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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
In today's NYT Pips easy, Ian Livengood delivers a true confidence-builder — a vertical sum-12 region instantly forces a triple of the same digit, and a four-cell equals cage cascades into a uniform band without any forks. The domino count is tiny, so you'll scan, place, and solve in a smooth flow.
Livengood's medium raises the stakes with two large equals quartets dominating the center-left. The bottleneck is spotting how the double dominoes anchor each cage; once you place the [1,1] and [5,5], the rest unwinds logically. One greater-than cell and an empty cell break up the symmetry just enough to avoid confusion.
Rodolfo Kurchan's hard is the day's real workout. A rare 'less 1' region forces a column of zeros, a sum-12 pair demands two 6's from separate dominoes, and a six-cell unequal region keeps you vigilant. The clues interlock tightly — expect a deliberate, deeply satisfying solve that rewards methodical pip tracking.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Hint 1: Target the fixed sum
Look for a region with a sum constraint that stands out — its total restricts the possible digits so severely that only one number can occupy every cell inside it.
💡 Hint 2: The vertical sum column
The vertical region in column 2 (cells [0,2], [1,2], [2,2]) requires a sum of 12. Given the available set of dominoes, this forces all three cells to hold the same pip. Find the domino that can supply a pair of that digit.
💡 Hint 3: The full solution
Place the [4,4] domino to cover [0,2] and [1,2]. [2,2] then needs a 4; use the [3,4] domino with 4 at [2,2] and 3 at [2,1] (satisfying the sum-3 there). The big equals region must all be 2: place [4,2] at [3,0]-[3,1] (4 in the empty cell, 2 in the equals), [2,3] at [3,2]-[3,3] (2 in equals, 3 in the sum-3 cell [3,3]), and finish with [2,2] at [4,1]-[4,2].
💡 Hint 1: Hug the equals cages
Identify the regions that force uniformity — the large 'equals' cages will dictate large swaths of the grid. Pay attention to how the limited domino set can supply identical pairs to each quartet.
💡 Hint 2: Anchor the two quartets
The left equals region ([2,0],[2,1],[3,0],[3,1]) and the central one ([2,2],[2,3],[3,2],[3,3]) both need four matching numbers. The domino list contains two double-number dominos — they'll each claim two cells inside these cages and fix the repeated value.
💡 Hint 3: The full solution
Place [1,1] vertically at [2,0]-[3,0] to start the 1's cage. Complete the 1's with [5,1] at [2,2]=5, [2,1]=1, and [1,6] for [3,1]=1, [4,1]=6 (satisfying the greater-3 there). The central cage becomes all 5's: use [5,5] at [3,2]-[3,3], and [2,5] at [1,3]=2, [2,3]=5. The equals pair [1,4],[2,4] gets [4,3] at [0,4]=4 (greater-3), [1,4]=3, and [3,0] at [2,4]=3, [3,4]=0 (empty).
💡 Hint 1: Find the tightest restriction
Start by locating the most restrictive region: a 'less than 1' cage. That condition means every cell inside can hold only one possible pip value — zero. Use that anchor to crack the left side of the grid.
💡 Hint 2: Zero column and sum-5 neighbor
The 'less 1' region occupies [5,0], [6,0], and [7,0] at bottom-left. The [0,0] domino will cover two of them, while the third zero must come from a domino that carries a zero partner, such as [0,5]. The cell [4,0] sits just above and has a sum-5 constraint — it will pair with [5,0] via that same [0,5] domino, forcing 5 above and 0 below.
💡 Hint 3: The sum-12 pair and top-right dominoes
After placing [0,5] (5 at [4,0], 0 at [5,0]) and [0,0] at [6,0]-[7,0], move to the top right. The sum-12 pair [0,4],[0,5] must both be 6, but no [6,6] exists. Instead, use [5,6] to give 6 to [0,5] and 5 to [1,5], and [3,6] to give 6 to [0,4] and 3 to [1,4]. That 3 satisfies the sum-3 cell at [1,4].
