Revision as of 20:41, 13 May 2024 by Admin (Created page with "One possible solution , after reading online descriptions of recursive merge sort, e.g. Dasgupta : <syntaxhighlight lang="C"> // to compile: gcc -Wall rmergesort.c -lm -o rmergesort /* ============================================================================ Name : rmergesort.c Author : Anon Version : 0.1 Copyright : (C)2013 under CC-By-SA 3.0 License Description : Recursive Merge Sort, Ansi-style =============================================...")
Exercise
ABy Admin
May 13'24
Answer
One possible solution , after reading online descriptions of recursive merge sort, e.g. Dasgupta :
// to compile: gcc -Wall rmergesort.c -lm -o rmergesort
/*
============================================================================
Name : rmergesort.c
Author : Anon
Version : 0.1
Copyright : (C)2013 under CC-By-SA 3.0 License
Description : Recursive Merge Sort, Ansi-style
============================================================================
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
//const int MAX = 200;
const int MAX = 20000000;
int *b;
int printOff = 0;
// this debugging print out of the array helps to show
// what is going on.
void printArray(char* label, int* a, int sz) {
int h = sz/ 2;
int i;
if (printOff) return;
printf("\n%s:\n", label);
for (i = 0; i < h; ++i ) {
printf("%d%c", a[i],
// add in a newline every 20 numbers
( ( i != 0 && i % 20 == 0 )? '\n': ' ' ) );
}
printf(" | ");
for (;i < sz; ++i) {
printf("%d%c", a[i],
( ( i != 0 && i % 20 == 0 )? '\n': ' ' ) );
}
putchar('\n');
}
void mergesort(int* a, int m ) {
printArray("BEFORE", a, m);
if (m > 2) {
// if greater than 2 elements, then recursive
mergesort(a, m / 2);
mergesort(a + m / 2, m - m / 2);
} else if (m == 2 && a[0] > a[1]) {
// if exactly 2 elements and need swapping, swap
int t = a[1];
a[1] = a[0];
a[0] = t;
goto end;
}
// 1 or greater than 2 elements which have been recursively sorted
// divide the array into a left and right array
// merge into the array b with index l.
int n = m/2;
int o = m - n;
int i = 0, j = n;
int l = 0;
// i is left, j is right ;
// l should equal m the size of the array
while (i < n) {
if ( j >= m) {
// the right array has finished, so copy the remaining left array
for(; i < n; ++i) {
b[l++] = a[i];
}
// the merge operation is to copy the smaller current element and
// increment the index of the parent sub-array.
} else if( a[i] < a[j] ) {
b[l++] = a[i++];
} else {
b[l++] = a[j++];
}
}
while ( j < m) {
// copy the remaining right array , if any
b[l++] = a[j++];
}
memcpy(a, b, sizeof(int) * l );
end:
printArray("AFTER", a, m);
return;
}
void rand_init(int* a, int n) {
int i;
for (i = 0; i < n; ++i ) {
a[i] = rand() % MAX;
}
}
int main(void) {
puts("!!!Hello World!!!"); /* prints !!!Hello World!!! */
// int N = 20;
// int N = 1000;
// int N = 1000000;
int N = 100000000; // still can't make a stack overflow on ubuntu,4GB, phenom
printOff = 1;
int *a;
b = calloc( N, sizeof(int));
a = calloc( N, sizeof(int));
rand_init(a, N);
mergesort(a, N);
printOff = 0;
printArray("LAST", a, N);
free(a);
free(b);
return EXIT_SUCCESS;
}
/* Having failed to translate my concept of non-recursive merge sort,
* I tackled the easier case of recursive merge sort.
* The next task is to translate the recursive version to a non-recursive
* version. This could be done by replacing calls to mergesort, with
* pushes onto a stack of
* tuples of ( <array start address>, <number of elements to process> )
*/
/* The central idea of merging, is that two sorted lists can be
* merged into one sorted list, by comparing the top of each list and
* moving the lowest valued element onto the end of the new list.
* The other list which has the higher valued element keeps its top
* element unchanged. When a list is exhausted, copy the remaining other list
* onto the end of the new list.
*/
/* The recursive part, is to defer any work in sorting an unsorted list,
* by dividing it into two lists until there is only 1 or two elements,
* and if there are two elements, sort them directly by swapping if
* the first element is larger than the second element.
*
* After returning from a recursive call, merge the lists, which will
* begin with one or two element sorted lists. The result is a sorted list
* which will be returned to the parent of the recursive call, and can
* be used for merging.
*/
/* The following is an imaginary discussion about what a programmer
* might be thinking about when programming:
*
* Visualising recursion in terms of a Z80 assembly language, which
* is similar to most assembly languages, there is a data stack (DS) and
* a call stack (CS) pointer, and each recursive call to mergesort
* pushes the return address , which is the program address of the instruction
* after the call , onto the stack pointed to by CS and CS is incremented,
* and the address of the array start and integer which is the subarray length
* onto the data stack pointed to by DS, which will be incremented twice.
*
* If the number of recursive , active calls exceed the allowable space for either the call stack
* or the data stack, then the program will crash , or a process space protection
* violation interrupt signal will be sent by the CPU, and the interrupt vector
* for that signal will jump the processor's current instruction pointer to the
* interrupt handling routine.
*/