# Floyd's algorithm source code in C

## Algorithm

The Floyd–Warshall algorithm compares all possible paths through the graph between each pair of vertices. It is able to do this with only Θ(|*V*|^{3}) comparisons in a graph. This is remarkable considering that there may be up to Ω(|*V*|^{2}) edges in the graph, and every combination of edges is tested. It does so by incrementally improving an estimate on the shortest path between two vertices, until the estimate is optimal.

Consider a graph *G* with vertices *V*, each numbered 1 through *N*. Further consider a function shortestPath(*i*, *j*, *k*) that returns the shortest possible path from *i* to *j* using vertices only from the set {1,2,...,*k*} as intermediate points along the way. Now, given this function, our goal is to find the shortest path from each *i* to each *j* using only vertices 1 to *k* + 1.

There are two candidates for each of these paths: either the true shortest path only uses vertices in the set {1, ..., *k*}; or there exists some path that goes from *i* to *k* + 1, then from *k* + 1 to *j* that is better. We know that the best path from *i* to *j* that only uses vertices 1 through *k* is defined by shortestPath(*i*, *j*, *k*), and it is clear that if there were a better path from *i* to *k* + 1 to *j*, then the length of this path would be the concatenation of the shortest path from *i* to *k* + 1 (using vertices in {1, ..., *k*}) and the shortest path from *k* + 1 to *j*(also using vertices in {1, ..., *k*}).

If *w*(*i*,*j*) is the weight of the edge between vertices *i* and *j*, we can define shortestPath(*i*, *j*, *k*) in terms of the following recursive formula: the base case is

*i*,

*j*,0) =

*w*(

*i*,

*j*)

and the recursive case is

This formula is the heart of the Floyd–Warshall algorithm. The algorithm works by first computing shortestPath(*i*, *j*, *k*) for all (*i*, *j*) pairs for *k* = 1, then *k* = 2, etc. This process continues until *k* = *n*, and we have found the shortest path for all (*i*, *j*) pairs using any intermediate vertices.

## Floyd's Algorithm

/*C PROGRAM TO IMPLEMENT ALL PAIR SHORTEST PATH USING FLOYDS ALGORITHM INPUT: N VALUE FOR NUMBER OF VERTICES OUTPUT: ADJACENCY MATRIX SHORTEST DISTANCE MATRIX */ #include<stdio.h> #include<stdlib.h> void AdjacencyMatrix(int a[][100],int n){ //To generate adjacency matrix for given nodes int i,j; for(i = 0;i < n; i++) { for(j = 0;j <= n; j++) { a[i][j] = 0; } } for(i = 1; i < n; i++) { for(j=0;j<i;j++) { a[i][j] = rand()%10; a[j][i] = 99; } } } int min(int a,int b){ if(a < b) return a; else return b; } void floyds(int a[][100],int n){ int i,j,k; for(k = 0;k < n ; k++) { for(i = 0;i < n; i++) { for(j = 0;j < n ; j++) { a[i][j] = min (a[i][j], a[i][k] + a[k][j] ); } } } } int main() { int a[100][100],n,i,j; FILE *fp = fopen("floyds.dot","w"); /* FOR DIGRAPH */ fprintf(fp,"digraph A {\n"); /* WRITE TO FILE */ printf("Enter the vertices of the digraph\n"); scanf("%d",&n); AdjacencyMatrix(a,n); printf("\t\tAdjacency Matrix of the graph\n"); /* PRINT ADJACENCY MATRIX */ for(i = 0;i < n; i++) { for(j = 0;j < n; j++) { printf("\t%d",a[i][j]); if(a[i][j] != 0) { fprintf(fp,"%d -> %d\n",i,j); } } printf("\n"); } floyds(a,n); printf("\n Shortest distance matrix\n"); /*PRINT SHORTEST DISTANCE MATRIX*/ for(i = 0;i < n; i++) { for(j = 0;j < n; j++) { printf("\t%d",a[i][j]); } printf("\n"); } fprintf(fp,"}\n\n"); fclose(fp); return 0; }

## Comments

good.... helps in lab

Concerning floyds(int a[][100],int n).

What does 'a' and represent and what does each of the two dimensions of a represent?

What does 'n' represent?

I have a list of locations, with a list of connections between those locations and the distance associated with those connections. I need to find shortest path for any given two locations - but need to understand how to apply "floyds(int a[][100],int n)" to it.

i wanted code for floyd algorithm using openGL