Shortest Path Problem: Difference between revisions
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Time complexity: <math> O(V^2 + VE) </math> | Time complexity: <math> O(V^2 + VE) </math> | ||
* Sparse: <math> O(V^2) </math> | * Sparse: <math> O(V^2) </math> | ||
* Dense: <math> O(V^3) </math> | * Dense: | ||
* <math> O(V^3) </math> | |||
<pre> | <pre> | ||
Revision as of 02:03, 28 February 2024
Definitions
A path is a sequence of nodes such that for all consecutive nodes, there exist an edge
Let there be a weight assigned to each edge.
Single Source Shortest Path (SSSP)
Given a graph , source node , outupt the shortest path from the source
Variants
- Single destination problem: shortest path from all nodes to a single destination
- Single pair problem: Shortest path between input pair
Implementation: Bellman Ford
All shortest path must have edges. If this condition is not satisfied, there is a cycle in in the path, and therefore it is not the shortest.
OPT(a, n-1) = min(w(u,a) + OPT(n-1, u) for all (u, a) in E)
Time complexity:
- Sparse:
- Dense:
// G: Graph with vertices (V) and edges (E)
// w[e]: weight of edge e
// S: starting node
Bellman-Ford(G, w, S)
V, E = G
pi[v] = null for all v // traceback
// initialize all shortest path algo
d[v] = infty for all v
d[s] = 0
for i from 1 to |V| - 1:
for all (u,v) in E
if d[v] > d[u] + w(u, v):
d[v] = d[u] + w(u,v)
pi[v] = u;
for all (u, v) in E:
if d[v] > d[u] + w(u, v):
// has negative cycle
return false
This accounts for negative edges but not negative cycles. Bellman Ford can detect it by running an extra time, since if there is a negative cycle, the run time will improve.
