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Shortest Path
Question- 1) Explain Single Source Shortest Paths in a directed acyclic graph?
Answer: A directed Acyclic graph is a graph with directed edges and no cycles. We can calculate single source shortest distances in O(V+E) time for Directed Acyclic Graphs where V is the number of nodes and E is the number of edges present in the graph. The idea is to use topological sorting.
Topological sorting for Directed Acyclic Graph (DAG) is a linear ordering of vertices such that for every directed edge ab, vertex a comes before b in the ordering.
Consider an example:
Consider the above diagram. The objective is to find the minimum distance between point A and H. To achieve this, we will use topological sorting.
∞ | ∞ | ∞ | ∞ | ∞ | ∞ | ∞ | ∞ |
A | B | C | D | E | F | G | H |
Now start with A and explore all the possible nodes and choose the minimum edge weight node. Initially like from A we can explore B and C and among them B have the minimum weight edges so we will choose that.
0 | 3 | ∞ | ∞ | ∞ | ∞ | ∞ | ∞ |
A | B | C | D | E | F | G | H |
Now we can observe that from B as well as from A we can explore c so we will choose minimum as A-C is 6 and A-B-C is 7 so we will choose c
0 | 3 | 6 | ∞ | ∞ | ∞ | ∞ | ∞ |
A | B | C | D | E | F | G | H |
Repeat the process until you reach the end point i.e. H
0 | 3 | 6 | 7 | ∞ | ∞ | ∞ | ∞ |
A | B | C | D | E | F | G | H |
0 | 3 | 6 | 7 | 3 | ∞ | ∞ | ∞ |
A | B | C | D | E | F | G | H |
0 | 3 | 6 | 7 | 3 | 12 | ∞ | ∞ |
A | B | C | D | E | F | G | H |
0 | 3 | 6 | 7 | 3 | 12 | 9 | ∞ |
A | B | C | D | E | F | G | H |
0 | 3 | 6 | 7 | 3 | 12 | 9 | 11 |
A | B | C | D | E | F | G | H |
So, the single source shortest paths between A and H in a directed acyclic graph give above is of weight 11.
i.e. A->B->D->G->H
Code implementation:
// C++ program to find single source shortest paths for Directed Acyclic Graphs
#include<iostream>
#include <bits/stdc++.h>
#define INF INT_MAX
usingnamespacestd;
// Graph is represented using adjacency list. Every node of adjacency list
// contains vertex number of the vertex to which edge connects. It also
// contains weight of the edge
classAdjListNode
{
intv;
intweight;
public:
AdjListNode(int_v, int_w) { v = _v; weight = _w;}
intgetV() { returnv; }
intgetWeight() { returnweight; }
};
// Class to represent a graph using adjacency list representation
classGraph
{
intV; // No. of vertices'
// Pointer to an array containing adjacency lists
list<AdjListNode> *adj;
// A function used by shortestPath
voidtopologicalSortUtil(intv, boolvisited[], stack<int>&Stack);
public:
Graph(intV); // Constructor
// function to add an edge to graph
voidaddEdge(intu, intv, intweight);
// Finds shortest paths from given source vertex
voidshortestPath(ints);
};
Graph::Graph(intV)
{
this->V = V;
adj = newlist<AdjListNode>[V];
}
voidGraph::addEdge(intu, intv, intweight)
{
AdjListNodenode(v, weight);
adj[u].push_back(node); // Add v to u's list
}
voidGraph::topologicalSortUtil(intv, boolvisited[], stack<int>&Stack)
{
// Mark the current node as visited
visited[v] = true;
// Recur for all the vertices adjacent to this vertex
list<AdjListNode>::iterator i;
for(i = adj[v].begin(); i != adj[v].end(); ++i)
{
AdjListNode node = *i;
if(!visited[node.getV()])
topologicalSortUtil(node.getV(), visited, Stack);
}
// Push current vertex to stack which stores topological sort
Stack.push(v);
}
// The function to find shortest paths from given vertex. It uses recursive
// topologicalSortUtil() to get topological sorting of given graph.
