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09.14.2005 at 04:55PM PDT, ID: 21561958
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7.4

How to list all possible number combinations from some given single digit numbers

Zone: C Programming Language
Tags: combinations, number, possible, all

Hi,

Trying to get my head around how to generate all possible number combinations from only four unique numbers. For example: 2, 1, 8, 3 in all possible combinations: 1238, 2138, 3812 and so on.

What is the best way to go about doing this?

Thanks

 
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Question Stats
Zone: Programming
Question Asked By: Elastex
Solution Provided By: slyong
Participating Experts: 4
Solution Grade: A
Views: 208
Translate:
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09.14.2005 at 05:22PM PDT, ID: 14885955

Rank: Wizard

ozo:

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09.14.2005 at 05:54PM PDT, ID: 14886073
slyong:

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09.14.2005 at 11:25PM PDT, ID: 14887081
billtouch:

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09.15.2005 at 09:36AM PDT, ID: 14890880
doraiswamy:

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09.15.2005 at 10:52AM PDT, ID: 14891566
Elastex:

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09.15.2005 at 12:23PM PDT, ID: 14892495
billtouch:

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09.16.2005 at 11:50AM PDT, ID: 14900456
billtouch:

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09.14.2005 at 05:22PM PDT, ID: 14885955

Rank: Wizard

ozo:
Is this homework?
One way would be to use three nested loops to rearrange an array
 
09.14.2005 at 05:54PM PDT, ID: 14886073
slyong:
There are quite a few permutation algorithm out there...:

1 - Recursive permutation algorithm like ozo said
2 - A permutation algorithm based on a string permutation algorithm
3 - An ordered (lexicographic) permuation algorithm

Google up "number permutation algorithm c" will land you into those pages.
Accepted Solution
 
09.14.2005 at 11:25PM PDT, ID: 14887081
billtouch:
The simplest algorithm would be to treat each number as a symbol and assign it a numeric value: 0, 1, 2, 3.  Each symbol will have two bits. 4 symbols is 8 bits or one unsigned char.

Therefore - all possible permutations are represented by all the values of unsigned char, i.e. 0-255. To see what the symbols are represented, split the unsigned char into 4 2bit ints and use that as an index into your symbol table.

The only remaining trick is to filter out every instance where a symbol is shown multiple times. All that is left is the desired result.

Your algorithm will have a for(;;) loop and a one line filter. It would be cool if you could post your results here.

Bill
 
09.15.2005 at 09:36AM PDT, ID: 14890880
doraiswamy:
Preferable to use recursion. It will print all permutations. Pseudocode given below:

permute(string1, string2)
{
if length(string2) = 1 then
  print string1 + string2
else
  for i = 1 to length(string2)
     string3 = string1 + first_character_of(string2)
     string4 = 2nd character of string2 to end of string
     permute(string3, string4)
     rotate string 2 left by 1 character, and put the original 1st character to end of string
   next
endif
return
}

main()
{
permute("", string)
}
 
09.15.2005 at 10:52AM PDT, ID: 14891566
Elastex:


Thanks slyong.

That's what I needed to know: permutation algorithm.


ozo
No it is not homework, just trying to learn how to program in C...

Bill
Below is code that does what I want...

#include <stdio.h>

void out(const int *s, const int size)
{
  if (s != 0) {
    for (int i = 0; i < size; i++) {
      printf("%1d", s[i] );
    }
    printf("\n");
  }
}


void permute(int *s, const int start, const int n)
{  
  if (start == n-1) {
    out(s, n);
  }
  else {
    for (int i = start; i < n; i++) {
      int tmp = s[i];
     
      s[i] = s[start];
      s[start] = tmp;
      permute(s, start+1, n);
      s[start] = s[i];
      s[i] = tmp;
    }
  }
}

main()
{
  int s[] = {1, 2, 3, 8};
  permute(s, 0, sizeof(s)/sizeof(int));
}
 
09.15.2005 at 12:23PM PDT, ID: 14892495
billtouch:
In computer science classes, understanding recursion is extremely important and a great achievement. Being able to find a recursive solution for a problem is something that only those with a real understanding of software engineering can do.

Given that, and having been writing embedded software for umpteen years (where recursion is not allowed because of its unbounded nature), I discovered that it is even more elegant to take the recursive solution and create a non-recursive solution. When that is done. you will find that ultimately, the final solution will be even more efficient.

Take the following code snippit:
 
char S[4] = "2183";
for (i=0; i<256; i++) {
   if ( /* filter code not included in case this is homework */ )
      print ( "%c, %c, %c, %c\n", S[i&0x03], S[(i>>2)&0x03], ...etc... );
}

For the filter code, think of using the parts of i as an index into a boolean array placed in a for(;;) loop.

As you can see, the elegance of simplicity outweighs the elegance of recursion. (IMHO)

Once the solution is posted, I will post the filter code. It is as simple as the above snippit.

Keep on coding!
Bill
 
09.16.2005 at 11:50AM PDT, ID: 14900456
billtouch:
A shorter one pass solution...

=====================================
#include <stdio.h>

char S[4] = "2183";

main()
{
        int i,j;
        int b[4] = {0,0,0,0};

        for ( i=0; i<256; i++) {
                b[0]=b[1]=b[2]=b[3]=0;                     // zero the flags
                for ( j=0; j<4; j++ ) {                    // Filter
                        if ( b[(i>>(j*2))&0x03] ) break;   // test if the flag indexed by 2 bit value is set
                        b[(i>>(j*2))&0x03] = 1;            // Was not set - so set it
                }
                if ( j==4 )      // this means we went thru all 4 iterations with no dulplicates
                        printf( "%c, %c, %c, %c\n",
                                S[(i)&0x03], S[(i>>2)&0x03], S[(i>>4)&0x03], S[(i>>6)&0x03] );
        }
}
 
 
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