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🎲 Today's Puzzle Overview
Ian Livengood keeps today's easy tight — five dominoes and one key observation. A three-cell equals region runs across the middle row, and before you place anything, it pays to simply count how many times each pip value appears across your set. Only one value shows up exactly three times, and once you spot it, two dominos snap into place simultaneously. Everything after that is one smooth pass.
Livengood's medium uses a different kind of filter. Two 'less than 2' constraints sit at the bottom corners of a seven-domino board, and they immediately narrow your options: only two of your tiles carry a pip low enough to qualify. Dropping those anchors in place locks pip 5 into two separate equals regions at once, and the cascade works its way upward to a greater-than constraint at the top that ties it all together.
Rodolfo Kurchan brings the hard puzzle, a twelve-domino layout built around a striking spine of single-cell sums. Seven individual cells across the edges and center of the grid each carry an exact sum target, and together they form a logical chain — each placement forces the next, from blank faces near the middle outward to the equals regions at the top and bottom rows. Big grid, tight logic.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Hint 1: Count before you place
The three-cell equals region in the middle row needs all three cells to show the same pip. Before placing anything, go through all five dominoes and tally how many times each pip value appears — one number will stand out as the only one that shows up exactly three times.
💡 Hint 2: Three sixes, two dominos
Pip 6 appears exactly three times across your set: once on [1|6] and twice on [6|6]. That's the value that fills the three-cell equals region. [6|6] covers the right two cells horizontally, and [1|6] drops its 6-face into the leftmost cell of that row via a vertical placement — with the 1-face landing in the empty cell above.
💡 Full answer
① [6|6] horizontal row 1 cols 1→2: pip 6 at [1,1] and [1,2]. ② [1|6] vertical col 0: pip 1 at [0,0] (empty ✓), pip 6 at [1,0] (equals: 6=6=6 ✓). ③ [5|3] vertical col 1: pip 5 at [2,1] (empty ✓), pip 3 at [3,1]. ④ [3|1] horizontal row 3 cols 2→3: pip 3 at [3,2] (equals: 3=3 ✓), pip 1 at [3,3]. ⑤ [2|2] horizontal row 2 cols 3→4: pip 2 at [2,3] and [2,4] (2+2+1=5 ✓).
💡 Hint 1: The bottom corners are the most restricted cells
Two cells at the bottom of the board each carry a 'less than 2' constraint — they can only hold pip 0 or pip 1. Before working out the equals regions, figure out which of your seven dominos actually have a face low enough to reach those spots.
💡 Hint 2: Only two dominos qualify
Scan your tiles: [2|0] has a 0-face, and [5|1] has a 1-face. Nothing else comes close. Place [2|0] vertically so its 0 lands at the bottom-left, and [5|1] vertically so its 1 lands at the bottom-right. That immediately sets pip 5 in two cells, which feeds both equals regions above.
💡 Full answer
① [2|0] vertical col 0: pip 2 at [2,0] (empty ✓), pip 0 at [3,0] (<2 ✓). ② [5|1] vertical col 3: pip 5 at [2,3], pip 1 at [3,3] (<2 ✓). ③ [5|6] vertical col 3 rows 0→1: pip 5 at [1,3], pip 6 at [0,3]. ④ [5|5] horizontal row 2 cols 4→5: pip 5 at [2,4] (equals: 5=5=5 ✓), pip 5 at [2,5] (empty ✓). ⑤ [3|3] horizontal row 2 cols 1→2: pip 3 at [2,1] and [2,2]. ⑥ [4|3] vertical col 2: pip 4 at [0,2], pip 3 at [1,2] (equals: 3=3=3 ✓; 4+6=10>9 ✓). ⑦ [2|4] horizontal row 1 cols 1→0: pip 2 at [1,1] (empty ✓), pip 4 at [1,0] (>2 ✓).
💡 Hint 1: Seven solo constraints, start at the most extreme
This board has seven single-cell sum constraints scattered down its edges. The ones with the smallest targets — sum=0 and sum=1 — are the most constrained. A cell that must sum to zero can hold exactly one pip value, and figuring that out points you to the right domino immediately.
💡 Hint 2: Zero and one anchor the middle rows
Sum=0 at [2,1] means that cell must be pip 0. Only [0|0] can supply a blank face there — and its partner lands in the adjacent empty cell. Sum=1 at [2,0] then forces pip 1, which comes from [1|0] placed vertically, dropping a 0 into the empty cell above. Two dominos placed before you've touched any of the equals regions.
💡 Hint 3: Four more single-cell sums fall in sequence
From the middle, work outward. Sum=2 at [3,0] and sum=3 at [4,0] are adjacent — a single domino [3|2] satisfies both at once, vertically in column 0. Then sum=2 at [4,2] forces [2|6], whose 6-face drops into [4,3] and feeds the sum=15 region. Sum=4 at [5,0] places [4|1] vertically, leaving pip 1 at [6,0].
