Double dabble

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In computer science, the double dabble algorithm is used to convert binary numbers into decimal (in particular, binary-coded decimal, or BCD, notation). The algorithm operates as follows:

Suppose the original number to be converted is stored in a register that is n bits wide. Reserve a scratch space wide enough to hold both the original number and its BCD representation; ceil(n/3) bits will be enough. It takes a maximum of 4 bits in binary to store each decimal digit.

Then partition the scratch space into BCD digits (on the left) and the original register (on the right). For example, if the original number to be converted is eight bits wide, the scratch space would be partitioned as follows:

100s Tens Ones   Original
0010 0100 0011   11110011

The diagram above shows the binary representation of 24310 in the original register, and the BCD representation of 243 on the left.

The scratch space is initialized to all zeros, and then the value to be converted is copied into the "original register" space on the right.

0000 0000 0000   11110011

The algorithm then iterates n times. On each iteration, the entire scratch space is left-shifted one bit. However, before the left-shift is done, any BCD digit which is greater than 4 is incremented by 3. The increment ensures that a value of 5, incremented and left-shifted, becomes 16, thus correctly "carrying" into the next BCD digit.

The double-dabble algorithm, performed on the value 24310, looks like this:

0000 0000 0000   11110011   Initialization
0000 0000 0001   11100110   Shift
0000 0000 0011   11001100   Shift
0000 0000 0111   10011000   Shift
0000 0000 1010   10011000   Add 3 to ONES, since it was 7
0000 0001 0101   00110000   Shift
0000 0001 1000   00110000   Add 3 to ONES, since it was 5
0000 0011 0000   01100000   Shift
0000 0110 0000   11000000   Shift
0000 1001 0000   11000000   Add 3 to TENS, since it was 6
0001 0010 0001   10000000   Shift
0010 0100 0011   00000000   Shift
   2    4    3
       BCD

Now eight shifts have been performed, so the algorithm terminates. The BCD digits to the left of the "original register" space display the BCD encoding of the original value 243.

Another example for the double dabble algorithm - value 6524410.

 104  103  102   101  100    Original binary
0000 0000 0000 0000 0000   1111111011011100   Initialization
0000 0000 0000 0000 0001   1111110110111000   Shift left (1st)
0000 0000 0000 0000 0011   1111101101110000   Shift left (2nd)
0000 0000 0000 0000 0111   1111011011100000   Shift left (3rd)
0000 0000 0000 0000 1010   1111011011100000   Add 3 to 100, since it was 7
0000 0000 0000 0001 0101   1110110111000000   Shift left (4th)
0000 0000 0000 0001 1000   1110110111000000   Add 3 to 100, since it was 5
0000 0000 0000 0011 0001   1101101110000000   Shift left (5th)
0000 0000 0000 0110 0011   1011011100000000   Shift left (6th)
0000 0000 0000 1001 0011   1011011100000000   Add 3 to 101, since it was 6
0000 0000 0001 0010 0111   0110111000000000   Shift left (7th)
0000 0000 0001 0010 1010   0110111000000000   Add 3 to 100, since it was 7
0000 0000 0010 0101 0100   1101110000000000   Shift left (8th)
0000 0000 0010 1000 0100   1101110000000000   Add 3 to 101, since it was 5
0000 0000 0101 0000 1001   1011100000000000   Shift left (9th)
0000 0000 1000 0000 1001   1011100000000000   Add 3 to 102, since it was 5
0000 0000 1000 0000 1100   1011100000000000   Add 3 to 100, since it was 9
0000 0001 0000 0001 1001   0111000000000000   Shift left (10th)
0000 0001 0000 0001 1100   0111000000000000   Add 3 to 100, since it was 9
0000 0010 0000 0011 1000   1110000000000000   Shift left (11th)
0000 0010 0000 0011 1011   1110000000000000   Add 3 to 100, since it was 8
0000 0100 0000 0111 0111   1100000000000000   Shift left (12th)
0000 0100 0000 1010 0111   1100000000000000   Add 3 to 101, since it was 7
0000 0100 0000 1010 1010   1100000000000000   Add 3 to 100, since it was 7
0000 1000 0001 0101 0101   1000000000000000   Shift left (13th)
0000 1011 0001 0101 0101   1000000000000000   Add 3 to 103, since it was 8
0000 1011 0001 1000 0101   1000000000000000   Add 3 to 101, since it was 5
0000 1011 0001 1000 1000   1000000000000000   Add 3 to 100, since it was 5
0001 0110 0011 0001 0001   0000000000000000   Shift left (14th)
0001 1001 0011 0001 0001   0000000000000000   Add 3 to 103, since it was 6
0011 0010 0110 0010 0010   0000000000000000   Shift left (15th)
0011 0010 1001 0010 0010   0000000000000000   Add 3 to 102, since it was 6
0110 0101 0010 0100 0100   0000000000000000   Shift left (16th)
   6    5    2    4    4
            BCD

