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🎲 Today's Puzzle Overview
Ian Livengood's easy grid for April 3rd is sparse — ten cells arranged in a loosely connected layout across four rows. The [5|5] double is the only double in today's easy set, and it sits right at the center of the puzzle's logic: the sum constraint near the top of the grid can only be satisfied if the double lands there. From that one placement, a chain of sums leads all the way to a tidy equals check in the bottom corner.
Livengood's medium puzzle also opens with a deduction that won't take long to spot. Two separate greater-than constraints share the same threshold — and with no 5-pip domino anywhere in today's set, both constrained cells can only hold one possible value. That single observation locks down a three-cell equals region in the center of the board, and once those three cells agree on their shared value, every remaining domino resolves without any backtracking.
Rodolfo Kurchan's hard puzzle for today scatters three doubles across an eight-row grid — and there are exactly three equals regions waiting to receive them. Each double anchors one region, but the challenge is working out which double belongs where. Once those placements click into place, a web of sum constraints tightens along the right edge of the board, and a sprawling six-cell unequal region in the middle demands six distinct values from six specific dominoes.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 The only double has one job to do
There's a two-cell sum constraint near the top of the grid. Today's easy set contains exactly one double — a domino whose two faces show the same pip count. That double is the only tile that can anchor one of those cells and still leave the other cell a reachable total.
💡 Place the double, then read off the neighbor
Once the double is in place, the sum constraint tells you the exact pip value of the adjacent cell. There's only one domino in today's set with that pip face that can reach that position — its orientation is forced, and it in turn feeds the next sum region.
💡 Full solution
Place [5|5] vertically at (0,1) and (1,1) — both cells show 5. Sum-6 at (1,1)+(1,2)=6: (1,1)=5, so (1,2)=1. The [3|1] domino orients with 1 at (1,2) and 3 at (2,2). Sum-7 at (2,2)+(2,3)=7: (2,2)=3, so (2,3)=4. The [4|0] domino orients with 4 at (2,3) and 0 at (3,3). Sum-3 at (2,0)+(3,0)+(3,1)=3: the [0|0] double fills (2,0)=0 and (3,0)=0, so (3,1)=3. The [3|0] domino orients with 3 at (3,1) and 0 at (3,2). Equals check at (3,2) and (3,3): 0=0 ✓. Puzzle complete.
💡 Two constraints, one value
Two cells in this puzzle carry greater-than constraints with the same threshold. Check the domino set — there's a pip value that appears in the set but only just barely clears that bar, and another value that clears it comfortably. Figure out which values are even possible, and you'll find only one that works for both cells.
💡 The equals region in the center unlocks
Once both greater-than cells are settled, look at the three-cell equals region touching one of them. One of those three cells is already fixed. That means all three must share the same value — and from there, two more dominoes fall into place automatically.
💡 Full solution
Both greater-than cells must exceed 4. No 5-pip domino exists in today's set, so both cells must be 6. The [4|6] domino covers (1,1) and (0,1) vertically: (0,1)=6, (1,1)=4. The [6|3] domino covers (2,0) and (3,0) vertically: (2,0)=6, (3,0)=3. Equals region (1,1),(1,2),(2,1): (1,1)=4, so all three must be 4. The [2|4] domino covers (1,3) and (1,2) horizontally: (1,2)=4, (1,3)=2. The [4|3] domino covers (2,1) and (3,1) vertically: (2,1)=4, (3,1)=3. Equals region (1,3),(1,4): (1,3)=2, so (1,4)=2. The [2|2] double covers (0,4) and (1,4) vertically: (0,4)=2, (1,4)=2 ✓. Sum-6 at (2,2)+(2,3)=6: the [2|0] domino covers (3,2) and (2,2): (3,2)=2, (2,2)=0. Then (2,3)=6. The [6|1] domino covers (2,3) and (2,4): (2,3)=6, (2,4)=1. Sum-8: 3+3+2=8 ✓. Puzzle complete.
💡 Count the doubles, count the equals regions
Before placing anything, do two counts: how many doubles are in today's hard set, and how many equals regions appear on the board. The numbers match — and that tells you exactly where each double needs to go.
💡 Start with the largest equals region
The four-cell equals region in the upper-left corner has the tightest constraints. Only one double in the set can anchor two adjacent cells there with the correct value. Place it, and three other cells in that region lock down immediately — which then cascades into the cells directly next to the region.
