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🎲 Today's Puzzle Overview
March 20, 2026 brings a trio of Pips puzzles that balance elegance with a satisfying crunch. Ian Livengood handles the easy and medium grids — his signature style is on full display: tight constraint networks where each new placement unlocks the next, leaving almost no room for guessing.
The easy grid uses a mix of greater-than, equals, and sum regions across just five dominoes. One constraint immediately narrows the field, and the solution unfolds naturally from there. Medium scales up to eight dominoes with a clean chain of logic starting from a single forced cell in the middle of the board.
For the hard puzzle, constructor Rodolfo Kurchan steps in with something beautifully uniform: an 8×6 grid dominated almost entirely by sum-6 regions, with a single equals constraint anchoring the top-left corner. Every sum-6 you resolve unlocks the next, creating a satisfying cascade — but knowing where to start is the real challenge.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 One constraint points straight to a domino
Find the region that sets a strict lower bound on a cell's value. Only one domino in your set can satisfy it — that's your starting point.
💡 Bottom-left corner is almost locked in
The greater-than constraint in the lower-left forces a specific pip value. Think about what the highest available pip count is, and which domino carries it.
💡 Full answer
The greater-than region needs a value above 5 — place [6–0] vertically in the bottom-left with the 6 at row 3. Its 0 lands at row 2, column 0. The equals row (row 1, columns 0–2) is all 3s: [3–3] fills the first two cells, and the [3–5] domino's 3-end drops into the third. [3–5] runs vertically into row 2. The last two cells in row 2 take [0–0] horizontally. That leaves [5–5] to fill the top two cells of the right column, completing the sum-10 region: 5 + 5 + 0 = 10.
💡 One cell tells you everything
There's a single-cell sum region somewhere in the grid — whatever the target is, that cell's value is already decided. Start there.
💡 A equals region at the top anchors a whole domino
The top-left equals constraint spans three cells across two rows. Once you figure out which pip value they all share, one double-pip domino drops right into place.
💡 Full answer
The single-cell sum=6 at row 2, column 3 forces that cell to 6 — place [0–6] vertically with 6 up and 0 at row 3. That 0 feeds the equals region at [3,2]–[3,3]: both must be 0, so [0–1] runs vertically there (1 at row 2). Now [2,1]+[2,2]=3 with [2,2]=1, so [2,1]=2 — that's [2–5] going up to put 5 at [1,1]. The top-left equals region needs all three cells equal to 5: [5–5] fills [0,0]–[1,0]. Single-cell sum=4 at [5,3] forces 4 there: [3–4] runs horizontally. Sum=3 at [5,1]+[5,2] with [5,2]=3 means [5,1]=0: [0–2] drops vertically, 2 at [4,1]. That 2 pins the equals region [3,0],[4,0],[4,1] as all 2s: [2–2] fills the top two cells. Last domino [6–6] seals the bottom equals region.
💡 The top-left corner is more constrained than it looks
There's an equals constraint hiding in the top-left — three cells that must all share the same value. Think about which pip count could realistically fill all three.
💡 Single-cell sum regions give free information
Several cells are isolated in their own sum regions, which means their value is handed to you outright. Find them first and work outward from those anchors.
💡 The equals constraint and its neighbor interact directly
The equals region in the top-left and the adjacent single-cell sum region share a cell. Once you resolve the equals constraint, the neighboring sum is also forced.
💡 The mid-left section has a sum pair and an equals pair stacked side by side
Columns 0 and 1 in the middle rows hold a sum=6 pair and an equals pair running parallel. The values in one column constrain the other — look for a domino that satisfies both at once.
💡 Full answer
Top-left equals region takes [0–0]: [0,0]=0, [1,0]=0. The adjacent single-cell sum=6 at [0,1] forces 6 there — [0–6] places 0 at [1,1] (equals confirmed) and 6 at [0,1]. Top-right sum=6 pairs: [5–1] at row 0 (5+1=6), [3–3] at row 1 (3+3=6). Mid-left: [5–4] goes horizontally with 5 at [3,1] and 4 at [3,0]; the sum pair [3,0]+[4,0]=6 needs 2 at [4,0], so [2–5] places 2 there and 5 at [4,1] — matching the equals constraint [3,1]=[4,1]=5. The 2×2 sum=6 block: [3–1] covers [4,4]=3 and [3,4]=1; [1–1] covers [3,5]=1 and [4,5]=1 (1+1+3+1=6). Bottom-left triple: [2–6] places 2 at [6,0] and 6 at [6,1] (sum=6 confirmed); [2–2] fills [7,0]=2 and [7,1]=2 (2+2+2=6). Bottom-right: [6–1] gives 6 at [6,5] and 1 at [6,4]; [5–6] gives 5 at [7,4] and 6 at [7,5]. All sum=6 regions confirmed.
🎨 Pips Solver
Mar 20, 2026
Click a domino to place it on the board. You can also click the board, and the correct domino will appear.
✅ 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: Let the greater-than region lead
The bottom-left cell sits inside a greater-than region — its value must exceed 5. With pip counts capped at 6, only a 6 will work. Scan your five dominoes: [6–0] is the only one holding a 6. Place it vertically in column 0, rows 2–3, with the 6 facing down at row 3.
2
Step 2: Resolve the equals row
Placing [6–0] drops a 0 into row 2, column 0. Now look at the equals region stretching across row 1, columns 0–2. All three cells must share the same pip count. The domino [3–3] slots into columns 0–1 at row 1, giving both cells a 3. For the region to hold, column 2 must also show a 3 — that cell belongs to a different domino.
