. The only way to cover it is with a domino that carries a 3, and its partner cell will be on row 0.
💡 Hint 3 (Final)
Place domino [2] (3|6) horizontally at [0,2]-[0,3], with 3 at [0,2] and 6 at [0,3]. The equals region spanning [1,0], [2,0], [2,1] must all be 3: place domino [4] (3|3) vertically at [1,0]-[2,0], then domino [0] (0|3) horizontally at [2,1]-[2,2] (3 at [2,1], 0 at [2,2]). The sum-8 region gets domino [3] (4|4) vertically at [2,3]-[3,3] for two 4s. Finally, the less-6 cell [0,1] takes domino [1] (5|6) with 5, and the empty [0,0] gets the 6.
💡 Hint 1
An 'unequal' region is a rarity — when a 2×2 block demands four different values, it acts as a final gatekeeper. Look for it late in the solve, but let the column constraints drive your start.
💡 Hint 2
The left column has an equals region spanning rows 1–3. With the available double dominos, only one pip value can appear three times, forcing that column to a specific number and dictating which double you use.
💡 Hint 3 (Final)
Place domino [8] (3|3) vertically at [2,0]-[3,0] and domino [6] (6|3) vertically at [0,0]-[1,0] (6 at [0,0], 3 at [1,0]) to satisfy the equals column. The sum-11 pair [0,0]-[0,1] gets 5 at [0,1] via domino [0] (5|1) horizontally to [0,2] (1 at [0,2]). The sum-5 region then demands 2+2, so domino [3] (2|2) across [0,3]-[0,4]. Single sum-4 cell [0,5] forces domino [5] (4|4) across [0,5]-[0,6], and the equals pair [0,6]-[0,7] gets domino [7] (4|5) with 4 at [0,7], 5 at [1,7]. Right column sum-11 [1,7]-[2,7] takes domino [4] (0|6) vertically with 6 at [2,7], 0 at [3,7]. Finally, the unequal region [2,3]-[3,4]: domino [1] (1|3) vertically [2,3]-[3,3] (3,1) and domino [2] (0|2) vertically [2,4]-[3,4] (2,0).
💡 Hint 1
Scout for the sum-1 constraints — single cells that must be exactly 1, and a sum-1 pair that must be 0+1. They are the puzzle's skeleton; everything else hangs off them.
💡 Hint 2
The left edge has two consecutive sum-1 cells at [3,0] and [3,1]. Their forced value of 1 will feed into the sum-11 pair above them, narrowing the possibilities for the vertical domino that crosses the boundary.
💡 Hint 3
Start with [3,0]=1. That cell must pair with a domino containing a 1; the only candidate that reaches it vertically is the 5|1 domino. Placing it puts a 5 at [2,0], which immediately forces the sum-11 partner [1,0] to be 6, pulling in the 3|6 domino.
💡 Hint 4
After fixing the left column, the top-row sum-11 gets a 3 from the 3|6 domino at [0,0], so [0,1] and [0,2] must sum to 8. Only a double-4 can do that. Meanwhile, the sum-1 at [3,1] will pair with 2|1 domino to its right, and the sum-11 stripe in the middle [1,4]-[2,4] will demand a 6|6 horizontally above and a 5|2 vertically below.
💡 Hint 5 (Final)
Full placement: Domino [12] (5|1) vertically [2,0]-[3,0] (5,1). Domino [6] (3|6) vertically [0,0]-[1,0] (3,6). Domino [2] (4|4) horizontally [0,1]-[0,2] (4,4). Domino [10] (0|0) horizontally [0,7]-[1,7] (0,0). Domino [8] (6|6) horizontally [0,4]-[1,4] (6,6). Domino [5] (5|2) vertically [2,4]-[3,4] (5,2). Domino [4] (3|4) horizontally [3,5]-[3,6] (4,3). Domino [0] (3|0) vertically [2,7]-[3,7] (0,3). Domino [7] (2|1) horizontally [3,1]-[3,2] (1,2). Domino [3] (5|0) horizontally [5,3]-[5,4] (0,5). Domino [9] (1|1) horizontally [6,3]-[7,3] (1,1). Domino [11] (2|6) vertically [5,5]-[6,5] (6,2). Domino [1] (5|5) vertically [8,3]-[9,3] (5,5). Domino [13] (2|4) horizontally [7,4]-[7,5] (4,2).
🎨 Pips Solver
Jun 11, 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
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
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: Crack the equals region
The equals region spanning [1,0], [2,0], [2,1] requires three identical pips. Among the available dominoes, only value 3 appears on enough tiles: the double-3 (domino 4) and the 0-3 (domino 0). This forces the region to be all 3s.
2
Step 2: Place the double-3 anchor
Domino 4 (3|3) must cover two of the equals cells. The geometry allows vertical placement at [1,0]-[2,0], setting both to 3.
3
Step 3: Extend the 3 and crack the sum-8
The third equals cell [2,1] must get its 3 from domino 0 (0|3), placed horizontally at [2,1]-[2,2]. That puts a 3 at [2,1] and a 0 at [2,2], which lies in the sum-8 region. With 0 fixed, the other two cells [2,3] and [3,3] must sum to 8 — so both must be 4. Domino 3 (4|4) fits vertically there.
