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🎲 Today's Puzzle Overview
April 13's Easy grid comes from Ian Livengood and is built around a design that might be the most direct Pips puzzle possible: almost every cell sits alone in its own region, and every one of those solo regions carries a sum constraint that names the pip count outright. Five dominoes, ten cells, five fixed values. The puzzle is less about deduction and more about assignment — read the labels, match the halves, and the lone double in the set tells you where both its identical faces go.
Rodolfo Kurchan contributes both the Medium and Hard grids. The Medium is a fully packed 4×4 board where every region uses an equals constraint, with a single less-than cell in the corner as the only break from that pattern. Three doubles sit in the set — [5|5], [0|0], and [1|1] — and each one anchors a different equals region. Tracking where each double lands, and what value it then forces through the rest of its region, drives the entire solve.
Kurchan's Hard puzzle spans a sparse 5×5 frame with only 16 of 25 cells occupied. The constraint mix is sharp at both extremes: a sum of exactly 2 in one corner and a sum-greater-than-9 in the opposite corner pull in opposite directions at once. Between them, a chain of sum-9 pairs, equals pairs, a less-than, and another greater-than link up so tightly that the second deduction flows directly from the first, and so on all the way to the last tile.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Every cell announces its own value
Look at the regions: almost every occupied cell sits alone with its own sum constraint. That constraint isn't a total to chase — it's a direct label. Read off each required pip count before you touch a single domino.
💡 The one double places itself
Two cells must both equal 2 and they sit side by side. The only domino that can fill two adjacent cells with identical pip counts is a double. Today's set contains exactly one double. Find it and slot it in — that's your first guaranteed placement.
💡 Full solution
Single-cell sums force five values immediately: (0,2)=1, (0,3)=2, (2,3)=3, (3,0)=5, (3,2)=4. Place [2|2] vertically at (0,3) and (1,3): both show 2. Place [0|1] horizontally at (0,1) and (0,2): (0,2)=1 lands the 1-pip there, so (0,1)=0. Place [1|4] horizontally at (3,1) and (3,2): (3,2)=4 lands the 4-pip there, so (3,1)=1. Place [3|4] vertically at (2,3) and (3,3): (2,3)=3 lands the 3-pip there, so (3,3)=4. Place [1|5] vertically at (2,0) and (3,0): (3,0)=5 lands the 5-pip there, so (2,0)=1. All constraints satisfied. Puzzle complete.
💡 Three doubles are in the set — track them down
Today's domino set contains three doubles: [5|5], [0|0], and [1|1]. Each one must go inside an equals region where two adjacent cells share the same value. Figure out which equals region each double belongs to before you try to place anything else.
💡 The [5|5] double sets off a chain through the top-right block
One equals region spans four cells in the top-right corner of the grid. The [5|5] double belongs there — place it in the two vertically adjacent cells, both showing 5. That forces the entire four-cell region to 5, which in turn tells you the orientation of the domino beside it.
💡 Full solution
Place [5|5] vertically at (0,3) and (1,3): both = 5. The four-cell top-right equals region forces (0,2)=5 and (1,2)=5. Place [5|0] vertically at (1,2) and (2,2): (1,2)=5, so (2,2)=0. The four-cell bottom-right equals region forces all of (2,1), (2,3), and (3,3) to 0. Place [0|0] vertically at (2,3) and (3,3): both = 0. Place [3|0] vertically at (1,1) and (2,1): (2,1)=0, so (1,1)=3. Place [4|5] horizontally at (0,1) and (0,2): (0,2)=5, so (0,1)=4. The two-cell top-left equals region forces (0,0)=4. Place [4|1] vertically at (0,0) and (1,0): (0,0)=4, so (1,0)=1. The four-cell left equals region forces (2,0), (3,0), and (3,1) all to 1. Place [1|1] vertically at (2,0) and (3,0): both = 1. Place [2|1] horizontally at (3,2) and (3,1): (3,1)=1, so (3,2)=2. Less-than check: 2 < 3 ✓. Puzzle complete.
💡 Sum = 2 is almost impossibly small
Two cells in the top-left corner must total exactly 2. With pip values running 0 through 6, the options are extremely narrow. Work out which domino halves can land in those two cells — then the rest of the top row falls into place from there.
