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🎲 Today's Puzzle Overview
February 24, 2026 has Ian Livengood on both easy and medium. The easy is a compact 3×3 grid with just four dominoes — a single-cell sum with an unusually low target locks in the first placement immediately and lets you unpack everything else in sequence. Medium stretches to six rows and introduces two sum-zero cells, which act as free moves that feed the larger sum-12 region above them.
Rodolfo Kurchan's hard puzzle takes a thematic approach: almost every region targets the same value. That uniformity removes most of the sum arithmetic and shifts the challenge to a mix of less-than, greater-than, and equals constraints scattered across the board. Two large equals regions in particular demand that you figure out what value fills multiple cells simultaneously — and those placements ripple outward through the rest of the grid.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 One region has an unusually low target
Find the single-cell sum region with the smallest target on the board. There's only one pip value that satisfies it — and knowing that value tells you which domino belongs there and which end faces which direction.
💡 That low-target cell triggers the bottom-row pair
Once the small-target cell is filled, the other half of that domino lands in the bottom-row pair region. You already know its value — use it to calculate what the remaining cell in that pair must be.
💡 Full answer
The single-cell sum=1 at [2,2] must hold exactly 1. Place [1–6] horizontally with 1 at [2,2] and 6 at [2,1]. The pair sum=6 at [2,0]+[2,1] now has [2,1]=6, so [2,0]=0. Place [4–0] vertically with 4 at [1,0] and 0 at [2,0]. The equals region at [0,0]–[1,0] requires [0,0]=4 as well. Place [2–4] horizontally with 2 at [0,1] and 4 at [0,0]. One domino left: [2–2] fills the L-shaped sum=6 region at [0,1]+[0,2]+[1,2]. [0,1]=2 is already in place — so [0,2]+[1,2] must equal 4. Both cells need 2. Place [2–2] vertically with 2 at [0,2] and 2 at [1,2]. 2+2+2=6 ✓.
💡 Two free moves are hiding in the grid
Two cells each carry a sum constraint targeting zero. A sum of zero can only mean one thing. Find those cells — they cost you nothing and immediately reveal which dominoes and which ends go there.
💡 The zeros feed the large sum region above them
Once you know where the zero-pip ends land, one of them sits inside the three-cell sum=12 region. With that cell's value confirmed, the remaining two cells in that region need to make up the rest of 12 — and that heavily constrains which dominoes go there.
💡 Full answer
Sum=0 at [0,2] forces 0 — place [0–5] horizontally with 0 at [0,2] and 5 at [0,1]. Sum=0 at [5,2] forces 0 — place [1–0] vertically with 1 at [5,1] and 0 at [5,2]. Sum=12 at [0,1]+[1,0]+[1,1]: [0,1]=5 already; remaining [1,0]+[1,1] must equal 7. Place [3–5] vertically with 3 at [1,2] (empty ✓) and 5 at [1,1]. Place [2–2] vertically with 2 at [1,0] and 2 at [2,0]. Check: 5+2+5=12 ✓. Equals at [2,0]–[3,0]: [2,0]=2, so [3,0]=2 — place [2–6] vertically with 2 at [3,0] and 6 at [4,0] (sum=6 ✓). Single-cell sum=6 at [4,3]: place [6–1] vertically with 6 at [4,3] and 1 at [3,3]. Equals at [1,3],[2,3],[3,3]: [3,3]=1, all must be 1 — place [1–1] vertically with 1 at [1,3] and 1 at [2,3]. Equals at [3,1]–[3,2]: place [4–4] horizontally, both cells showing 4.
💡 Almost every region shares the same target
Look at the sum regions across the board — most of them have identical targets. That removes most of the arithmetic. The real challenge comes from the polarity constraints and the two equals regions, which force specific pip values in key spots.
💡 Single-cell sum regions give you three free pip values
There are several lone sum cells in this puzzle, each immediately revealing a specific pip count. Find them all — they anchor their surrounding regions and give you the footholds needed to unpack the polarity constraints nearby.
💡 The two equals regions each span four cells
Two large equals regions sit near the bottom and right of the grid. All four cells in each region must share the same pip count. Think about which value a four-cell equals region can realistically hold given the domino options available.