💡 The right edge resolves through a chain of sums
After the upper-left region is placed, follow the chain down the right side of the grid: an equals pair, then a sum-11, then a sum-12. Each result feeds the next constraint, and two more doubles land in their regions along the way.
💡 The six-cell unequal region needs six distinct values
A large region in the center of the board requires every cell to show a different pip count. By the time you reach it, several surrounding values are already fixed. Use those anchors to determine which of the remaining dominoes can fit each position without repeating a value.
💡 Full solution
Four-cell equals at [0,0],[0,1],[1,0],[2,0]: the [2|2] double covers (0,0) and (1,0) vertically — both show 2, so (0,1)=2 and (2,0)=2. The [0|2] domino covers (0,2) and (0,1): (0,2)=0, (0,1)=2 ✓. Equals at (0,2),(0,3): (0,3)=0. The [0|4] domino: (0,3)=0, (0,4)=4. Equals at (0,4),(1,4): (1,4)=4. The [4|6] domino: (1,4)=4, (1,5)=6. Sum-11: (1,5)=6, so (2,5)=5. The [5|6] domino: (2,5)=5, (3,5)=6. Sum-12: (3,5)=6, so (4,5)=6. The [1|6] domino: (4,5)=6, (5,5)=1. Three-cell equals [4,0],[5,0],[6,0]: the [5|5] double covers (4,0) and (5,0) — both show 5, so (6,0)=5. The [3|5] domino: (7,0)=3, (6,0)=5 ✓. Sum-4 at (3,0)=4: the [2|4] domino covers (2,0) and (3,0): (2,0)=2 ✓, (3,0)=4. Unequal region: [0|3] at (3,2),(2,2): (3,2)=0, (2,2)=3. [1|2] at (4,2),(5,2): (4,2)=1, (5,2)=2. [4|5] at (4,3),(3,3): (4,3)=4, (3,3)=5. Check: {3,0,5,1,4,2} all distinct ✓. Four-cell equals [5,5],[6,4],[6,5],[7,4]: (5,5)=1. The [1|1] double covers (6,4) and (6,5) — both show 1 ✓. The [1|4] domino: (7,4)=1, (7,3)=4. Sum-4 at (7,3)=4 ✓. The [6|6] double covers (7,1) and (7,2): both show 6. Equals ✓. Puzzle complete.
🎨 Pips Solver
Apr 3, 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 double is the only tile that fits the sum region
The sum-6 region spans (1,1) and (1,2). The [5|5] double placed vertically at (0,1) and (1,1) puts 5 in cell (1,1). That forces (1,2) to be 1 — no other arrangement reaches the sum-6 target without going over. The double is the logical starting point because it's the only tile in today's set that can sit in that column without creating a conflict.
2
Step 2: Sum-6 forces the orientation of the next domino
With (1,1)=5, the sum-6 constraint requires (1,2)=1. The only domino with a 1-pip face that can reach (1,2) is [3|1]. Its orientation is fixed: 1 at (1,2) and 3 at (2,2), placed vertically.
3
Step 3: Sum-7 places the third domino
The region at (2,2) and (2,3) must total 7. Cell (2,2)=3 is now set, so (2,3) must be 4. The [4|0] domino orients with 4 at (2,3) and 0 at (3,3), placed vertically.
4
Step 4: Sum-3 and the equals check close the puzzle
The three-cell region at (2,0), (3,0), and (3,1) must sum to 3. The [0|0] double fills (2,0)=0 and (3,0)=0, contributing nothing to the total. So (3,1) must be exactly 3. The [3|0] domino places 3 at (3,1) and 0 at (3,2). Final check: the equals region at (3,2) and (3,3) requires matching values — (3,2)=0 and (3,3)=0. They match. All constraints satisfied.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Greater-than constraints name the pip value
Two cells have greater-than-4 constraints: (0,1) and (2,0). Scan the domino list — the only pip values above 4 in today's medium set are the two 6-pip faces (no 5-pip domino appears). So both cells must be exactly 6. The [4|6] domino covers (1,1) and (0,1) vertically: (0,1)=6, (1,1)=4. The [6|3] domino covers (2,0) and (3,0) vertically: (2,0)=6, (3,0)=3.
2
Step 2: The three-cell equals region locks the center
The equals region spans (1,1), (1,2), and (2,1). Cell (1,1)=4 is already fixed, so all three cells must be 4. The [2|4] domino covers (1,3) and (1,2) horizontally: (1,2)=4 satisfies the equals constraint, and (1,3)=2. The [4|3] domino covers (2,1) and (3,1) vertically: (2,1)=4 ✓, (3,1)=3.
