Question

In C++, please build a simple Distance Vector program that will communicate with N partners and do the following. (NOTE: All connections between a server and clients should be TCP/IP socket.)

1. You will run N instances on your machine
2. Each instance will run on a different port (instance 1 will run on port 18181, instance 2 on port 18182, instance 3 on port 18183, etc)
3. The program will start by reading in the appropriate neighbors file and vector file.
4. The program will ONLY read vectors where the fromNode is equal to that node. For instance, node 1 will only read in vectors where node1 is the fromNode.
5. The program will support the following requests from clients:
   1. Show files at the server’s current directory: ‘print’
      1. This will print the current vector table on a node.
      2. ONLY print valid vectors (don’t print uninitialized ones)
   2. Download files: ‘refresh’
      1. This will force the current node to send its vector table to all its neighbors.
   3. Upload files: ‘update’

1. This will be of the form: Update fromNode toNode cost
2. Update checks that the toNode is the current node. If it is not it is ignored
3. Update then checks if the toNode-fromNode exists in vector table. If it does not it is added and neighbors are notified
4. If the pair is in the table, and if the new cost is less than the old cost, that cost is stored and the neighbors are notified.

Answer 1

#include<stdio.h>

#include<iostream.h>

Using namespace std;

Struct node

{

unsigned dist[6];

unsigned from[6];

}

DVR[10];

int main()

{

cout<<"\n\n-------------------- Distance Vector Routing Algorithm----------- ";

int costmat[6][6];

int nodes, i, j, k;

cout<<"\n\n Enter the number of nodes : ";

cin>>nodes; //Enter the nodes

cout<<"\n Enter the cost matrix : \n" ;

for(i = 0; i < nodes; i++)

{

for(j = 0; j < nodes; j++)

{

cin>>costmat[i][j];

costmat[i][i] = 0;

DVR[i].dist[j] = costmat[i][j]; //initialise the distance equal to cost matrix

DVR[i].from[j] = j;

}

}

for(i = 0; i < nodes; i++)

for(j = i+1; j < nodes; j++)

for(k = 0; k < nodes; k++)

if(DVR[i].dist[j] > costmat[i][k] + DVR[k].dist[j])

{ //We calculate the minimum distance

DVR[i].dist[j] = DVR[i].dist[k] + DVR[k].dist[j];

DVR[j].dist[i] = DVR[i].dist[j];

DVR[i].from[j] = k;

DVR[j].from[i] = k;

}

for(i = 0; i < nodes; i++)

{

cout<<"\n\n For router: "<<i+1;

for(j = 0; j < nodes; j++)

cout<<"\t\n node "<<j+1<<" via "<<DVR[i].from[j]+1<<" Distance "<<DVR[i].dist[j];

}

cout<<" \n\n ";

return 0;

}