💡 Hint 4: Sum=15 with one cell already set
[4,3]=6 from the previous step. The three-cell sum=15 region needs [3,4]+[4,4]=9. Only one combination works with your remaining dominos: [3|5] contributes 3 at [3,4] and 5 at [2,4], while [1|6] contributes 6 at [4,4] and 1 at [5,4]. That 5 at [2,4] feeds the sum=10 pair just above.
💡 Full answer
① [0|0] horizontal row 2 cols 1→2: pip 0 at [2,1] (sum=0 ✓) and [2,2] (empty ✓). ② [1|0] vertical col 0 rows 2→1: pip 1 at [2,0] (sum=1 ✓), pip 0 at [1,0] (empty ✓). ③ [3|2] vertical col 0: pip 2 at [3,0] (sum=2 ✓), pip 3 at [4,0] (sum=3 ✓). ④ [2|6] vertical col 2: pip 2 at [4,2] (sum=2 ✓), pip 6 at [4,3]. ⑤ [4|1] vertical col 0: pip 4 at [5,0] (sum=4 ✓), pip 1 at [6,0]. ⑥ [3|5] vertical col 4: pip 3 at [3,4], pip 5 at [2,4]. ⑦ [1|6] vertical col 4: pip 6 at [4,4] (3+6+6=15 ✓), pip 1 at [5,4] (empty ✓). ⑧ [5|2] vertical col 4: pip 5 at [1,4] (5+5=10 ✓), pip 2 at [0,4]. ⑨ [2|2] horizontal row 0 cols 2→3: pip 2 at [0,2] and [0,3] (equals: 2=2=2 ✓). ⑩ [4|4] horizontal row 0 cols 0→1: pip 4 at [0,0] and [0,1] (equals: 4=4 ✓). ⑪ [3|3] horizontal row 6 cols 3→4: pip 3 at [6,3] and [6,4]. ⑫ [4|3] horizontal row 6 cols 1→2: pip 4 at [6,1] (1+4=5 ✓), pip 3 at [6,2] (equals: 3=3=3 ✓).
🎨 Pips Solver
Mar 19, 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 pips to crack the equals region
The equals region at [1,0], [1,1], and [1,2] demands all three cells match. Go through your five dominoes — [3|1], [1|6], [5|3], [6|6], [2|2] — and tally each value: pip 1 appears twice, pip 2 twice, pip 3 twice, pip 5 once, and pip 6 three times (from [1|6] and [6|6]). Only pip 6 occurs exactly three times. Those three cells must all show 6.
2
Step 2: Lock in the two six-bearing dominos
[6|6] goes horizontal in row 1, covering [1,1] and [1,2] — both faces are already 6. [1|6] goes vertical in column 0: pip 1 lands at [0,0] (empty constraint, no issue) and pip 6 lands at [1,0] (equals: 6=6=6 ✓). Two tiles, three cells solved in one move.
3
Step 3: The two-cell equals region points to pip 3
Remaining dominos: [3|1], [5|3], [2|2]. The equals pair at [3,1] and [3,2] needs matching pips. Pip 3 appears on both [3|1] and [5|3] — that's the value. [5|3] vertical in column 1: pip 5 at [2,1] (empty constraint ✓), pip 3 at [3,1]. [3|1] horizontal in row 3: pip 3 at [3,2] (equals: 3=3 ✓), pip 1 at [3,3].
4
Step 4: Sum=5 completes with [2|2]
The sum=5 region covers [2,3], [2,4], and [3,3]. [3,3]=1 is already placed. So [2,3]+[2,4] must equal 4. The only domino left is [2|2], and 2+2=4 exactly. Horizontal in row 2: pip 2 at [2,3], pip 2 at [2,4] (2+2+1=5 ✓). Done.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Two 'less than 2' cells reveal the anchors
[3,0] and [3,3] both carry less-than-2 constraints — each must be pip 0 or pip 1. Run through your seven dominos and look for faces with those values: [2|0] has a 0-face, [5|1] has a 1-face. Nothing else qualifies. These two tiles are the anchors for the entire puzzle.
2
Step 2: Place the anchors and read off the cascade
[2|0] goes vertical in column 0: pip 2 at [2,0] (empty ✓), pip 0 at [3,0] (<2 ✓). [5|1] goes vertical in column 3: pip 5 at [2,3], pip 1 at [3,3] (<2 ✓). Now [2,3]=5. The equals region {[1,3],[2,3],[2,4]} must all match, which means [1,3]=5 and [2,4]=5.