💡 Hint 4: Equals chain and low-value cages
With [1,5]=5, the equals region [1,5],[2,5],[3,5] becomes all 5's; use [5,5] at [2,5]-[3,5]. The 'less 7' region [2,3],[2,4],[3,4] needs numbers below 7 — the [2,2] domino supplies a pair of 2's at [2,4]-[3,4], and later [5,2] puts a 2 at [2,3] and a 5 at [2,2] (sum-5). The 'less 4' cage [3,2],[4,1],[4,2] gets 1's from [1,1] at [4,1]-[4,2] and [3,1]'s 1 at [3,2] (its 3 at [3,3] in the less-13 zone).
💡 Hint 5: The complete hard solution
Place [0,5] at [4,0]=5,[5,0]=0; [0,0] at [6,0]-[7,0]=0; [5,6] at [1,5]=5,[0,5]=6; [3,6] at [1,4]=3,[0,4]=6; [5,5] at [2,5]-[3,5]=5; [2,2] at [2,4]-[3,4]=2; [1,1] at [4,1]-[4,2]=1; [3,1] at [3,2]=1,[3,3]=3; [5,2] at [2,2]=5,[2,3]=2; [3,4] at [5,2]=3,[5,1]=4; [4,4] at [6,1]-[7,1]=4; [3,3] at [4,3]-[5,3]=3; [3,5] at [7,4]=3,[6,4]=5; [4,2] at [5,5]=4,[6,5]=2; [6,0] at [4,6]=6,[5,6]=0. All pips respect every less, sum, equals, and unequal constraint.
🎨 Pips Solver
May 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: Forcing the sum-12 column
The vertical region at cells [0,2], [1,2], and [2,2] must sum to 12. With pip values 0–6 and the available domino set ([4,2], [3,4], [2,3], [4,4], [2,2]), the only way to hit 12 with three cells is 4+4+4. So all three cells will be 4.
2
Step 2: Placing the double-4
The only domino that can supply two 4's is [4,4]. The cell [0,2] has only one adjacent cell in the grid — [1,2] — so the [4,4] domino must cover [0,2] and [1,2] (both become 4). This locks two of the three sum-12 cells, leaving [2,2] needing a 4.
3
Step 3: Completing the column and the sum-3 clue
[2,2] is adjacent to [2,1] (which has its own sum-3 region) and to [3,2] (part of the big equals region). The only remaining domino with a 4 is [3,4]. Placing it with its 4 at [2,2] forces its 3 into [2,1], perfectly satisfying that single-cell sum-3.
4
Step 4: Settling the equals cage and leftovers
The equals region ([3,1],[3,2],[4,1],[4,2]) now demands a uniform value. [3,1] is adjacent to the empty cell [3,0]; use [4,2] there, putting 4 in the empty cell and 2 in [3,1]. This sets the equals value to 2. Then [2,3] covers [3,2] and the sum-3 cell [3,3]: 2 in equals, 3 for the sum. Finally [2,2] fills [4,1] and [4,2] with 2s. All constraints are satisfied.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Spot the double domino anchors
Two large equals regions each demand four identical numbers: one at [2,0]-[3,1] and another at [2,2]-[3,3]. The only double-value dominoes in the set are [1,1] and [5,5]. Therefore the left quartet will be all 1's (forced by [1,1]) and the central quartet all 5's (forced by [5,5]).
2
Step 2: Start the 1's cage
Place the [1,1] domino vertically over [2,0] and [3,0] — the only adjacent pair in that cage that doesn't conflict with other constraints yet. Now [2,0] and [3,0] are 1, leaving [2,1] and [3,1] of the equals cage needing 1's.
