Graph Algorithms and Sorting Techniques in C
1. Kruskal’s Algorithm
This C code implements Kruskal’s algorithm to find the minimum spanning tree of a graph:
#include<stdio.h>
#define INF 999
#define MAX 100
int p[MAX], c[MAX][MAX], t[MAX][2];
int find(int v) {
while (p[v]) {
v = p[v];
}
return v;
}
void union1(int i, int j) {
p[j] = i;
}
void kruskal(int n) {
int i, j, k, u, v, min, res1, res2, sum = 0;
for (k = 1; k < n; k++) {
min = INF;
for (i = 1; i < n - 1; i++) {
for (j = 1; j <= n; j++) {
if (i == j) continue;
if (c[i][j] < min) {
u = find(i);
v = find(j);
if (u != v) {
res1 = i;
res2 = j;
min = c[i][j];
}
}
}
}
union1(res1, find(res2));
t[k][1] = res1;
t[k][2] = res2;
sum = sum + min;
}
printf("\nCost of spanning tree is=%d", sum);
printf("\nEdges of spanning tree are:\n");
for (i = 1; i < n; i++)
printf("%d -> %d\n", t[i][1], t[i][2]);
}
int main() {
int i, j, n;
printf("\nEnter the n value:");
scanf("%d", &n);
for (i = 1; i <= n; i++)
p[i] = 0;
printf("\nEnter the graph data:\n");
for (i = 1; i <= n; i++)
for (j = 1; j <= n; j++)
scanf("%d", &c[i][j]);
kruskal(n);
return 0;
}
2. Prim’s Algorithm
This C code implements Prim’s algorithm to find the minimum spanning tree of a graph:
#include<stdio.h>
#define INF 999
int prim(int c[10][10], int n, int s) {
int v[10], i, j, sum = 0, ver[10], d[10], min, u;
for (i = 1; i <= n; i++) {
ver[i] = s;
d[i] = c[s][i];
v[i] = 0;
}
v[s] = 1;
for (i = 1; i <= n - 1; i++) {
min = INF;
for (j = 1; j <= n; j++) {
if (v[j] == 0 && d[j] < min) {
min = d[j];
u = j;
}
}
v[u] = 1;
sum = sum + d[u];
printf("\n%d -> %d sum=%d", ver[u], u, sum);
for (j = 1; j <= n; j++) {
if (v[j] == 0 && c[u][j] < d[j]) {
d[j] = c[u][j];
ver[j] = u;
}
}
}
return sum;
}
void main() {
int c[10][10], i, j, res, s, n;
printf("\nEnter n value:");
scanf("%d", &n);
printf("\nEnter the graph data:\n");
for (i = 1; i <= n; i++)
for (j = 1; j <= n; j++)
scanf("%d", &c[i][j]);
printf("\nEnter the source node:");
scanf("%d", &s);
res = prim(c, n, s);
printf("\nCost=%d", res);
}
3A. Floyd-Warshall Algorithm
This C code implements the Floyd-Warshall algorithm to find the shortest paths between all pairs of vertices in a graph:
#include<stdio.h>
#define INF 999
int min(int a, int b) {
return (a < b) ? a : b;
}
void floyd(int p[][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] = min(p[i][j], p[i][k] + p[k][j]);
}
int main() {
int a[10][10], n, i, j;
printf("\nEnter the n value:");
scanf("%d", &n);
printf("\nEnter the graph data:\n");
for (i = 1; i <= n; i++)
for (j = 1; j <= n; j++)
scanf("%d", &a[i][j]);
floyd(a, n);
printf("\nShortest path matrix\n");
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++)
printf("%d ", a[i][j]);
printf("\n");
}
return 0;
}
3B. Warshall’s Algorithm
This C code implements Warshall’s algorithm to find the transitive closure of a graph:
#include<stdio.h>
void warsh(int p[][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] = p[i][j] || (p[i][k] && p[k][j]);
}
int main() {
int a[10][10], n, i, j;
printf("\nEnter the n value:");
scanf("%d", &n);
printf("\nEnter the graph data:\n");
for (i = 1; i <= n; i++)
for (j = 1; j <= n; j++)
scanf("%d", &a[i][j]);
warsh(a, n);
printf("\nResultant path matrix\n");
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++)
printf("%d ", a[i][j]);
printf("\n");
}
return 0;
}
4. Dijkstra’s Algorithm
