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🎲 Today's Puzzle Overview
August 23, 2025 — puzzles #16, #17, and #18, all by Heidi Erwin.
Easy is a compact 3×3 grid where a pair sum=1 forces the first placement almost immediately — a total of just 1 across two cells means one must be 0 and one must be 1. That cascades into an equals row and a pair sum below it.
Medium introduces a sum=12 vertical pair — two cells totaling 12 both require a 6. A less-than-2 constraint at the top and two equals regions of different sizes complete the puzzle.
The hard puzzle has a distinctive feature: four separate single-cell sum=6 regions scattered across the board, each handing you a free 6. Between them, a four-cell sum=4 row and a four-cell equals region (all 1s) form the connective tissue. The less-than-6 column constraint on the right edge is satisfied almost as a side effect once the main pieces fall into place.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 The pair sum=1 is the tightest constraint on the board
Two cells must total just 1 — the only combination is 0 and 1. One of your dominoes has a 0 on both ends, which can deliver a zero to the pair sum region. Once that zero is placed, the other cell is forced to be 1.
💡 The cell forced to 1 sits in the equals row above it
The domino that fills the second cell of the pair sum also has its other end reach into the three-cell equals row. That reveals the value all three cells must share — and the remaining dominos fall right into place.
💡 Full answer
Sum=1 at [0,1]+[0,2]: place [0–0] horizontally with 0 at [0,0] (empty ✓) and 0 at [0,1]. Now [0,2] must be 1. Place [3–1] vertically with 3 at [1,2] and 1 at [0,2] ✓. Equals row [1,0],[1,1],[1,2]: [1,2]=3, so all must be 3 — place [3–3] horizontally with 3 at [1,0] and 3 at [1,1]. Last domino: [4–0]. Sum=4 at [2,0]+[2,1] — place [4–0] horizontally with 4 at [2,0] and 0 at [2,1]. 4+0=4 ✓.
💡 Sum=12 in a vertical pair means both cells must be 6
Two cells stacked in the left column must total 12. The only way to get there is with two 6s — the maximum pip value. Look for the two dominoes that each carry a 6 and can reach those cells.
💡 Less-than-2 at the top filters immediately to 0 or 1
The top cell of the right column is capped below 2 — it must hold a 0 or a 1. That cell anchors a three-cell equals region, so whatever value you place there becomes the shared value for all three cells. Figure out which domino satisfies the less-than and delivers the right number.
💡 Full answer
Less<2 at [0,1] forces 0 or 1 — place [1–5] horizontally with 1 at [0,1] ✓ and 5 at [0,2]. Equals [0,2],[1,1],[1,2]: [0,2]=5, all three must be 5 — place [5–5] horizontally with 5 at [1,1] and 5 at [1,2]. Sum=12 at [2,0]+[3,0]: place [6–3] vertically with 6 at [2,0] and 3 at [1,0] (empty ✓); place [4–6] horizontally with 4 at [3,1] and 6 at [3,0]. 6+6=12 ✓. Equals [3,1],[3,2],[4,1]: [3,1]=4 anchors it — place [1–4] vertically with 1 at [4,2] (empty ✓) and 4 at [3,2]; place [3–4] horizontally with 3 at [4,0] (empty ✓) and 4 at [4,1]. All three cells equal 4 ✓.
💡 Four scattered single-cell sum=6 regions give away four free 6s
Look across the grid for lone sum=6 cells — there are four of them. Each one forces a 6 at that position. The domino carrying each 6 also places a second pip elsewhere, which feeds into the adjacent constraints.
💡 The four-cell sum=4 row sits between two of those 6s
A horizontal region in row 1 covers four cells and must total just 4. Two of those cells get their values from the second ends of the sum=6 dominoes. With those values known, the remaining two cells have very little left to contribute.
💡 A four-cell equals region in the lower-left is anchored at 1
The equals region covering four cells forms an irregular block. Its value is constrained by the sum=4 row above it — once you figure out what pip appears in the cell that links the two regions, the equals region fills itself in.
💡 The less-than-6 column on the right satisfies itself
A four-cell column region requires every cell to be less than 6. This is a relatively loose constraint — values 0 through 5 all qualify. The dominoes that handle this column carry small pip values that fit naturally once the main regions are resolved.