3
Step 3: Complete the equals chain
The only remaining domino with a 3 is [3–5]. Orient it vertically in column 2: the 3 goes at row 1 (satisfying the equals constraint) and the 5 lands at row 2. Three cells, three 3s — the equals region is satisfied.
4
Step 4: Fill in row 2
Columns 0, 2 in row 2 are now occupied. Columns 3 and 4 are open and both fall outside any region constraint. The [0–0] domino fits perfectly here horizontally — place it across [2,3] and [2,4], both showing 0.
5
Step 5: Close out the sum-10 column
The right column (column 4) has a sum-10 constraint across rows 0–2. Row 2 already shows a 0 from [0–0]. That means rows 0 and 1 together need to sum to 10. The only domino left is [5–5]: drop it vertically into rows 0–1 of column 4. 5 + 5 + 0 = 10. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Claim the free cell
Spot the single-cell sum region in column 3, row 2 — its target is 6, so that cell's value is already decided: 6. Look through your dominoes for one that has a 6 and can reach that cell. The [0–6] domino fits: stand it vertically with 6 at row 2 and 0 dropping down to row 3, column 3.
2
Step 2: Propagate through equals
That 0 at row 3, column 3 now falls inside an equals region covering [3,2] and [3,3]. Both cells must be equal, so [3,2] is also 0. The [0–1] domino handles column 2: 0 at row 3, 1 reaching up to row 2.
3
Step 3: Work the sum-3 region upward
The sum-3 region at [2,1] and [2,2] now has its right cell known — [2,2]=1. The left cell needs to make up the rest: 3 − 1 = 2, so [2,1]=2. The [2–5] domino places 2 there and stretches up to leave 5 at row 1, column 1.
4
Step 4: Lock the top-left equals region
That 5 at [1,1] sits inside the equals region spanning [0,0], [1,0], and [1,1]. All three must be equal to 5. The [5–5] double domino drops vertically into column 0, rows 0–1. All three cells confirmed as 5.
5
Step 5: Decode the single-cell sum-4
Row 5, column 3 holds a lone sum region targeting 4 — that cell must be 4. The [3–4] domino places 3 at [5,2] and 4 at [5,3], running horizontally.
6
Step 6: Finish the sum-3 in row 5
The sum-3 region at [5,1] and [5,2] now has its right cell resolved: [5,2]=3. So [5,1] needs to be 0. The [0–2] domino runs vertically: 0 at row 5 and 2 up at row 4, column 1.
7
Step 7: Trigger the mid-left equals region
That 2 at [4,1] belongs to the equals region covering [3,0], [4,0], and [4,1]. All three must be equal to 2. The [2–2] double domino slots into column 0, rows 3–4. Done.
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Step 8: The last domino seals it
One domino and one region remain: [6–6] and the equals region at the bottom, [6,2]–[6,3]. An equals constraint with a double-pip domino — both cells become 6. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Identify the unique equals anchor
Unlike everything else in this grid, the top-left corner holds an equals constraint — cells [0,0], [1,0], and [1,1] must all share the same pip value. What single value could fill all three? The [0–0] domino covers [0,0] and [1,0], both showing 0. That forces [1,1]=0 as well.
2
Step 2: Resolve the adjacent single-cell sum
Cell [0,1] is its own sum region with a target of 6, so it must hold a 6. The [0–6] domino connects [1,1] and [0,1]: its 0-end lands at [1,1] (already confirmed as 0) and its 6-end fills [0,1]. Clean fit.
3
Step 3: Fill the top-right sum pairs
Two sum-6 regions sit side by side in the top-right: [0,4]+[0,5] and [1,4]+[1,5]. The [5–1] domino gives 5+1=6 and covers row 0. The [3–3] domino gives 3+3=6 and covers row 1. Both regions satisfied in one move each.
4
Step 4: Untangle the mid-left column pair
In the middle section, columns 0 and 1 carry a sum-6 pair ([3,0]+[4,0]) and an equals pair ([3,1]+[4,1]) stacked side by side. The [5–4] domino runs horizontally: 5 at [3,1] and 4 at [3,0]. For the sum to reach 6, [4,0] must be 2 — the [2–5] domino delivers exactly that, placing 2 at [4,0] and 5 at [4,1]. The equals pair [3,1]=[4,1] checks out: both are 5.
5
Step 5: Crack the 2×2 sum block
The four-cell sum-6 region at [3,4],[3,5],[4,4],[4,5] needs to total 6 across two dominos. The [3–1] domino goes vertically in column 4: 1 at row 3 and 3 at row 4. The [1–1] domino goes vertically in column 5: 1 at row 3 and 1 at row 4. Total: 1+1+3+1=6. Both dominoes placed.
6
Step 6: Work through the bottom-left triple
The sum-6 region at the bottom-left covers three cells: [6,0], [7,0], [7,1]. The adjacent single-cell sum=6 at [6,1] locks that cell to 6 immediately. The [2–6] domino connects [6,0]=2 and [6,1]=6 — the single-cell constraint is met. The [2–2] domino fills the remaining two cells: [7,0]=2 and [7,1]=2. Check: 2+2+2=6. ✓
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Step 7: Close out the bottom-right corner
Four cells remain, all in sum-6 regions. The [6–1] domino places 6 at [6,5] and 1 at [6,4] — matching the single-cell sum=6 at [6,5] and contributing 1 to the pair at [6,4]+[7,4]. The [5–6] domino places 5 at [7,4] and 6 at [7,5]: 1+5=6 for the column pair, and [7,5]=6 confirms its own single-cell region. Every constraint 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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