Four chairs are arranged in a circle as shown here...They are designated A, B, C, D.Each chair may have an occupant (logic 1),or may be empty (logic 0).Create both the |M...) and Em(...)(Maxterm and Minterm) expressions forthe following functions...a) F(D, C, B, A) = 1 iff there are no adjacent empty chairsb) G(D, C, B, A) =1 iff there are at least 3 adjacent empty chairsc) H(D, C, B, A) = 1 iff there are more people sitting in chairs A, D than in chairs B, C.Hint: create a truth table of all possibilities and fill it in with the logic from your understanding ofthe problem sentence. Then create the minterm and maxterms expressions for each.

Since A, B, C, D are four variables, so the truth table contain 24 = 16 rows. (a) Given F(D, C, B,…

Answered: Four chairs are arranged in a circle as…

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

See similar textbooks

Related questions

Question

Four chairs are arranged in a circle as shown here...
They are designated A, B, C, D.
Each chair may have an occupant (logic 1),
or may be empty (logic 0).
Create both the |M...) and Em(...)
(Maxterm and Minterm) expressions for
the following functions...
a) F(D, C, B, A) = 1 iff there are no adjacent empty chairs
b) G(D, C, B, A) =1 iff there are at least 3 adjacent empty chairs
c) H(D, C, B, A) = 1 iff there are more people sitting in chairs A, D than in chairs B, C.
Hint: create a truth table of all possibilities and fill it in with the logic from your understanding of
the problem sentence. Then create the minterm and maxterms expressions for each.

Expert Solution

Step 1

Since A, B, C, D are four variables, so the truth table contain 2⁴ = 16 rows.

(a) Given F(D, C, B, A) = 1 iff there are no adjacent empty chairs. The truth table is:

A	B	C	D	F	Minterms	Maxterms
0	0	0	0	0		(A+B+C+D)
0	0	0	1	0		(A+B+C+D')
0	0	1	0	0		(A+B+C'+D)
0	0	1	1	0		(A+B+C'+D')
0	1	0	0	0		(A+B'+C+D)
0	1	0	1	1	A'BC'D
0	1	1	0	0		(A+B'+C'+D)
0	1	1	1	1	A'BCD
1	0	0	0	0		(A'+B+C+D)
1	0	0	1	0		(A'+B+C+D')
1	0	1	0	1	AB'CD'
1	0	1	1	1	AB'CD
1	1	0	0	0		(A'+B'+C+D)
1	1	0	1	1	ABC'D
1	1	1	0	1	ABCD'
1	1	1	1	1	ABCD

Now to get desired SOP expression add the minterms for the output 1's.
F=A'BC'D + A'BCD + AB'CD' + AB'CD + ABC'D + ABCD' + ABCD = $\sum_{}$ (5, 7, 10, 11, 13, 14, 15)

Now to get desired POS expression multiply the maxterms for the output 0's.
F= (A+B+C+D)(A+B+C+D')(A+B+C'+D)(A+B+C'+D')(A+B'+C+D)(A+B'+C'+D)(A'+B+C+D)(A'+B+C+D')(A'+B'+C+D)
= $\prod_{}$ (0, 1, 2, 3, 4, 6, 8, 9, 12)