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🎲 Today's Puzzle Overview
Ian Livengood put together today's easy, and it's built around a single elegant insight. A four-cell equals region sits across the middle of the grid — unusually large for this difficulty level. The puzzle is asking you to find the one pip value that appears four times across your six-domino set. Once you spot it, three tiles lock into place simultaneously, and the rest of the puzzle completes in one smooth pass.
Livengood's medium is a step up in complexity. Eight dominoes, a 4×6 grid, and a bottom row that needs all four cells to hold the same value. The surface-level entry point is a sum constraint near the top, but the real challenge is tracing how it propagates downward through an equals pair, a three-cell equals region, and finally into the all-zero bottom row. Each deduction is short and clean — the puzzle rewards methodical thinking over guessing.
Rodolfo Kurchan returns for the hard, a 9×7 grid with sixteen dominoes. The standout feature is a four-cell column that must sum to exactly zero — meaning every pip in that column is a blank face. Three dominoes supply those zeros, and once placed they trigger a chain through a sum=24 region (all sixes), two sum=10 constraints, and a sum=4 column that forces all ones. The board is large but the logic is tight.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Hint 1: One region is bigger than it looks
There's a region spanning four cells that must all show the same pip. Before placing anything, count how many times each pip value appears across your entire domino set — one number will stand out as the only one that can fill all four spots.
💡 Hint 2: Four sixes, three dominos
The four-cell equals region runs through the middle rows. The pip 6 appears on three of your dominoes: 6|5, 6|6, and 6|3. Together those three tiles carry exactly four 6-pips — perfect for a four-cell equals constraint. Once those go in, the 3-pip leftover from 6|3 satisfies the sum=3 at [2,5] for free, and the last three dominos each have just one open spot.
💡 Full answer
① [6|5] horizontal row 1 cols 2→1: 6-pip at [1,2], 5-pip at [1,1] (empty). ② [6|6] horizontal row 1 cols 3→4: pip 6 at both [1,3] and [1,4]. ③ [6|3] horizontal row 2 cols 4→5: 6-pip at [2,4] (equals: 6=6=6=6 ✓), 3-pip at [2,5] (sum=3 ✓). ④ [2|3] horizontal row 2 cols 1→0: 2-pip at [2,1] (empty), 3-pip at [2,0] (sum=3 ✓). ⑤ [3|5] horizontal row 0 cols 4→5: 3-pip at [0,4] (empty), 5-pip at [0,5] (>4 ✓). ⑥ [4|4] horizontal row 0 cols 0→1: pip 4 at both (equals: 4=4 ✓).
💡 Hint 1: Start at the top
There's a single-cell sum constraint near the top of the grid. It forces one specific pip value at that position, which tells you exactly which domino goes there and how it's oriented. That first placement starts a chain that leads all the way down to the bottom row.
💡 Hint 2: Sum=3 starts the cascade
Sum=3 at [0,2] means [0,2] must be pip 3. That locks in domino 3|0 placed vertically, dropping a 0 into [1,2]. Then look at the three-cell equals region in the middle and the four-cell bottom row — zero is going to appear a lot here. The bottom row's equals constraint forces all four cells to the same value, and once you see what that value must be, four dominos place themselves.
💡 Full answer
① [3|0] vertical col 2 rows 0→1: 3-pip at [0,2] (sum=3 ✓), 0-pip at [1,2] (empty). ② [1|1] vertical col 1 rows 1→2: pip 1 at both [1,1] and [2,1] (equals ✓). ③ [5|5] vertical col 3 rows 1→2: pip 5 at both [1,3] and [2,3]. ④ [0|5] vertical col 2 rows 3→2: 0-pip at [3,2], 5-pip at [2,2] (equals: 5=5=5 ✓). ⑤ [0|0] horizontal row 3 cols 0→1: pip 0 at both. ⑥ [4|0] horizontal row 3 cols 4→3: 4-pip at [3,4], 0-pip at [3,3] (equals: 0=0=0=0 ✓). ⑦ [4|3] vertical col 5 rows 3→2: 4-pip at [3,5] (equals: 4=4 ✓), 3-pip at [2,5] (sum=3 ✓). ⑧ [4|2] horizontal row 1 cols 4→5: 4-pip at [1,4] (>3 ✓), 2-pip at [1,5] (empty).
