C++ Program to Implement Booth’s Multiplication Algorithm for Multiplication of 2 signed Numbers

Booth’s algorithm is a multiplication algorithm that multiplies two signed binary numbers in 2’s compliment notation. Booth used desk calculators that were faster at shifting than adding and created the algorithm to increase their speed.

Algorithm

Begin
   Put multiplicand in BR and multiplier in QR
      and then the algorithm works as per the following conditions:
   1. If Qn and Qn+1 are same i.e. 00 or 11 perform arithmetic shift by 1 bit.
   2. If Qn Qn+1 = 10 do A= A + BR and perform arithmetic shift by 1 bit.
   3. If Qn Qn+1 = 01 do A= A – BR and perform arithmetic shift by 1 bit.
End

Example Code

#include<iostream>
using namespace std;
void add(int a[], int x[], int q);
void complement(int a[], int n) {
   int i;
   int x[8] = { NULL };
   x[0] = 1;
   for (i = 0; i < n; i++) {
      a[i] = (a[i] + 1) % 2;
   }
   add(a, x, n);
}
void add(int ac[], int x[], int q) {
   int i, c = 0;
   for (i = 0; i < q; i++) {
      ac[i] = ac[i] + x[i] + c;
      if (ac[i] > 1) {
         ac[i] = ac[i] % 2;
         c = 1;
      }else
         c = 0;
      }
   }
   void ashr(int ac[], int qr[], int &qn, int q) {
      int temp, i;
      temp = ac[0];
      qn = qr[0];
      cout << "\t\tashr\t\t";
      for (i = 0; i < q - 1; i++) {
         ac[i] = ac[i + 1];
         qr[i] = qr[i + 1];
      }
      qr[q - 1] = temp;
   }
   void display(int ac[], int qr[], int qrn) {
      int i;
      for (i = qrn - 1; i >= 0; i--)
         cout << ac[i];
      cout << " ";
      for (i = qrn - 1; i >= 0; i--)
         cout << qr[i];
   }
   int main(int argc, char **argv) {
      int mt[10], br[10], qr[10], sc, ac[10] = { 0 };
      int brn, qrn, i, qn, temp;
      cout << "\n--Enter the multiplicand and multipier in signed 2's
      complement form if negative--";
      cout << "\n Number of multiplicand bit=";
      cin >> brn;
      cout << "\nmultiplicand=";
      for (i = brn - 1; i >= 0; i--)
         cin >> br[i]; //multiplicand
      for (i = brn - 1; i >= 0; i--)
         mt[i] = br[i];
      complement(mt, brn);
      cout << "\nNo. of multiplier bit=";
      cin >> qrn;
      sc = qrn;
      cout << "Multiplier=";
      for (i = qrn - 1; i >= 0; i--)
         cin >> qr[i];
         qn = 0;
         temp = 0;
         cout << "qn\tq[n+1]\t\tBR\t\tAC\tQR\t\tsc\n";
         cout << "\t\t\tinitial\t\t";
         display(ac, qr, qrn);
         cout << "\t\t" << sc << "\n";
         while (sc != 0) {
            cout << qr[0] << "\t" << qn;
            if ((qn + qr[0]) == 1) {
               if (temp == 0) {
                  add(ac, mt, qrn);
                  cout << "\t\tsubtracting BR\t";
                  for (i = qrn - 1; i >= 0; i--)
                     cout << ac[i];
                  temp = 1;
               }
            else if (temp == 1) {
               add(ac, br, qrn);
               cout << "\t\tadding BR\t";
               for (i = qrn - 1; i >= 0; i--)
                  cout << ac[i];
                  temp = 0;
            }
            cout << "\n\t";
            ashr(ac, qr, qn, qrn);
         }
         else if (qn - qr[0] == 0)
            ashr(ac, qr, qn, qrn);
            display(ac, qr, qrn);
            cout << "\t";
            sc--;
            cout << "\t" << sc << "\n";
   }
   cout << "Result=";
   display(ac, qr, qrn);
}

Output

--Enter the multiplicand and multipier in signed 2's complement form if
negative--
Number of multiplicand bit=5
multiplicand=0 1 1 1 1
No. of multiplier bit=5
Multiplier=1 0 1 1 1
qn q[n+1] BR AC QR sc
initial 00000 10111 5
1 0 subtracting BR 10001
ashr 11000 11011 4
1 1 ashr 11100 01101 3
1 1 ashr 11110 00110 2
0 1 adding BR 01101
ashr 00110 10011 1
1 0 subtracting BR 10111
ashr 11011 11001 0
Result=11011 11001

Nancy Den

Updated on 30-Jul-2019 22:30:25

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