Subsequence calculator. Can i make it more efficient?

Question

The idea of subsequences is explained very well in this post:Generate subsequences

But i didnt understand the answers on that question because im a beginner.

What i wanted to know is if i could make my C program any more efficient while still keeping it simple and understandable and without using functions?

#include <stdio.h>
#define NUM 123456

int main(void) {

    int i,num=NUM,x,y,mask,digits=0,max=1;

    while ( num != 0 ) {  //calculate digits
        num /= 10;
        digits++;
    }

    for ( i = 1; i <= digits; i++ ) {  //calculate max number of subsequences
        max *= 2;     
    }
    max=max-2;

    printf("Subsequences are:\n");
    for ( i = 1; i <= max ; i++ ) {
        mask = i;     //digit selector
        x = 1;        //multiplier
        num = NUM;
        y=0;          //subsequence value

        while ( num != 0 ) {
            if ( mask % 2 == 1 ) {
                y += num % 10 * x;
                x *= 10; 
            }
            num /= 10;
            mask /= 2;
        }
        printf("%d \n" , y);
    } 

    return 0;
}

Note that when we define NUM as a number such as 5111 or 100 some of the subsequences appear twice. Is there any simple way to fix that? Thanks!

Answer 1

The root of the reason that certain subsequences appear more than once with some numbers is because those numbers have repetitions of the same digit.

That repetition could be eliminated by saving each subsequence in an array and checking that array to see if the specific subsequence is already in the array. If already in the array, do not print. Otherwise, add subsequence to array contents and print

Answer 2

The problem can be divided into two tasks: (1) find all subsequences of an array of digits and (2) pack and unpack integers into digits.

Let's consider the subsequences of the array {a, b, c}. You can generate them by walking through the array from left to right and follow two paths: One where you include the current element in a subsequence and one where you don't.

That leads to a recursive approach tat we can represent as a tree:

                  {}
               /      \
         {}               {a}
       /    \           /     \
    {}        {b}     {a}      {ab}
   /  \      /   \   /   \    /    \
  {}  {c}  {b} {bc} {a} {ac} {ab} {abc}

When we branch left, we skip the current element and when we go right, we include the element. The elements themselves are the depth of the tree: On the first level we treat element a, on the next band on the last c.

The bottom row has all subsequences. This includes the empty sequence and the full sequence, which you don't want. But let's include them for now. (The arrays in the bottom row are usually called a power set, which is a nice web-searchable term.)

I mentioned recursion, which entails recursive functions, and functions are out.

So we need to tackle the problem another way. Let's turn the tree on its side. A dash denotes that the element was skipped. The notation on the right uses another notation: 0 means the element was skipped, 1 means the element was included:

- - -        000
- - c        001
- b -        010
- b c        011
a - -        100
a - c        101
a b -        110
a b c        111

I hope the codes on the right look familiar, because that's how you count from 000 to 111 in binary. That's a nice way to enumerate our elements. Now we need a way to tell which bits are set in each number.

The rightmost bit is set when the number is odd. We can find out about the other bits by repeatedly dividing the number by two, which in binary is a shift to the right, discarding the rightmost bit.

Now how to extract the digits from the original number? That number is a decimal number; it's in base 10. We can use the same approach as for finding the bits in the binary number, because the bits 0 and 1 are the binary digits.

Start with the number. The last digit is the result of taking the remainder after a division by 10. Then divide the number by ten until it is zero. This code yields the digits from right to left. So does the code for finding the bits, which means we can find whether a bit is set and which digit to print in a single loop, always taking the rightmost bit and if it is set, print the rightmost digit of the original number.

The empty and the full subsequences are the first and last items in the enumeration. If you don't want them, skip them.

That leaves the problem of the duplicated subsequences if the digit has repeated digits. I don' see an easy way to solve this except user3629249's suggestion to create the subsequence anyway and later check whether is has already been printed.

An easy way to do this is to keep an array of the subsequences. That array has max entries. After you have filled that array, sort it and then print it, but skip entries that are equal to the previous entry.

Here's an example implementation that uses an array of digits so that the original number doesn't have to be decomposed each time. It uses the sorting function qsort from <stdlib.h>, which requires a sorting function:

#include <stdlib.h>
#include <stdio.h>

#define NUM 412131

typedef unsigned int uint;

int uintcmp(const void *pa, const void *pb)
{
    const uint *a = pa;
    const uint *b = pb;

    return (*a > *b) - (*a < *b);
}

int main(void)
{
    uint digit[20];             // array of digits
    size_t ndigit = 0;          // length of array
    uint num = NUM;
    uint max = 1;
    size_t i;

    while (num) {
        digit[ndigit++] = num % 10;
        num /= 10;
        max *= 2;
    }

    uint res[max];              // array of subsequences

    for (i = 0; i < max; i++) {
        uint mask = i;          // mask for bit detection
        uint j = ndigit;        // index into digit array
        uint s = 0;

        while (j--) {
            if (mask % 2) s = s*10 + digit[j];
            mask /= 2;
        }

        res[i] = s;
    }

    qsort(res, max, sizeof(*res), uintcmp);

    for (i = 1; i < max - 1; i++) {
        if (res[i] != res[i - 1]) printf("%u\n", res[i]);
    }

    return 0;
}