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Which are the essential prime implicants of the following Boolean function?

$f(a, b, c)= a' c+ ac'+b' c$

  1. $a' c$ and $ac'$
  2. $a' c$ and $b' c$
  3. $a' c$ only.
  4. $ac'$ and $bc'$
in Digital Logic by Veteran (52.2k points)
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2 Answers

+6 votes
Best answer

$f(a,b,c) = a'c+ac'+b'c$

We can write these product of sum terms into canonical product of sum form.

$f(a,b,c) = \underbrace{a'b'c}_{001}+\underbrace{a'bc}_{011}+\underbrace{ab'c'}_{100}+\underbrace{abc'}_{110}+\underbrace{ab'c}_{101}+\underbrace{a'b'c}_{001}$

$f(a,b,c) = \sum(1,3,4,5,6)$

Now, we can draw the k-map for these minterms.

  • Prime implicant of $f$ is an implicant that is minimal - that is, the removal of any literal from product term results in a non-implicant for $ f$.
  • Essential prime implicant is an prime implicant that cover an output of the function that no combination of other prime implicants is able to cover.

Prime implicants are$:a'c,b'c,ab',ac'$

Essential prime implicants are$:a'c,ac'\:\text{(green color)}$.

References:

by Veteran (54.8k points)
selected by
+1
Perfect explanation.
0

@Lakshman Patel RJIT @techbd123 

No. of Implicants = No. of Minterms = 5

Please, correct me if i am wrong.

+1

@ayushsomani

It is not true.

+1

@ayushsomani

Number of implicants $=$ number of minterms $+$ number of subcubes of size $2$

Number of implicants $= 5 + 4 = 9$

PS: Number of implicants $=$  number of subcubes of size $2^{0} \:+$  number of subcubes of size $2^{1} +$  number of subcubes of size $2^{2} +$ number of subcubes of size $2^{3}$

+19 votes
Answer : A.

Using K map $f = ac' + a'c$
by Loyal (8.6k points)
edited by
+1
yes you are right, b'c is selective prime implicant.
+3

Just to see it visually ;)

Clearly $A\bar{C}$ and $\bar{A}C$ cover cells that are not covered by any others.

0

Essential prime implicant: A prime implicant that includes one or more distinguished one cells. Essential prime implicants are important because a minimal sum contains all essential prime implicants.

Reference:

+1
Someone please edit the answer. f is not a$\overline{c}$ + $\overline{a}$c

f = a$\overline{c}$ + $\overline{a}$c + $\overline{b}$c

or

f = a$\overline{c}$ + $\overline{a}$c + a$\overline{b}$

 

EPIs are a$\overline{c}$ , $\overline{a}$c
+1
Please check my solution.

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