recycle in permutation

[rahs08]

hello, dear all. i'have small problem. here is my algorithm

#include <stdio.h>

/**
Read a number, n, from standard input and print the
permutations.
*/

void write( int *x, int n)
{
if (x != 0) {
for (int i = 0; i < n; i++) {
printf("%4d", x[i] );
}
printf("\n");
}
} // print


void swap(int *x, int i, int j)
{
int t;
t = x[i];
x[i] = x[j];
x[j] = t;
}


void recycle(int *x, int start, int n)
{
int tmp = x[start+1];
for (int i = start+1; i <n-2; ++i) {
x[i] = x[n-i-1];
x[n-i-1] = tmp;
}
}


void starterequiv(int *x, int start, int n)
{
write(x, n);
if (start < n) {
int i, j;
for (i = n-1; i > start; i--) {
for (j = i + 1; j < n; j++) {
swap(x, i, j);
starterequiv(x, i, n);
}
recycle(x, i, n);
}
}
}


void initiate(int *x, int n)
{
for (int i = 0; i <n; i++) {
x[i] = i+1;
}
} // init


int main()
{
char buf[100];
int n;
printf("Enter the number of elements: ");
fgets(buf, sizeof(buf), stdin );
sscanf_s(buf, "%d", &n);

if (n > 0 && n <= 100) {
int *x = new int[n];
initiate(x, n);
starterequiv(x, 0, n);
}
}

The problem is
i didn;t get an output same as follows (n=4)
1234
1432 (recycle)
1423 (swap)
1324 (recycle)
1342 (swap)
1243. ( recycle).
Its based combinatoric concept. how to fix it?

[exception]

What exactly do you want to do, what is the output and what would you expect it to be? Do you want *all* permutations of some input digits? (That's what the text says, however, for 4 you have 4! = 24 permutations, you only list 5). A little bit more explanation and a little bit less not-working source code would perhaps be helpful.

[rahs08]

ok. thanks for the reply me. my algorithm output as follows: n= 4
1234
1243
1423
1432
1234
1243.

i didn;t need all 24 permutation. i just need first 6 of 24 permutation which begin with element '1'. i knew that we can solve using swap and recursion.

i need to program which swap and reverse is run by side by side

recycle and swap in loop. i think the problem is at starterequiv function.

[jsmith]

Alternatively the STL provides all of this for you:

#include <algorithm>
#include <iostream>
#include <string>

int main() {
    // Assume this is what you want to permute:
    std::string numbers( "4321" );

    // next_permutation requires the first permutation to be in sorted order
    std::sort( numbers.begin(), numbers.end() );

    // The sorted sequence is the first permutation
    std::cout << numbers << std::endl;

    // Loop through all permutations...
    while( std::next_permutation( numbers.begin(), numbers.end() ) )
        std::cout << numbers << std::endl;
}

[rahs08]

ok.its help me so much. but i have to use my own algorithm in spite of next-permutation. Maybe its quite difficult to do loop swap and reverse. that's ok. here is my output from the algorithm. i just generate 6.
1234
1243
1324
1342
1423
1432.

I just need there first three. how to use filter/erase/delete function to delete the last three.then my output will become as follows:

1234
1243
1324

[rahs08]

void starterequiv(int *arr,  int start, int SIZE)
{	int count = 1;
	if (start > SIZE) 
	 print(arr, SIZE);
	else
	{ int i, j; 
		starterequiv(arr,start+1 ,SIZE);
    for (i = SIZE-1; i > start; i--) { 
      for (j = i+1 ; j <= SIZE-1; j++) {
	    swap(arr, i, j);
		starterequiv(arr, i, SIZE);
		count++ ;

      } 
	 shiftleft(arr,i,SIZE); 
		}
	if (count == 4) exit(0);
	}
}

my output:
Enter the number of elements: 4
1234
1243
1243
1324
1342
Press any key to continue . . .

the are two problem:
(i) repeated permutation.
and the second one is
if (count == 4) exit(0); is not function. for to list the first 4 of the permutation.

[Umz]

try using a loop in the main() function that calls the function, that way it will only loop 4 times and finish.

int main() {

for (int i = 0; i <= 4; i++) {
      starterequiv(value,value,value);
}

[rahs08]

thanks, when i put this function as your suggestion,

for (int i = 0; i <= 4; i++) {
      starterequiv(arr,0,SIZE);
}

then my output as follows:

Enter the number of elements: 4
1234
1243
1243
1324
1342
1423
1432
1234
1243
1243
1324
1342
1423
1432
1234
1243
1243
1324
1342
1423
1432
1234
1243
1243
1324
1342
1423
1432
1234
1243
1243
1324
1342
1423
1432
no. permutation :31
Press any key to continue . . .

list 31 or list the first 4.

[Umz]

Sorry, misunderstood what your program did. If thats the case just move the for() inside the starterequiv instead.

 void starterequiv(int *arr,  int start, int SIZE)
{	
                int count = 0;
                for (count; count <=4; i++) {

	if (start > SIZE) {
	    print(arr, SIZE);
	} else {
	        int i, j; 
                        starterequiv(arr,start+1 ,SIZE);
                        for (i = SIZE-1; i > start; i--) { 
                        for (j = i+1 ; j <= SIZE-1; j++) {
	        swap(arr, i, j); } }
                        starterequiv(arr, i, SIZE);
	        count++ ;
                }

	shiftleft(arr,i,SIZE); 
} 

This way the loop is inside the void function and only goes 4 times, this should work.