💡 Greater-than constraints at the edges point to specific pip values
The top-right corner holds a greater-than constraint with a high threshold. Combined with the adjacent single-cell sum region, that threshold pins the value for that cell almost completely. Work outward from there along the top edge.
💡 Full answer
Three single-cell sum=4 regions: place [3–4] horizontally with 3 at [0,2] (less<4 ✓) and 4 at [0,1] ✓. Place [6–4] horizontally with 6 at [2,0] (greater>4 ✓) and 4 at [2,1] ✓. Place [4–4] vertically with 4 at [1,6] ✓ and 4 at [2,6]. Greater>4 at [0,6]: place [6–2] horizontally with 6 at [0,6] ✓ and 2 at [0,5]. Sum=4 at [0,4]+[0,5]: [0,5]=2, so [0,4]=2 — place [0–2] vertically with 0 at [1,4] and 2 at [0,4]. Sum=4 across [1,4]+[2,4]+[2,5]+[2,6]: [1,4]=0, [2,6]=4; remaining two cells need 0+0 — place [0–0] horizontally with 0 at [2,4] and 0 at [2,5]. Check: 0+0+0+4=4 ✓. Sum=4 at [1,2]+[2,2]: place [4–0] vertically with 0 at [1,2] and 4 at [2,2]. Left-column polarity: [3–5] vertically with 3 at [0,0] (less<4 ✓) and 5 at [1,0] (greater>4 ✓). Bottom-left equals [5,0],[6,0],[6,1] must all be 3: place [1–3] vertically with 1 at [4,0] and 3 at [5,0]; place [3–3] horizontally with 3 at [6,0] and 3 at [6,1] ✓. Sum=4 at [4,0]+[4,1]+[4,2]+[5,2]: [4,0]=1; need three more 1s — place [1–1] horizontally with 1 at [4,1] and 1 at [4,2]; place [1–5] vertically with 1 at [5,2] and 5 at [6,2] (greater>4 ✓). 1+1+1+1=4 ✓. Right equals [4,5],[4,6],[5,6],[6,6] all 5: place [5–5] vertically with 5 at [4,6] and 5 at [5,6]; place [2–5] horizontally with 2 at [4,4] and 5 at [4,5]; place [5–6] vertically with 5 at [6,6] and 6 at [6,5] (greater>4 ✓). Sum=4 at [4,4]+[5,4]: [4,4]=2, so [5,4]=2 — place [2–3] vertically with 2 at [5,4] and 3 at [6,4] (less<4 ✓). 2+2=4 ✓.
🎨 Pips Solver
Feb 24, 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: The sum-1 cell unlocks the first placement
The cell at [2,2] carries a single-cell sum=1 constraint — it must hold exactly 1. Of your four dominoes, [1–6] is the only one with a 1. Place it horizontally: 1 at [2,2] and 6 going left to [2,1].
2
Step 2: The bottom-row pair falls into place
The pair sum=6 at [2,0] and [2,1] now has [2,1]=6 confirmed. Subtract: [2,0] must be 0. The [4–0] domino delivers: place it vertically with 4 at [1,0] and 0 dropping to [2,0].
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Step 3: The equals region links the left column
The equals constraint at [0,0] and [1,0] requires both cells to match. [1,0]=4, so [0,0] must also be 4. The [2–4] domino threads the top row: place it horizontally with 2 at [0,1] and 4 at [0,0].
4
Step 4: Last domino closes the L-shaped sum region
One domino and one region remain. The sum=6 at [0,1]+[0,2]+[1,2] has [0,1]=2 already placed. The remaining two cells need to sum to 4 — that means both must be 2. Place [2–2] vertically with 2 at [0,2] and 2 at [1,2]. Final check: 2+2+2=6 ✓. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Collect the two zero-pip free moves
Cell [0,2] is a single-cell sum=0 region — it must hold 0. The [0–5] domino places 0 there and 5 going right to [0,1]. Set that aside and find the other zero: cell [5,2] is also a sum=0 region. The [1–0] domino places 1 at [5,1] and 0 at [5,2].
2
Step 2: Unpack the sum-12 triangle
The sum=12 region covers [0,1], [1,0], and [1,1]. [0,1]=5 from step 1, so the other two cells need to total 7. The [3–5] domino contributes: place it vertically with 3 at [1,2] (empty ✓) and 5 at [1,1]. Now [1,0] needs to be 2. The [2–2] domino fits: place it vertically with 2 at [1,0] and 2 at [2,0]. Sum check: 5+2+5=12 ✓.
