Midterm 1 Practice Problem 1 CSE140
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!A!B
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!AB
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AB
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A!B
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!C!D
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x
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!CD
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1
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x
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CD
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1
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1
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x
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x
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C!D
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1
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x
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x
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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?
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!A!B
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!AB
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AB
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A!B
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!C!D
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x
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!CD
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1 *
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x
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CD
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1
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1
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x
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x
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C!D
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1 *
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x
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x
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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)
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CD \ AB
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00
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01
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11
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10
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00
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1
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1
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x
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1
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01
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1 *
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x
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1
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11
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x
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x
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10
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1 *
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x
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x
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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’
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D \ BC
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00
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01
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11
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10
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0
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0
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0
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0
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1
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0
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Y = (B+C)(B’+D)
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D \ BC
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00
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01
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11
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10
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0
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1
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1
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1
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1
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1
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