💡 Hint 1: One region rules the bottom half
Somewhere in this grid there's a four-cell region with a sum constraint so extreme that there's only one pip value that can satisfy it. Find that region first — it determines three domino placements and kicks off nearly everything else in the lower half.
💡 Hint 2: Sum=0 forces four blank faces
The column region at rows 5–8 of col 4 must sum to zero. Pip values can't be negative, so every one of those four cells must show pip 0. Three dominos supply those blanks: 4|0, 0|0, and 5|0. Placing all three gives you the bottom half's backbone.
💡 Hint 3: Sum=9 and sum=24 follow immediately
With [5,3]=4 set by domino 4|0, the sum=9 pair at {[4,3],[5,3]} tells you [4,3]=5. That value feeds into the four-cell sum=24 region — six times four is twenty-four, so every cell there must be pip 6. That forces 6|5, 6|6, and 6|3 all at once.
💡 Hint 4: Two sum=10 constraints, then the right column
After the six-pip block, work the lower-right: sum=10 at {[7,3],[8,3]} gives you [7,3]=5, which places 5|3. Sum=10 at {[7,2],[8,1],[8,2]} resolves with 4|3. Then sum=11 at {[1,6],[2,6],[3,6]} unravels from [3,6]=3 upward, and the sum=4 column on the left forces all ones across four cells.
💡 Full answer
① [4|0] horiz row 5 cols 3→4: 4 at [5,3], 0 at [5,4] (sum=0 col start). ② [0|0] vert col 4 rows 6→7: pip 0 at both [6,4] and [7,4]. ③ [5|0] horiz row 8 cols 3→4: 5 at [8,3], 0 at [8,4] (sum=0: 0+0+0+0=0 ✓). ④ [6|5] horiz row 4 cols 4→3: 6 at [4,4], 5 at [4,3] (5+4=9 ✓). ⑤ [6|6] vert col 5 rows 4→5: 6 at [4,5], 6 at [5,5]. ⑥ [6|3] vert col 6 rows 4→3: 6 at [4,6] (6+6+6+6=24 ✓), 3 at [3,6]. ⑦ [5|3] horiz row 7 cols 3→2: 5 at [7,3] (5+5=10 ✓), 3 at [7,2]. ⑧ [4|3] horiz row 8 cols 2→1: 4 at [8,2], 3 at [8,1] (3+3+4=10 ✓). ⑨ [3|2] vert col 5 rows 8→7: 3 at [8,5] (sum=3 ✓), 2 at [7,5] (empty). ⑩ [4|5] horiz row 2 cols 6→5: 4 at [2,6], 5 at [2,5] (5+5=10 ✓). ⑪ [4|2] vert col 6 rows 1→0: 4 at [1,6] (4+4+3=11 ✓), 2 at [0,6] (empty). ⑫ [5|2] horiz row 2 cols 4→3: 5 at [2,4] (sum=10 ✓), 2 at [3,4] (empty). ⑬ [1|5] vert col 2 rows 1→0: 1 at [1,2] (sum=4 start), 5 at [0,2] (sum=5 ✓). ⑭ [3|1] horiz row 2 cols 3→2: 3 at [2,3] (sum=3 ✓), 1 at [2,2]. ⑮ [1|1] vert col 2 rows 3→4: 1 at [3,2], 1 at [4,2] (1+1+1+1=4 ✓). ⑯ [2|2] vert col 0 rows 7→8: 2 at [7,0] (empty), 2 at [8,0] (empty).
🎨 Pips Solver
Mar 18, 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: Count your pip values to crack the equals region
The equals constraint spans four cells: [1,2], [1,3], [1,4], and [2,4]. All four must show the same pip. Go through your six dominoes and tally each value: you get one 2, one 3, two 4s, one 5, and — critically — four 6s (from 6|5, 6|6, and 6|3). Only 6 appears exactly four times. Those three dominos must orient with their 6-face into the equals region.
2
Step 2: Place the three six-pip dominos
6|5 goes horizontal at row 1: 6-pip at [1,2], 5-pip at [1,1] (the empty cell, no constraint). 6|6 goes horizontal at row 1: 6-pip at [1,3] and 6-pip at [1,4]. 6|3 goes horizontal at row 2: 6-pip at [2,4] (equals: 6=6=6=6 ✓), 3-pip at [2,5]. That 3-pip lands directly in the sum=3 constraint at [2,5] — satisfied for free.
