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🎲 Today's Puzzle Overview
In today's NYT Pips easy, Ian Livengood builds a tight grid around a sum-11 region that forces a heavy double-five placement, immediately triggering an equals cascade through column 1 that locks a zero across three cells. The deduction graph is sparse but deterministic, with only five dominoes covering ten cells.
Livengood's medium puzzle expands the constraint web with a sum-4 fulcrum that parasitically feeds on a double-zero domino, branching into a chain of equals pairings that propagate zeros, twos, and fours across the grid. The solving path is linear but requires careful inventory management as each domino exacts a precise value handoff.
Rodolfo Kurchan's hard grid is a masterclass in constraint density. Two sum-12 regions anchor opposite corners, a sum-0 trio in row 2 acts as a zero sink that forces three distinct zero-bearing dominoes, and a four-cell equals chain in row 3 synchronizes with a parallel three-cell equals in column 1. The unequal region on the right then resolves a six-cell mix with no repeats, tying together the remaining dominoes in a satisfying final press.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Search for heavy hitters
Identify a sum region that requires a very high total; in parallel, notice an equals region spanning multiple cells that will demand a repeated small digit.
💡 Zero in on the column
The sum-11 region in the top right can only be satisfied by the [5,5] domino covering cells [1,2]-[1,3]. That leaves a single cell in the sum region needing 1, and forces the equals region in column 1 (cells [1,1], [2,1], [3,1]) to converge on 0.
💡 Full chain revealed
Place [5,5] horizontally in [1,2]-[1,3]. The sum-11 residual cell [0,2] must be 1, so the [1,1] domino goes in [0,1]-[0,2] (values 1,1). To make the equals region work, place the [2,0] domino in [1,0]-[1,1] giving 2 and 0 (the 0 at [1,1]). Then the [0,3] domino goes in [2,1]-[2,2] with 0 at [2,1] and 3 at [2,2] (satisfying greater-than-2). Finally, place [0,1] in [3,1]-[3,2] with 0 at [3,1] and 1 at [3,2] (satisfying less-than-2).
💡 Find the sum that demands zeros
Look for a sum region whose target forces the use of a domino with two identical low numbers; also track a region that requires a specific single high value.
💡 Sum-4 in row 1
The sum-4 region at [1,1]-[1,2]-[1,3] forces the [0,0] domino to place its two zeros in [1,1] and [1,2], leaving [1,3] to take the 4 from a [4,3] domino. The greater-than-0 cell [0,3] then gets the companion 3.
💡 Assembly complete
Place [0,0] horizontally in [1,1]-[1,2] (0,0). Put [4,3] vertically in [1,3]-[0,3] with [1,3]=4, [0,3]=3. The sum-5 cell [2,0] needs a 5, so place [2,5] horizontally in [2,0]-[2,1] with 5 and 2. The equals pair [2,1]-[2,2] forces [2,2]=2, so place [4,2] horizontally in [2,2]-[2,3] (2,4). The equals pair [2,3]-[3,3] forces [3,3]=4, so [4,6] goes horizontally in [3,2]-[3,3] with 6 and 4. The less-1 cell [3,0] forces 0, so [0,6] goes horizontally in [3,0]-[3,1] (0,6). Finally, the equals pair [0,0]-[1,0] takes [1,1] vertically with 1,1.
💡 Spot the forced sums
Examine sum regions that demand extreme values, especially sum-12 pairs and a sum-0 trio. Also notice a long equals chain running horizontally through row 3 and another vertical equals in column 1.
💡 Corner anchors
The sum-12 region at [0,4]-[0,5] forces double 6, so the [6,6] domino must go there horizontally. The left sum-12 region [3,0]-[4,0] forces both cells to be 6, but they aren’t covered together; you’ll need to split a 6 between them using a domino with a 6.
