Question

Show how bitwise operations on bit strings can be used to find these combinations of\(A = \{ a,b,c,d,e\},\)\(B = \{ b,c,d,g,p,t,v\},\)\(C = \{ c,e,i,o,u,x,y,z\} ,\) and \(D = \{ d,e,h,i,n,o,t,u,x,y\} \)

a) \(A \cup B\)

b) \(A \cap B\)

c) \((A \cup D) \cap (B \cup C)\)

d) \(A \cup B \cup C \cup D\)

Short answer

Given:

\(A = \{ a,b,c,d,e\} ,\)

\(B = \{ b,c,d,g,p,t,v\} ,\)

\(C = \{ c,e,i,o,u,x,y,z\} ,\)

\(D = \{ d,e,h,i,n,o,t,u,x,y\} \)

If the \(ith\) bit in the string is a \(1\), then the \(ith\) letter of the alphabet is in the set.

If the \(ith\) bit in the string is a \(0\), then the \(ith\) letter of the alphabet is Not in the set.

The alphabet contains \(26\) letters; thus, each string needs to contain \(26\) bits.

\(A:11111\)\(00000\)\(00000\)\(00000\)\(00000\)\(0\)

\(B:01110\)\(01000\)\(00000\)\(10001\)\(01000\)\(0\)

\(C:00101\)\(00010\)\(00001\)\(00000\)\(10111\)\(0\)

\(D:00011\) \(00110\) \(00011\) \(00001\) \(10011\) \(0\)

Step-by-step solution

Step 1

(a). Union \(A \cup B\): all elements that are either in \(A\) OR in \(B\)

If either string contains a\(1\)on the\(ith\)bit, then\(A \cup B\)contains a\(1\)on the\(ith\)bit as well.

\(A \cup B:\) \(11111\) \(01000\) \(00000\) \(10001\) \(01000\) \(0\)

The string then corresponds with the set:

\(A \cup B = \{ a,b,c,d,e,g,p,t,v\} \)

Step 2

(b). Intersection \(A \cap B\): all elements that are both in \(A\) AND in \(B\)

If both strings contain a\(1\)on the\(ith\)bit, then\(A \cap B\)contains a\(1\)on the\(ith\)bit as well.

\(A \cup B:\) \(01110\) \(00000\) \(00000\) \(00000\) \(00000\) \(0\)

This string then corresponds with set:

\(A \cap B = \{ b,c,d\} \)

Step 3

(c). If either string contains a \(1\) on the \(ith\) bit, then the union contains a \(1\) on the \(ith\) bit as well.

\(A \cup D:\) \(11111\) \(00110\) \(00011\) \(00001\) \(10011\) \(0\)

\(B \cup C:\) \(01111\) \(01010\) \(00001\) \(10001\) \(11011\) \(1\)

This string then corresponds with the set:

\((A \cup D) \cap (B \cup C):\{ b,c,d,e,i,o,t,u,x,y\} \)

Step 4

(d). If either string contains a \(1\) on the \(ith\) bit, then the union contains a \(1\) on the \(ith\) bit as well.

\(A \cup D:\) \(11111\) \(00110\) \(00011\) \(00001\) \(10011\) \(0\)

\(B \cup C:\) \(01111\) \(01010\) \(00001\) \(10001\) \(11011\) \(1\)

Since the union is commutative, \(A \cup B \cup C \cup D\)is equal to\((A \cup D) \cap (B \cup C)\)

\(A \cup B \cup C \cup D\): \(11111\) \(01110\) \(00011\) \(10001\) \(11011\) \(1\)

This string then corresponds with the set:

\(A \cup B \cup C \cup D = \{ a,b,c,d,e,g,h,i,n,o,p,t,u,v,x,y,z\} \)

Thus, we conclude that

A) \(\{ a,b,c,d,e,g,p,t,v\} \)

B) \(\{ b,c,d\} \)

C) \(\{ b,c,d,e,i,o,t,u,x,y\} \)

D) \(\{ a,b,c,d,e,g,h,i,n,o,p,t,u,v,x,y,z\} \)