Sixteen shifts have been performed, so the algorithm terminates. The BCD digits is: 6*104 + 5*103 + 2*102 + 4*101 + 4*100 = 65244.

C implementation[edit]

The double dabble algorithm might look like this when implemented in C. Notice that this implementation is designed to convert an "input register" of any width, by taking an array as its parameter and returning a dynamically allocated string. Also notice that this implementation does not store an explicit copy of the input register in its scratch space, as the description of the algorithm did; copying the input register into the scratch space was just a pedagogical device.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
 
/*
   This function takes an array of n unsigned integers,
   each holding a value in the range [0, 65535],
   representing a number in the range [0, 2**(16n)-1].
   arr[0] is the most significant "digit".
   This function returns a new array containing the given
   number as a string of decimal digits.
 
   For the sake of brevity, this example assumes that
   calloc and realloc will never fail.
*/
void double_dabble(int n, const unsigned int *arr, char **result)
{
    int nbits = 16*n;         /* length of arr in bits */
    int nscratch = nbits/3;   /* length of scratch in bytes */
    char *scratch = calloc(1 + nscratch, sizeof *scratch);
    int i, j, k;
    int smin = nscratch-2;    /* speed optimization */
 
    for (i=0; i < n; ++i) {
        for (j=0; j < 16; ++j) {
            /* This bit will be shifted in on the right. */
            int shifted_in = (arr[i] & (1 << (15-j)))? 1: 0;
 
            /* Add 3 everywhere that scratch[k] >= 5. */
            for (k=smin; k < nscratch; ++k)
              scratch[k] += (scratch[k] >= 5)? 3: 0;
 
            /* Shift scratch to the left by one position. */
            if (scratch[smin] >= 8)
              smin -= 1;
            for (k=smin; k < nscratch-1; ++k) {
                scratch[k] <<= 1;
                scratch[k] &= 0xF;
                scratch[k] |= (scratch[k+1] >= 8);
            }
 
            /* Shift in the new bit from arr. */
            scratch[nscratch-1] <<= 1;
            scratch[nscratch-1] &= 0xF;
            scratch[nscratch-1] |= shifted_in;
        }
    }
 
    /* Remove leading zeros from the scratch space. */
    for (k=0; k < nscratch-1; ++k)
      if (scratch[k] != 0) break;
    nscratch -= k;
    memmove(scratch, scratch+k, nscratch+1);
 
    /* Convert the scratch space from BCD digits to ASCII. */
    for (k=0; k < nscratch; ++k)
      scratch[k] += '0';
 
    /* Resize and return the resulting string. */
    *result = realloc(scratch, nscratch+1);
    return;
}
 
/*
   This test driver should print the following decimal values:
   246
   16170604
   1059756703745
*/
int main(void)
{
    unsigned int arr[] = { 246, 48748, 1 };
    char *text = NULL;
    int i;
    for (i=0; i < 3; ++i) {
        double_dabble(i+1, arr, &text);
        printf("%s\n", text);
        free(text);
    }
    return 0;
}

Historical[edit]

In the 1960s, the term double dabble was also used for the mental algorithm used by programmers to convert a binary number to decimal. It is performed by reading the binary number from left to right, doubling if the next bit is zero, and doubling and adding one if the next bit is one. In the example above, 11110011, the thought process would be: "one, three, seven, fifteen, thirty, sixty, one hundred twenty-one, two hundred forty-three," the same result as that obtained above.