Writing Merge Sort Pseudo-Code Procedure in C - c

Question

I have been going through Introduction to Algorithms, and have been trying to implement the MERGE-SORT algorithm in C programming language to gain a better understanding of it.
The book presents two pseudo-codes:
and
While I do understand the above procedures, I must be missing something during the implementation.
I must be missing something from the pseudo-code but cannot figure it out yet. Any suggestions as to why this is happening would be appreciated.
EDIT: Updated Code and Output
/* C program for Merge Sort */
#include<stdlib.h>
#include<stdio.h>
void MERGE(int [], int , int , int );
void printArray(int [], int );
void MERGE_SORT(int [], int , int );
int main(void)
{
int A[] = { 12, 11, 13, 5, 6, 7, 2, 9 };
int arr_size = sizeof(A) / sizeof(A[0]);
printf("Given array is \n");
printArray(A, arr_size);
MERGE_SORT(A, 0, arr_size); //Fixed: Index to start from zero
printf("\nSorted array is \n");
printArray(A, arr_size);
return 0;
}
void MERGE(int A[], int p, int q, int r)
{
int i = 0;
int j = 0;
int n1 = q - p + 1; //Computing length of sub-array 1
int n2 = r - q; //Computing length of sub-array 2
int *L = malloc((n1 + 2) * sizeof(*L + 1)); //Creating Left array
int *R = malloc((n2 + 2) * sizeof(*R + 1)); //Creating Right array
for (int i = 0; i <= n1; i++) { //Fixed: <=, i start from 0
L[i] = A[p + i - 1];
}
for (int j = 0; j <= n2; j++) { //Fixed: <=, i start from 0
R[j] = A[q + j];
}
L[n1 + 1] = 99; //Placing Sentinel at the end of array
R[n2 + 1] = 99;
i = 1;
j = 1;
/*Prior to the first iteration k = p, so the subarray is empty.
Both L[i] and R[j] are the smallest elements of their arrays and have not
been copied back to A*/
for (int k = p; k <= r; k++) { //Fixed: <=
if (L[i] <= R[j]) {
A[k] = L[i];
i++;
}
else { //Fixed: Assignment and not condition check for A[k]
A[k] = R[j];
j++;
}
}
free(L);
free(R);
}
void MERGE_SORT(int A[], int p, int r)
{
//During first iteration p = 1 & r = 8
if (p < r) {
int q = (p + r) / 2;
MERGE_SORT(A, p, q);
MERGE_SORT(A, q + 1, r);
MERGE(A, p, q, r);
}
}
/* Function to print an array */
void printArray(int Arr[], int size)
{
int i;
for (i = 0; i < size; i++)
printf("%d ", Arr[i]);
printf("\n");
}

Answer 1

Looked in the pseudo code and found out that some things have been mistakenly written wrong.
1. You need to be careful with the array index to start from 0 or 1
2. Merge last part in for loop is actually an assignment instead for conditional check.
Edit: Have updated the code to fix for the error Stack around the variable A was corrupted
Please find the corrected code here(Lookout for //Fixed for fixes)
/* C program for Merge Sort */
#include<stdlib.h>
#include<stdio.h>
void MERGE(A, p, q, r);
void printArray(Arr, size);
void MERGE_SORT(A, p, r);
int main(void)
{
int A[] = { 12, 11, 13, 5, 6, 7, 2, 9 };
int arr_size = sizeof(A) / sizeof(A[0]);
printf("Given array is \n");
printArray(A, arr_size);
MERGE_SORT(A, 0, arr_size - 1); //Fixed: Index to start from zero, arr_size - 1
printf("\nSorted array is \n");
printArray(A, arr_size);
return 0;
}
void MERGE(int A[], int p, int q, int r)
{
int i = 0;
int j = 0;
int n1 = q - p + 1; //Computing length of sub-array 1
int n2 = r - q; //Computing length of sub-array 2
int *L = malloc((n1+1) * sizeof(*L+1)); //Creating Left array
int *R = malloc((n2+1) * sizeof(*R+1)); //Creating Right array
for (int i = 0; i < n1; i++) { //Fixed: i start from 0
L[i] = A[p + i];
}
// int arr_size = sizeof(A) / sizeof(A[0]);
for (int j = 0; j < n2; j++) { //Fixed: j start from 0
R[j] = A[q + j + 1];
}
L[n1] = 99; //Placing Sentinel at the end of array
R[n2] = 99;
i = 0; //Fixed: i and j to start from 0
j = 0;
/*Prior to the first iteration k = p, so the subarray is empty.
Both L[i] and R[j] are the smallest elements of their arrays and have not
been copied back to A*/
for (int k = p; k <= r; k++) { //Fixed: <=
if (L[i] <= R[j]) {
A[k] = L[i];
i++;
}
else { //Fixed: Assignment and not condition check for A[k]
A[k] = R[j];
j++;
}
}
free(L);
free(R);
}
void MERGE_SORT(int A[], int p, int r)
{
//During first iteration p = 1 & r = 8
if (p < r) {
int q = (p + r) / 2;
MERGE_SORT(A, p, q);
MERGE_SORT(A, q + 1, r);
MERGE(A, p, q, r);
}
}
/* Function to print an array */
void printArray(int Arr[], int size)
{
int i;
for (i = 0; i < size; i++)
printf("%d ", Arr[i]);
printf("\n", size);
}
Hope it helps.
Revert for any doubts.