3
Step 3: Equals in the left column propagates downward
The equals region at [2,0] and [3,0] needs both cells to match. [2,0]=2, so [3,0]=2. The [2–6] domino slots in vertically: 2 at [3,0] and 6 at [4,0]. The single-cell sum=6 at [4,0] confirms: 6=6 ✓.
4
Step 4: The right column chain locks in three cells
The single-cell sum=6 at [4,3] forces 6 there. Place [6–1] vertically: 6 at [4,3] and 1 at [3,3]. That 1 anchors the three-cell equals region at [1,3], [2,3], and [3,3] — all must be 1. Place [1–1] vertically with 1 at [1,3] and 1 at [2,3]. ✓
5
Step 5: Last domino seals the middle equals pair
One domino and one region remain: [4–4] and the equals region at [3,1]–[3,2]. A double domino in an equals region — it's a perfect match. Place [4–4] horizontally: 4 at [3,1] and 4 at [3,2]. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Three single-cell sums give away three dominoes
Three lone sum=4 cells sit across the board, each requiring exactly 4. Cell [0,1]: place [3–4] horizontally with 3 at [0,2] (less-than-4 ✓) and 4 at [0,1]. Cell [2,1]: place [6–4] horizontally with 6 at [2,0] (greater-than-4 ✓) and 4 at [2,1]. Cell [1,6]: place [4–4] vertically with 4 at [1,6] and 4 at [2,6].
2
Step 2: Decode the top-right corner
Cell [0,6] has a greater-than-4 constraint — it needs 5 or 6. The [6–2] domino places 6 at [0,6] (✓) and 2 at [0,5]. The pair sum=4 at [0,4]+[0,5] now has [0,5]=2, so [0,4]=2. Place [0–2] vertically with 0 at [1,4] and 2 at [0,4].
3
Step 3: Fill the four-cell sum region in the top-right block
The sum=4 region at [1,4]+[2,4]+[2,5]+[2,6] already has [1,4]=0 and [2,6]=4. The remaining two cells [2,4] and [2,5] must total 0 — both are 0. Place [0–0] horizontally: 0 at [2,4] and 0 at [2,5]. Sum check: 0+0+0+4=4 ✓.
4
Step 4: Handle the vertical sum pair and the left-column polarity
Sum=4 at [1,2]+[2,2]: place [4–0] vertically with 0 at [1,2] and 4 at [2,2]. 0+4=4 ✓. Left column: [0,0] needs less-than-4 and [1,0] needs greater-than-4. The [3–5] domino threads both: 3 at [0,0] (less-than-4 ✓) and 5 at [1,0] (greater-than-4 ✓).
5
Step 5: Solve the bottom-left equals region
The equals region at [5,0], [6,0], and [6,1] needs all three cells to match. Working backward from available dominoes: place [1–3] vertically with 1 at [4,0] and 3 at [5,0]. Place [3–3] horizontally with 3 at [6,0] and 3 at [6,1]. All three cells equal 3. ✓
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Step 6: Clear the bottom-left four-cell sum
The sum=4 region covers [4,0], [4,1], [4,2], and [5,2]. [4,0]=1 from step 5, so the other three cells need 3 total — all must be 1. Place [1–1] horizontally with 1 at [4,1] and 1 at [4,2]. Place [1–5] vertically with 1 at [5,2] and 5 at [6,2] (greater-than-4 ✓). Sum check: 1+1+1+1=4 ✓.
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Step 7: Lock the four-cell equals region on the right
The equals region at [4,5], [4,6], [5,6], and [6,6] needs all four cells to share the same value. Place [5–5] vertically with 5 at [4,6] and 5 at [5,6]. Place [2–5] horizontally with 2 at [4,4] and 5 at [4,5]. Place [5–6] vertically with 5 at [6,6] and 6 at [6,5] (greater-than-4 ✓). Equals check: all four cells show 5. ✓
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Step 8: The last two placements close the puzzle
Sum=4 at [4,4]+[5,4]: [4,4]=2 from step 7, so [5,4]=2. Place [2–3] vertically with 2 at [5,4] and 3 at [6,4] (less-than-4 ✓). 2+2=4 ✓. 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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