3
Step 3: Sum=3 at [2,0] points to 2|3
Cell [2,0] needs sum=3, so [2,0]=3. Among the remaining three dominos (2|3, 4|4, 3|5), only 2|3 has a 3-face available. Place it horizontal at row 2: 3-pip at [2,0] (sum=3 ✓), 2-pip at [2,1] (empty cell).
4
Step 4: Greater>4 at [0,5] assigns 3|5
Cell [0,5] must hold pip 5 or 6. The two dominos left are 3|5 and 4|4. Only 3|5 has a pip above 4. Place it horizontal at row 0: 3-pip at [0,4] (empty), 5-pip at [0,5] (>4 ✓).
5
Step 5: 4|4 closes the board
One domino remains — 4|4 — and one slot remains: [0,0] and [0,1]. The equals constraint there needs both cells equal. 4|4 is a double, so both faces are 4. Horizontal at row 0: 4-pip at [0,0], 4-pip at [0,1] (equals: 4=4 ✓). Done.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Sum=3 at [0,2] determines the first domino
Cell [0,2] carries a sum=3 constraint — it must hold pip 3 on its own. Two dominos have a 3-face: 4|3 and 3|0. Only 3|0 can orient to place 3 at [0,2] via a vertical placement at col 2 rows 0→1: 3-pip at [0,2] (sum=3 ✓), 0-pip at [1,2] (the empty cell below).
2
Step 2: Equals pair at {[1,1],[2,1]} takes 1|1
The equals region spanning [1,1] and [2,1] needs both cells to show the same pip. Any domino with matching faces could theoretically work, but only 1|1 is a true double with a low enough value. Place it vertically at col 1 rows 1→2: pip 1 at [1,1], pip 1 at [2,1] (equals ✓).
3
Step 3: Three-cell equals region needs all 5s
The equals region {[1,3],[2,2],[2,3]} must be all the same value. Domino 5|5 placed vertically at col 3 rows 1→2 gives 5-pip at [1,3] and 5-pip at [2,3]. For [2,2] to also equal 5, domino 0|5 must orient with its 5-face at [2,2]. Place 0|5 vertically at col 2 rows 3→2: 0-pip at [3,2], 5-pip at [2,2] (equals: 5=5=5 ✓).
4
Step 4: Four-cell bottom equals needs all zeros
The region {[3,0],[3,1],[3,2],[3,3]} requires all four cells to match. [3,2]=0 is already set from the previous step. So all four must be 0. Domino 0|0 handles two of them — horizontal at row 3 cols 0→1: pip 0 at [3,0] and [3,1].
5
Step 5: 4|0 fills the last zero and sets [3,4]=4
[3,3] still needs to be 0. Domino 4|0 goes horizontal at row 3 cols 4→3: 4-pip at [3,4], 0-pip at [3,3] (equals: 0=0=0=0 ✓). Note that [3,4]=4 will matter in the next step.
6
Step 6: Equals {[3,4],[3,5]} and sum=3 at [2,5] combine
[3,4]=4 is locked. The equals region {[3,4],[3,5]} means [3,5] must also be 4. Domino 4|3 goes vertically at col 5 rows 3→2: 4-pip at [3,5] (equals: 4=4 ✓), 3-pip at [2,5] (sum=3 ✓). Two constraints resolved in one placement.
7
Step 7: Greater>3 at [1,4] places the last domino
One domino left: 4|2. Cell [1,4] must be greater than 3, so it needs pip 4, 5, or 6. The 4-face of 4|2 is the match. Horizontal at row 1 cols 4→5: 4-pip at [1,4] (>3 ✓), 2-pip at [1,5] (empty). Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Sum=0 column forces four blank faces
The region {[5,4],[6,4],[7,4],[8,4]} must sum to 0. Since pip values are never negative, every single cell in that column must show pip 0. Three dominos supply those zeros: 4|0 (one face), 0|0 (both faces), and 5|0 (one face). These three placements form the foundation of the entire board.