This C code implements Dijkstra’s algorithm to find the shortest paths from a source vertex to all other vertices in a graph:
#include<stdio.h>
#define INF 999
void dijkstra(int c[10][10], int n, int s, int d[10]) {
int v[10], min, u, i, j;
for (i = 1; i <= n; i++) {
d[i] = c[s][i];
v[i] = 0;
}
v[s] = 1;
for (i = 1; i <= n; i++) {
min = INF;
for (j = 1; j <= n; j++) {
if (v[j] == 0 && d[j] < min) {
min = d[j];
u = j;
}
}
v[u] = 1;
for (j = 1; j <= n; j++) {
if (v[j] == 0 && (d[u] + c[u][j]) < d[j]) {
d[j] = d[u] + c[u][j];
}
}
}
}
int main() {
int c[10][10], d[10], i, j, s, n;
printf("\nEnter n value:");
scanf("%d", &n);
printf("\nEnter the graph data:\n");
for (i = 1; i <= n; i++)
for (j = 1; j <= n; j++)
scanf("%d", &c[i][j]);
printf("\nEnter the source node:");
scanf("%d", &s);
dijkstra(c, n, s, d);
for (i = 1; i <= n; i++)
printf("\nShortest distance from %d to %d is %d", s, i, d[i]);
return 0;
}
5. Topological Sort
This C code implements topological sort to find a linear ordering of vertices in a directed acyclic graph (DAG) such that for every directed edge (u, v), vertex u comes before vertex v in the ordering:
#include<stdio.h>
int temp[10], k = 0;
void sort(int a[][10], int id[], int n) {
int i, j;
for (i = 1; i <= n; i++) {
if (id[i] == 0) {
id[i] = -1;
temp[++k] = i;
for (j = 1; j <= n; j++) {
if (a[i][j] == 1 && id[j] != -1) {
id[j]--;
}
}
i = 0;
}
}
}
void main() {
int a[10][10], id[10], n, i, j;
printf("\nEnter the n value:");
scanf("%d", &n);
for (i = 1; i <= n; i++)
id[i] = 0;
printf("\nEnter the graph data:\n");
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++) {
scanf("%d", &a[i][j]);
if (a[i][j] == 1) {
id[j]++;
}
}
}
sort(a, id, n);
if (k != n) {
printf("\nTopological ordering not possible");
} else {
printf("\nTopological ordering is:");
for (i = 1; i <= k; i++)
printf("%d ", temp[i]);
}
}
6. 0/1 Knapsack Problem (Recursive)
This C code solves the 0/1 knapsack problem using recursion. Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible:
#include<stdio.h>
int w[10], p[10], n;
int max(int a, int b) {
return a > b ? a : b;
}
int knap(int i, int m) {
if (i == n)
return w[i] > m ? 0 : p[i];
if (w[i] > m)
return knap(i + 1, m);
return max(knap(i + 1, m), knap(i + 1, m - w[i]) + p[i]);
}
int main() {
int m, i, max_profit;
printf("\nEnter the no. of objects:");
scanf("%d", &n);
printf("\nEnter the knapsack capacity:");
scanf("%d", &m);
printf("\nEnter profit followed by weight:\n");
for (i = 1; i <= n; i++)
scanf("%d %d", &p[i], &w[i]);
max_profit = knap(1, m);
printf("\nMax profit=%d", max_profit);
return 0;
}
7. Fractional Knapsack Problem (Greedy)
This C code solves the fractional knapsack problem using a greedy approach:
#include <stdio.h>
#define MAX 50
int p[MAX], w[MAX], x[MAX];
double maxprofit;
int n, m, i;
void greedyKnapsack(int n, int w[], int p[], int m) {
double ratio[MAX];
for (i = 0; i < n; i++) {
ratio[i] = (double)p[i] / w[i];
}
for (i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
if (ratio[i] < ratio[j]) {
double temp = ratio[i];
ratio[i] = ratio[j];
ratio[j] = temp;
int temp2 = w[i];
w[i] = w[j];
w[j] = temp2;
temp2 = p[i];
p[i] = p[j];
p[j] = temp2;
}
}
}
int currentWeight = 0;
maxprofit = 0.0;
for (i = 0; i < n; i++) {
if (currentWeight + w[i] <= m) {
x[i] = 1;
currentWeight += w[i];
maxprofit += p[i];
} else {
x[i] = (m - currentWeight) / (double)w[i];
maxprofit += x[i] * p[i];
break;
}
}
