# 辅导 CSE140 Midterm 1 Practice Problem 1讲解 R编程

Midterm 1 Practice Problem 1               CSE140

 !A!B !AB AB A!B !C!D x !CD 1 x CD 1 1 x x C!D 1 x x

utput = 1 if input is a prime number   !A!BC!D = 2; !A!BCD = 3; !AB!CD = 5; !ABCD = 7

utput = don’t care if output > 9

How many prime implicants?    !A!BC + !ABD + !ACD          BD + !BC

How many distinguished 1-cells?

How many essential prime implicants?

How many minimal sums?

 !A!B !AB AB A!B !C!D x !CD 1 * x CD 1 1 x x C!D 1 * x x

utput = 1 if input is a prime number   !A!BC!D = 2; !A!BCD = 3; !AB!CD = 5; !ABCD = 7

utput = don’t care if output > 9

How many prime implicants? 3  dark blue center square, lighter blue split square, blue highlight line

How many distinguished 1-cells? 2  (denoted as usual with *; x are never distinguished)

How many essential prime implicants? 2 dark and light blue squares; blue highlight line is not essential

How many minimal sums? just one: BD + CB’  (don’t need CD – this has a consensus flavor to it)

MIdterm 1 Practice Problem 2

utput = 1 if number is not a prime

How many prime Implicants ?

How many distinguished 1 cells?

How many essential prime impilcants?

How many minimal sums?

utput = 1 if number is not a prime

How many prime Implicants?   4            Includes A=1 half-plane and CD=00 top row

How many distinguished 1 cells? 2 (*, as always)

How many essential prime impilcants? 2

How many minimal sums? 1     B’C’ + BD’   (interesting: + A buys us nothing)

 CD  \  AB 00 01 11 10 00 1 1 x 1 01 1 * x 1 11 x x 10 1 * x x

Problem 3: finding minimal product of sums

1. Fill out Kmap for Y=0

2. Find minimal sum for Y=0

3. Apply DeMorgan twice

Y’ = B’C’ + BD’

group Y=0 terms: Y’ = B’C’ + BD’

invert both sides: Y = (B’C’ + BD’)’

apply Demorgan to the OR: Y = (B’C’)’(BD’)’

apply Demorgan to each AND: Y = (B+C)(B’+D)

Y’ = B’C’ + BD’

 D \ BC 00 01 11 10 0 0 0 0 1 0

Y = (B+C)(B’+D)

 D \ BC 00 01 11 10 0 1 1 1 1 1