Concept

Dijkstra's algorithm — where it appears

A shortest-path method that settles vertices in order of distance, whose complexity class is decided by which priority queue it is given. Its complexity class is decided by the priority queue it is given, which the pseudocode names as a queue and nothing more.

Named by 12 essays across one field — each of them below, with the objects they name alongside it.

10010³10³10⁴10⁵10⁶10⁷Vcounted workBreadth-firstDijkstra, binary heapDijkstra, all V queuedBellman–Ford, all passesV from 64 to 2048, sparse, fixed average degreework = scans + visits + relaxations + queue comparisons

Counting on a graph

An instrumented array counts comparisons, swaps, reads and writes, and none of those is what a graph algorithm spends its time on. Three new primitives are needed — an adjacency scanned, a vertex first reached, an edge relaxed — and once they exist, breadth-first and depth-first search turn out to be the same algorithm by every count kept on arrays.

graphs · Graph
sparse, fixed average degree10010³10³10⁴10⁵10⁶VworkDijkstra, binary heap — E log EDijkstra, all V queued — V^2dense, fixed density10010⁴10⁵VworkDijkstra, binary heap — V + EDijkstra, all V queued — Ethe class is a property of the sweep as much as of the algorithmwork = scans + visits + relaxations + queue comparisons

Two parameters, one bound, no order

With one size parameter the candidate classes are ordered — n beats n log n beats n², always, and comparing two bounds is reading them. With two, E log V and V² have no order at all, and which one is smaller is a property of the graph. Sweeping V at fixed degree and at fixed density are different experiments, and the same algorithm fits different classes in the two.

graphs · Graph
adjacency scansrelaxationsqueue comparisonsvisitsBreadth-first14,336Dijkstra, all V queued2,118,656Dijkstra, binary heap56,973Bellman–Ford, all passes50,309,120Prim64,884Kruskal74,008V = 2048, E = 6,144, sparse, fixed average degreeevery segment counted exactly

The queue decides the class, and the pseudocode does not name it

Dijkstra's algorithm is eleven lines of pseudocode with a priority queue in the middle of them. Which queue is not stated, and it is the difference between 56,973 units of work and 2,118,656 on the same graph. Two of the three queues here also fail to fit the class they are famous for, in a regime each.

graphs · Graph
10010³10³10⁴10⁵10⁶10⁷Vcounted workTopological order, one passDijkstra, binary heapBellman–Ford, all passesV from 64 to 2048, directed, acyclicwork = scans + visits + relaxations + queue comparisons

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.

graphs · Graph
No estimate543 cells expanded · path 58Straight-line estimate325 cells expanded · path 58Estimate doubled71 cells expanded · path 64V = 900, E = 895, every edge costs onethe estimate is a function of the vertex, supplied by the caller

The precondition on a function the caller writes

Dijkstra expands 1,582 cells to find a path of 98 across a fifty-square grid. The same loop, with the straight-line distance to the goal added to each key, expands 405 and finds the same 98. The estimate has to be a function the caller supplies, and the guarantee holds only while that function never overestimates — a condition on somebody else's code, not on the graph.

graphs · Graph
10010³10³10⁴10⁵10⁶10⁷Vcounted workBreadth-firstDijkstra, binary heapBellman–FordBellman–Ford, all passesV from 64 to 2048, sparse, fixed average degreework = scans + visits + relaxations + queue comparisons

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.

graphs · Graph
cells expandedreads building the estimateNo estimate1,572 expanded · path 356Straight-line estimate1,550 expanded · path 356A*, landmarks252 expanded · path 356 · 6,328 reads to buildV = 2,500, steps cost one to nine, seed 20260910shortest path 356

An estimate borrowed from an easier problem

On a grid where every step costs one, the straight-line distance to the goal cuts a search from 543 cells to 325. On terrain where steps cost between one and nine it cuts 1,572 to 1,550, because it still believes every step costs one. Four exact distance tables, computed once, cut the same search to 252 — and cost 6,328 reads to build, so they pay for themselves on the fifth query.

graphs · Graph
cells expandedNo estimate543 expanded · path 58Straight-line estimate325 expanded · path 58Dijkstra, reduced costs325 expanded · path 58 · 0 arcs priced below zeroV = 900, unit steps, seed 20260910shortest path 58

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.

graphs · Graph
stop where they meetextra path length42321stop when keys reach itextra path length110203040grid, by seedV = 900, 28% blocked, steps cost one to nine5 of 40 wrong under the meeting rule

Where two searches should stop

Search from both ends of a shortest-path query at once and the two frontiers meet somewhere in the middle, having expanded about two thirds of what one search would. Stop at the first vertex both searches have finished, and on five of forty weighted grids the path returned is longer than the shortest. The rule that is always right stops on a different condition, and on one of those grids it also stops sooner.

graphs · Graph
steps cost one to ninesteps cost one01,0002,0003,000cells expanded, mean over the gridsA* from one endalways shortestalways shortesttwo-ended Dijkstraalways shortestalways shortesttwo A*, separate estimateswrong on 12always shortesttwo A*, stop on either keyalways shortestalways shortesttwo A*, averaged potentialalways shortestalways shortest40 grids a bar, 2,500 cellsestimate: straight-line cells

Two estimates that must agree

Run A* from both ends of a query at once, each search guided by its own straight-line estimate, and stop by the rule that is correct for two-ended Dijkstra. On 40 weighted grids it expands 1,002 cells on average and returns a longer path than the shortest on 12 of them. Give both searches one potential, half of one estimate minus half of the other, and the same rule is correct again — on all 40 grids, for 1,041 cells. Two estimates that measure different things cannot share a stopping rule until they are made to measure the same thing.

graphs · Graph
24816326412825510⁶10⁷average out-degreecounted workBellman–Ford from every sourceJohnson's reweightingFloyd–Warshall256 vertices, every answer comparedwork: relaxations + heap comparisons

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.

graphs · Graph
00.50011.50arcs redrawn between queriesstored potential ÷ Bellman–Ford per query0.02%0.1%0.5%2%5%20%recomputed from nothingmended from the broken arcs256 vertices, 128 queriesarc costs redrawn

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.

graphs · Graph

Named alongside it

The objects these essays reach for when they reach for this one.

Shortest pathSearch frontierHeuristic searchNegative weightRelaxationAdmissibilityBellman–FordDensityPotential functionPreconditionReweightingSparse graph

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