printf("Optimal solution for greedy method: %.1f\n", maxprofit);
printf("Solution vector for greedy method: ");
for (i = 0; i < n; i++)
printf("%d\t", x[i]);
}
int main() {
printf("Enter the number of objects: ");
scanf("%d", &n);
printf("Enter the objects' weights: ");
for (i = 0; i < n; i++)
scanf("%d", &w[i]);
printf("Enter the objects' profits: ");
for (i = 0; i < n; i++)
scanf("%d", &p[i]);
printf("Enter the maximum capacity: ");
scanf("%d", &m);
greedyKnapsack(n, w, p, m);
return 0;
}
8. Sum of Subsets Problem
This C code solves the sum of subsets problem using backtracking. Given a set of integers and a target sum, determine if there is a subset of the given set with sum equal to the given sum:
#include<stdio.h>
#define MAX 10
int s[MAX], x[MAX], d;
void sumofsub(int p, int k, int r) {
int i;
x[k] = 1;
if ((p + s[k]) == d) {
for (i = 1; i <= k; i++)
if (x[i] == 1)
printf("%d ", s[i]);
printf("\n");
} else if (p + s[k] + s[k + 1] <= d) {
sumofsub(p + s[k], k + 1, r - s[k]);
}
if ((p + r - s[k] >= d) && (p + s[k + 1] <= d)) {
x[k] = 0;
sumofsub(p, k + 1, r - s[k]);
}
}
int main() {
int i, n, sum = 0;
printf("\nEnter the n value:");
scanf("%d", &n);
printf("\nEnter the set in increasing order:");
for (i = 1; i <= n; i++)
scanf("%d", &s[i]);
printf("\nEnter the max subset value:");
scanf("%d", &d);
for (i = 1; i <= n; i++)
sum = sum + s[i];
if (sum < d || s[1] > d) {
printf("\nNo subset possible");
} else {
sumofsub(0, 1, sum);
}
return 0;
}
9. Selection Sort and Time Complexity
This C code implements selection sort and measures its execution time. It also includes Python code to visualize the time complexity:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
void selectionSort(int arr[], int n) {
int i, j, min_idx;
for (i = 0; i < n - 1; i++) {
min_idx = i;
for (j = i + 1; j < n; j++) {
if (arr[j] < arr[min_idx]) {
min_idx = j;
}
}
int temp = arr[min_idx];
arr[min_idx] = arr[i];
arr[i] = temp;
}
}
void generateRandomNumbers(int arr[], int n) {
for (int i = 0; i < n; i++) {
arr[i] = rand() % 10000;
}
}
int main() {
int n;
printf("Enter number of elements: ");
scanf("%d", &n);
if (n <= 5000) {
printf("Please enter a value greater than 5000\n");
return 1;
}
int *arr = (int *)malloc(n * sizeof(int));
if (arr == NULL) {
printf("Memory allocation failed\n");
return 1;
}
generateRandomNumbers(arr, n);
clock_t start = clock();
selectionSort(arr, n);
clock_t end = clock();
double time_taken = ((double)(end - start)) / CLOCKS_PER_SEC;
printf("Time taken to sort %d elements: %f seconds\n", n, time_taken);
free(arr);
return 0;
}
Python code for visualization:
import matplotlib.pyplot as plt
n_values = [6000, 7000, 8000, 9000, 10000]
time_taken = [0.031000, 0.034000, 0.047000, 0.052000, 0.077000] # Replace with actual times recorded
plt.plot(n_values, time_taken, marker='o')
plt.title('Selection Sort Time Complexity')
plt.xlabel('Number of Elements (n)')
plt.ylabel('Time taken (seconds)')
plt.grid(True)
plt.show()
10. Quick Sort and Time Complexity
This C code implements quick sort and measures its execution time. It also includes Python code to visualize the time complexity:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
void swap(int *a, int *b) {
int t = *a;
*a = *b;
*b = t;
}
int partition(int arr[], int low, int high) {
int pivot = arr[high];
int i = (low - 1);
for (int j = low; j <= high - 1; j++) {
if (arr[j] < pivot) {
i++;
swap(&arr[i], &arr[j]);
}
