Shortest path

GRAPH WITH LENGTHDijkstra’s Algorithm

Edge with Length

Distance / time / resource ,etc..

Pattaya

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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

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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

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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 + w(u, v).– If v's alarm is set for later than T + w(u, v), then reset it to

this earlier time.

Dijkstra’s Algo from BFSprocedure dijkstra(G, w, s)//Input: Graph G = (V,E), directed or undirected; vertex s V; positive edge lengths w// 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] + w(v, u) dist[u] = dist[v] + w(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, w, s)//Input: Graph G = (V,E), directed or undirected; vertex s V; positive edge lengths w// 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] + w(v,u) dist[u] = dist[v] + w(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)• Beware!!! Can be O (V2 lg V) • if the graph is not sparse

Might be V2

Good when the graph is

sparse

Fibonacci Heap

• Using Fibonacci Heap– 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 through a node closer than v

• Shortest path to A pass through B which is… in BFS sense… is further than A

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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] + w(v,z)) dist[z] = dist[v] + w(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) + w(u,v) ) over all edges (u,v)

• Initial d(*,s) = 0, d(0,v) =

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, w, s)//Input: Graph G = (V,E), directed; vertex s V; edge lengths w (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] + w(a,b) dist[b] = dist[a] + w(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] + w(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] + w(a,b) printf(“negative cycle\n”)

Shortest Path in DAG• Path in DAG appears in linearized order

procedure dag-shortest-path(G, w, s)//Input: DAG G = (V,E), vertex s V; edge lengths w (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] + w(u,v) dist[v] = dist[u] + w(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) + w(k,b) ) over all edges (k,b)

• Initial d0(a,a) = 0, d0 (a,b) = , d1(a,b) = w(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 w (may be negative), no negative cycle// Output: dist[u,v] is the shortest distance from u to v.

dist = w

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)