3
Step 3: Finish the 1's and inject a 5
[3,1] is adjacent to [4,1] (a greater-than-3 cell) and [3,0] (already 1). The [1,6] domino fits perfectly: place its 1 at [3,1] and its 6 at [4,1], satisfying the greater-3. For [2,1], which borders the central cage at [2,2], use the [5,1] domino: put 1 at [2,1] and 5 at [2,2]. Now the central 5's cage has its first 5.
4
Step 4: Build the 5's quartet
With [2,2]=5, the rest of the central equals cage must be 5's. Place [5,5] vertically at [3,2]-[3,3]. The remaining cell [2,3] needs 5; it is adjacent to the empty cell [1,3]. Use the [2,5] domino there: 2 at [1,3] (empty, no constraint) and 5 at [2,3], completing the 5's cage.
5
Step 5: Handle the final equals pair and remainder
The equals region [1,4]-[2,4] needs equal values. [2,4] is adjacent to the empty [3,4], so a domino covering both must assign one value to the equals cell and one to the empty. The [3,0] domino provides 3 and 0: put 3 at [2,4], 0 at [3,4]. Then [1,4] must also be 3; it pairs with [0,4] (greater-than-3). The [4,3] domino has 3 and 4: place 3 at [1,4] and 4 at [0,4] to satisfy greater-3. All dominoes are used, all constraints met.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Zero column and sum-5
The 'less 1' region at [5,0],[6,0],[7,0] forces these three cells to be 0. The [0,0] double-zero domino covers two of them — logically [6,0] and [7,0] because [5,0] pairs upward with [4,0], which has a sum-5 constraint. The only 0+5 domino is [0,5]; place it with 5 at [4,0] and 0 at [5,0]. Now the column's zeros are set.
2
Step 2: Top-right sum-12 and sum-3
The region [0,4],[0,5] must sum to 12, forcing two 6's. No single domino provides a 6+6 pair, so each 6 must come from a separate domino with a 6. [5,6] places 6 at [0,5] and 5 at [1,5]; [3,6] places 6 at [0,4] and 3 at [1,4]. That 3 fulfills the sum-3 requirement at [1,4]. Now [1,5] is 5, triggering an equals chain.
3
Step 3: Equals-5 column
The equals region at [1,5],[2,5],[3,5] forces all three to become 5. With [1,5] already 5, the double-5 domino [5,5] naturally covers the adjacent pair [2,5] and [3,5], giving both the needed 5's. This tower of 5's is now complete.
4
Step 4: Less-7 and less-4 cages
The less-7 region ([2,3],[2,4],[3,4]) must contain numbers below 7. The [2,2] domino fits here — place it over [2,4] and [3,4], both becoming 2. The less-4 cage ([3,2],[4,1],[4,2]) requires numbers less than 4; the [1,1] domino delivers 1's to [4,1] and [4,2]. Then [3,2] needs a 1, supplied by the [3,1] domino with its 1 at [3,2] and its 3 moving to [3,3] (part of the less-13 zone). To finish the less-7's remaining cell [2,3] with a 2, use the [5,2] domino: 5 at [2,2] (satisfying the sum-5 cell there) and 2 at [2,3].
5
Step 5: Less-13 regions and the unequal finale
The large less-13 region [3,3],[4,3],[5,2],[5,3] is satisfied by the low pips already in motion: [3,3] has 3, [4,3] and [5,3] get the [3,3] domino (3 and 3), and [5,2] becomes 3 from the [3,4] domino, which places its 4 at [5,1] (inside the less-13 column). The remaining less-13 column [5,1],[6,1],[7,1] gets 4's: [4,4] covers [6,1]-[7,1] with 4's, and [5,1] already has 4. Finally, the unequal region spans six cells — use [4,2] for 4 at [5,5] and 2 at [6,5]; [6,0] for 6 at [4,6] and 0 at [5,6]; and [3,5] for 5 at [6,4] and 3 at [7,4]. Every domino and constraint now matches.
💡 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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