}
swap(&arr[i + 1], &arr[high]);
return (i + 1);
}
void quickSort(int arr[], int low, int high) {
if (low < high) {
int pi = partition(arr, low, high);
quickSort(arr, low, pi - 1);
quickSort(arr, pi + 1, high);
}
}
void generateRandomNumbers(int arr[], int n) {
for (int i = 0; i < n; i++) {
arr[i] = rand() % 100000;
}
}
int main() {
int n;
printf("Enter number of elements: ");
if (scanf("%d", &n) != 1) {
printf("Invalid input\n");
return 1;
}
if (n <= 5000) {
printf("Please enter a value greater than 5000\n");
return 1;
}
int *arr = (int *)malloc(n * sizeof(int));
if (arr == NULL) {
printf("Memory allocation failed\n");
return 1;
}
srand(time(0));
generateRandomNumbers(arr, n);
clock_t start = clock();
quickSort(arr, 0, n - 1);
clock_t end = clock();
double time_taken = ((double)(end - start)) / CLOCKS_PER_SEC;
printf("Time taken to sort %d elements: %f seconds\n", n, time_taken);
free(arr);
return 0;
}
Python code for visualization:
import matplotlib.pyplot as plt
n_values = [10000, 20000, 30000, 35000, 50000]
time_taken = [0.0000, 0.015000, 0.011000, 0.003000, 0.015000] # Replace with actual times recorded
plt.plot(n_values, time_taken, marker='o')
plt.title('Quick Sort Time Complexity')
plt.xlabel('Number of Elements (n)')
plt.ylabel('Time taken (seconds)')
plt.grid(True)
plt.show()
11. Merge Sort and Time Complexity
This C code implements merge sort and measures its execution time. It also includes Python code to visualize the time complexity:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
void merge(int arr[], int left, int mid, int right) {
int i, j, k;
int n1 = mid - left + 1;
int n2 = right - mid;
int *L = (int *)malloc(n1 * sizeof(int));
int *R = (int *)malloc(n2 * sizeof(int));
for (i = 0; i < n1; i++)
L[i] = arr[left + i];
for (j = 0; j < n2; j++)
R[j] = arr[mid + 1 + j];
i = 0;
j = 0;
k = left;
while (i < n1 && j < n2) {
if (L[i] <= R[j]) {
arr[k] = L[i];
i++;
} else {
arr[k] = R[j];
j++;
}
k++;
}
while (i < n1) {
arr[k] = L[i];
i++;
k++;
}
while (j < n2) {
arr[k] = R[j];
j++;
k++;
}
free(L);
free(R);
}
void mergeSort(int arr[], int left, int right) {
if (left < right) {
int mid = left + (right - left) / 2;
mergeSort(arr, left, mid);
mergeSort(arr, mid + 1, right);
merge(arr, left, mid, right);
}
}
void generateRandomArray(int arr[], int n) {
for (int i = 0; i < n; i++)
arr[i] = rand() % 100000;
}
int main() {
int n;
printf("Enter the number of elements: ");
scanf("%d", &n);
if (n <= 5000) {
printf("Please enter a value greater than 5000\n");
return 1;
}
int *arr = (int *)malloc(n * sizeof(int));
if (arr == NULL) {
printf("Memory allocation failed\n");
return 1;
}
generateRandomArray(arr, n);
clock_t start = clock();
for (int i = 0; i < 1000; i++) {
mergeSort(arr, 0, n - 1);
}
clock_t end = clock();
double time_taken = ((double)(end - start)) / CLOCKS_PER_SEC / 1000.0;
printf("Time taken to sort %d elements: %f seconds\n", n, time_taken);
free(arr);
return 0;
}
Python code for visualization:
import matplotlib.pyplot as plt
n_values = [6000, 7000, 8000, 9000, 10000, 11000, 12000, 13000, 15000]
time_taken = [0.000709, 0.000752, 0.000916, 0.001493, 0.001589, 0.002562, 0.001944, 0.002961, 0.003563] # Replace with actual times recorded
plt.plot(n_values, time_taken, marker='o')
plt.title('Merge Sort Time Complexity')
plt.xlabel('Number of Elements (n)')
plt.ylabel('Time taken (seconds)')
plt.grid(True)
plt.show()