💡 Zero vacuum in row 2
The sum-0 cells [2,1], [2,2], [2,3] each must be 0. Since no double-zero exists, you’ll place three different zero-bearing dominoes: [0,4] vertically in [2,1]-[3,1] (0,4), [0,2] vertically in [2,2]-[3,2] (0,2), and [0,1] horizontally in [2,3]-[2,4] (0,1). The 4 at [3,1] sets the equals region [3,1]-[4,1]-[4,2] to 4.
💡 Equals chains solidify
The four-cell equals at [3,2]-[3,5] now has [3,2]=2 from the [0,2] domino, so all four become 2. This forces [3,4]-[3,5] to use a [2,2] domino ([3,3]=2, [3,4]=2) and [3,5]=2 from the [3,2] domino placed vertically with [2,5]=3. The sum-1 cell [2,0] gets its 1 from the [6,1] domino, whose 6 lands at [3,0] to complete the left sum-12.
💡 Final domino positions
Place [6,6] horizontally in [0,4]-[0,5] (6,6). Place [6,1] vertically in [3,0]-[2,0] (6,1). For the sum-0 row, place [0,4] vertically in [2,1]-[3,1] (0,4), [0,2] vertically in [2,2]-[3,2] (0,2), [0,1] horizontally in [2,3]-[2,4] (0,1). The equals chain [3,2]-[3,5] gets [2,2] domino horizontally in [3,3]-[3,4] (2,2) and [3,2] domino vertically in [2,5]-[3,5] (3,2). The equals region [5,0]-[5,2] takes [5,5] horizontally (5,5) with [5,3]=0 from [3,0] domino placed vertically in [4,3]-[5,3] (3,0). Left sum-12 leftover [4,0]=6 from [6,4] placed horizontally in [4,0]-[4,1] (6,4); the 4 at [4,1] aligns with the equals-three, and [4,2] gets 4 from [4,5] domino placed vertically in [4,2]-[5,2] (4,5). The unequal rightmost columns resolve with [3,4] vertically in [3,8]-[4,8] (3,4), [6,2] horizontally in [4,7]-[5,7] (6,2), [1,5] horizontally in [5,6]-[4,6] (1,5), and the [1,4] domino vertically in [1,4]-[1,5] (1,4) completing the sum-2 and sum-7 regions.
🎨 Pips Solver
May 20, 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: Anchor on sum-11
The region [0,2],[1,2],[1,3] must sum to 11. The only way to pack that much value with available dominoes is to use [5,5] covering [1,2]-[1,3] horizontally. That contributes 10, leaving [0,2] needing 1.
2
Step 2: Fill the top row gap
With [0,2] forced to 1, the [1,1] domino must cover it and its only adjacent neighbor [0,1] to give a pair of 1s. Place [1,1] horizontally in [0,1]-[0,2] (1,1).
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Step 3: Resolve the equals column
The equals region [1,1],[2,1],[3,1] now has [1,1] free. The domino [2,0] (2-0) can place its 0 at [1,1] and its 2 at [1,0] vertically (or horizontally? the adjacency is vertical). That satisfies the empty cell [1,0]=2 and sets the equals value to 0.
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Step 4: Place the remaining dominoes
With the equals value locked at 0, [2,1] must be 0; the [0,3] domino fits there horizontally into [2,1]-[2,2], giving 0 at [2,1] and 3 at [2,2] (fulfilling the greater-2 constraint). Then [3,1] needs 0, so the [0,1] domino goes in [3,1]-[3,2] with 0 at [3,1] and 1 at [3,2] (fulfilling the less-2 constraint).
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Exploit the sum-4 region
The sum-4 region [1,1],[1,2],[1,3] forces a total of 4 across three cells. The only domino that can contribute two zeros is [0,0]; placing it horizontally in [1,1]-[1,2] (0,0) leaves a residual of 4 at [1,3].
2
Step 2: Branch to the greater-than-zero cell
[1,3]=4 now requires a domino that supplies a 4, so the [4,3] domino must be placed vertically in [1,3]-[0,3] with [1,3]=4 and [0,3]=3. That satisfies the greater-0 constraint on [0,3].
