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воскресенье, 20 апреля 2014 г.

Multiplication of big numbers represented as strings

This is the usual task in interview: solve multiplication of integer numbers presented as strings, for ex., "25" * "25" = "625". Length of these strings may be any (not 16 bits or 32 bits only).
Implementation will use school way of multiplication of numbers (sign is ignored), see:
   1856
   1856
   ----
  11136
  9280
14848
1856
=======
3444736
So, "1856" * "1856" = "3444736", we multiply each digit of second number on each digit of first one, shift result number by one to left, summarize them and don't forget about carry. Carry is of 2 kinds: carry of multiplication and carry of sum. Algorithm is:
  • multiply last digit of 2nd number by last digit of 1st one
  • write result (only low digit) in result string at LAST POSITION
  • if resulting number is > 10, save carry (divided by 10)
  • make the same with the 2nd digit of 2nd number, but instead of saving resulting digit, make summation: with carry of multiplication AND carry of sum
  • don't forget to make shifting to left when go to next digit of 1nd number
So, each "saving" of result digit should make summation and there are not number 9280 in middle result but sum of 11136 and 9280 ("103936"). Implementation on C looks like this:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include "algutils.h"

#define ASCII2I(CH) ((CH)?((CH)-'0'):0)
#define I2ASCII(I) ((I)+'0')

char* mul(const char *s1, const char *s2, char *res, int reslen)
{
    int r, // index in result string
        i, // index in s1
        _i, // offset (to left) of next middle number
        j, // index in s2
        len1 = strlen(s1), len2 = strlen(s2);
    int d1, // digit in s1
        d2, // digit in s2
        ts, // temporary summ
        c, // curry in multiplication
        cs; // curry in summation
    memset(res, 0, reslen);
    for (i=len2-1, _i=0, cs=0; i>=0; i--, _i++) {
        d2 = ASCII2I(s2[i]);
        r = reslen - 2 - _i;
        for (j=len1-1, c=0; j>=0; j--) {
            d1 = ASCII2I(s1[j]);
            d1 = d1*d2 + c;
            ts = ASCII2I(res[r]) + (d1%10) + cs;
            res[r] = I2ASCII(ts%10);
            cs = ts/10;
            c = d1/10;
            r--;
        }
        do {
            // do rest of sum (high digits)
            ts = ASCII2I(res[r]) + (c%10) + cs;
            if (!ts) break;
            res[r] = I2ASCII(ts%10);
            cs = ts/10;
            r--;
        } while (cs);
    }
    return (res + r + 1);
}

///////////////////////////////////////////////
//              Main: tests
///////////////////////////////////////////////

#define RND(N) (srand(time(NULL)), rand()%(N))

int main(void)
{
    char _res[32], *res;
    char s1[32] = {0};
    char s2[32] = {0};
    int n1, n2, rn, resn, chk;

    //printf("%s\n", mul("1856", "1856", _res, sizeofarray(_res)));
    //return 0;

    for (int i=0; i<100; i++) {
        sprintf(s1, "%d", RND(9999999));
        sprintf(s2, "%d", 1+RND(9999999));
        printf("STR: %s * %s = %s\n", s1, s2, (res=mul(s1, s2, _res, sizeofarray(_res))));
        n1 = atoi(s1); n2 = atoi(s2); rn = n1*n2; resn = atoi(res);
        printf("CTR: %d * %d = %d\n", n1, n2, rn);
        printf("verify <%d>: %d (%d ? %d)\n", i, (chk=(resn==rn)), rn, resn);
        if (!chk) { printf("FAILED!\n"); return (1); }
        else printf("\n\n");
    }
    printf("All tests are passed\n");
    return (0);
}
Tests are cycle of multiplication of random numbers. If test is failed, cycle is break.

четверг, 17 апреля 2014 г.

