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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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🎲 Today's Puzzle Overview
Easy, by Ian Livengood, is a confidence-builder — a tiny 5×3 grid where a three-cell equals region immediately dictates the opening. Spot that equal-value column and the rest dominoes into place with zero guesswork.
Medium, from Rodolfo Kurchan, lifts the challenge with a sparse but clever layout. A three-cell equals chain on the right forces an early commitment to zeros, then a sum-11 region and a parallel equals in row 1 create a tight cross-constraint. Once you lock the rightmost column, the left side unravels smoothly.
Today's NYT Pips hard by Rodolfo Kurchan is a rich tangle of sum and equals regions across a full grid. A deceptively simple single-cell sum-2 at the bottom right acts as the key — it fixes a digit that echoes through two separate four-cell equals regions, one in the center and one along the edge. Navigating the interconnected greater-than and sum constraints after that feels like untangling a knot, but the logical path is profoundly satisfying.
💡 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 type that forces all its cells to have the same value. That’s often the key to unlock an easy puzzle.
💡 Hint 2
The three-cell equals region in the left column can only be one specific number. Check which numbers appear on at least two different dominoes.
💡 Hint 3 (Full Answer)
The left-column equals must be 1s. Place the [1|1] domino vertically at [0,0]–[1,0]. Place [6|1] at [2,0]–[2,1] so [2,0]=1 joins the equals and [2,1]=6. The less-4 cell at [0,2] takes 3 from [5|3] at [0,1]–[0,2]. Then the sum-2 region (cells [3,2]–[4,2]) gets 2+0 from [6|2] at [2,2]–[3,2] and [0|4] at [4,2]–[4,1].
💡 Hint 1
Anchor on a multi-cell equals region. The longest chain of equal cells is your starting point — consider which numbers can fill it using the dominoes you have.
💡 Hint 2
The three-cell equals in column 6 (rows 1–3) can only be 0 or 5. Look at the dominoes with [0|0] and [5|5], then see how the sum-11 region in row 2 demands a 5 and a 6; that will help you decide.
💡 Hint 3 (Full Answer)
Column 6 equals is 0. Place [0|0] at [2,6]–[3,6] and [0|1] at [1,6]–[0,6] (1 lands at [0,6], 0 at [1,6]). Row 1’s equals [1,3]–[1,4] must be 5, so [5|5] goes at [1,3]–[2,3] (both 5). Then [2,2] must be 6 to sum 11; place [6|3] at [2,2]–[3,2] (6 and 3). Finish with [6|5] at [0,4]–[1,4] (6,5), and the sum-3 region [1,0]–[1,1] with [6|1] at [0,0]–[1,0] (6,1) and [4|2] at [2,1]–[1,1] (4,2).
💡 Hint 1
A single-cell region is your Rosetta stone. It forces a precise value that will ripple across the board.
💡 Hint 2
The sum-2 region at [6,5] must be exactly 2. That cell needs a domino with a 2; look at which dominoes can cover it and what the neighboring cell’s region requires.
💡 Hint 3
Cover [6,5] with the [1|2] domino, placing the 2 there and the 1 at [6,6]. That instantly makes the four-cell equals in column 6 all 1s, so the [1|1] domino will cover two of them. The central four-cell equals (at [3,4],[4,2],[4,3],[4,4]) becomes all zeros because of the [0|0] domino.
💡 Hint 4
After locking 1s in column 6, place [1|1] at [4,6]–[5,6] and [3|1] at [2,6]–[3,6] to finish that edge. For the central zero equals, put [0|0] at [4,3]–[4,4], then [0|4] at [4,2]–[3,2] (satisfying less-5 with 4) and [0|5] at [3,4]–[2,4]. Row 0’s sum-12 regions require 6+6 pairs — [6|2] covers [0,1]–[0,0], [6|5] covers [0,2]–[0,3], [1|6] covers [0,4]–[0,5], and [6|3] covers [0,6]–[1,6]. The greater-10 region gets [6|6], and the greater-13 region gets three 5s from [5|5] and [5|3].
💡 Hint 5 (Full Answer)
Full answer: Start with [1|2] at [6,5]–[6,6] (2,1). Place [1|1] at [4,6]–[5,6] and [3|1] at [2,6]–[3,6] (3,1). Equals-2 at left: [4|2] at [2,0]–[1,0] (4,2) and [6|2] at [0,1]–[0,0] (6,2). Central zeros: [0|0] at [4,3]–[4,4], [0|4] at [4,2]–[3,2] (0,4), [0|5] at [3,4]–[2,4] (0,5). Row 0: [6|5] at [0,2]–[0,3] (6,5), [1|6] at [0,4]–[0,5] (1,6), [6|3] at [0,6]–[1,6] (6,3). Greater-10: [6|6] at [2,2]–[2,3]. Greater-13: [5|5] at [5,0]–[6,0] (5,5), [5|3] at [6,1]–[6,2] (5,3). Empty row 6: [3|3] at [6,3]–[6,4].
🎨 Pips Solver
Jun 23, 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: Focus on the equals column
The three-cell region in the left column ([0,0],[1,0],[2,0]) forces all values equal. Check the available dominoes: the only number that appears on two different dominoes is 1 (on [1|1] and [6|1]). Therefore, every cell in that column must be 1.
