3b. Warshal's algorithm.

 #include<stdio.h>

    #include<math.h>

    int max(int, int);

    void warshal(int p[10][10], int n) {

        int i, j, k;

        for (k = 1; k <= n; k++)

            for (i = 1; i <= n; i++)

                for (j = 1; j <= n; j++)

                    p[i][j] = max(p[i][j], p[i][k] && p[k][j]);

    }

    int max(int a, int b) {

        ;

        if (a > b)

            return (a);

        else

            return (b);

    }

    void main() {

        int p[10][10] = { 0 }, n, e, u, v, i, j;

        printf("\n Enter the number of vertices:");

        scanf("%d", &n);

        printf("\n Enter the number of edges:");

        scanf("%d", &e);

        for (i = 1; i <= e; i++) {

            printf("\n Enter the end vertices of edge %d:", i);

            scanf("%d%d", &u, &v);

            p[u][v] = 1;

        }

        printf("\n Matrix of input data: \n");

        for (i = 1; i <= n; i++) {

            for (j = 1; j <= n; j++)

                printf("%d\t", p[i][j]);

            printf("\n");

        }

        warshal(p, n);

        printf("\n Transitive closure: \n");

        for (i = 1; i <= n; i++) {

            for (j = 1; j <= n; j++)

                printf("%d\t", p[i][j]);

            printf("\n");

        }

      

    }

OUTPUT:

 Enter the number of vertices:5

 Enter the number of edges:11

 Enter the end vertices of edge 1:1 1

 Enter the end vertices of edge 2:1 4

 Enter the end vertices of edge 3:3 2

 Enter the end vertices of edge 4:3 3

 Enter the end vertices of edge 5:3 4

 Enter the end vertices of edge 6:4 2

 Enter the end vertices of edge 7:4 4

 Enter the end vertices of edge 8:5 2

 Enter the end vertices of edge 9:5 3

 Enter the end vertices of edge 10:5 4

 Enter the end vertices of edge 11:5 5