voidGraph::shortestPath(ints)
{
stack<int> Stack;
intdist[V];
// Mark all the vertices as not visited
bool*visited = newbool[V];
for(inti = 0; i< V; i++)
visited[i] = false;
// Call the recursive helper function to store Topological Sort
// starting from all vertices one by one
for(inti = 0; i< V; i++)
if(visited[i] == false)
topologicalSortUtil(i, visited, Stack);
// Initialize distances to all vertices as infinite and distance
// to source as 0
for(inti = 0; i< V; i++)
dist[i] = INF;
dist[s] = 0;
// Process vertices in topological order
while(Stack.empty() == false)
{
// Get the next vertex from topological order
intu = Stack.top();
Stack.pop();
// Update distances of all adjacent vertices
list<AdjListNode>::iterator i;
if(dist[u] != INF)
{
for(i = adj[u].begin(); i != adj[u].end(); ++i)
if(dist[i->getV()] >dist[u] + i->getWeight())
dist[i->getV()] = dist[u] + i->getWeight();
}
}
// Print the calculated shortest distances
for(inti = 0; i< V; i++)
(dist[i] == INF)? cout<< "INF ": cout<<dist[i] << " ";
}
// Driver program to test above functions
intmain()
{
// Create a graph given in the above diagram. Here vertex numbers are
// 0, 1, 2, 3, 4, 5 with following mappings:
// 0=r, 1=s, 2=t, 3=x, 4=y, 5=z
Graph g(6);
g.addEdge(0, 1, 5);
g.addEdge(0, 2, 3);
g.addEdge(1, 3, 6);
g.addEdge(1, 2, 2);
g.addEdge(2, 4, 4);
g.addEdge(2, 5, 2);
g.addEdge(2, 3, 7);
g.addEdge(3, 4, -1);
g.addEdge(4, 5, -2);
ints = 1;
cout<< "Following are shortest distances from source "<< s <<" n";
g.shortestPath(s);
return0;
}
Question- 2) Explain Negative Weight Edges ?
Answer: An edge is (together with vertices) one of the two basic units out of which graphs are constructed. Each edge has two (or in hypergraphs, more) vertices to which it is attached, called its endpoints. Edges may be directed or undirected, Positive or negative. Negative edges are used to represent deprivation or the negative aspect of a graph.
Negative edges represent loss in real world scenario.
Consider an example, a person visited multiple cities. The edges represent the profit or loss the person undergoes while visiting one city from another depending on various factors.
Consider the value of graph in thousands. From the above graph we can say that if the person travels from A to B he gets a profit of 8000 whereas if a person moves from G to C it suffers a loss of 1000.
So, we can say that negative edges signify something opposite to the natural meaning.
Question- 3) Explain Dijkstra's Algorithm? Advantages and Disadvantages.
Answer: Dijkstra's Algorithm is a shortest path algorithm used to find the shortest path from source to all the possible vertex present in the graph. The algorithm used greedy approach to solve this problem. It is one of the most efficient algorithms but is not efficient to deal with negative weighted graph
The time complexity for all V vertices is V * (E*logV) i.eO(VElogV).
Pseudo code :
Input: Graph G, source vertex s, w (the weight of the edge)
Output: Shortest distance between source s and destination
Notation:
G =Weighted directed graph
S= Set of vertices
w (u, v) = the weight of the edge between u and v
s=source vertex
DIJKSTRA (G, w, s)
Initialize-single source (G, s)
S=Î¦
Q=G.V
While Q !=Î¦
U=EXTRACT-MIN(Q)
S=S U {u}
for each vertex v â‚¬ G.Adj[u]
RELAX(u,v,w)
Consider the below graph:
We will be using Dijkstra's algorithm to find the shortest distance from vertex V1 to all other.
V1 | V2 | V3 | V4 | V5 | V6 |
0 | ∞ | ∞ | ∞ | ∞ | ∞ |
As our starting vertex is V1 so choose V1 and explore all the possible vertex that can be accessed through V1
V2 | V3 | V4 | V5 | V6 |
2 | 4 | ∞ | ∞ | ∞ |
Now among these the smallest one is V2 so select V2. And explore all vertex and if it can relax any vertex with less value than replace that value. We can observe that V3 can be reached through v2 with total weight 2+1=3
V2 | V3 | V4 | V5 | V6 |
2 | 3 | 9 | ∞ | ∞ |
Now select V2
V2 | V3 | V4 | V5 | V6 |
2 | 3 | 9 | 6 | ∞ |
Now Select the minimum in all unselected vertex.
V2 | V3 | V4 | V5 | V6 |
2 | 3 | 8 | 6 | 11 |
Now select V4
V2 | V3 | V4 | V5 | V6 |
2 | 3 | 8 | 6 | 11 |
Now select V4
V2 | V3 | V4 | V5 | V6 |
2 | 3 | 8 | 6 | 9 |
Now select 6
V2 | V3 | V4 | V5 | V6 |
2 | 3 | 8 | 6 | 9 |
So, the result is
Vertex | Distance |
V2 | 2 |
V3 | 3 |
V4 | 8 |
V5 | 6 |
V6 | 9 |
Advantages: -
1) Implemented in many used applications like Google Maps.
2) It is used in finding Shortest Path.
3) To find locations of Map which refers to vertices of graph.
4) Distance between the location refers to edges.
5) It is used in IP routing to find Open shortest Path First.
6) It is used in the telephone network.
Disadvantages: -
1) It does blind search so wastes lot of time while processing.
2) It cannot handle negative edges.
3) This leads to acyclic graphs and most often cannot obtain the right shortest path.
Code Implementation:
// A C++ program for Dijkstra's single source shortest path algorithm.