2
Step 2: Place the double-1 domino
You need two 1s from [1|1] and a third 1 from [6|1]. The [1|1] domino can only cover two adjacent cells in the column; place it vertically at [0,0]–[1,0] (both 1).
3
Step 3: Supply the third 1
The remaining cell [2,0] must get its 1 from the [6|1] domino. The only adjacent cell is [2,1], so place [6|1] horizontally at [2,0]–[2,1] giving [2,0]=1 and [2,1]=6.
4
Step 4: Finish the grid
The equals region [2,1]–[2,2] now forces [2,2]=6; place [6|2] at [2,2]–[3,2] with [3,2]=2. The sum-2 region ([3,2]–[4,2]) then demands [4,2]=0, so place [0|4] at [4,2]–[4,1] (0 and 4). The less-4 cell [0,2] gets 3 from the last domino [5|3] at [0,1]–[0,2].
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Solve the three-cell equals in column 6
The column‑6 equals (cells [1,6],[2,6],[3,6]) can only be 0 or 5. Using 5 would require the [5|5] and [6|5] dominoes, but [6|5] is needed for sum-11 later. Instead, choose 0: place the double‑0 domino [0|0] at [2,6]–[3,6] and the [0|1] domino at [1,6]–[0,6] with [1,6]=0, [0,6]=1.
2
Step 2: Deduce the equals region in row 1
The sum‑11 region at [2,2]–[2,3] must be 5 and 6. Row 1’s equals [1,3]–[1,4] sits directly above and, with [5|5] available, must be 5. Place [5|5] at [1,3]–[2,3] so both cells are 5. Then [2,2] is forced to 6; place [6|3] at [2,2]–[3,2] (6 and 3).
3
Step 3: Place the [6|5] domino
Now [1,4]=5 needs a partner. The empty cell [0,4] is adjacent; place [6|5] at [0,4]–[1,4] with [0,4]=6, [1,4]=5.
4
Step 4: Handle the sum‑3 region
The sum‑3 region ([1,0],[1,1]) must total 3. The remaining dominoes are [6|1] and [4|2]. [0,0] is empty. Place [6|1] at [0,0]–[1,0] giving 6 and 1, then [1,1] needs 2 to reach sum 3, so place [4|2] at [2,1]–[1,1] (4 and 2). The less‑6 on [2,1] is satisfied by the 4.
5
Step 5: Verify
All cells are covered and constraints satisfied: the right column equals are 0, row-1 equals are 5, sum-11 is 6+5, sum-3 is 1+2. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Anchor on the single-cell sum-2
The tiny sum‑2 region at [6,5] forces that cell to be exactly 2. Look at dominoes containing a 2. The [1|2] domino is the natural choice; placing it at [6,5]–[6,6] gives [6,5]=2 and [6,6]=1. That 1 immediately tells you the four‑cell equals region in column 6 ([3,6],[4,6],[5,6],[6,6]) must all be 1.
2
Step 2: Resolve the right‑edge equals
To fill four equal 1s, use the double‑1 domino [1|1] covering [4,6]–[5,6] and a second domino to supply a 1 for [3,6]. The [3|1] domino works nicely when placed at [2,6]–[3,6], giving [3,6]=1 and [2,6]=3. That satisfies the smaller equals [1,6],[2,6] which then becomes 3, so [1,6] must also be 3 (handled later by [6|3] at [0,6]–[1,6]).
3
Step 3: Zero in on the central equals
The four‑cell equals region in the center ([3,4],[4,2],[4,3],[4,4]) must be all zeros because the double‑0 domino [0|0] is available and no other number can fill four identical cells without conflict. Place [0|0] at [4,3]–[4,4] (both 0). The remaining two zeros come from [0|4] and [0|5]. Place [0|4] at [4,2]–[3,2] — the 0 satisfies the equals and the 4 at [3,2] satisfies the less‑5 region. Place [0|5] at [3,4]–[2,4] — the 0 satisfies the equals and the 5 fills the empty cell [2,4].
4
Step 4: Row 0’s sum cascades
Row 0 has two sum‑12 regions and one sum‑6. [0,1]–[0,2] must sum to 12, so both cells are 6. Use [6|2] at [0,1]–[0,0] (6 and 2) and [6|5] at [0,2]–[0,3] (6 and 5). That gives [0,3]=5, which pairs with [0,4] in the sum‑6 region; [0,4] must be 1, so place [1|6] at [0,4]–[0,5] (1 and 6). The remaining sum‑12 [0,5]–[0,6] now has [0,5]=6, so [0,6] must be 6 as well — place [6|3] at [0,6]–[1,6] (6 and 3). This also sets [1,6]=3, matching the equals-3 we anticipated.
5
Step 5: Left-side equals and the heavy sums
The equals‑2 region [0,0],[1,0] already has [0,0]=2 from [6|2]; [1,0] needs 2, so place [4|2] at [2,0]–[1,0] (4 and 2). That sets [2,0]=4, and the three‑cell equals [2,0],[3,0],[4,0] must all be 4, so place [4|4] at [3,0]–[4,0]. The greater‑10 region [2,2]–[2,3] (sum>10) forces two 6s; place [6|6] there. The greater‑13 region [5,0],[6,0],[6,1] needs sum>13 — fill it with three 5s: [5|5] at [5,0]–[6,0] and [5|3] at [6,1]–[6,2] (5 and 3). Finally, the empty cells [6,3]–[6,4] take [3|3].
💡 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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