Ternary Counting In C - C Snipplr Social Repository

Revision: 61181

at December 2, 2012 12:41 by rtperson

Initial Code

int countInTernary(int numdigits) {
    char num[numdigits+1];
    memset(num, '0', sizeof num);
    num[numdigits] = 0;
    char *curIndex = &num[numdigits-1];
    char *end = &num[numdigits-1];
    char *start = &num[0];
    bool done = false;
    
    while (!done) {
        printf("%s\n", num);
        if (*curIndex < '2') {
            (*curIndex)++;
        } else {
            /* move left until you find a number < 2 */
            while (curIndex >= start && *curIndex == '2') {
                --curIndex;
            }
            if (curIndex < start) {
                done = true;
            } else {
                (*curIndex)++;
                while (curIndex < end) {
                    curIndex++;
                    *curIndex = '0';
                }                
            }
        }
    }  
    return EXIT_SUCCESS;
}

Initial URL

Initial Description

Here's a fairly useless but fun piece of code: counting in base 3. This takes in a number of  digits and counts up in ternary. Be warned, though. The algorithm is (for obvious reasons) O(n^3), where n is the number of digits. Actually probably theta(n^3), since n^3 is both the upper and lower bound. Either way, it spits out 531,441 possibilities for 12 digits, so be careful not to run it on too many digits.

Initial Title

Ternary Counting In C

Initial Tags

Initial Language

Choose a language for easy browsing: