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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
This NYT Pips easy puzzle by Ian Livengood is a confidence-builder: a compact grid with three equals regions and a few sum-6 targets that lock in early. The bottleneck is the triple-cell equals at the top left — once you realize it forces a specific identical-pip domino, the entire puzzle unravels in a straightforward chain. No branching, just clean deductions.
Medium from Rodolfo Kurchan presents a 4×9 landscape where the standout is a sum-16 column on the far right. That bottleneck narrows the pip options severely and fixes two dominos at once. From there, a series of sum-11, sum-9, and greater-than constraints flow logically, with the greater-4 cell demanding a 6 and the sum-4 pairs tidying up. It's a balanced mid-level solve that rewards spotting the column early.
Hard by Rodolfo Kurchan is a dense 10×6 puzzle with multiple microscopic constraints: a singleton sum-1, several sum-2 cells, a less-3 pair, and a triple equals. The puzzle opens with the sum-1 dictating an equals chain, then the single sum-6 and sum-4 forces create a cascade. Managing the interaction between the 6-6 placement and the sum-8 duo is the toughest nut to crack. Expect a methodical but satisfying hour of Pips.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Scan for equals regions
Start by identifying all regions that force cells to have identical pip values. They'll be key to forcing specific dominos.
💡 The triple equals corner
The three-cell equals region at positions [0,0], [0,1], and [1,0] must all show the same pip. Which available identical-pip dominos can cover two of those cells while allowing the third to match? The answer drives the entire puzzle.
💡 The Solve
Place the [4,4] domino on [0,0] and [0,1] to satisfy the triple equals, making [1,0]=4. The two single-cell sum-6 regions at [2,1] and [2,2] each require a 6, so place [6,6] there. The right-hand equals [0,2] and [1,2] must be equal, so [1,1] goes there. The sum-6 pair [2,0] and [3,0] needs a total of 6: place [5,3] with its 3 on [3,0] and 5 on [3,1]; then [3,4] fills [2,0] with 3 and [1,0] with 4, completing all constraints.
💡 Hunt for extreme sums
Look for a region with an unusually high sum target. That will dramatically narrow your pip choices early on.
💡 The sum-16 column
The column on the far right, containing cells [0,8], [1,8], and [2,8], must sum to 16. With pip values up to 6, the only workable combination is 6, 5, and 5. Which dominos can deliver that?
💡 The Solve
Place [5,5] on [0,8] and [1,8] (5+5) and [1,6] on [2,8]/[3,8] with the 6 on [2,8] and the 1 on [3,8] to hit 16 and satisfy the less-3 at [3,8]. The greater-4 at [3,4] demands a 6, so put [6,6] on [2,4]/[3,4]. For sum-11, [4,5] goes on [0,4]/[1,4] (4 and 5) to add with the 6 at [2,4]. Sum-9 uses [6,4] on [0,6]/[1,6] (6) and [1,3] on [2,6]/[3,6] (3+1) to make 6+3=9. Then sum-4 at [2,0]-[2,1] gets [2,2] at [1,0]/[2,0] (2+2) and [3,2] at [3,1]/[2,1] (3+2). Finally the sum-4 single at [1,2] takes [4,3] with 4 there and 3 on [2,2].
💡 Zero in on tiny sums
This puzzle has several regions with sum targets of 1 or 2, and a less-3 pair. Spotting where a 1 is forced will unlock the first domino placement.
💡 The sum-1 anchor
The cell at [7,3] must be exactly 1 (sum-1). It shares a domino with an adjacent cell that belongs to a large equals region. That means the other pip on that domino will determine the value for that entire equals chain.
💡 Seeding the equals
Place the [2,1] domino with its 1 on [7,3] and its 2 on [8,3]. This forces the equals region [8,3], [9,3], [9,4] to all be 2. Use the [2,2] domino on [9,3] and [9,4] to finish it.
💡 Conquer the corner singles
The sum-6 single at [8,5] must be 6; the adjacent sum-2 at [9,5] needs a 2. The [2,6] domino fits perfectly. The sum-4 pair [8,0]-[8,1] requires 3+1, so place [3,1] there. The less-3 pair [6,0]-[6,1] takes [0,1] with 0 and 1.
💡 The Solve
Place [2,1] with 1 at [7,3] (sum-1) and 2 at [8,3]; then [2,2] on [9,3]/[9,4] completes the equals region of 2s. The sum-6 at [8,5] gets [2,6] with 6 there and 2 at [9,5] (sum-2). Sum-4 at [8,0]-[8,1] uses [3,1] (3,1). The less-3 pair [6,0]-[6,1] takes [0,1] (0,1). The triple equals in column 3 forces [4,4] on [1,3]/[2,3] and [0,4] on [0,2]/[0,3] with 4 at [0,3]. Sum-2 at [1,0] gets 2 from [4,2] (4 at [2,0], 2 at [1,0]). Greater-3 at [1,5] gets 5 via [0,5] (0 at [1,6], 5 at [1,5]), satisfying the less-3 there. The sum-6 pair [5,3]-[5,4] takes [3,2] giving 3 and 2 (with sum-2 at [5,5] taking the 2), then [3,6] on [5,3]/[6,3] puts 3 and 6 (sum-6 at [6,3] gets 6). Equals at [3,6]/[4,6] takes [6,6]. Sum-8 at [3,8]/[4,8] uses [3,5] (3,5). Finally, [0,2] fills [7,4]=0, [7,5]=2 (sum-2).
🎨 Pips Solver
Jun 27, 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: Identify the triple equals
The region comprising [0,0], [0,1], and [1,0] must have the same pip in all three cells. The only way to achieve this with a single domino is to use one with identical pips covering two cells; the third cell will then inherit that value. The available identical-pip dominos are [1,1], [4,4], and [6,6]. Because of the sum-6 singles and the right-side equals later, [4,4] is the only one that fits without conflict. Place it on [0,0] and [0,1], setting [1,0] to 4 automatically.
