Suppose we find the product of the binary numbers 100 0111 and 111 0111 using the algorithm discussed in lecture and shown below: binary Product (a, u): c = [] #empty container 8=0 for i = 0: k-1: if u i = 1: for j = 0:len (c): returns 100 0111 Select all the partial products that would be elements in the container c. 1000 1110 s = binaryAdd (s, c_i) 1 0001 1100 x = a + "i zeros" c.append (x) 10 0011 1000 100 0111 0000 1000 1110 0000 10001 1100 0000 10 0011 1000 0000 100 0111 0000 0000 10 0001 0000 0001

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
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Suppose we find the product of the binary numbers 100 0111 and 111 0111 using the algorithm discussed in lecture and shown below:
binary Product (a, u):
c = [] #empty container
S = 0
for i = 0: k-1:
if u i = 1:
for j = 0:1en (c):
return s
100 0111
Select all the partial products that would be elements in the container c.
1000 1110
s = binaryAdd (s, c_i)
1 0001 1100
10 0011 1000
x = a + "i zeros"
c.append (x)
100 0111 0000
1000 1110 0000
1 0001 1100 0000
10 0011 1000 0000
100 0111 0000 0000
10 0001 0000 0001
Transcribed Image Text:Suppose we find the product of the binary numbers 100 0111 and 111 0111 using the algorithm discussed in lecture and shown below: binary Product (a, u): c = [] #empty container S = 0 for i = 0: k-1: if u i = 1: for j = 0:1en (c): return s 100 0111 Select all the partial products that would be elements in the container c. 1000 1110 s = binaryAdd (s, c_i) 1 0001 1100 10 0011 1000 x = a + "i zeros" c.append (x) 100 0111 0000 1000 1110 0000 1 0001 1100 0000 10 0011 1000 0000 100 0111 0000 0000 10 0001 0000 0001
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