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🎲 Today's Puzzle Overview
Ian Livengood's easy grid for April 8th is stripped down to just four dominoes and four constraints — the smallest board you'll see this week. Two of those four tiles are doubles, and the opening deduction comes from the grid's only inequality: a greater-than-8 constraint that can only be satisfied by one specific arrangement of the available faces. Place those two tiles correctly at the bottom and every other constraint falls in a single chain upward through the grid.
Rodolfo Kurchan's medium puzzle is ruled by a trio of sum=8 regions that stack across the top of the board. Three separate domino pairs all need to total the same number — and with no matching pip values to anchor from, the entry point is hidden in the bottom-left corner: a two-cell sum=1 region that is the tightest constraint on the board. Only one combination of tiles can supply a total of 1, and placing them immediately triggers a sum=10 chain upward that unlocks two of the three sum=8 pairs in sequence. The two doubles in today's set are the last pieces seated, not the first.
Rodolfo Kurchan's hard puzzle for April 8th spreads across eight rows and features three distinct equals regions, each anchored by one of the three doubles in today's set. The opening move comes not from the doubles but from two mirrored sum=3 single-cell constraints at the far corners of the top row — each names its domino immediately. Those two placements hand off values to an equals pair and a four-cell equals region that cascades all the way down the center of the board in zeros. A five-cell equals region below that locks to a different value entirely, and a three-cell equals region at the lower-left closes the board from the bottom up.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 The inequality at the bottom rules out most arrangements
Today's easy grid has just four dominoes and four constraints. Start with the most restrictive: the greater-than-8 region at the bottom of the grid requires two cells to sum to at least 9. Check the pip faces available across all four tiles — only one pairing of two domino halves can actually reach that threshold. Find it, and both bottom cells are fixed.
💡 The bottom locks the top — one chain from there
Once the greater-than-8 constraint is resolved, the value in the bottom-left cell feeds directly into the sum=6 region above it. That nails the pip above, which in turn feeds sum=3 at the top. The last domino — the one covering the top-right — is fully determined by what's already been placed. No guessing required.
💡 Full solution
Greater-than-8 at (2,1)+(2,2): the [4|4] double places 4 at (2,0) and 4 at (2,1). The [0|5] domino places 0 at (1,2) and 5 at (2,2). Sum: 4+5=9>8 ✓. Sum=6 at (1,0)+(2,0): (2,0)=4, so (1,0)=2. The [2|2] double covers (0,0) and (1,0) vertically: 2 at (0,0), 2 at (1,0) ✓. Sum=3 at (0,0)+(0,1): (0,0)=2, so (0,1)=1. The [1|4] domino covers (0,1) and (0,2) horizontally: 1 at (0,1) ✓, 4 at (0,2). Sum=4 at (0,2)+(1,2): (0,2)=4, (1,2)=0 → 4+0=4 ✓. All four constraints satisfied. Puzzle complete.
💡 Sum=1 is the tightest constraint on the board — start there
Three regions in today's medium puzzle all require a sum of 8, which sounds hard to crack. Before touching any of them, look at the bottom-left: a two-cell region that must total just 1. A sum of 1 means one cell holds 0 and the other holds 1. Find the two dominoes in today's set that can supply those faces, and both bottom-left slots are immediately determined.
💡 Sum=10 climbs upward and cracks two of the sum=8 pairs
Once the bottom-left is placed, one of those tiles deposits a pip value into the start of the sum=10 region above it. That value forces the other end of sum=10 — and that end feeds directly into one of the three sum=8 pairs. A second sum=8 pair resolves from the same chain. By the time you're done, only the three-cell equals region and the bottom-right equals pair need attention.