3
Step 3: The equals pair and the double
The two-cell equals region at (1,3) and (1,4): (1,3)=2, so (1,4) must also be 2. The [2|2] double covers (0,4) and (1,4) vertically: (0,4)=2 and (1,4)=2 ✓. Single-cell sum-2 at (0,4): (0,4)=2 ✓.
4
Step 4: Sum-6 and the remaining two dominoes
Sum-6 at (2,2)+(2,3)=6. The remaining dominoes are [2|0] and [6|1]. The [2|0] domino covers (3,2) and (2,2): (3,2)=2, (2,2)=0. Then (2,3) must be 6. The [6|1] domino covers (2,3) and (2,4): (2,3)=6, (2,4)=1. Cell (2,4) is an empty-type region — no constraint to satisfy. Sum-8 at (3,0)+(3,1)+(3,2): 3+3+2=8 ✓. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Match each double to its equals region
Three doubles in today's set: [2|2], [5|5], [6|6]. Three equals regions on the board: four cells at [0,0],[0,1],[1,0],[2,0]; three cells at [4,0],[5,0],[6,0]; two cells at [7,1],[7,2]. Each double anchors one region by supplying two adjacent equal-value cells. The assignment follows from the surrounding constraints — start with the largest region.
2
Step 2: Four-cell equals region at the upper-left — all must be 2
The region [0,0],[0,1],[1,0],[2,0] requires one shared value. The [2|2] double placed vertically at (0,0) and (1,0) gives both cells the value 2. That forces (0,1)=2 and (2,0)=2 as well. The entire upper-left corner is now settled.
3
Step 3: Adjacent constraints cascade across the top row
The [0|2] domino covers (0,2) and (0,1): (0,1)=2 is fixed, so (0,2)=0. Equals at (0,2),(0,3): (0,3)=0. The [0|4] domino covers (0,3) and (0,4): (0,3)=0, (0,4)=4. Equals at (0,4),(1,4): (1,4)=4. The [4|6] domino covers (1,4) and (1,5): (1,4)=4, (1,5)=6.
4
Step 4: Sum-11 and sum-12 resolve the right edge
Sum-11 at (1,5)+(2,5)=11: (1,5)=6, so (2,5)=5. The [5|6] domino covers (2,5) and (3,5): (2,5)=5, (3,5)=6. Sum-12 at (3,5)+(4,5)=12: (3,5)=6, so (4,5)=6. The [1|6] domino covers (5,5) and (4,5): (4,5)=6, (5,5)=1.
5
Step 5: Three-cell equals at the left edge — all must be 5
The region [4,0],[5,0],[6,0] requires one shared value. The [5|5] double placed vertically at (4,0) and (5,0) gives both cells 5, forcing (6,0)=5. The [3|5] domino covers (7,0) and (6,0): (6,0)=5 ✓, (7,0)=3. Cell (7,0) is an empty-type region — no constraint.
6
Step 6: Single-cell sum-4 at (3,0)
Cell (3,0) must equal 4. The [2|4] domino covers (2,0) and (3,0) vertically: (2,0)=2 is already confirmed from the upper-left equals region, so (3,0)=4 ✓.
7
Step 7: The six-cell unequal region fills with six distinct values
The region [2,2],[3,2],[3,3],[4,2],[4,3],[5,2] requires all six cells to differ. The [0|3] domino covers (3,2) and (2,2): (3,2)=0, (2,2)=3. The [1|2] domino covers (4,2) and (5,2): (4,2)=1, (5,2)=2. The [4|5] domino covers (4,3) and (3,3): (4,3)=4, (3,3)=5. Unequal check: {3, 0, 5, 1, 4, 2} — all six values distinct ✓.
8
Step 8: Four-cell equals at the bottom-right — all must be 1
The region [5,5],[6,4],[6,5],[7,4] requires one shared value. Cell (5,5)=1 is already fixed, so all four must be 1. The [1|1] double covers (6,4) and (6,5): both show 1 ✓. The [1|4] domino covers (7,4) and (7,3): (7,4)=1 ✓, (7,3)=4. Sum-4 at (7,3)=4 ✓.
9
Step 9: The last double closes the puzzle
The [6|6] double covers (7,1) and (7,2) horizontally: both show 6. Equals constraint at (7,1),(7,2) ✓. All twelve constraints across the board are satisfied. Puzzle complete.
💡 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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