3
Step 3: Supply both fives in the equals region
[5|6] goes vertical, column 3: pip 5 at [1,3] (equals region, first cell set), pip 6 at [0,3]. [5|5] goes horizontal in row 2 cols 4→5: pip 5 at [2,4] (equals: 5=5=5 ✓), pip 5 at [2,5] (empty ✓). The right-side equals region is sealed.
4
Step 4: The other equals region needs all threes
Remaining dominos: [2|4], [4|3], [3|3]. The equals region {[1,2],[2,1],[2,2]} needs a shared value. Pip 3 appears on [4|3] (one face) and [3|3] (both faces) — three threes for three cells. [3|3] horizontal in row 2 cols 1→2: pip 3 at [2,1] and [2,2]. [4|3] vertical in column 2: pip 4 at [0,2], pip 3 at [1,2] (equals: 3=3=3 ✓). The greater-than-9 constraint at [0,2],[0,3] gives 4+6=10 ✓ for free.
5
Step 5: Last domino satisfies the greater-than constraint
One tile remains — [2|4] — and two open cells: [1,0] and [1,1]. Cell [1,0] must be greater than 2. Orient [2|4] so its 4-face reaches [1,0]: pip 2 at [1,1] (empty ✓), pip 4 at [1,0] (>2 ✓). Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Sum=0 means one thing — blank face
The single-cell sum=0 constraint at [2,1] forces pip 0. There is no other possibility. Domino [0|0] is the only tile that can place a blank face at [2,1] while keeping the other face legal. Put it horizontal in row 2: pip 0 at [2,1] (sum=0 ✓), pip 0 at [2,2] (empty constraint ✓).
2
Step 2: Sum=1 at [2,0] places the second anchor
[2,0] must hold pip 1. Domino [1|0] goes vertical in column 0: pip 1 at [2,0] (sum=1 ✓), pip 0 at [1,0] (the empty-constraint cell above). Two neighboring sum constraints, two dominos placed.
3
Step 3: Sum=2 and sum=3 share a single domino
[3,0] must be pip 2 (sum=2) and [4,0] must be pip 3 (sum=3). These two cells sit in the same column one row apart, and domino [3|2] satisfies both at once: vertical in column 0, pip 2 at [3,0] (sum=2 ✓), pip 3 at [4,0] (sum=3 ✓). Two constraints for the price of one placement.
4
Step 4: Sum=2 at [4,2] drops a six into the sum=15 region
[4,2] must be pip 2. Domino [2|6] goes vertical in column 2: pip 2 at [4,2] (sum=2 ✓), pip 6 at [4,3]. That 6 is now locked into one of the three cells of the sum=15 region — and it's about to do a lot of work.
5
Step 5: Sum=4 at [5,0] — domino [4|1] finishes the left edge
[5,0] must be pip 4. Domino [4|1] vertical in column 0: pip 4 at [5,0] (sum=4 ✓), pip 1 at [6,0]. The entire left-edge column is now resolved.
6
Step 6: Sum=15 with [4,3]=6 already known
The sum=15 region covers [3,4], [4,3], and [4,4]. [4,3]=6 is set. The remaining two cells must sum to 9. Check your available dominos: [3|5] contributes either a 3 or a 5, and [1|6] contributes either a 1 or a 6. Only 3+6=9 works. So [3,4]=3 and [4,4]=6. Place [3|5] vertical in column 4: pip 3 at [3,4], pip 5 at [2,4]. Place [1|6] vertical in column 4: pip 6 at [4,4] (3+6+6=15 ✓), pip 1 at [5,4] (empty ✓).
7
Step 7: Sum=10 at {[1,4],[2,4]} — [2,4]=5 is already set
[2,4]=5 from Step 6. Sum=10 means [1,4]=5. Domino [5|2] goes vertical in column 4: pip 5 at [1,4] (5+5=10 ✓), pip 2 at [0,4].
8
Step 8: Both top equals regions resolve from [0,4]=2
[0,4]=2 is now locked. The three-cell equals region {[0,2],[0,3],[0,4]} must all match, so [0,2]=2 and [0,3]=2. Domino [2|2] horizontal in row 0 cols 2→3: pip 2 at both (equals: 2=2=2 ✓). The other top equals region {[0,0],[0,1]} takes the remaining double — [4|4] horizontal in row 0 cols 0→1: pip 4 at both (equals: 4=4 ✓).
9
Step 9: Bottom row seals everything
Two dominos remain: [3|3] and [4|3]. The three-cell equals region at the bottom {[6,2],[6,3],[6,4]} needs matching pips. Domino [3|3] provides two threes at [6,3] and [6,4]. Domino [4|3] goes horizontal in row 6: pip 4 at [6,1], pip 3 at [6,2] (equals: 3=3=3 ✓). Sum=5 at {[6,0],[6,1]}: [6,0]=1 and [6,1]=4 — 1+4=5 ✓. 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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