In-place non-recursive Merge Sort

Merge sorting is a good known algorithm, but usually is used recursive implementation with additional array. This is an another version: without recursive and without usage of additional array.
Merge sorting idea is to sort small parts (pairs) than longer (4 items) and etc, but each next sort phase is MERGE of previous sorted parts. Merging of 2 sorted parts should saves order of items:
1,5 merging 2,3 is 1,2,3,5.
If we avoid recursive implementation then we should generate indexes of all parts (2 items, 4 items...). This sub-task will be done in mergesort() function. Another sub-task is to merge 2 sorted parts into new one, but sorted too. This is implemented in _merge() function, which avoid additional array of result by using of swapping of items: we has 2 arrays with 2 indexes and compare both items (from one array and another, indexed by these 2 indexes). If they need reordering then we swap items. Index of lesser item is incremented (see code below).
Main problem is to calculate middle index (for splitting input array into 2 parts for merging). Next formula is used:
mid = lev*2 + pielen*pie;
if (pielen > len || mid > len-1) mid = lastleft;
where lev is the level of pseudo-recursion, pielen is the length of pie (part) = {2, 4, 8...}. len is the length of input array, lastleft is the index of left bound of previous "level". So, we should to get something like this:
lev=0 left=0   mid=0   right=2
lev=0 left=2   mid=2   right=4
lev=0 left=4   mid=4   right=6
lev=0 left=6   mid=6   right=8
lev=0 left=8   mid=8   right=10
lev=0 left=10 mid=10 right=12
lev=1 left=0   mid=2   right=4
lev=1 left=4   mid=6   right=8
lev=1 left=8   mid=10 right=12
lev=2 left=0   mid=4   right=8
lev=2 left=8   mid=8   right=12
lev=3 left=0   mid=8   right=12
So, this is the code:
#include <stdio.h>
#include "algutils.h"

template<typename>
static inline void _merge(T *arr, int left, int mid, int right)
{
    T tmp, // tmp item needed for inserting if head position on shift
      *larr, // left subarray
      *rarr; // right subarray
    int llen, // length of left subarray
        rlen; // length of right subarray
    int l, r; // indexes of compared items in left, right subarrays

    l = r = 0;
    larr = &arr[left];
    if (mid==left) {
        // When len==2 swapping is more preferable
        rarr = &arr[mid+1];
        llen = mid - left + 1;
        rlen = right - mid - 1;
    } else {
        // Common case
        rarr = &arr[mid];
        llen = mid - left;
        rlen = right - mid;
    }
    // Index in left subarray is limited by "right" index
    while ((larr+l < rarr+r) && r < rlen) {
        if (larr[l] > rarr[r]) {
            tmp = rarr[r];
            RSHIFT_ARR(larr+l, llen-(l-r), 1);
            larr[l] = tmp;
            r++;
        }
        l++;
    }
}

template<typename>
int mergsort(T* arr, int len)
{
    int lev, // level of "recursion"
        pie, // number of subarray (of part)
        pielen, // subarray length: 2, 4, 8...
        left, right, // bounds (right is not included)
        mid, // middle point for splitting - is the same as left bound of subarray but on prev. level
        lastleft; // last left index

    if (!arr||!len) return (0);

    lev = 0;
    pielen = 2;
    while (pielen < 2*len) {
        pie = 0;
        do {
            left = pie*pielen;
            right = left + pielen;
            if (right > len) right = len;
            mid = lev*2 + pielen*pie;
            if (pielen > len || mid > len-1) mid = lastleft; //mid = len-2;
            _merge(arr, left, mid, right);
            pie++;
        } while (right < len-1);
        pielen *= 2;
        lev++;
        lastleft = left;
    }
    return (1);
}


/*****************************************
                   main
 ****************************************/
int main(void)
{
    int arr1[] = {10, 19, 1, 2, 10, 450, 6, 7, 8, 100, 120, 121};
    mergsort(arr1, sizeofarray(arr1));
    PRINTARR(arr1, "%d");
    return (0);
}
Macros RSHIFT_ARR() is trivial:
/// Shift array @a ARR of length @a LEN by @a N positions to right
#define RSHIFT_ARR(ARR, LEN, N) do { \
    register int _rshift_arr_i_; \
    for (_rshift_arr_i_=(N)+(LEN)-1; \
         _rshift_arr_i_>(N)-1; \
         _rshift_arr_i_--) \
    { \
        *((ARR)+_rshift_arr_i_) = *((ARR)+(_rshift_arr_i_-(N))); \
    } \
} while (0)