recursive algorithm

This is a discussion on recursive algorithm within the C Programming forums, part of the General Programming Boards category; i'm having a problem with below recursive code. there are 4 chars stored in a array, 'a', 'b', 'c', 'd'. ...

  1. #1
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    Jul 2004
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    3

    recursive algorithm

    i'm having a problem with below recursive code.
    there are 4 chars stored in a array, 'a', 'b', 'c', 'd'. input is between 1 and 16, and output should look like this:

    when input is 1:
    a
    b
    c
    d

    when iput is 2:
    aa
    ab
    ac
    ad
    ba
    bb
    bc
    bd
    ca
    cb
    cc
    cd
    da
    db
    dc
    dd

    and so on...

    here is the code i'v been working:
    Code:
    void f(int x, int num)
    {
    	int i;
            char a[4] = {'a', 'b', 'c', 'd'};
    	char temp[16];
    
    	if(num == 0)
    		return;
    
    	else
    		for(i = 0; i < 4; i++)
    		{
    			temp[x] = a[i];
    
    			f(x + 1, num - 1);
    
    			temp[x + 1] = '\0';
    
    			printf("%s\n", temp);
    		}
    }
    where should i change to make it work?

  2. #2
    Super Moderator
    Join Date
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    Posts
    4,913
    What isn't working now? You need to be more specific about that so we know where to look.

  3. #3
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    3
    Quote Originally Posted by sean_mackrory
    What isn't working now? You need to be more specific about that so we know where to look.
    ok...
    when u run the above code, u get this result:
    Code:
    $ ./test.out
    Enter a number between 1 and 16: 1
    a
    b
    c
    d
    
    $ ./test.out
    Enter a number between 1 and 16: 2
     a
     b
     c
     d
    a
     a
     b
     c
     d
    b
     a
     b
     c
     d
    c
     a
     b
     c
     d
    d
    here is the full code:
    Code:
    #include <stdio.h>
     
    void f(int, int, int);
     
    int main(void)
    {
            int x = 0, num;
     
            printf("Enter a number between 1 and 16: ");
            scanf("%d", &num);
     
            if (num < 1 || num > 16)
                    printf("-> %d is not a valid number.\n", num);
     
            else
                    f(x, num, num);
    }
     
    void f(int x, int n, int num)
    {
            int i;
            char a[4] = {'a', 'b', 'c', 'd'};
            char temp[16];
     
            if(num == 0)
                    return;
     
            else
                    for(i = 0; i < 4; i++)
                    {
                            temp[x] = a[i];
     
                            f(x + 1, n, num - 1);     // here?
     
                            temp[x + 1] = '\0';
     
                            printf("%s\n", temp);     // and maybe here?
                    }
    }

  4. #4
    & the hat of GPL slaying Thantos's Avatar
    Join Date
    Sep 2001
    Posts
    5,681
    I like these types of problems. Always good for a little dusting of the skills
    Here is my solution done up as much like yours as possible. Things I would notice: When I do the print, how I declared the array (note you don't have to do yours dynamically I did just for kicks)
    Code:
    #include <stdio.h>
    #include <stdlib.h>
    void f(const int num, const int pos)
    {
      static const char a[4] = {'a', 'b', 'c', 'd'};
      static char *array;
      int counter;
      if ( pos == 0 )
        array = malloc (num + 1);
    
      if ( pos >= num )
      {
        array[num] = '\0';
        puts(array);
        return;
      }
    
      for (counter=0; counter<sizeof a; counter++)
      {
        array[pos]=a[counter];
        f(num, pos+1);
      }
    
      if ( pos == 0)
        free(array);
    
    }
    
    int main(void) {
      f(3,0);
      putchar('\n');
      return 0;
    }

  5. #5
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    3

    Thumbs up

    thanks alot, Thantos.

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