Check a topological-order certificate edge by edge — Proof Before You Proceed
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Operator-authored reference; no visitor notes are included.
Operator-authored worked example.
Claim: the finite graph with vertices A,B,C,D and edges A->B, A->C, B->D, C->D is acyclic. Assume the supplied list is complete and vertices are distinct. Certificate: [A,B,C,D].
Assign positions A=0, B=1, C=2, D=3. Check EVERY edge: A->B: 0<1; A->C: 0<2; B->D: 1<3; C->D: 2<3. A cycle would force positions to increase strictly back to their starting position, impossible.
Offline Python 3 checker:
def verify(vertices, edges, order):
if len(order) != len(vertices) or len(set(order)) != len(order) or set(order) != set(vertices):
return False
p = {v:i for i,v in enumerate(order)}
return all(u in p and v in p and p[u]<p[v] for u,v in edges)
vs = ['A','B','C','D']
es = [('A','B'),('A','C'),('B','D'),('C','D')]
assert verify(vs, es, ['A','B','C','D'])
assert not verify(vs, es, ['A','D','B','C'])
assert not verify(vs, es, ['A','B','C','C'])
assert not verify(vs, es+[('D','A')], ['A','B','C','D'])
Invalid order [A,D,B,C] fails B->D (2<1 is false) and C->D (3<1 is false). Rejecting it does not prove the graph cyclic: the original order is valid. Adding D->A creates a cycle witnessed by A->B->D->A. A certificate covers only supplied dependencies; missing edges are outside its claim.