2
Step 2: Place the three zero-supply dominos
4|0 horizontal at row 5 cols 3→4: 4-pip at [5,3], 0-pip at [5,4] (sum=0 column, first zero). 0|0 vertical at col 4 rows 6→7: pip 0 at [6,4] and [7,4]. 5|0 horizontal at row 8 cols 3→4: 5-pip at [8,3], 0-pip at [8,4] (sum=0: 0+0+0+0=0 ✓). The column is sealed.
3
Step 3: Sum=9 at {[4,3],[5,3]} gives [4,3]=5
[5,3]=4 is locked from Step 2. The sum=9 constraint means [4,3]=5. Domino 6|5 goes horizontal at row 4 cols 4→3: 6-pip at [4,4], 5-pip at [4,3] (5+4=9 ✓).
4
Step 4: Sum=24 forces four sixes
Region {[4,4],[4,5],[4,6],[5,5]} must sum to 24. With four cells and maximum pip 6, the only possibility is 6+6+6+6. [4,4]=6 is already set. So [4,5]=6, [4,6]=6, [5,5]=6. Domino 6|6 vertical at col 5 rows 4→5: 6-pip at [4,5] and [5,5]. Domino 6|3 vertical at col 6 rows 4→3: 6-pip at [4,6] (sum=24: 6+6+6+6 ✓), 3-pip at [3,6].
5
Step 5: Sum=10 at {[7,3],[8,3]} resolves cleanly
[8,3]=5 from Step 2. Sum=10 means [7,3]=5. Domino 5|3 horizontal at row 7 cols 3→2: 5-pip at [7,3] (5+5=10 ✓), 3-pip at [7,2].
6
Step 6: Sum=10 at {[7,2],[8,1],[8,2]} — three cells, one deduction
[7,2]=3 is locked from Step 5. So [8,1]+[8,2] must equal 7. Domino 4|3 placed horizontal at row 8 cols 2→1: 4-pip at [8,2], 3-pip at [8,1] (3+3+4=10 ✓).
7
Step 7: Sum=3 at {[8,5]} pins 3|2
Cell [8,5] alone carries the sum=3 constraint, so [8,5]=3. Domino 3|2 goes vertical at col 5 rows 8→7: 3-pip at [8,5] (sum=3 ✓), 2-pip at [7,5] (empty constraint).
8
Step 8: Sum=10 at {[2,4],[2,5]} and sum=11 at the right column
Domino 4|5 horizontal at row 2 cols 6→5: 4-pip at [2,6], 5-pip at [2,5]. Now sum=10 at {[2,4],[2,5]}: [2,5]=5, so [2,4]=5. Domino 5|2 horizontal at row 2 cols 4→3: 5-pip at [2,4] (5+5=10 ✓), 2-pip at [3,4] (empty). Also from [2,6]=4: sum=11 region {[1,6],[2,6],[3,6]} has [3,6]=3 and [2,6]=4, so [1,6]=4.
9
Step 9: 4|2 closes the right column and sum=5 at [0,2]
[1,6]=4 is determined. Domino 4|2 vertical at col 6 rows 1→0: 4-pip at [1,6] (sum=11: 4+4+3=11 ✓), 2-pip at [0,6] (empty).
10
Step 10: Sum=4 column forces all ones — domino 1|5 starts it
Region {[1,2],[2,2],[3,2],[4,2]} must sum to 4 across four cells. The only way is 1+1+1+1. So every cell must be pip 1. Domino 1|5 vertical at col 2 rows 1→0: 1-pip at [1,2] (sum=4 start), 5-pip at [0,2] (sum=5 ✓).
11
Step 11: 3|1 brings sum=3 and another one into the column
Domino 3|1 horizontal at row 2 cols 3→2: 3-pip at [2,3] (sum=3 ✓ at {[2,3]}), 1-pip at [2,2] (sum=4 continues).
12
Step 12: 1|1 closes the sum=4 column
Domino 1|1 vertical at col 2 rows 3→4: 1-pip at [3,2], 1-pip at [4,2] (sum=4: 1+1+1+1=4 ✓). All four column cells are pip 1.
13
Step 13: 2|2 fills the last two empty cells
One domino remains — 2|2 — and two empty cells: [7,0] and [8,0], both in empty-constraint regions. Domino 2|2 vertical at col 0 rows 7→8: 2-pip at [7,0] (empty ✓), 2-pip at [8,0] (empty ✓). 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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