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)