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🎲 Today's Puzzle Overview
Today's NYT Pips features a pair of Ian Livengood grids and a demanding Rodolfo Kurchan puzzle. The easy 4×4 opens with a sum‑4 quad that acts as a single‑value funnel, forcing a cascade of specific domino placements through a less‑than, a sum‑6, and an equals region.
The medium, a 5×6 layout, is driven by a network of small sum constraints—most notably a sum‑0 trio that seeds three zeros and dictates the entire bottom‑left architecture. The top‑right sum‑7 and sum‑4 pairs then pull dominoes in a chain that leaves little room for deviation.
Kurchan's hard 5×8 is a constraint‑rich web: two less‑than cells anchor the left and right edges, while a sum‑7 column and a sum‑9 column create interlocking dependencies. An equals region, a sum‑5 vertical, and multiple greater‑than restrictions tighten until only one arrangement survives.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Hint 1: Spot the binding constraints
Search for a region that forces all cells to be identical, and another that sums a 2×2 block to a very small number—these will drive the early placements.
💡 Hint 2: Zero in on the lower equals pair
The equals region lives in row 3, columns 1–2. The only domino with two equal high numbers will fit here, while the 2×2 sum‑4 block in the top right corner will need the [1,1] and [1,0] dominoes.
💡 Hint 3: Complete solution
Place [5,5] horizontally at (3,1)-(3,2). The 2×2 sum‑4 block uses [1,1] across (2,2)-(2,3) and [1,0] vertically with 1 at (1,3) and 0 at (0,3) (satisfying less‑than‑5). The sum‑6 pair on the left gets [1,3] vertically with 3 at (1,1) and 1 at (1,2), then [3,6] vertically with 3 at (2,1) and 6 at (2,0) for the greater‑than‑5 cell.
💡 Hint 1: Find the zero forcing
A sum‑0 region containing three cells will lock in the only double‑zero domino and demand a zero from an adjacent domino.
💡 Hint 2: Focus on the bottom‑left corner
The sum‑0 region spans (3,0), (4,0), and (4,1). That forces the [0,0] domino vertically on the left edge and reserves a zero at (4,1) from the [1,0] domino. The sum‑7 pair at top right then needs the [5,6] and [2,3] dominoes.
💡 Hint 3: Full grid layout
Place [0,0] at (3,0)-(4,0). Put [5,6] horizontally with 5 at (0,2), 6 at (0,1); then [0,4] vertically with 0 at (1,1), 4 at (2,1). [2,3] goes vertically 2 at (0,3), 3 at (1,3). [1,1] horizontal at (2,2)-(2,3) gives 1/1. [1,0] vertical with 1 at (4,2), 0 at (4,1); [2,0] horizontal with 0 at (4,3), 2 at (4,4). [0,3] vertical with 0 at (4,5), 3 at (3,5).
💡 Hint 1: Find the most restrictive cell
Two single‑cell less‑than constraints at the top edge limit the pips to 1 or 0. The less‑than‑2 cell in column 3 is especially narrow.
💡 Hint 2: Lock the top‑right corner
Cell (0,3) must be 1 because the only smaller pip is 0, but that domino choice forces a partner. The [4,1] domino fits vertically, placing 1 at (0,3) and 4 at (1,3).
💡 Hint 3: Work down column 0
The less‑than‑3 at (0,0) forces a 1, which takes the [1,1] domino horizontally. Then the sum‑7 column (0) with (1,0)=1 demands a 6 and a 0 below, pulling in the [6,3] and [0,5] dominoes in that order.
💡 Hint 4: Resolve the sum‑9 column and equals region
Column 3 already has 4 at top; to sum 9 across three cells, the remaining two must total 5. The [2,1] domino provides 2 at (2,3) and 1 at (2,2). The equals pair at (2,5)-(2,6) must both be 1, so the [1,3] domino puts 1 at (2,5) and 3 above, while [5,1] fills (2,7) with 5 and (2,6) with 1.
💡 Hint 5: Full hard solution
Place [4,1] vertical at (1,3)-(0,3); [1,1] horizontal at (0,0)-(1,0); [6,3] vertical at (2,0)-(2,1); [0,5] vertical at (3,0)-(4,0); [2,1] horizontal at (2,3)-(2,2); [3,5] vertical at (3,3)-(4,3); [1,3] vertical at (2,5)-(1,5); [5,1] horizontal at (2,7)-(2,6); [0,0] vertical at (3,7)-(4,7); [4,4] horizontal at (3,5)-(4,5).
🎨 Pips Solver
Aug 4, 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: Fill the sum‑4 quad with ones
The sum‑4 region spans (1,2), (1,3), (2,2), (2,3). Only a value of 1 in every cell sums to 4 with four cells. The sole double‑1 domino, [1,1], must occupy two of these. Place it horizontally covering (2,2) and (2,3).