💡 Each constraint hands you the next
Once the top-left corner is set, the sum-9 region directly beside it gives you another value for free. That value feeds the equals pair above it, which feeds the greater-than region across the grid, and so on. This puzzle runs like a single unbroken chain.
💡 Full solution
Sum=2 at (0,0)+(0,1): place [2|5] vertically at (0,0) and (1,0) with 2 at (0,0), giving (1,0)=5; place [0|3] horizontally at (0,1) and (0,2) with 0 at (0,1), giving (0,2)=3. Sum check: 2+0=2 ✓. Sum=9 at (0,2)+(0,3): (0,2)=3, so (0,3)=6. Place [6|0] horizontally at (0,3) and (0,4): (0,3)=6, (0,4)=0. Equals at (0,4)/(1,4): (0,4)=0 forces (1,4)=0. Place [0|5] vertically at (1,4) and (2,4): (1,4)=0, (2,4)=5. Greater-than-9 at (2,4)+(3,4): (2,4)=5, so (3,4) must exceed 4 — only 6 works. Place [4|6] vertically at (4,4) and (3,4): (3,4)=6, (4,4)=4. Sum=11>9 ✓. Sum=9 at (1,0)+(2,0): (1,0)=5, so (2,0)=4. Place [4|1] vertically at (2,0) and (3,0): (2,0)=4, (3,0)=1. Less-than-3 at (3,0)+(4,0): (3,0)=1, so (4,0) must be at most 1. Place [1|2] vertically at (4,0) and (4,1): (4,0)=1, (4,1)=2. Sum=2<3 ✓. Equals at (4,1)/(4,2): (4,1)=2 forces (4,2)=2. Place [4|2] horizontally at (4,3) and (4,2): (4,2)=2, (4,3)=4. Equals at (4,3)/(4,4): 4=4 ✓. Puzzle complete.
🎨 Pips Solver
Apr 13, 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: Five single-cell regions name their values outright
Every region in this puzzle is a single cell carrying a sum constraint. That means no addition is needed — each constraint is just the required pip count. Read them off in order: (0,2) must show 1, (0,3) must show 2, (2,3) must show 3, (3,0) must show 5, (3,2) must show 4. Write those five values down. The other five cells — (0,1), (1,3), (2,0), (3,1), (3,3) — are empty-type and carry no constraint of their own.
2
Step 2: The [2|2] double fills the two cells that must both be 2
Cells (0,3) and (1,3) must both equal 2, and they are adjacent vertically. The only domino that can place the same pip count in two touching cells is a double. Today's set contains exactly one double: [2|2]. Place it vertically at (0,3) and (1,3), both showing 2. Both the sum-2 constraint and the empty constraint at (1,3) are satisfied.
3
Step 3: Every remaining domino has at least one end forced
Place [0|1] horizontally at (0,1) and (0,2): cell (0,2) must be 1, so the 1-pip lands there and (0,1)=0. Place [1|4] horizontally at (3,1) and (3,2): cell (3,2) must be 4, so the 4-pip lands there and (3,1)=1. Place [3|4] vertically at (2,3) and (3,3): cell (2,3) must be 3, so the 3-pip lands there and (3,3)=4. Place [1|5] vertically at (2,0) and (3,0): cell (3,0) must be 5, so the 5-pip lands there and (2,0)=1. All constraints satisfied. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: The [5|5] double anchors the top-right equals region
The four-cell region at (0,2), (0,3), (1,2), and (1,3) requires all cells to show the same pip count. Cells (0,3) and (1,3) are adjacent vertically, so a double can fill them both with one placement. The [5|5] double is the only tile in today's set that can anchor two adjacent cells in an equals region at a value high enough to be consistent with the full 4×4 grid. Place [5|5] vertically at (0,3) and (1,3): both show 5. The equals constraint now forces (0,2)=5 and (1,2)=5.
2
Step 2: (2,2) falls to 0, locking the bottom-right equals region
The [5|0] domino connects (1,2) and (2,2) vertically. Cell (1,2)=5 is now fixed, so the 0-pip must go to (2,2). The four-cell equals region at (2,1), (2,2), (2,3), and (3,3) must all share one value — and (2,2)=0 fixes that value to 0. All four cells must be 0. Place [0|0] vertically at (2,3) and (3,3): both show 0, satisfying (2,3) and (3,3). Cell (2,1) must also be 0.