💡 Full solution
Sum=1 at (2,0)+(2,1)=1: [6|0] places 0 at (2,0) and 6 at (3,0). [5|1] places 5 at (2,2) and 1 at (2,1). Sum=10 at (1,2)+(2,2)=10: (2,2)=5, so (1,2)=5. [5|2] covers (1,2) and (1,1): 5 at (1,2) ✓, 2 at (1,1). Sum=8 at (0,1)+(1,1)=8: (1,1)=2, so (0,1)=6. [4|6] covers (0,2) and (0,1): 4 at (0,2), 6 at (0,1) ✓. Sum=8 at (0,2)+(0,3)=8: (0,2)=4, so (0,3)=4. [3|4] covers (1,3) and (0,3): 3 at (1,3), 4 at (0,3) ✓. Sum=3 at (1,3)=3 ✓. Sum=8 at (0,0)+(1,0)=8: the [4|4] double placed vertically — 4+4=8 ✓. Equals at (3,0),(3,1): (3,0)=6, so (3,1)=6. [6|2] covers (3,1) and (3,2): 6 at (3,1) ✓, 2 at (3,2). Three-cell equals at (2,3),(3,2),(3,3): (3,2)=2, so all three must equal 2. [2|2] double covers (2,3) and (3,3): 2 at (2,3) ✓, 2 at (3,3) ✓. All eight constraints satisfied. Puzzle complete.
💡 Two mirrored corners open the board
The top row carries a sum=3 single-cell constraint at the left corner and another at the right corner — both name the same exact pip value. Scan today's hard set for dominoes that can deliver a 3-pip face to each of those positions. There is exactly one tile for each corner, and placing both immediately reveals values in two neighboring regions.
💡 Four zeros cascade through the center
From the two corner placements, an equals pair resolves and hands off a zero to the start of a four-cell equals region that spans cells in two different rows. All four cells in that region must share the same pip count — and the value turns out to be zero. Three additional dominoes each contribute a 0-pip face to lock those cells, while their other halves seed the sum constraints directly below.
💡 Full solution
Sum=3 at (0,0): [3|6] places 3 at (0,0) ✓, 6 at (1,0). Sum=3 at (0,6): [3|1] places 3 at (0,6) ✓, 1 at (1,6). Equals at (1,5),(1,6): (1,6)=1, so (1,5)=1. [1|0] covers (1,5) and (1,4): 1 at (1,5) ✓, 0 at (1,4). Four-cell equals (1,4),(2,2),(2,3),(2,4)=0: (1,4)=0, so all four cells must be 0. [6|0] covers (3,4) and (2,4): 6 at (3,4), 0 at (2,4) ✓. [0|4] covers (2,3) and (3,3): 0 at (2,3) ✓, 4 at (3,3). [5|0] covers (3,2) and (2,2): 5 at (3,2), 0 at (2,2) ✓. Sum=5 at (3,2)=5 ✓. Five-cell equals (3,3),(4,2),(4,3),(5,3),(6,3)=4: (3,3)=4. [4|4] double covers (4,2) and (4,3): 4 at (4,2) ✓, 4 at (4,3) ✓. [4|1] covers (5,3) and (5,2): 4 at (5,3) ✓, 1 at (5,2). [2|4] covers (7,3) and (6,3): 2 at (7,3), 4 at (6,3) ✓. Sum=11 at (3,4)+(4,4)=11: (3,4)=6, so (4,4)=5. [1|5] covers (5,4) and (4,4): 1 at (5,4), 5 at (4,4) ✓. Three-cell equals (5,2),(6,1),(6,2)=1: (5,2)=1. [1|1] double covers (6,1) and (6,2): 1 at (6,1) ✓, 1 at (6,2) ✓. Sum=6 at (5,4)+(6,4)=6: (5,4)=1, so (6,4)=5. [3|5] covers (6,5) and (6,4): 3 at (6,5), 5 at (6,4) ✓. Equals at (6,5),(7,5): (6,5)=3, so (7,5)=3. [3|2] covers (7,5) and (7,4): 3 at (7,5) ✓, 2 at (7,4). Equals at (7,3),(7,4): (7,4)=2, so (7,3)=2 ✓ (already placed). Sum=11 at (1,0)+(1,1)=11: (1,0)=6, so (1,1)=5. [5|5] double covers (1,1) and (1,2): 5 at (1,1) ✓, 5 at (1,2). Sum=10 at (1,2)+(1,3)=10: (1,2)=5, so (1,3)=5. [5|2] covers (1,3) and (0,3): 5 at (1,3) ✓, 2 at (0,3) (empty ✓). Sum=9 at (7,1)+(7,2)=9: [5|4] places 5 at (7,1) and 4 at (7,2) → 5+4=9 ✓. All fifteen constraints satisfied. Puzzle complete.