2
Step 2: Extend the ones with the zero‑one domino
Cells (1,2) and (1,3) remain unfilled and must be 1. The [1,0] domino can supply a 1 to (1,3) and a 0 above. Place it vertically with 1 at (1,3) and 0 at (0,3). The less‑than‑5 constraint at (0,3) accepts 0, locking the placement.
3
Step 3: Satisfy the sum‑6 pair and greater‑than cell
The sum‑6 pair at (1,1) and (2,1) needs a total of 6. The [1,3] domino is the only source of a 3 that can reach (1,1); place it vertically with 3 at (1,1) and 1 at (1,2) (already 1). The second 3 for (2,1) comes from [3,6], placed vertically with 3 at (2,1) and 6 at (2,0). The single‑cell greater‑than‑5 at (2,0) forces exactly 6, completing the chain.
4
Step 4: Lock the equals region
The equals region at (3,1) and (3,2) requires identical values. The only remaining unused domino is [5,5]; place it horizontally across those two cells.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Seed zeros with the double‑zero domino
The sum‑0 region spans (3,0), (4,0), and (4,1). All three cells must be 0. The [0,0] domino covers two of them; place it vertically at (3,0) and (4,0). The third zero will come later.
2
Step 2: Crack the top‑right sum‑7 and its neighbor
The sum‑7 pair at (0,2)-(0,3) demands 5 and 2. The only 5 is in the [5,6] domino. Place it horizontally with 5 at (0,2) and 6 at (0,1). The adjacent sum‑6 region (0,1)-(1,1) now has 6, so (1,1) must be 0. The [0,4] domino supplies that 0 and a 4 for (2,1) when placed vertically.
3
Step 3: Complete the sum‑7 and the sum‑4 chain
The remaining sum‑7 cell (0,3) needs 2. Place [2,3] vertically: 2 at (0,3), 3 at (1,3). The sum‑4 region (1,3)-(2,3) then totals 3+1, so (2,3) must be 1. The [1,1] domino fits horizontally at (2,2)-(2,3), giving a 1 at (2,2) and completing the sum‑5 pair (2,1)-(2,2) as 4+1.
4
Step 4: Close the bottom‑left zero region and sum‑1
The sum‑1 pair at (4,2)-(4,3) requires 1 and 0. Place [1,0] vertically: 1 at (4,2) and 0 at (4,1)—fulfilling the missing zero from the sum‑0 trio. Then place [2,0] horizontally: 0 at (4,3) and 2 at (4,4) to satisfy the sum‑2 single cell.
5
Step 5: Place the final vertical in column 5
The sum‑3 single cell at (3,5) needs 3. The [0,3] domino supplies it vertically: 3 at (3,5) and 0 at (4,5). The unconstrained cell (4,5) accepts the 0 harmlessly.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Force the top‑right less‑than cell
Cell (0,3) obeys a less‑than‑2 constraint, limiting it to 0 or 1. The available pips and partner constraints make 1 the only viable choice. Place [4,1] vertically with 1 at (0,3) and 4 at (1,3).
2
Step 2: Anchor the top‑left less‑than cell
Cell (0,0) is less‑than‑3, forcing a 1 (0 would block later sums). Place [1,1] horizontally across (0,0) and (1,0), giving both pips as 1.
3
Step 3: Solve the sum‑7 column and its bottom greater‑than
Column 0’s sum‑7 region (1,0), (2,0), (3,0) already has 1, so (2,0)+(3,0)=6. The domino [6,3] provides the 6: place it vertically with 6 at (2,0) and 3 at (2,1). Then (3,0) must be 0; place [0,5] vertically with 0 at (3,0) and 5 at (4,0), satisfying the greater‑than‑4 constraint at (4,0).
4
Step 4: Resolve the sum‑9 column and its partner
Column 3’s sum‑9 region (1,3)=4, (2,3), (3,3) needs 5 more. The [2,1] domino supplies 2 at (2,3) and 1 at (2,2), the latter fulfilling the sum‑4 pair (2,1)-(2,2) with 3+1. The remaining (3,3) demands 3, so place [3,5] vertically with 3 at (3,3) and 5 at (4,3), satisfying the greater‑than‑4 at (4,3).
5
Step 5: Untangle the equals region and sum‑3
The equals pair (2,5)-(2,6) requires identical pips. Domino [1,3] puts 1 at (2,5) and 3 at (1,5) (sum‑3 cell satisfied). Then (2,6) also needs 1, so place [5,1] horizontally with 5 at (2,7) and 1 at (2,6).
6
Step 6: Finish with the sum‑5 column and greater‑than region
The sum‑5 column at (2,7)=5, (3,7), (4,7) forces 0+0 below. Place [0,0] vertically at (3,7)-(4,7). The remaining greater‑than region at (3,5)-(4,5) takes the [4,4] domino horizontally, completing the grid.
💡 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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