3
Step 3: (2,1)=0 resolves the middle domino and propagates upward
The [3|0] domino connects (1,1) and (2,1) vertically. Cell (2,1)=0 is now fixed, so the 3-pip goes to (1,1). Next, the [4|5] domino connects (0,1) and (0,2) horizontally. Cell (0,2)=5 is fixed, so the 4-pip goes to (0,1). The two-cell equals region at (0,0) and (0,1) must share one value — (0,1)=4 forces (0,0)=4.
4
Step 4: The left column equals region locks down from (1,0)=1
The [4|1] domino connects (0,0) and (1,0) vertically. Cell (0,0)=4 is fixed, so the 1-pip goes to (1,0). The four-cell equals region at (1,0), (2,0), (3,0), and (3,1) must all share one value — (1,0)=1 sets that value to 1. Place the [1|1] double vertically at (2,0) and (3,0): both show 1, satisfying those two cells. Cell (3,1) must also be 1.
5
Step 5: The last domino satisfies the less-than constraint automatically
The only remaining domino is [2|1], covering (3,2) and (3,1) horizontally. Cell (3,1)=1 is fixed, so the 1-pip goes there and (3,2)=2. The less-than constraint at (3,2) requires it to be less than 3 — and 2 < 3 ✓. All constraints satisfied. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Sum=2 forces the top-left corner
Cells (0,0) and (0,1) must total exactly 2. This is an extremely tight constraint — each cell can be at most 2. Scanning the domino set for halves small enough: [2|5] offers a 2, and [0|3] offers a 0. With (0,0)=2 from [2|5] and (0,1)=0 from [0|3], the sum is 2 ✓. Place [2|5] vertically at (0,0) and (1,0): (0,0)=2, (1,0)=5. Place [0|3] horizontally at (0,1) and (0,2): (0,1)=0, (0,2)=3.
2
Step 2: Sum=9 at (0,2)+(0,3) resolves [6|0]
Cell (0,2)=3 is now set. The sum-9 region requires (0,2)+(0,3)=9, so (0,3)=6. The [6|0] domino connects (0,3) and (0,4) horizontally: (0,3)=6, (0,4)=0.
3
Step 3: Equals at (0,4)/(1,4) forces [0|5] into position
Cell (0,4)=0 is set. The equals region requires (1,4)=0 as well. The [0|5] domino connects (1,4) and (2,4) vertically: (1,4)=0 forces (2,4)=5.
4
Step 4: Greater-than-9 at (2,4)+(3,4) places [4|6]
Cell (2,4)=5 is set. The region requires (2,4)+(3,4)>9, so (3,4) must exceed 4. The only value available from the remaining dominoes is 6. Place [4|6] vertically at (4,4) and (3,4): (3,4)=6, (4,4)=4. Sum = 5+6 = 11 > 9 ✓.
5
Step 5: Sum=9 at (1,0)+(2,0) places [4|1]
Cell (1,0)=5 is set. The region requires (1,0)+(2,0)=9, so (2,0)=4. The [4|1] domino connects (2,0) and (3,0) vertically: (2,0)=4, (3,0)=1.
6
Step 6: Less-than-3 at (3,0)+(4,0) places [1|2]
Cell (3,0)=1 is set. The region requires (3,0)+(4,0)<3, so (4,0) must be at most 1. The [1|2] domino connects (4,0) and (4,1) vertically: placing 1 at (4,0) gives (4,1)=2. Sum = 1+1 = 2 < 3 ✓.
7
Step 7: Equals at (4,1)/(4,2) orients [4|2]
Cell (4,1)=2 is set. The equals region requires (4,2)=2 as well. The [4|2] domino connects (4,3) and (4,2) horizontally: (4,2)=2 forces (4,3)=4.
8
Step 8: Equals at (4,3)/(4,4) closes the puzzle
Cell (4,3)=4 and cell (4,4)=4 — both were determined in previous steps. The equals constraint requires them to match: 4 = 4 ✓. All constraints 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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