4
Step 4: Top row sweep
The isolated sum-3 cell [0,2] demands a 3, so domino 2 (3|6) goes horizontally [0,2]-[0,3] with 3 at [0,2] and 6 at [0,3]. Then the less-6 cell [0,1] needs a pip under 6; the only remaining place for domino 1 (5|6) is horizontally [0,1]-[0,0] with 5 at [0,1] and 6 at [0,0] (empty).
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Establish the left column equals
The equals region occupies [1,0], [2,0], [3,0]. Only value 3 can appear three times (double-3 domino 8, plus 3 from domino 6 or 1). So the column gets 3s. Place domino 8 (3|3) vertically at [2,0]-[3,0]. To give [1,0] a 3, use domino 6 (6|3) vertically at [0,0]-[1,0] with 6 at [0,0] and 3 at [1,0].
2
Step 2: Resolve the top sum-11 and sum-5
[0,0] is 6, so the sum-11 pair [0,0]-[0,1] needs 5 at [0,1]. Domino 0 (5|1) placed horizontally [0,1]-[0,2] gives 5 and 1. Now [0,2]=1 is part of the sum-5 region [0,2]-[0,4]; the other two cells must sum to 4, forcing both 2. Use domino 3 (2|2) horizontally [0,3]-[0,4].
3
Step 3: Navigate the 4s and equals on the right
A sum-4 singleton at [0,5] requires a 4. Domino 5 (4|4) covers [0,5]-[0,6] with two 4s, simultaneously satisfying the equals pair [0,6]-[0,7]. So [0,7] becomes 4. Domino 7 (4|5) then fits [0,7]-[1,7] with 4 at [0,7] and 5 at [1,7].
4
Step 4: Fill the rightmost column
The sum-11 pair [1,7]-[2,7] has 5 at [1,7], so [2,7] must be 6. Place domino 4 (0|6) vertically [2,7]-[3,7] with 6 at [2,7] and 0 at [3,7] (leftover empty cell).
5
Step 5: Satisfy the unequal 2×2
The region [2,3]-[3,4] needs four distinct values. With remaining dominoes 1 (1|3) and 2 (0|2), place domino 1 vertically [2,3]-[3,3] (3 at [2,3], 1 at [3,3]) and domino 2 vertically [2,4]-[3,4] (2 at [2,4], 0 at [3,4]). All values 3,2,1,0 — unique.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Unlock the left edge with sum-1 and sum-11
The sum-1 cell [3,0] must be 1. The only domino that can place a 1 there while covering upward is domino 12 (5|1), placed vertically [2,0]-[3,0] (5 above, 1 below). Then the sum-11 pair [1,0]-[2,0] now has 5 at [2,0], so [1,0] must be 6. Place domino 6 (3|6) vertically [0,0]-[1,0] (3 at [0,0], 6 at [1,0]).
2
Step 2: Complete the top-left sum-11
[0,0] is 3, so the sum-11 region [0,0]-[0,2] needs 8 more across [0,1] and [0,2]. The only domino that fits is 4|4, placed horizontally [0,1]-[0,2] (both 4). The empty corners [0,7]-[1,7] can be filled with domino 10 (0|0) horizontally.
3
Step 3: Handle the second sum-1 and the equals tail
The sum-1 at [3,1] forces a 1 there. Domino 7 (2|1) placed horizontally [3,1]-[3,2] gives 1 at [3,1] and 2 at [3,2] (empty). Then tackle the sum-11 strip in the middle: [1,4]-[2,4] must sum to 11. Place domino 8 (6|6) horizontally [0,4]-[1,4] to satisfy the greater-4 cell at [0,4] and put 6 at both; then [2,4] needs 5. Use domino 5 (5|2) vertically [2,4]-[3,4] (5 at [2,4], 2 at [3,4]).
4
Step 4: Resolve the sum-6 and equals pair
The sum-6 region [3,4]-[3,5] now has 2 at [3,4], so [3,5] must be 4. Domino 4 (3|4) placed horizontally [3,5]-[3,6] puts 4 at [3,5] and 3 at [3,6]. The equals region [3,6]-[3,7] then demands both 3, so [3,7] must be 3. Place domino 0 (3|0) vertically [2,7]-[3,7] with 3 at [3,7] and 0 at [2,7].
5
Step 5: Satisfy the sum-1 pair and adjacent sum-11
The sum-1 pair [5,3]-[6,3] needs 0 and 1. Domino 3 (5|0) horizontally [5,3]-[5,4] sets [5,3]=0 and [5,4]=5. Then domino 9 (1|1) horizontally [6,3]-[7,3] sets both 1, completing sum-1. The sum-11 region [5,4]-[5,5] gets [5,4]=5, so [5,5] must be 6. Place domino 11 (2|6) vertically [5,5]-[6,5] (6 at [5,5], 2 at [6,5]).
6
Step 6: Finish the vertical sum-11 and remaining sum-4 cells
The sum-11 region [7,3]-[9,3] has 1 at [7,3]; the other two cells need sum 10, so use domino 1 (5|5) vertically [8,3]-[9,3] (both 5). Finally, sum-4 region [6,5]-[7,5] has 2 at [6,5], so [7,5] needs 2. The sum-4 singleton [7,4] needs 4. Domino 13 (2|4) horizontally [7,4]-[7,5] places 4 at [7,4] and 2 at [7,5].
💡 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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