Simulation Border Gateway protocol Routing Algorithm CS1305- NETWORK LAB

Step by step BGP Algorithm
The number of nodes is obtained and stored in variable ‘n’
The distance between all nodes is accepted in the 4 two dimensional array ‘a’
It is values are printed back in the form of matrix
The shortest pat between any 2 nodes is found using the expression a[i][j]=a[i][k]+a[k][j]
The distance between nodes to itself is made ‘0’
The output Matrix is printed back to the user
Finally terminate Border Gateway protocol program

BGP Source code in C language programming CS1305- NETWORK LAB
#include<stdio.h>
main()
{
int i,n,j,k,a[10][10],b[10][10];
printf("ENTER THE NO OF NODES");
scanf("%d",&n);
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
printf("ENTER THE DISTANCE BETWEEN HOSTS %d,%d :",i+1,j+1);
scanf("%d",&a[i][j]);
}
}
printf("THE GIVEN INPUT\n");
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
printf("%d\t",a[i][j]);
}
printf("\n");
}
for(k=0;k<n;k++)
{
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
if(a[i][j]>a[i][k]+a[k][j])
{
a[i][j]=a[i][k]+a[k][j];
}
}
}
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
b[i][j]=a[i][j];
if(i==j)
b[i][j]=0;
}
}
printf("OUTPUT MATRIX \n");
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
printf("%d\t",b[i][j]);
}
printf("\n");
}
}

Output CS1305- NETWORK LAB
[it28@localhost studentwebsite]$cc protocol.c
[it28@localhost studentwebsite]$./a.out
Enter the no of nodes: 2
ENTER THE DISTANCE BETWEEN HOSTS 1 1:1
ENTER THE DISTANCE BETWEEN HOSTS 1 2:2
ENTER THE DISTANCE BETWEEN HOSTS 2 1:3
ENTER THE DISTANCE BETWEEN HOSTS 2 2:4
The given output
1 2
3 4
Output Matrix
0 2
3 0
Result
The shortest path between any two given nodes was found using BGP routing in CS1305- NETWORK LAB successfully.

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