बूथ का एल्गोरिथ्म एक गुणन एल्गोरिथ्म है जो दो हस्ताक्षरित बाइनरी नंबरों को 2 के कॉम्प्लिमेंट नोटेशन में गुणा करता है। बूथ ने डेस्क कैलकुलेटर का इस्तेमाल किया जो जोड़ने की तुलना में शिफ्टिंग में तेज़ थे और उन्होंने अपनी गति बढ़ाने के लिए एल्गोरिदम बनाया।
एल्गोरिदम
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
उदाहरण कोड
#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);
} आउटपुट
--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