SHORTEST PATHNattee Niparnan
Dijkstra’s Algorithm
Graph with Length
Edge with Length
Length functionl(a,b) = distance from a to b
Finding Shortest Path
BFS can give us the shortest path Just convert the length edge into unit
edge
However, this is very slowImagine a case when the length is 1,000,000
Alarm Clock Analogy No need to walk to every
node Since it won’t change
anything We skip to the “actual”
node Set up the clock at alarm at
the target node
Alarm Clock Algorithm
Set an alarm clock for node s at time 0.
Repeat until there are no more alarms: Say the next alarm goes off at time T, for
node u. Then: The distance from s to u is T. For each neighbor v of u in G:
If there is no alarm yet for v, set one for time T + l(u, v).
If v's alarm is set for later than T + l(u, v), then reset it to this earlier time.
Dijkstra’s Algo from BFSprocedure dijkstra(G, l, s)//Input: Graph G = (V;E), directed or undirected; vertex s V; positive edge lengths l// Output: For all vertices u reachable from s, dist[u] is setto the distance from s to u.
for all v V dist[v] = + prev[v] = nil
dist[s] = 0Priority_Queue H = makequeue(V) // enqueue all vertices (using dist as keys)while H is not empty v = H.deletemin() for each edge (v,u) E if dist[u] > dist[v] + l(v, u) dist[u] = dist[v] + l(v, u) prev[u] = v H.decreasekey(v,dist[v]) // change the value of v to dist[v]
Another Implementation of Dijkstra’s Growing from Known Region of
shortest path Given a graph and a starting node s
What if we know a shortest path from s to some subset S’ V?
Divide and Conquer Approach?
Dijktra’s Algo #2procedure dijkstra(G, l, s)//Input: Graph G = (V;E), directed or undirected; vertex s V; positive edge lengths l// Output: For all vertices u reachable from s, dist[u] is setto the distance from s to u.
for all u V : dist[u] = + prev[u] = nil
dist[s] = 0
R = {} // (the “known region”)while R ≠ V : Pick the node v R with smallest dist[] Add v to R for all edges (v,u) E: if dist[u] > dist[v] + l(v,u) dist[u] = dist[v] + l(v,u) prev[u] = v
Analysis
There are |V| ExtractMin Need to check all edges
At most |E|, if we use adjacency list Maybe |V2|, if we use adjacency matrix Value of dist[] might be changed
Depends on underlying data structure
Choice of DS
Using simple array Each ExtractMin uses O(V) Each change of dist[] uses O(1) Result = O(V2 + E) = O(V2)
Using binary heap Each ExtractMin uses O(lg V) Each change of dist[] uses O(lg V) Result = O( (V + E) lg V)
Can be O (V2 lg V)
Might be V2
Good when the graph is sparse
Fibonacci Heap
Using simple array Each ExtractMin uses O( lg V)
(amortized) Each change of dist[] uses O(1)
(amortized) Result = O(V lg V + E)
Graph with Negative Edge
Disjktra’s works because a shortest path to v must pass throught a node closer than v
Shortest path to A pass through B which is… in BFS sense… is further than A
Negative Cycle
A graph with a negative cycle has no shortest path The shortest.. makes no sense..
Hence, negative edge must be a directed
Key Idea in Shortest Path
Update the distance
if (dist[z] > dist[v] + l(v,z))
dist[z] = dist[v] + l(v,z)
This is safe to perform
Another Approach to Shortest Path A shortest path must has at most |V| - 1
edges Writing a recurrent d(n, v) = shortest distance from s to v
using at most n edges d(n,v) = min of
d(n – 1, v) min( d(n – 1, u) + l(u,v) ) over all
edges (u,v) Initial d(*,s) = 0, d(|V| - 1,v) = inf
Bellman-Ford Algorithm
Dynamic Programming Since d(n,*) use d(n – 1,*), we use
only 1D array to store D
Bellman-Ford Algorithmprocedure BellmanFord(G, l, s)//Input: Graph G = (V,E), directed; vertex s V; edge lengths l (may be negative), no negative cycle// Output: For all vertices u reachable from s, dist[u] is setto the distance from s to u.
for all u V : dist[u] = + prev[u] = nil
dist[s] = 0
repeat |V| - 1 times: for all edges (a,b) E: if dist[b] > dist[a] + l(a,b) dist[b] = dist[a] + l(a,b) prev[b] = a
Analysis
Very simple Loop |V| times Each loop takes |E| iterations
O(V E) Dense graph O(V3)
Bellman-Ford is slower but can detect negative cycle
Detecting Negative Cycle
After repeat the loop |V|-1 times If we can still do
dist[v] = dist[u] + l(y,v) Then, there is a negative cycle
Because there is a shorter path eventhough we have repeat this |V|-1 time
for all edges (a,b) E if dist[b] > dist[a] + l(a,b) printf(“negative cycle\n”)
Shortest Path in DAG Path in DAG appears in linearized
orderprocedure dag-shortest-path(G, l, s)//Input: DAG G = (V;E), vertex s V; edge lengths l (may be negative)// Output: For all vertices u reachable from s, dist[u] is setto the distance from s to u.
for all u V : dist[u] = + prev[u] = nil
dist[s] = 0Linearize GFor each u V , in linearized order: for all edges (u,v) E: if dist[v] > dist[u] + l(u,v) dist[v] = dist[u] + l(y,v) prev[b] = a
All Pair Shortest Path
All Pair Shortest Path Problem Input:
A graph Output:
A matrix D[1..v,1..v] giving the shortest distance between every pair of vertices
Approach
Standard shortest path gives shortest path from a given vertex s to every vertex Repeat this for every starting vertex s Dijkstra O(|V| * (|E| log |V|) ) Bellman-Ford O(|V| * (|V|3) )
Dynamic Algorithm Approach Floyd-Warshall
Dynamic Algorithm Approach Using the previous recurrent
d(n, v) = shortest distance from s to v using at most n edges
Change to dn(a,b) = shortest distance from a to b using at
most n edges
dn(a,b) = min of dn-1(a,b)
min(dn-1(a,k) + l(k,b) ) over all edges (k,b)
Initial d0(a,a) = 0, d0 (a,b) = inf , d1(a,b) = l(a,b)
Dynamic Algorithm Approach Again, A shortest path must has at
most |V| - 1 edges What we need to find is d|V|(a,b)
Let D{M} be the matrix dm(a,b) Start with D{1} and compute D{2}
Similar to computation of dist[] in Bellman-Ford
Repeat until we have D{|V|}
Floyd-Warshall
Use another recurrent dk(a,b) is a shortest distance from a
to b that can travel via a set of vertex {1,2,3,…,k}
d0(a,b) = l(a,b) because it can not go pass any vertex
d1(a,b) means a shortest distance that the path can have only vertex 1 (not including a and b)
Recurrent Relation
dk(a,b) = min of dk-1(a,b)
dk-1(a,k) + dk-1 (k,b)
Initial d0(a,b) = l(a,b)
Floyd-Warshall procedure FloydWarshall(G, l)//Input: Graph G = (V,E); vertex s V; edge lengths l (may be negative), no negative cycle// Output: dist[u,v] is the shortest distance from u to v.
dist = l
for k = 0 to |V| - 1 for i = 0 to |V| - 1 for j = 0 to |V| - 1 dist[i][j] = min (dist[i][j], dist[i][k] + dist[k][j])
Analysis
Very simple 3 nested loops of |V| O(|V|3)