Answer 2

here are some changes i have done to your code `
#include<stdlib.h>
#include<stdio.h>
void MERGE(int *A,int p,int q,int r);
void printArray(int *Arr,int size);
void MERGE_SORT(int *A,int p,int r);
int main(void){
int A[] = { 12, 11, 13, 5, 6, 7, 2, 9 };
int arr_size = sizeof(A) / sizeof(A[0]);
printf("Given array is \n");
printArray(A, arr_size);
MERGE_SORT(A, 0, arr_size -1); // pass the indices of the array
printf("\nSorted array is \n");
printArray(A, arr_size);
return 0;
}
void MERGE(int A[], int p, int q, int r){
int i = 0;
int j = 0;
int k; //declair it here
int n1 = q - p + 1; //Computing length of sub-array 1
int n2 = r - q; //Computing length of sub-array 2
int *L = malloc((n1) * sizeof(*L+1)); //Creating Left array
int *R = malloc((n2) * sizeof(*R+1)); //Creating Right array
for (int i = 0; i < n1; i++) { //start coping from zero
L[i] = A[p + i];
}
for (int j = 0; j < n2; j++) {
R[j] = A[q +1 + j];
}
// L[n1] = 99; we won't be needing these as to mark the end we already know the size of arrays
// R[n2] = 99;
// i = 1;
// j = 1;
/*Prior to the first iteration k = p, so the subarray is empty.
Both L[i] and R[j] are the smallest elements of their arrays and have not
been copied back to A*/
for (k = p; k < r+1 && i < n1 && j<n2; k++) {
//i & j checks weather the array has completed or not
if (L[i] <= R[j]) {
A[k] = L[i];
i++;
}
else {
A[k]=R[j];
j++;
}
}
// when one of the array is empty u can copy the rest of the array with out compairing
while(i<n1)
A[k++]=L[i++];
while(j<n2)
A[k++]=R[j++];
free(L);
free(R);
}
void MERGE_SORT(int A[], int p, int r)
{
//During first iteration p = 1 & r = 8
if (p < r) {
int q = (p + r) / 2;
MERGE_SORT(A, p, q);
MERGE_SORT(A, q + 1, r);
MERGE(A, p, q, r);
}
}
/* Function to print an array */
void printArray(int Arr[], int size){
int i;
for (i = 0; i < size; i++)
printf("%d ", Arr[i]);
printf("\n");
}`
first of all you were not passing the right parameters to the function.
then the concept of using infinity to indicate is not good as one can want to sort bigger number than that in that case you would have to increase infinity an alternative approach is given above.
Mean while i also solved the problem with your code here it was again with the array index were not rightly used check it out now its working:`
#include<stdlib.h>
#include<stdio.h>
void MERGE(A, p, q, r);
void printArray(Arr, size);
void MERGE_SORT(A, p, r);
int main(void)
{
int A[] = { 12, 11, 13, 5, 6, 7, 2, 9 };
int arr_size = sizeof(A) / sizeof(A[0]);
printf("Given array is \n");
printArray(A, arr_size);
MERGE_SORT(A, 1, arr_size);
printf("\nSorted array is \n");
printArray(A, arr_size);
return 0;
}
void MERGE(int A[], int p, int q, int r)
{
int i = 0;
int j = 0;
int n1 = q - p + 1; //Computing length of sub-array 1
int n2 = r - q; //Computing length of sub-array 2
int *L = malloc((n1+1) * sizeof(*L+1)); //Creating Left array
int *R = malloc((n2+1) * sizeof(*R+1)); //Creating Right array
for (int i = 1; i < n1; i++) {
L[i] = A[p + i - 1];
}
for (int j = 1; j < n2; j++) {
R[j] = A[q + j];
}
L[n1] = 99; //Placing Sentinel at the end of array
R[n2] = 99;
i = 1;
j = 1;
/*Prior to the first iteration k = p, so the subarray is empty.
Both L[i] and R[j] are the smallest elements of their arrays and have not
been copied back to A*/
for (int k = p; k < r; k++) {
if (L[i] <= R[j]) {
A[k] = L[i];
i++;
}
else if (A[k] == L[i])
j++;
}
free(L);
free(R);
}
void MERGE_SORT(int A[], int p, int r)
{
//During first iteration p = 1 & r = 8
if (p < r) {
int q = (p + r) / 2;
MERGE_SORT(A, p, q);
MERGE_SORT(A, q + 1, r);
MERGE(A, p, q, r);
}
}
/* Function to print an array */
void printArray(int Arr[], int size)
{
int i;
for (i = 0; i < size; i++)
printf("%d ", Arr[i]);
printf("\n");
}