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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 this NYT Pips set, Ian Livengood's easy offers a gentle warm-up with a single sum-1 anchor that cascades cleanly. Rodolfo Kurchan's medium challenges solvers to spot a zero-heavy structure where empty cells become 0, forcing a precise domino chain from a greater-4 trigger. His hard puzzle is the densest — an equals region, a less-3 cell, and interlocking sum-3 and sum-6 rings create a rewarding tangle that demands careful cross-referencing. Overall, a well-calibrated climb from beginner-friendly to deviously tight.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Hint 1: Spot the tightest sum
Look for a region whose sum target is impossibly small — that’s your starting anchor.
💡 Hint 2: The lone cell
The cell at row 2, col 2 belongs to a sum‑1 region. It must be a 1, and only one domino contains two 1s. That domino must cover this cell and its left neighbor.
💡 Hint 3: Full break-in
Place the [1,1] domino horizontally over [2,1] and [2,2]. The left sum‑5 region then forces the [2,2] domino vertically at [1,0]‑[1,1]. The top sum‑5 takes the [2,3] domino vertically at [1,3]‑[0,3] (2 and 3), and the remaining [0,3] domino fills [2,3]‑[2,4] with 0 and 3.
💡 Hint 1: Zero in on the empties
Many cells have no constraint — expect them to become 0s. Focus on the few cells that demand specific high values.
💡 Hint 2: Greater‑4 forces a 6
The cell at [4,1] requires a number greater than 4. The only domino with a 6 is [0,6]; its companion 0 will naturally sit in the empty cell [5,1] below it.
💡 Hint 3: Complete logic chain
Place [0,6] vertically: 6 at [4,1], 0 at [5,1]. The sum‑3 at [4,2] must be 3, so pair it with the 6 in the sum‑11 region via the [3,6] domino (3 at [4,2], 6 at [3,2]). That forces [3,1] to be 5, so [0,5] goes horizontally at [3,0]‑[3,1] (0,5). The sum‑9 at [2,4]‑[3,4] is fed by [0,4] vertically (0 at [3,3], 4 at [3,4]) and [3,5] horizontally (5 at [2,4], 3 at [2,5]). Finally, [0,3] covers [2,3]‑[1,3] (0,3) and [6,6] sits at [0,4]‑[1,4] to complete the upper sum‑9 and greater‑3.
💡 Hint 1: Equal parts and a less
Identify the equals region (two cells that must match) and the less‑than‑3 cell. Think about which values can satisfy the match while obeying the small‑digit limit nearby.
💡 Hint 2: Values snap into place
The equals pair at [4,4]‑[4,5] will eventually both be 4. The less‑3 cell at [0,2] has to be 1. These two facts constrain the entire top row and right‑side column.
💡 Hint 3: Top‑left domino tiling
With [0,2]=1, the sum‑6 region at [0,0]‑[0,1] needs a 5 and a 1. Place the [5,1] domino horizontally covering [0,1]=5 and [0,2]=1. Now [0,0] becomes 1, and the sum‑6 single cell at [1,0] demands a 6 — use the [1,6] domino vertically (1 at [0,0], 6 at [1,0]).
💡 Hint 4: Right edge resolved
The sum‑3 at [0,6] forces a 3 there, so [0,5] (greater‑3) must be at least 4. The [4,3] domino fits perfectly horizontally: [0,5]=4, [0,6]=3. Then [0,4] (greater‑3) takes 5 from the [5,0] domino, whose 0 starts the sum‑3 column at [1,4].
💡 Hint 5: Full grid assembly
With [1,4]=0, the column sum‑3 needs 3 more from [2,4] and [3,4]. Place [3,2] horizontally for [2,4]=3, [2,5]=2; then the sum‑3 on [2,5]‑[2,6] forces [2,6]=1. The [1,3] domino covers [2,6]=1 and [1,6]=3. The sum‑3 at [4,0]‑[4,1] takes [0,6] horizontally: [4,1]=0, [4,2]=6, so [4,0]=3 via [3,3] vertically (3,0). The sum‑6 at [2,0]‑[2,1]‑[3,0] is completed by [2,1] (2,1). Finally, the equals pair: [4,4] gets 4 from [4,0] (4,0) vertically, and [4,5]‑[4,6] get [4,4] horizontally for 4,4.
🎨 Pips Solver
Jul 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: The tiny sum‑1 cell
The region at [2,2] must sum to 1, so it contains a 1. The only way to place a 1 here is with the [1,1] domino, which must occupy [2,2] and its only adjacent cell, [2,1]. Place [1,1] horizontally: [2,1]=1, [2,2]=1.
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Step 2: Left sum‑5 solved
The sum‑5 region at [1,0], [1,1], [2,1] now has a known 1 at [2,1]. The remaining two cells must sum to 4 — the [2,2] domino (two 2s) fits vertically: [1,0]=2, [1,1]=2.
