Negative weight — where it appears
Named by 7 essays across one field — each of them below, with the objects they name alongside it.
The precondition that removes the queue
Dijkstra maintains a priority queue to discover which vertex is safe to finalise next, and on a directed acyclic graph 65% of its counted work goes into that queue. The order it is discovering is already known. Relaxing in topological order makes exactly one relaxation per arc — 1,536 arcs, 1,536 relaxations — with no queue at all, and negative weights are fine.
The bound with a precondition
Bellman–Ford is O(V·E), and on a graph of 2,048 vertices it stops after seven passes of the 2,047 the bound allows — a factor of 289 between the bound and the run. Dijkstra is faster and returns a wrong answer on four vertices if one arc is negative. Both facts are about the same clause: the qualifier at the end of the sentence.
A graph is as hard as its largest cycle
Negative arcs rule out Dijkstra's algorithm and leave Bellman–Ford, which on a thousand vertices does three million units of work. Stopping it when a pass changes nothing brings that to 78,496. Finding the strongly connected components first and running it inside each one brings it to 38,549 — and to a quarter of the early-exit cost when the components are small, because every cycle lives inside one.
An estimate is a reweighting
Reprice every arc by the estimate's drop across it and run plain Dijkstra, and it expands the same 325 cells A* does, in the same order, because the two are one algorithm. Replace the estimate with one that is still never too high but drops too fast between neighbours, and 215 arcs go below zero — and on a stated grid the search that refuses to reopen a finished cell returns a path of 178 where the shortest is 169.
One Bellman–Ford buys every Dijkstra
A directed graph of 256 vertices with a third of its arcs negative needs shortest paths between every pair. Running Bellman–Ford from every source costs 25.8 million counted operations on the densest graph drawn; running it once, repricing every arc by what it found, and then running Dijkstra from every source costs 13.1 million, and the one Bellman–Ford is under one per cent of that. Floyd–Warshall's 16.8 million is never the cheapest count on the plate. On the sparsest graphs the repeated Bellman–Ford wins, because its early exit makes nine passes rather than 255.
How long a reweighting stays true
Johnson's one Bellman–Ford run costs under one per cent of an all-pairs computation because it is divided over every source. Asked one query at a time it is divided over nothing, and it still repays itself after 2.7 queries — because the preparation is one Bellman–Ford and every query saves a third of another. What decides the trade is not the query count but whether the graph holds still: at half a per cent of arcs redrawn between queries the stored potential is worth exactly nothing, and its life is geometric at a per-arc failure rate of 6.6%.
A potential mended where it broke
A stored reweighting on a 256-vertex graph with negative arcs costs 10,045 relaxations to rebuild, and rebuilding it every time an update breaks it stops paying once half a per cent of arcs change between queries. Mending it from the arcs that broke costs 16 to 442 relaxations instead, and the stored potential stays at two thirds of the per-query cost at every rate of change. When the change is a vertex whose costs all move together, a repair reaches nearly every vertex. It still costs a third of a rebuild.
Named alongside it
The objects these essays reach for when they reach for this one.
Shortest pathBellman–FordDijkstra's algorithmRelaxationPotential functionReweightingEarly exitPreconditionAmortisationBreak-evenCounted primitiveDirected acyclic graph