💡 Full answer
Four sum=6 singles: place [6–2] vertically with 6 at [2,4] ✓ and 2 at [1,4]; place [6–3] horizontally with 6 at [3,4] ✓ and 3 at [3,3] (greater>2 ✓); place [1–6] vertically with 1 at [3,1] and 6 at [4,1] ✓; place [0–6] horizontally with 0 at [4,4] (empty ✓) and 6 at [4,3] ✓. Sum=4 at [1,1]+[1,2]+[1,3]+[1,4]: [1,4]=2; remaining three cells need 2. Place [1–2] vertically with 1 at [2,1] and 2 at [1,1]; place [0–0] horizontally with 0 at [1,2] and 0 at [1,3]. Check: 2+0+0+2=4 ✓. Equals [2,1],[3,0],[3,1],[4,0]: [2,1]=1 and [3,1]=1 anchor it — place [1–1] vertically with 1 at [3,0] and 1 at [4,0]. All four equal 1 ✓. Less-than-6 column [0,6],[1,6],[2,6],[3,6] and remaining sums: place [1–4] horizontally with 1 at [0,6] ✓ and 4 at [0,7] (empty ✓); place [1–3] horizontally with 1 at [1,6] ✓ and 3 at [1,7] (sum=3 ✓); place [1–0] vertically with 1 at [3,6] ✓ and 0 at [2,6] ✓. All four column cells less than 6 ✓. Puzzle complete.
🎨 Pips Solver
Aug 23, 2025
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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: Sum=1 pair forces the first two placements
The sum=1 region at [0,1]+[0,2] needs two cells totaling 1. Place [0–0] horizontally: 0 at [0,0] (empty ✓) and 0 at [0,1]. Now [0,2] must be 1. Place [3–1] vertically: 3 at [1,2] and 1 at [0,2] ✓.
2
Step 2: Equals row fills from the known value
The equals region at [1,0],[1,1],[1,2] has [1,2]=3 from step 1 — all three must be 3. Place [3–3] horizontally: 3 at [1,0] and 3 at [1,1]. ✓
3
Step 3: Last domino closes the bottom
One domino remains: [4–0]. The sum=4 at [2,0]+[2,1] needs two cells totaling 4 — place [4–0] horizontally: 4 at [2,0] and 0 at [2,1]. 4+0=4 ✓. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Less-than-2 and the top equals region
Cell [0,1] must be less than 2 — so 0 or 1. Place [1–5] horizontally: 1 at [0,1] ✓ and 5 at [0,2]. That 5 anchors the equals region at [0,2],[1,1],[1,2] — all three cells must be 5. Place [5–5] horizontally: 5 at [1,1] and 5 at [1,2].
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Step 2: Sum=12 pair needs two 6s
The sum=12 region at [2,0]+[3,0] requires 6+6. Place [6–3] vertically: 6 at [2,0] and 3 at [1,0] (empty ✓). Place [4–6] horizontally: 4 at [3,1] and 6 at [3,0]. 6+6=12 ✓.
3
Step 3: Bottom equals region fills from the known 4
The equals region at [3,1],[3,2],[4,1] has [3,1]=4 anchoring it — all three must be 4. Place [1–4] vertically: 1 at [4,2] (empty ✓) and 4 at [3,2]. Place [3–4] horizontally: 3 at [4,0] (empty ✓) and 4 at [4,1]. All three equal 4 ✓. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Two sum=6 singles in the middle — collect and cascade
Cell [2,4] is a lone sum=6 — place [6–2] vertically: 6 at [2,4] ✓ and 2 at [1,4]. Cell [3,4] is a lone sum=6 — place [6–3] horizontally: 6 at [3,4] ✓ and 3 at [3,3] (greater-than-2 ✓).
2
Step 2: Two more sum=6 singles in the lower section
Cell [4,1] is a lone sum=6 — place [1–6] vertically: 1 at [3,1] and 6 at [4,1] ✓. Cell [4,3] is a lone sum=6 — place [0–6] horizontally: 0 at [4,4] (empty ✓) and 6 at [4,3] ✓. All four sum=6 regions satisfied.
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Step 3: Sum=4 across row 1
The sum=4 region at [1,1]+[1,2]+[1,3]+[1,4] has [1,4]=2 from step 1. Remaining three cells need 2. Place [1–2] vertically: 1 at [2,1] and 2 at [1,1]. Place [0–0] horizontally: 0 at [1,2] and 0 at [1,3]. Check: 2+0+0+2=4 ✓.
4
Step 4: Four-cell equals region locks at 1
The equals region at [2,1],[3,0],[3,1],[4,0] has [2,1]=1 and [3,1]=1 already placed — all four must be 1. Place [1–1] vertically: 1 at [3,0] and 1 at [4,0]. ✓
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Step 5: Fill the less-than-6 column and close out
The right column [0,6],[1,6],[2,6],[3,6] needs every cell below 6. Place [1–4] horizontally: 1 at [0,6] ✓ and 4 at [0,7] (empty ✓). Place [1–3] horizontally: 1 at [1,6] ✓ and 3 at [1,7] (sum=3 ✓). Place [1–0] vertically: 1 at [3,6] ✓ and 0 at [2,6] ✓. All four column cells less than 6 ✓. 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.
💬 Community Discussion
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