# Simplify the following Boolean equations.

#### audreyeckman

Joined Jan 7, 2016
8
i. Simplify the following Boolean equations.

ii. Sketch a reasonably simple combinational circuit implementing the simpliﬁed equation.

iii. Compare the numbers of literals and operators versus the numbers of gates, nets, and pins in the schematic diagrams

A. a′bc′ + a′b′d + a′b′cd′ + bc′d + acd + abcd′ + ac′d

So far, I have gotten this. Does this seem correct? I am unsure how to sketch the circuit...can someone please help with this?

i.

= a’bc’(d + d’) + a’b’(c + c’)d + a’b’cd’ + (a + a’)bc’d + a(b + b’)cd + abcd’ + a(b + b’)c’d

= a’bc’d + a’bc’d’ + a’b’cd + a’b’c’d + a’b’cd’ + abc’d + a’bc’d + abcd + ab’cd + abcd’ + abc’d + ab’c’d

= a’bc’d + a’bc’d’ + a’b’cd + a’b’c’d + a’b’cd’ + abc’d + abcd + ab’cd + abcd’ + ab’c’d

#### shteii01

Joined Feb 19, 2010
4,644
You started with 7 pieces, your "solution" has 10 pieces. You just made things more complicated. Which is opposite of: simplify.

What techniques to simplify Boolean equations do you know?

#### WBahn

Joined Mar 31, 2012
25,928
i. Simplify the following Boolean equations.

ii. Sketch a reasonably simple combinational circuit implementing the simpliﬁed equation.

iii. Compare the numbers of literals and operators versus the numbers of gates, nets, and pins in the schematic diagrams

A. a′bc′ + a′b′d + a′b′cd′ + bc′d + acd + abcd′ + ac′d

So far, I have gotten this. Does this seem correct? I am unsure how to sketch the circuit...can someone please help with this?

i.

= a’bc’(d + d’) + a’b’(c + c’)d + a’b’cd’ + (a + a’)bc’d + a(b + b’)cd + abcd’ + a(b + b’)c’d

= a’bc’d + a’bc’d’ + a’b’cd + a’b’c’d + a’b’cd’ + abc’d + a’bc’d + abcd + ab’cd + abcd’ + abc’d + ab’c’d

= a’bc’d + a’bc’d’ + a’b’cd + a’b’c’d + a’b’cd’ + abc’d + abcd + ab’cd + abcd’ + ab’c’d
It appears you have tried to convert the given expression into the canonical sum-of-products form, which is generally NOT considered "simplified". Though note that the problem gives no metric by which to determine which of two expressions is "simpler", though it does at least as for comparisons.

As for how to implement a Boolean expression in a circuit, do you know how to implement the following:

A', AB, A+B

using a NOT, AND, and OR gate respectively?

If so, then you have all the building blocks, unless using XOR is an option.

#### audreyeckman

Joined Jan 7, 2016
8
I am in the first week of my CSE 140 course so this is all new to me.

I am doing this over and attempting to use Shannon's expansion:

Shannon’s expansion

Let f(a,b,c,d) = a′bc′ + a′b′d + a′b′cd′ + bc′d + acd + abcd′ + ac′d

f(1,b,c,d) = 0bc’ + 0b’d + 0b’cd’ + bc’d + 1cd + 1bcd’ + 1c’d
= bc’d + cd + bcd’ + c’d
= d(c’ + c) + b(c’d + cd’)
= c’d(1 + b) + cd + bcd’
= c’d + cd + bcd’

f(0,b,c,d) = 1bc’ + 1b’d + 1b’cd’ + bc’d + 0cd + 0bcd’ + 0c’d
= bc’ + b’d + b’cd’ + bc’d
= bc’(1 + d) + b’d + b’cd’
= bc’ + b’d + b’cd’

Apply Shannon’s expansion:
LHS = af(1,b,c,d)+ a’f(0,b,c,d)
= ac’d + acd + abcd’ + a’bc’ + a’b’d + a’b’cd’

How does this look so far?

#### Kermit2

Joined Feb 5, 2010
4,162
Look again at the 4th line down from the sentence SHANNON'S EXPANSION. You dropped a term.

• audreyeckman

#### audreyeckman

Joined Jan 7, 2016
8
Woops I had a strikethrough on that line in my word doc.. so it shouldn't be there at all.. it should be like this

Let f(a,b,c,d) = a′bc′ + a′b′d + a′b′cd′ + bc′d + acd + abcd′ + ac′d

f(1,b,c,d) = 0bc’ + 0b’d + 0b’cd’ + bc’d + 1cd + 1bcd’ + 1c’d
= bc’d + cd + bcd’ + c’d

= c’d(1 + b) + cd + bcd’
= c’d + cd + bcd’

f(0,b,c,d) = 1bc’ + 1b’d + 1b’cd’ + bc’d + 0cd + 0bcd’ + 0c’d
= bc’ + b’d + b’cd’ + bc’d
= bc’(1 + d) + b’d + b’cd’
= bc’ + b’d + b’cd’

Apply Shannon’s expansion:
LHS = af(1,b,c,d)+ a’f(0,b,c,d)
= ac’d + acd + abcd’ + a’bc’ + a’b’d + a’b’cd’​

#### audreyeckman

Joined Jan 7, 2016
8
Do I just continue applying shannon's expansion for other terms like c' to simplify more?

#### Kermit2

Joined Feb 5, 2010
4,162
Yes.

IF it results in a more simplified form.

#### RBR1317

Joined Nov 13, 2010
552
• audreyeckman

#### Glenn Holland

Joined Dec 26, 2014
705
I would use "Demorgan's Theorem" instead of a Karnaugh Map to simplify it.

If there is an inversion bar over an operator, remove only the bar above operator, then change the operator from AND to OR or vice versa.

#### shteii01

Joined Feb 19, 2010
4,644
I plotted the original Boolean function directly onto a Karnaugh map in order to track the progress of minimization. Note that each 4-variable term maps to one square on the K-map, while each 3-variable term maps to two squares, etc. (Some of the terms will overlap in a square because of logical redundancies.) The current state of your minimization is shown in red.
View attachment 98248
I am with Glenn, I don't think OP was taught Karnaugh Map yet. So the OP supposed to use Boolean Algebra rules. At least that is my understanding of the situation.

#### RBR1317

Joined Nov 13, 2010
552
So the OP supposed to use Boolean Algebra rules.
Note that the Karnaugh map was not suggested as the method for solution, merely as an aid to track progress, possibly as an indicator where further application of Boolean Algebra rules will lead to simplification.

#### WBahn

Joined Mar 31, 2012
25,928
But it can only serve that purpose if the TS understands what a Karnaugh map is and how to use it.

#### RBR1317

Joined Nov 13, 2010
552
...if the TS understands...
Quite so. But don't be too quick to assume what a TS is not capable of understanding. Writing/manipulating a K-map can be difficult, but reading one can be fairly easy and possibly learned with nothing more than a good example. I would hope that I provided a good example, but if not, this won't be the first wasted effort that led nowhere.