// The program is for adjacency matrix representation of the graph
#include <limits.h>
#include <stdio.h>
// Number of vertices in the graph
#define V 9
// A utility function to find the vertex with minimum distance value, from
// the set of vertices not yet included in shortest path tree
intminDistance(intdist[], boolsptSet[])
{
// Initialize min value
intmin = INT_MAX, min_index;
for(intv = 0; v < V; v++)
if(sptSet[v] == false&&dist[v] <= min)
min = dist[v], min_index = v;
returnmin_index;
}
// A utility function to print the constructed distance array
voidprintSolution(intdist[])
{
printf("Vertex \t\t Distance from Source\n");
for(inti = 0; i< V; i++)
printf("%d \t\t %d\n", i, dist[i]);
}
// Function that implements Dijkstra's single source shortest path algorithm
// for a graph represented using adjacency matrix representation
voiddijkstra(intgraph[V][V], intsrc)
{
intdist[V]; // The output array. dist[i] will hold the shortest
// distance from src to i
boolsptSet[V]; // sptSet[i] will be true if vertex i is included in shortest
// path tree or shortest distance from src to i is finalized
// Initialize all distances as INFINITE and stpSet[] as false
for(inti = 0; i< V; i++)
dist[i] = INT_MAX, sptSet[i] = false;
// Distance of source vertex from itself is always 0
dist[src] = 0;
// Find shortest path for all vertices
for(intcount = 0; count < V - 1; count++) {
// Pick the minimum distance vertex from the set of vertices not
// yet processed. u is always equal to src in the first iteration.
intu = minDistance(dist, sptSet);
// Mark the picked vertex as processed
sptSet[u] = true;
// Update dist value of the adjacent vertices of the picked vertex.
for(intv = 0; v < V; v++)
// Update dist[v] only if is not in sptSet, there is an edge from
// u to v, and total weight of path from srcto v through u is
// smaller than current value of dist[v]
if(!sptSet[v] && graph[u][v] &&dist[u] != INT_MAX
&&dist[u] + graph[u][v] <dist[v])
dist[v] = dist[u] + graph[u][v];
}
// print the constructed distance array
printSolution(dist);
}
// driver program to test above function
intmain()
{
/* Let us create the example graph discussed above */
intgraph[V][V] = { { 0, 4, 0, 0, 0, 0, 0, 8, 0 },
{ 4, 0, 8, 0, 0, 0, 0, 11, 0 },
{ 0, 8, 0, 7, 0, 4, 0, 0, 2 },
{ 0, 0, 7, 0, 9, 14, 0, 0, 0 },
{ 0, 0, 0, 9, 0, 10, 0, 0, 0 },
{ 0, 0, 4, 14, 10, 0, 2, 0, 0 },
{ 0, 0, 0, 0, 0, 2, 0, 1, 6 },
{ 8, 11, 0, 0, 0, 0, 1, 0, 7 },
{ 0, 0, 2, 0, 0, 0, 6, 7, 0 } };
dijkstra(graph, 0);
return0;
}
Question- 4) Explain Bellman-Ford Algorithm?
Answer: Bellman-Ford algorithm is another algorithm used to find the shortest distance from a source vertex to all the vertex present in the graph. Bellman-Ford algorithm is slower than Dijkstra algorithm but is considered as a versatile algorithm as it is capable of handling negative weighted graphs as well.
Pseudo code:
Input data:
graph with nodes V and arcs E with weights f(e); source node u. Output data: distances d(v) from the source node u to each node v âˆˆV.
for eachvâˆˆVdod(v) := âˆž
d(u) = 0
forifrom 1 to |V| - 1:
for eache = (w, v) âˆˆE:
ifd(v) >d(w) + f(e):
d(v) := d(w) + f(e)
Consider the below diagram:
We will be using Dijkstra’s algorithm to find the shortest distance from vertex V1 to all other.
Applying Bellman-Ford algorithm
Step1:
Count the number of vertices in the graph:
Now as in the diagram we can see there are 7 vertices so for this problem every vertex will undergo maximum n-1 relaxation i.e. 7-1=6 relaxation.
Relaxation formula:
If(d[u]+c(u,v)<d[v ])
Then
d[v]= d[u]+c(u,v)
where d[u], d[v] is the weight quantity associated with vertex u and v, c(u,v) is the edge weight between vertex u and v.
Step 2:
Enlist all the edges
(1,2) (1,3) (1,4) (2,5) (3,2) (3,5) (4,3) (4,6) (5,7) (6,7)
Initially all vertices are marked as infinity so
V1 | V2 | V3 | V4 | V5 | V6 | V7 |
0 | ∞ | ∞ | ∞ | ∞ | ∞ | ∞ |
Take the first pair (1,2):
So d[u]=d[V1]=0
c(u,v)=c(V1,V2)= 6 and d[v]=d[V1]=âˆž
Applying relaxation formula
d[u]+c(u,v)<d[v]) i.e. d[V1]+c(V1,V2)<d[V2])
0+6<âˆž so the condition satisfies hence replace d[v2] with d[V1]+c(V1,V2) so the table is
V1 | V2 | V3 | V4 | V5 | V6 | V7 |
0 | 6 | ∞ | ∞ | ∞ | ∞ | ∞ |
Now repeat the process for all the vertex taking the above table as the reference for updated value.
V1 | V2 |