🎨 Pips Solver
Apr 8, 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: Greater-than-8 pins two dominoes to the bottom
The region at (2,1) and (2,2) must sum to more than 8 — at least 9. Today's set has four tiles: [4|4], [1|4], [0|5], and [2|2]. The only way to reach a sum of 9 or more from these faces is to place [4|4] horizontally with 4 at (2,0) and 4 at (2,1), and [0|5] vertically with 0 at (1,2) and 5 at (2,2). Combined: 4+5=9>8 ✓.
2
Step 2: Sum=6 at the left column
The region at (1,0) and (2,0) must total 6. Cell (2,0)=4 is now fixed from Step 1, so (1,0) must be exactly 2. The only remaining domino with a 2-pip face is [2|2]. Placed vertically: 2 at (0,0) and 2 at (1,0) ✓.
3
Step 3: Sum=3 across the top row
The region at (0,0) and (0,1) must total 3. Cell (0,0)=2 is now fixed. So (0,1) must be 1. The only remaining domino is [1|4]. Placed horizontally: 1 at (0,1) ✓, 4 at (0,2).
4
Step 4: Sum=4 at the top-right confirms the final placement
The region at (0,2) and (1,2) must total 4. Cell (0,2)=4 and cell (1,2)=0 (from Step 1). 4+0=4 ✓. All four constraints across the board are satisfied. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Sum=1 at (2,0)+(2,1) demands pip-0 and pip-1
A total of 1 from two cells means one holds 0 and the other holds 1. Two dominoes in today's set carry a 0-pip face: [6|0] and [2|2]. Two carry a 1-pip face: [5|1] and no others fit here cleanly. [6|0] supplies 0 at (2,0) and 6 at (3,0). [5|1] supplies 1 at (2,1) and 5 at (2,2). Sum: 0+1=1 ✓.
2
Step 2: Sum=10 at (1,2)+(2,2)
Cell (2,2)=5 is now fixed. Sum=10 requires (1,2)=5. The [5|2] domino covers (1,2) and (1,1): 5 at (1,2) ✓, 2 at (1,1). Two values cascade from one step.
3
Step 3: Sum=8 at (0,1)+(1,1) feeds the next sum=8
Cell (1,1)=2 is now fixed. Sum=8 requires (0,1)=6. The [4|6] domino covers (0,2) and (0,1): 4 at (0,2), 6 at (0,1) ✓. Sum=8 at (0,2)+(0,3): (0,2)=4, so (0,3)=4. The [3|4] domino covers (1,3) and (0,3): 3 at (1,3), 4 at (0,3) ✓. Single-cell sum=3 at (1,3)=3 ✓. Three constraints resolved in two steps.
4
Step 4: The [4|4] double seals the third sum=8 pair
The region at (0,0) and (1,0) must total 8. Only one domino in today's set can hit 8 on its own from two faces: the [4|4] double. Place it vertically: 4 at (0,0), 4 at (1,0) → 4+4=8 ✓.
