Instructions: All answers must be proved: it is not sucient to simply state the
ID: 3843277 • Letter: I
Question
Instructions: All answers must be proved: it is not sucient to simply state the answer. All answers must be written in your own words.
Treat single arithmetic operations (addition, subtraction, multiplication, and division) as constant time operations.
Problem: In the Grand Tour problem, you are given a list of n cities along with their pairwise distances (i.e., you know the distance between every two cities). All distances are positive and symmetric (the distance between A and B is equal to the distance between B and A). You begin at a specied city C0, and must perform a ‘tour’, in which you visit every other city exactly once, and then return to C0. The goal is to minimize the total distance traveled.
Design a reasonable greedy algorithm for solving this problem. Does it always nd the correct answer (i.e., the shortest tour)? If yes, prove that it is always correct, and if no, provide a counterexample.
Note: I am posting this question second time and I did not satisy from first one. Please answer me clearly. Thank You!
Explanation / Answer
What you are asking is the example of Dijkstra's algorithm.
Dijkstra's Algorithm --
In Dijkstra’s algorithm we generate a SPT (shortest path tree) with given source as root. We maintain two sets, one set contains vertices included in shortest path tree, other set includes vertices not yet included in shortest path tree. At every step of the algorithm, we find a vertex which is in the other set (set of not yet included) and has minimum distance from source.
Below are the detailed steps used in Dijkstra’s algorithm to find the shortest path from a single source vertex to all other vertices in the given graph.
Algorithm
1) Create a set sptSet (shortest path tree set) that keeps track of vertices included in shortest path tree, i.e., whose minimum distance from source is calculated and finalized. Initially, this set is empty.
2) Assign a distance value to all vertices in the input graph. Initialize all distance values as INFINITE. Assign distance value as 0 for the source vertex so that it is picked first.
3) While sptSet doesn’t include all vertices
….a) Pick a vertex u which is not there in sptSetand has minimum distance value.
….b) Include u to sptSet.
….c) Update distance value of all adjacent vertices of u. To update the distance values, iterate through all adjacent vertices. For every adjacent vertex v, if sum of distance value of u (from source) and weight of edge u-v, is less than the distance value of v, then update the distance value of v.
JAVA PROGRAM --
import java.util.*;
import java.lang.*;
import java.io.*;
class ShortestPath
{
static final int V=9;
int minDistance(int dist[], Boolean sptSet[])
{
int min = Integer.MAX_VALUE, min_index=-1;
for (int v = 0; v < V; v++)
if (sptSet[v] == false && dist[v] <= min)
{
min = dist[v];
min_index = v;
}
return min_index;
}
void printSolution(int dist[], int n)
{
System.out.println("Vertex Distance from Source");
for (int i = 0; i < V; i++)
System.out.println(i+" "+dist[i]);
}
void dijkstra(int graph[][], int src)
{
int dist[] = new int[V];
Boolean sptSet[] = new Boolean[V];
for (int i = 0; i < V; i++)
{
dist[i] = Integer.MAX_VALUE;
sptSet[i] = false;
}
dist[src] = 0;
for (int count = 0; count < V-1; count++)
{
int u = minDistance(dist, sptSet);
// Mark the picked vertex as processed
sptSet[u] = true;
for (int v = 0; v < V; v++)
if (!sptSet[v] && graph[u][v]!=0 &&
dist[u] != Integer.MAX_VALUE &&
dist[u]+graph[u][v] < dist[v])
dist[v] = dist[u] + graph[u][v];
}
printSolution(dist, V);
}
public static void main (String[] args)
{
int graph[][] = new int[][]{{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}
};
ShortestPath t = new ShortestPath();
t.dijkstra(graph, 0);
}
}