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🎲 Today's Puzzle Overview
August 21, 2025 — puzzles #10, #11, and #12, all constructed by Heidi Erwin.
The easy puzzle is a column-running chain of pair sums, with a solo sum anchoring each end. The entire board flows from top to bottom in one sequence — each region's result feeds the next.
Medium brings back the unequal region type and combines it with an unusually large five-cell sum region that must total just 5. With five cells to fill and only 5 to distribute among them, most values have to be very small.
The hard puzzle is the most structurally elegant so far: two diamond-shaped five-cell equals regions sit at opposite ends of the board, one targeting 5 and one targeting 1. A four-cell sum=0 row runs between them as a dividing line. Solve the zero row first to anchor both diamonds, then each one unlocks the remaining constraints around it.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 All sum regions run down a single column
Every constraint in this puzzle forms a chain from top to bottom. The solo sum at [0,0] and the solo sum at [5,2] are the two endpoints — find them, pin those values, and the three pair sums in between each unlock from the known end.
💡 The solo sum at the top cascades downward
Once you place the domino that satisfies [0,0], its partner end lands in the first pair sum region. That gives you one cell of the pair for free — subtract from the target and you know the other cell too, which becomes the known end of the next pair sum below.
💡 Full answer
Sum=2 at [0,0] forces 2 — place [2–3] horizontally with 2 at [0,0] and 3 at [0,1]. Sum=7 at [0,1]+[1,1]: [0,1]=3, so [1,1]=4. Place [4–3] vertically with 4 at [1,1] and 3 at [2,1]. Sum=7 at [2,1]+[3,1]: [2,1]=3, so [3,1]=4. Place [4–5] vertically with 4 at [3,1] and 5 at [4,1]. Sum=9 at [4,1]+[5,1]: [4,1]=5, so [5,1]=4. Place [4–2] vertically with 4 at [5,1] and 2 at [5,2] (sum=2 ✓).
💡 Start with the solo sum=5 — it feeds the large region
The single-cell sum=5 at [3,3] hands you a value for free. Its domino partner falls inside the five-cell sum region, immediately contributing to that region's total of 5. With that cell known, figure out how little the remaining four cells can add up to.
💡 Five cells, total of 5 — most values must be very small
Once you know what the solo-sum domino contributes to the five-cell region, the rest of the cells have very little left to share. Look for dominoes with zeros and ones — they're the only way to keep the total that low.
💡 Full answer
Sum=5 at [3,3] forces 5 — place [5–0] horizontally with 5 at [3,3] and 0 at [3,2]. Five-cell sum=5 at [3,1]+[3,2]+[4,1]+[4,2]+[5,2]: [3,2]=0 placed; remaining four cells need 5. Place [3–2] vertically with 3 at [2,1] and 2 at [3,1] — remaining [4,1]+[4,2]+[5,2]=3. Place [1–5] horizontally with 1 at [4,1] and 5 at [4,0] (empty ✓). Place [1–1] vertically with 1 at [4,2] and 1 at [5,2]. Sum check: 2+0+1+1+1=5 ✓. Unequal region [0,1],[1,1],[2,1]: [2,1]=3 already placed; place [6–5] vertically with 6 at [0,1] and 5 at [1,1]. All three different: 6≠5≠3 ✓.
💡 Two diamond-shaped equals regions frame the board
A five-cell diamond equals region sits at the top of the grid and another at the bottom. Both share the same cross shape — one center cell and four neighbors. A four-cell sum=0 row runs between them. Start by zeroing out that middle row, and the values that anchor each diamond will emerge.
💡 The sum=0 row forces all four cells to zero — and reveals the diamonds' values
A four-cell region in row 3 must total zero, so every cell is 0. Three dominoes contribute to this row, and each one has a non-zero other end that falls outside it. Those non-zero ends anchor the two diamonds above and below.
💡 The top diamond is all 5s, the bottom diamond is all 1s
Once the zero row is filled, one cell of the top diamond and one cell of the bottom diamond get their values from the domino partners of the zeros. Both values are confirmed immediately — and from there the rest of each diamond's cells follow the equals constraint.
💡 The sum=10 pair and the sum=3 and sum=4 cells are the last pieces
With both diamonds filled in, three constraints remain: a two-cell sum=10 near the top-right, a single-cell sum=3, and a two-cell sum=4 at the bottom. Each is resolved by the domino ends already adjacent to the filled regions.