3
Step 3: Use the single-cell sum-5
The region at [2,0] demands a value of 5. The only domino containing a 5 is [2,5]; placing it horizontally in [2,0]-[2,1] gives [2,0]=5 and [2,1]=2.
4
Step 4: Propagate equals pairs
The equals pair [2,1],[2,2] forces [2,2]=2. The domino [4,2] (4-2) must cover [2,2]-[2,3] horizontally with [2,2]=2 and [2,3]=4. Then the equals pair [2,3],[3,3] forces [3,3]=4, so the [4,6] domino goes horizontally in [3,2]-[3,3] giving 6 and 4.
5
Step 5: Mop up the remaining cells
The less-1 cell [3,0] demands 0, so the [0,6] domino goes horizontally in [3,0]-[3,1] with 0 and 6. The equals pair [0,0],[1,0] forces both to be 1, so the [1,1] domino is placed vertically in [0,0]-[1,0] (1,1). All cells are filled and constraints satisfied.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Top-right sum-12 lock
The region [0,4],[0,5] must sum to 12, so both cells must be 6. The only available dual-6 domino is [6,6]; place it horizontally in [0,4]-[0,5] (6,6).
2
Step 2: Left sum-12 and sum-1 tie
The region [3,0],[4,0] also sums to 12, requiring values 6 and 6. The cell [2,0] has a sum-1 constraint, forcing 1. The [6,1] domino can supply 6 to [3,0] and 1 to [2,0] with a vertical placement in [3,0]-[2,0] (6,1). This covers one of the left 6s.
3
Step 3: Sum-0 zero chain in row 2
The region [2,1],[2,2],[2,3] sums to 0, so all must be 0. No double-zero exists, so three zero-bearing dominoes are required. Place [0,4] vertically in [2,1]-[3,1] (0,4), [0,2] vertically in [2,2]-[3,2] (0,2), and [0,1] horizontally in [2,3]-[2,4] (0,1). This feeds 4 into [3,1] and 2 into [3,2].
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Step 4: Equals cascades in row 3
The equals region [3,1],[4,1],[4,2] now receives [3,1]=4, so [4,1] and [4,2] both become 4. The four-cell equals at [3,2]-[3,5] inherits [3,2]=2, so all become 2. To achieve this, place [2,2] domino (2-2) horizontally in [3,3]-[3,4] and the [3,2] domino (3-2) vertically in [2,5]-[3,5] ([3,5]=2, [2,5]=3). This also resolves the sum-7 region [1,5]-[2,5] (3+4) – the 4 will come later.
5
Step 5: Bottom-left equals and zero
The equals region [5,0],[5,1],[5,2] forces three 5s, so the [5,5] domino goes horizontally in [5,0]-[5,1] and the 5 at [5,2] will need a partner. The sum-0 cell [5,3] must be 0, so place [3,0] domino vertically in [4,3]-[5,3] (3,0) – satisfying the greater-2 cell [4,3]=3. The left sum-12 needed [4,0]=6; that comes from the [6,4] domino placed horizontally in [4,0]-[4,1] (6,4), giving 6 and 4 (matching the equals 4).
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Step 6: Complete the unequal right side
The unequal region spans six cells requiring all values distinct. The remaining dominoes [4,5], [3,4], [6,2], [1,5], and [1,4] must fill them. Place [4,5] vertically in [4,2]-[5,2] (4,5) – giving [5,2]=5 matching the equals. Place [3,4] vertically in [3,8]-[4,8] (3,4). Place [6,2] horizontally in [4,7]-[5,7] (6,2). Place [1,5] horizontally in [5,6]-[4,6] (1,5). Finally, [1,4] goes vertically in [1,4]-[1,5] (1,4), completing the sum-2 (1+1) and sum-7 (4+3) regions. All constraints are met and the unequal region contains digits 3,4,1,5,6,2 with no repeats.
💡 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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