Help please

[k_w_s_t_a_s]

I'm trying to write a function :void permutations(int m,int n) to print all the permutations of the numbers 1,...,m per n.For example , permutations(4,2) should print:
(1,2)
(1,3)
(1,4)
(2,1)
(2,3)
(2,4)
(3,1)
(3,2)
(3,4)
(4,1)
(4,2)
(4,3)

I've tried many things with no results.Can you help me?
Recursive and non-recursive functions are both acceptable.

[alpha]

code?

and code tags...

no we won't write the program for you, but we are glad to help you with trouble you have.

[XSquared]

Its quite easily doable using recursion.

You need to show us some code before we will help you though.

[vasanth]

why is thge 2nd number needed thanb.. you just require 4 to generate it.. what is the job of 2 here.. is it saying 2 numbers or sometin...


you can use a loop say
the number is 3

1,2
1,3
2,1
2,3
3,1
3,2

I am helping ytou because you seem new here.. never repeat the same mistake again of only askin questions... put some effort..



the program would be..

# include <iostream.h>


int main()
{
int num;
cout<<"\n Enter the limit > ";
cin>>num;

for(int i=1;i<=num;i++)
{

for(int j=1;j<=num;j++)
         {
if(i!=j)
              {
                cout<<i<<","<<j<<"\n";
               }

         }

}



return 0;
}

this should do the work...

[k_w_s_t_a_s]

I am new here indeed.But I tried some things with no result.
Your program would be perfect if we knew from the start the numbers m and n in permutations(m,n),but we don't.I want the function to work for every pair m,n no matter how large they are.The second argument n is required.If we had :
permutations(5,4) it should print:
(1,2,3,4)
(1,2,3,5)
(1,2,4,3)
(1,2,4,5)
...
...
...
One thought i made was similar to yours but more general.I thought of making n nested for loops.But I also wanted inside the last nested loop to know all the values of the variables used in the above loops(i,j,k,...).I have no idea how to achieve this non-recursively.Do you?
I tried to make it recursively something like that:

#include <iostream>
#include <vector>

using namespace std;

void permutations(int m,int n) {
    
    static vector<int> a(n);
    static int N=n;
    bool check=true;
    a[n-1]=n;
    for(int i=1;i<=m;i++)
        if(n==1) {
            for(int j=0;j<N-1;j++)
                for(int k=j+1;k<N;k++)
                    if(a[j]==a[k]) check=false;
            if(check) {
                cout<<'(';
                for(int j=0;j<N;j++)
                    cout<<a[j]<<',';
                cout<<')'<<endl;
            }
        }
        else permutations(m,n-1);
}

int main()
{

permutations(5,3);

 cin.get();

}

But the result is:

(1,2,3,)
(1,2,3,)
(1,2,3,)
(1,2,3,)
(1,2,3,)
...

Any helpful ideas?

[k_w_s_t_a_s]

Something else:
Why doesn't the post include the spaces i've put in my program?
When i write it is ok,but when i post it all the spaces are eliminated.

[knave]

Go here: http://cboard.cprogramming.com/misc....&action=bbcode

Use the "vB Code" tags to format your future posts.

[PJYelton]

You were close, but you had two major mistakes that was causing it to screw up.

First off, checked gets set to true only at the beginning of the function, when instead it needs to be set to true before every time you look for duplicates.

Second, I'm not sure why you set a[n-1]=n. Wouldn't this make it so a[4] always equals 5, and a[3] always equals 4 etc? I think you meant to have that within the for loop and making it so a[n-1]=i. So your new code would look something like this and should work (although I haven't tested it!):

#include <iostream>
#include <vector>

using namespace std;

void permutations(int m,int n) {
    
    static vector<int> a(n);
    static int N=n;
    bool check=true;
    for(int i=1;i<=m;i++)
        a[n-1]=i;
        if(n==1) {
            checked=true;
            for(int j=0;j<N-1;j++)
                for(int k=j+1;k<N;k++)
                    if(a[j]==a[k]) check=false;
            if(check) {
                cout<<'(';
                for(int j=0;j<N;j++)
                    cout<<a[j]<<',';
                cout<<')'<<endl;
            }
        }
        else permutations(m,n-1);
}

int main()
{

permutations(5,3);

 cin.get();

}

[k_w_s_t_a_s]

I tested it and it doesn't print anything.

[hk_mp5kpdw]

You can use the next_permutation function defined in <algorithm> perhaps. Unless I made a typo:

#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

void do_permut(int i1, int i2)
{
    vector<int> vec;
    int skip = 0;

    // Exit if i2 > i1.

    if( i2 > i1 ) return;

    // Initialize vector.

    for( int i = 1; i < i1; ++i ) vec.push_back(i);

    // Output the permutations.

    do
    {
        if( skip++ % i2 == 0 )
        {
            cout << '(';
            for( int j = 0; j < i2; ++j )
            {
                cout << vec[j];
                if( j < (i2-1) ) cout << ',';
            }
            cout << ')' << endl;
        }
    } while( next_permutation(vec.begin(),vec.end()) );

}

int main()
{
    do_permut(4,2);
    return 0;
}

[edit]Tried it with do_permut(4,3) and it didn't work as expected, it does work with (4,2) but the algorithm needs a bit more work I guess. [/edit]

[XSquared]

#include <iostream>
#include <assert.h>

using namespace std;

#define MAX_SIZE 50

int cp[ MAX_SIZE ] = { 0 };

void permutations( int m, int n ) {

	int i;
	int l0 = -1;
	for( i = 0; i < n; i++ ) {
		l0 = cp[i] == 0 ? i : l0;
		if( l0 == i ) break;
	}
	if( l0 == -1 ) {
		for( i = 0; i < n; i++ )
			cout<<( i == 0 ? '(' : ',' )<<cp[i]<<( i == n - 1 ? ")\n" : " \b" );
		return;
	}
	int fnnu = 1;
	while( fnnu <= m ) {
		for( i = 0; i < l0; i++ )
			if( cp[i] == fnnu ) { fnnu++; goto eol; }
		cp[l0] = fnnu;
		permutations( m, n );
		fnnu++;
		eol:;
	}
	cp[l0] = 0;

}

int main( void ) {

	int m, n;
	cout<<"m: ";
	cin>>m;
	cout<<"n: ";
	cin>>n;
	assert( m >= n );
	permutations( m, n );

	return 0;

}

[PJYelton]

I tested it and it doesn't print anything.

Oops, didn't notice that you didn't put brackets around your for statement. Enclose the entire for statement in brackets and it should work.

[k_w_s_t_a_s]

Thank you all for your help.One last thing.Are there any non-recursive suggestions-ideas in how this could be done,but without the use of next_permutation in <algorithm>?

[k_w_s_t_a_s]

I am sorry I insist but isn't there anybody who can give some advice for a non-recursive solution?