A triangle cannot use two colorsComplete assignment space · 8 candidates
The problem
Decide and count color assignments satisfying every graph edge.
The checked result
Satisfiable: no; Solution count: 0.
Every assignment either yields a fully valid coloring or an edge witnessing failure. Counting includes distinct color labels, so symmetric assignments remain separate.
Why the checker accepts it
- Enumerate one color value for each vertex.
- Check every edge’s unequal-color constraint.
- Count the complete solution space.
- Validate a coloring witness, or require enumeration evidence when no model exists.
Formal specification
{
"vertices": 3,
"edges": [
[
0,
1
],
[
1,
2
],
[
2,
0
]
],
"colors": 2
}Claim and evidence
{
"claim": {
"satisfiable": false,
"solution_count": 0
},
"witness": {
"method": "exhaustive enumeration"
}
}Dataset construction
Deterministic finite fixture; full enumeration or witness replay as stated. Structured JSON; field meanings are stated in the specification. Acceptance covers 8 checker units for this record; the unit type is stated in its verification scope.
Complexity and limits
k colors on n vertices produce kⁿ assignments. Color-label permutations can create symmetric solutions.
This is a finite coloring instance; the solution count is not reduced by graph or color symmetries.
A boundary to investigate
Failing to find a solution is not equivalent to proving there is none. Add symmetry reduction and checked propagation or conflict explanations.