5
Step 5: The bottom equals pair and the three-cell equals region close the puzzle
Equals at (3,0),(3,1): (3,0)=6 from Step 1, so (3,1) must also be 6. The [6|2] domino covers (3,1) and (3,2): 6 at (3,1) ✓, 2 at (3,2). Three-cell equals at (2,3),(3,2),(3,3): (3,2)=2, so all three must equal 2. The [2|2] double covers (2,3) and (3,3): 2 at (2,3) ✓, 2 at (3,3) ✓. All eight constraints satisfied. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Two mirrored sum=3 corners name two dominoes
Cell (0,0) must equal exactly 3. Dominoes in today's set with a 3-pip face: [3|1], [3,5], [3,6], [3,2]. The [3|6] domino fits (0,0) and (1,0) vertically: 3 at (0,0) ✓, 6 at (1,0). Cell (0,6) must also equal 3. The [3|1] domino fits (0,6) and (1,6) vertically: 3 at (0,6) ✓, 1 at (1,6).
2
Step 2: Equals pair at (1,5),(1,6) and the four-cell equals region begin
Cell (1,6)=1 from Step 1. The equals region at (1,5) and (1,6) forces (1,5)=1. The [1|0] domino covers (1,5) and (1,4): 1 at (1,5) ✓, 0 at (1,4). The four-cell equals region (1,4),(2,2),(2,3),(2,4) now has its anchor: (1,4)=0, so all four cells must equal 0.
3
Step 3: Three dominoes fill the four-cell zero region
Each of the three remaining zero cells needs a domino with a 0-pip face at that position. [6|0] covers (3,4) and (2,4): 6 at (3,4), 0 at (2,4) ✓. [0|4] covers (2,3) and (3,3): 0 at (2,3) ✓, 4 at (3,3). [5|0] covers (3,2) and (2,2): 5 at (3,2), 0 at (2,2) ✓. Single-cell sum=5 at (3,2)=5 ✓.
4
Step 4: Five-cell equals region locks to 4
The five-cell region (3,3),(4,2),(4,3),(5,3),(6,3) requires one shared value. Cell (3,3)=4 is fixed from Step 3. The [4|4] double covers (4,2) and (4,3): both 4 ✓. The [4|1] domino covers (5,3) and (5,2): 4 at (5,3) ✓, 1 at (5,2). The [2|4] domino covers (7,3) and (6,3): 2 at (7,3), 4 at (6,3) ✓.
5
Step 5: Sum=11 at (3,4)+(4,4) and the three-cell equals region
Cell (3,4)=6 from Step 3. Sum=11 requires (4,4)=5. The [1|5] domino covers (5,4) and (4,4): 1 at (5,4), 5 at (4,4) ✓. Three-cell equals (5,2),(6,1),(6,2)=1: (5,2)=1 from Step 4. The [1|1] double covers (6,1) and (6,2): 1 at (6,1) ✓, 1 at (6,2) ✓.
6
Step 6: Sum=6 and the two equals pairs at the bottom
Sum=6 at (5,4)+(6,4): (5,4)=1, so (6,4)=5. The [3|5] domino covers (6,5) and (6,4): 3 at (6,5), 5 at (6,4) ✓. Equals at (6,5),(7,5): (6,5)=3, so (7,5)=3. The [3|2] domino covers (7,5) and (7,4): 3 at (7,5) ✓, 2 at (7,4). Equals at (7,3),(7,4): (7,4)=2, so (7,3)=2 — already confirmed from Step 4 ✓.
7
Step 7: Sum=11 at the top row and sum=10 beside it
Sum=11 at (1,0)+(1,1): (1,0)=6, so (1,1)=5. The [5|5] double covers (1,1) and (1,2): 5 at (1,1) ✓, 5 at (1,2). Sum=10 at (1,2)+(1,3): (1,2)=5, so (1,3)=5. The [5|2] domino covers (1,3) and (0,3): 5 at (1,3) ✓, 2 at (0,3) — (0,3) is an empty cell, any value is valid ✓.
8
Step 8: Sum=9 at the bottom-left closes the puzzle
The final region at (7,1) and (7,2) must total 9. The only remaining domino is [5|4]. Place it horizontally: 5 at (7,1), 4 at (7,2) → 5+4=9 ✓. All fifteen 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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