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Step 3: Top and right sum‑5 interplay
The top sum‑5 ([0,2],[0,3]) and the right sum‑5 ([1,3],[2,3]) share column 3. The [2,3] domino provides 2 and 3. Place it vertically: [1,3]=2, [0,3]=3. Both sums are now satisfied.
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Step 4: Fill the gap
The only uncovered cells are [2,3] and [2,4]. The [0,3] domino completes the grid horizontally: [2,3]=3, [2,4]=0. The empty cell at [2,4] accepts 0 without issue.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Greater‑4 anchor
Cell [4,1] requires a number >4. The only such digit in the domino set is 6, carried by the [0,6] domino. Place [0,6] vertically: [4,1]=6, [5,1]=0 (the empty cell at [5,1] takes the 0).
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Step 2: Sum‑3 meets sum‑11
The sum‑3 cell at [4,2] must be exactly 3. The neighboring cell [3,2] is part of a sum‑11 region with [3,1]. To pair 3 with a digit that can sum to 11 with its partner, [3,2] must be 6. Use the [3,6] domino vertically: [4,2]=3, [3,2]=6.
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Step 3: Sum‑11 completed
Now [3,1] must be 5. The only domino offering a 5 with an available companion is [0,5]. Place it horizontally: [3,0]=0 (empty), [3,1]=5. The sum‑11 is satisfied.
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Step 4: Zero down the middle
The empty cell [3,3] will hold a 0. Pair it with a 4 to start the sum‑9 at [2,4]‑[3,4]. The [0,4] domino fits vertically: [3,3]=0, [3,4]=4. Then [2,4] needs 5 to reach 9; the [3,5] domino covers that: [2,4]=5, [2,5]=3 (the greater‑0 cell at [2,5] welcomes 3).
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Step 5: Top‑right wrap‑up
The sum‑9 at [1,3]‑[1,4] requires 3 and 6. The [0,3] domino supplies 0 and 3: place it with 0 at [2,3] (empty) and 3 at [1,3]. Finally, the [6,6] domino places 6 at both [0,4] (greater‑3) and [1,4] to complete the puzzle.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Less‑3 forces the top left
The cell [0,2] is less than 3, so it must be 1 (0 is possible but would block the sum‑6). The sum‑6 at [0,0]‑[0,1] then requires 5 and 1. Place the [5,1] domino horizontally: [0,1]=5, [0,2]=1. Now [0,0] is forced to 1, making the sum‑6 cell [1,0]=6. The [1,6] domino covers both: [0,0]=1, [1,0]=6 vertically.
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Step 2: Rightmost column takes shape
The sum‑3 at [0,6] must be 3. With greater‑3 cells to its left, the [4,3] domino (4 and 3) fits: place it horizontally at [0,5]=4, [0,6]=3. The remaining greater‑3 cell [0,4] gets 5 from the [5,0] domino, whose 0 lands in [1,4] to start the column’s sum‑3 chain: [0,4]=5, [1,4]=0.
3
Step 3: Column of three
The sum‑3 region covering [1,4], [2,4], [3,4] now has a 0 at the top. The remaining 3 must be split between [2,4] and [3,4]. The [3,2] domino (3,2) can occupy [2,4] and [2,5] to satisfy both the column and the sum‑3 at [2,5]‑[2,6]: [2,4]=3, [2,5]=2. Then [2,6] becomes 1, and the sum‑3 cell at [1,6] gets its 3 from the [1,3] domino vertically: [2,6]=1, [1,6]=3.
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Step 4: Bottom‑left knot
The sum‑3 region at [4,0]‑[4,1] interacts with the sum‑6 cell [4,2] and the already‑known [4,1] from earlier steps. Place the [0,6] domino horizontally: [4,1]=0, [4,2]=6. Then [4,0] needs a 3 to sum to 3 with [4,1]=0. The [3,3] domino covers this and the sum‑6 across [2,0]‑[2,1]‑[3,0]: place [3,3] vertically at [3,0]=3, [4,0]=3.
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Step 5: Equal and finish
The sum‑6 on [2,0]‑[2,1]‑[3,0] now has two 3s, so [2,0] and [2,1] must sum to 0? Wait: the region sum is 6, so with [3,0]=3, the remaining 3 goes to [2,0]+[2,1]. The [2,1] domino (2,1) fits horizontally: [2,0]=1, [2,1]=2. Finally, the equals pair: [4,4]‑[4,5] need 4s. The [4,4] domino handles [4,5]=4, [4,6]=4 (greater‑3), and the [4,0] domino places 4 at [4,4] with 0 at [3,4] to close the sum‑3 column.
💡 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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