2
Step 2: Fill the forced 6s
There are two single-cell regions at [2,1] and [2,2], each with a sum target of 6. That means each cell must contain exactly 6. The [6,6] domino is the only one that can supply two 6s, so it must be placed covering [2,1] and [2,2] (any orientation works because both are 6).
3
Step 3: Resolve the right-hand equals
The region [0,2] and [1,2] is an equals, so both cells must have the same pip. The remaining identical-pip domino is [1,1], so place it there, giving [0,2]=1 and [1,2]=1.
4
Step 4: Satisfy the sum-6 pair and finish
The region [2,0] and [3,0] must sum to 6. We have already used 6s and 1s elsewhere; the only way to make 6 with remaining dominos is 3+3. Place [5,3] with its 3 on [3,0] and its 5 on [3,1] (empty). Now [2,0] must be 3, which is provided by the [3,4] domino: put its 3 on [2,0] and its 4 on [1,0]. That 4 matches the triple equals region, completing the grid.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Crack the sum-16 column
The rightmost column has cells [0,8], [1,8], and [2,8] that must sum to 16. With pip values at most 6, the only feasible combination is 6+5+5. The dominos [5,5] and [1,6] achieve this: place [5,5] vertically on [0,8] and [1,8] (both 5), and place [1,6] horizontally with its 6 on [2,8] and its 1 on [3,8]. The less-3 region at [3,8] is satisfied with the 1.
2
Step 2: Greater-4 and sum-11
The cell [3,4] has a 'greater 4' constraint, meaning it must be 5 or 6. Since 5 is used elsewhere, it must be 6. The [6,6] domino can cover [2,4] and [3,4] (both 6). Now the sum-11 region [1,4] and [2,4] already has a 6 at [2,4], so [1,4] needs a 5. Place the [4,5] domino on [0,4] (4) and [1,4] (5). That also satisfies the greater-1 at [0,4].
3
Step 3: Sum-9 in row 1-2
The region [1,6] and [2,6] must sum to 9. The adjacent [3,6] is empty. Place [1,3] on [3,6] (1) and [2,6] (3) to give [2,6]=3. Then [6,4] goes on [0,6] (4) and [1,6] (6) to make [1,6]=6, so 6+3=9.
4
Step 4: Sum-4 in the bottom left
The region [2,0] and [2,1] must sum to 4. Available dominos include [2,2] and [3,2]. Place [2,2] on [1,0] and [2,0] (both 2), giving [2,0]=2 and [1,0]=2 (empty). Then place [3,2] on [3,1] (3) and [2,1] (2), so [2,1]=2 and [3,1]=3 (greater-2 satisfied). Now [2,0]+[2,1]=2+2=4.
5
Step 5: Final single sum-4
The cell [1,2] has a sum-4 single, so it must be exactly 4. The remaining domino [4,3] gives that: place it on [1,2] (4) and [2,2] (3), where [2,2] is empty. All regions satisfied.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Anchor on sum-1
Cell [7,3] is a sum-1 region, forcing a pip of 1. It borders cell [8,3] which is part of an equals region. The [2,1] domino (pips 2 and 1) can link them: place the 1 on [7,3] and the 2 on [8,3]. This seeds the equals region [8,3], [9,3], [9,4] with a 2. To complete that region, place [2,2] on [9,3] and [9,4] — both become 2.
2
Step 2: Isolated forcing cells
Several single-cell constraints demand immediate values. The sum-6 at [8,5] forces a 6; the adjacent sum-2 at [9,5] needs a 2. The [2,6] domino fits perfectly, with 6 on [8,5] and 2 on [9,5]. The sum-4 pair at [8,0]-[8,1] requires a total of 4: place [3,1] with 3 and 1. The less-3 pair at [6,0]-[6,1] must be <3: use [0,1] with 0 and 1 there.
3
Step 3: The triple equals in column 3
The region [0,3], [1,3], [2,3] all must be identical. With the 2s and 6s placed elsewhere, the only possible identical-value domino that can cover two of these cells is [4,4]. Place it on [1,3] and [2,3] (both 4). The remaining cell [0,3] still needs a 4, so the [0,4] domino goes on [0,2] (empty) with 0 and [0,3] with 4. Now the triple equals is satisfied with all 4s.
4
Step 4: Top-row constraints
The sum-2 at [1,0] demands a 2. It's adjacent to [2,0] which has a greater-2 constraint (needs >2). The [4,2] domino gives 2 to [1,0] and 4 to [2,0], satisfying both. Next, the greater-3 at [1,5] needs a value >3; the domino [0,5] can place its 5 there and its 0 on [1,6] (less-3). This completes the top right.
5
Step 5: Central sum-6 and sum-2 dance
The region [5,3] and [5,4] must sum to 6, but [5,5] is a sum-2 single (needs 2). The [3,2] domino can cover [5,4] (3) and [5,5] (2), satisfying [5,5] and giving [5,4]=3. Then [5,3] must be 3 to complete sum-6. The [3,6] domino places its 3 on [5,3] and its 6 on [6,3] (sum-6 single). Now the equals region [3,6] and [4,6] requires equal values; the [6,6] domino fits there with both 6s.
6
Step 6: Bottom and final fill
The sum-8 pair at [3,8] and [4,8] needs a total of 8. Use [3,5] with 3 on [4,8] and 5 on [3,8]. Lastly, the only unused domino is [0,2]; place it on [7,4] (empty, 0) and [7,5] (sum-2, 2) to finish the grid. All constraints now hold.
💡 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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