💡 Full answer
Sum=0 at [3,1]+[3,2]+[3,3]+[3,4]: place [0–0] horizontally with 0 at [3,2] and 0 at [3,3]; place [0–5] vertically with 0 at [3,1] and 5 at [2,1]; place [1–0] vertically with 1 at [4,4] and 0 at [3,4]. All four: 0+0+0+0=0 ✓. Top diamond equals: [2,1]=5 anchors it — all five cells must be 5. Place [5–5] vertically with 5 at [0,1] and 5 at [1,1]. Place [2–5] vertically with 2 at [2,0] (empty ✓) and 5 at [1,0]. Place [5–6] horizontally with 5 at [1,2] and 6 at [1,3]. All five confirmed as 5 ✓. Sum=10 at [0,3]+[1,3]: [1,3]=6, so [0,3]=4 — place [4–4] horizontally with 4 at [0,2] (empty ✓) and 4 at [0,3]. 4+6=10 ✓. Bottom diamond equals: [4,4]=1 anchors it — all five must be 1. Place [1–1] horizontally with 1 at [5,3] and 1 at [5,4]. Place [6–1] vertically with 6 at [4,5] (empty ✓) and 1 at [5,5]. Place [4–1] horizontally with 4 at [6,3] and 1 at [6,4]. All five confirmed as 1 ✓. Sum=3 at [5,2]: place [0–3] vertically with 0 at [6,2] and 3 at [5,2] ✓. Sum=4 at [6,2]+[6,3]: 0+4=4 ✓. Puzzle complete.
🎨 Pips Solver
Aug 21, 2025
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: Solo sum=2 opens the chain
Cell [0,0] is a lone sum=2 region — it holds exactly 2. Place [2–3] horizontally: 2 at [0,0] and 3 at [0,1]. The 3 lands in the first pair sum region, immediately giving you one of the two values needed.
2
Step 2: Three pair sums cascade downward
Sum=7 at [0,1]+[1,1]: [0,1]=3, so [1,1]=4. Place [4–3] vertically: 4 at [1,1] and 3 at [2,1]. Sum=7 at [2,1]+[3,1]: [2,1]=3, so [3,1]=4. Place [4–5] vertically: 4 at [3,1] and 5 at [4,1]. Sum=9 at [4,1]+[5,1]: [4,1]=5, so [5,1]=4.
3
Step 3: Last domino closes the bottom
Place [4–2] vertically: 4 at [5,1] ✓ and 2 at [5,2] (sum=2 ✓). The entire column chain is resolved. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Solo sum=5 reveals the first two cells
Cell [3,3] is a lone sum=5 — it holds exactly 5. Place [5–0] horizontally: 5 at [3,3] and 0 at [3,2]. The 0 now occupies one cell of the five-cell sum region.
2
Step 2: Fill the left side of the five-cell region
Five-cell sum=5 with [3,2]=0 placed: remaining [3,1]+[4,1]+[4,2]+[5,2]=5. Place [3–2] vertically: 3 at [2,1] and 2 at [3,1]. Remaining [4,1]+[4,2]+[5,2]=3 — all three need to sum to 3.
3
Step 3: Finish the five-cell region with small values
Place [1–5] horizontally: 1 at [4,1] and 5 at [4,0] (empty ✓). Place [1–1] vertically: 1 at [4,2] and 1 at [5,2]. Full sum check: 2+0+1+1+1=5 ✓.
4
Step 4: Satisfy the unequal column
The unequal region at [0,1],[1,1],[2,1] requires three distinct values. [2,1]=3 is already placed. Place [6–5] vertically: 6 at [0,1] and 5 at [1,1]. Check: 3≠5≠6 — all different ✓. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Zero out row 3
The sum=0 region at [3,1],[3,2],[3,3],[3,4] forces all four cells to 0. Place [0–0] horizontally: 0 at [3,2] and 0 at [3,3]. Place [0–5] vertically: 0 at [3,1] and 5 at [2,1]. Place [1–0] vertically: 1 at [4,4] and 0 at [3,4]. All four cells zeroed ✓.
2
Step 2: Fill the top diamond — all five cells equal 5
The equals diamond at [0,1],[1,0],[1,1],[1,2],[2,1] has [2,1]=5 from step 1. All five must be 5. Place [5–5] vertically: 5 at [0,1] and 5 at [1,1]. Place [2–5] vertically: 2 at [2,0] (empty ✓) and 5 at [1,0]. Place [5–6] horizontally: 5 at [1,2] and 6 at [1,3]. All five cells confirmed as 5 ✓.
3
Step 3: Sum=10 pair resolves from the known 6
The sum=10 region at [0,3]+[1,3] has [1,3]=6 from step 2. So [0,3] must be 4. Place [4–4] horizontally: 4 at [0,2] (empty ✓) and 4 at [0,3]. 4+6=10 ✓.
4
Step 4: Fill the bottom diamond — all five cells equal 1
The equals diamond at [4,4],[5,3],[5,4],[5,5],[6,4] has [4,4]=1 from step 1. All five must be 1. Place [1–1] horizontally: 1 at [5,3] and 1 at [5,4]. Place [6–1] vertically: 6 at [4,5] (empty ✓) and 1 at [5,5]. Place [4–1] horizontally: 4 at [6,3] and 1 at [6,4]. All five confirmed as 1 ✓.
5
Step 5: Solo sum=3 and the bottom pair close the puzzle
One domino remains: [0–3]. Cell [5,2] is a lone sum=3 — it must hold 3. Place [0–3] vertically: 0 at [6,2] and 3 at [5,2] ✓. Sum=4 at [6,2]+[6,3]: [6,2]=0 and [6,3]=4 from step 4. 0+4=4 ✓. 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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