Dijkstra's Algorithm: Difference between revisions

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[[Category:Algorithms]]
{{Infobox Algorithm|runtime=O(VE) or O(V^2) depending on data structure|class=[[Graph Algorithm]], [[Greedy Algorithm]]}}
{{Infobox Algorithm|class=[[Graph Algorithms]] <br> [[Shortest Path Problem]]}}<aside class="portable-infobox noexcerpt pi-background pi-theme-default pi-layout-stacked">
 
== Dijkstra's Algorithm ==
'''Dijkstra's algorithm''' is a solution to the [[Shortest Path Problem|shortest path problem]] for graphs with positively-weighted edges.
</aside><span></span>
 
== Dijkstra's Algorithm ==
= Approach: Greedy =
Problem: [[Shortest Path Problem]].
 
Dijkstra's algorithm centers around this property: ''of all the edges leaving the source, the shortest edge must be in the shortest path.''


The weight of edges must be positive.
To reach the shortest path, Dijkstra always picks the shortest edge
leaving the known nodes. In each iteration, it finalizes the distance to
a node by picking the shortest one.


= Approach =
Test
= Implementation =
= Implementation =
<pre>
Dijkstra(G, w, s) {
    // initialize
    for all v in V:
        d[v] = infty
        pi[v] = null
    d[s] = 0
    Q = V  // priority queue of nodes
    while Q is not empty:
        u = Q.extractMin()
        for all v in adj[u]:
            if d[v] > d[u] + w(u,v):
                d[v] = d[u] + w(u,v)
                pi[v] = u
</pre>
[[Category:Algorithms]]

Latest revision as of 05:22, 20 March 2024

Dijkstra's algorithm is a solution to the shortest path problem for graphs with positively-weighted edges.

Approach: Greedy

Dijkstra's algorithm centers around this property: of all the edges leaving the source, the shortest edge must be in the shortest path.

To reach the shortest path, Dijkstra always picks the shortest edge leaving the known nodes. In each iteration, it finalizes the distance to a node by picking the shortest one.

Implementation

Dijkstra(G, w, s) {
    // initialize
    for all v in V:
        d[v] = infty
        pi[v] = null
    d[s] = 0
    Q = V   // priority queue of nodes
    while Q is not empty:
        u = Q.extractMin()
        for all v in adj[u]:
            if d[v] > d[u] + w(u,v):
                d[v] = d[u] + w(u,v)
                pi[v] = u