CPSC 359: Tutorial 2
PhD Student
Fall 2018
Today
More exercises related to algebraic minimization
So what are the tools we have for minimization?
Theorems
| a) | b) | |
|---|---|---|
| Postulate 2 | x+0 = x | x*1 = x |
| Postulate 5 | x+x' = 1 | x*x' = 0 |
| Theorem 1 | x + x = x | x*x = x |
| Theorem 2 | x+1 = 1 | x*0 = 0 |
| Theorem 3 | (x')' = x | |
| Theorem 4 | x+y=y+x | xy=yx |
| Theorem 4 | x+(y+z)=(x+y)+z | (xy)z=x(yz) |
| Theorem 5 | (x+y)' = x'y' | (xy)' = x' + y' |
| Theorem 6 | x+xy = x | x(x+y) = x |
An approach to minimization is to factor out terms like (x' + x) or x'x
Boolean Functions
\(f=x+y\cdot z\)
The goal of this tutorial is to get you familliar with boolean functions, i.e.
\(x\)
\(z\)
\(y\)
\(f\)
We'll look at a couple of these as examples
Theorems
In the original slide deck I forgot the absorption law
\(x + (x\cdot y) = x\) (OR Absorption Law)
\(x\cdot (x + y) = x\) (AND Absorption Law)
Algebraic Simplification
Recall the function from last tutorial
\(f=x+y^\prime+z^\prime+yz\)
We reasoned that \(f=1\) by the truth table, but how do we see it via algebraic manipulation?
Exercises
Algebraically minimize the following function, then create a truth table for both functions to show they're equivalent
\(f=x+x'y\)
| a) | b) | |
|---|---|---|
| Postulate 2 | x+0 = x | x*1 = x |
| Postulate 5 | x+x' = 1 | x*x' = 0 |
| Theorem 1 | x + x = x | x*x = x |
| Theorem 2 | x+1 = 1 | x*0 = 0 |
| Theorem 3 | (x')' = x | |
| Theorem 4 | x+y=y+x | xy=yx |
| Theorem 4 | x+(y+z)=(x+y)+z | (xy)z=x(yz) |
| Theorem 5 | (x+y)' = x'y' | (xy)' = x' + y' |
| Theorem 6 | x+xy = x | x(x+y) = x |
Exercises
Algebraically minimize the following function, then create a truth table for both functions to show they're equivalent
\(f=xy + x'z+yz\)
| a) | b) | |
|---|---|---|
| Postulate 2 | x+0 = x | x*1 = x |
| Postulate 5 | x+x' = 1 | x*x' = 0 |
| Theorem 1 | x + x = x | x*x = x |
| Theorem 2 | x+1 = 1 | x*0 = 0 |
| Theorem 3 | (x')' = x | |
| Theorem 4 | x+y=y+x | xy=yx |
| Theorem 4 | x+(y+z)=(x+y)+z | (xy)z=x(yz) |
| Theorem 5 | (x+y)' = x'y' | (xy)' = x' + y' |
| Theorem 6 | x+xy = x | x(x+y) = x |
Complement of a Function
\(f=xy + x'z+yz\)
At times, it's useful to take the compliment of a function, i.e.
\(f'=(xy + x'z+yz)'\)
Two "methods" for distributing the negation, DeMorgan's and the short cut method.
Short Cut Method
\(f=xy + x'z+yz\)
DeMorgan's Theorem is straight forward to use, but a bit tedious, so we have the "short cut method". i.e
(x+y)(x'+z)(y+z)
- Form the dual (ands become ors, ors become ands)
- Negate each literal
f=(x'+y')(x+z')(y'+z')
Exercises
Use both DeMorgan's rule and the shortcut method to calculate the complement of
\(f=(x'yz' + x'y'z)'\)
| a) | b) | |
|---|---|---|
| Postulate 2 | x+0 = x | x*1 = x |
| Postulate 5 | x+x' = 1 | x*x' = 0 |
| Theorem 1 | x + x = x | x*x = x |
| Theorem 2 | x+1 = 1 | x*0 = 0 |
| Theorem 3 | (x')' = x | |
| Theorem 4 | x+y=y+x | xy=yx |
| Theorem 4 | x+(y+z)=(x+y)+z | (xy)z=x(yz) |
| Theorem 5 | (x+y)' = x'y' | (xy)' = x' + y' |
| Theorem 6 | x+xy = x | x(x+y) = x |
Exercises
Use both DeMorgan's rule and the shortcut method to calculate the complement of
\(f=(x[y'z' +yz])\)'
| a) | b) | |
|---|---|---|
| Postulate 2 | x+0 = x | x*1 = x |
| Postulate 5 | x+x' = 1 | x*x' = 0 |
| Theorem 1 | x + x = x | x*x = x |
| Theorem 2 | x+1 = 1 | x*0 = 0 |
| Theorem 3 | (x')' = x | |
| Theorem 4 | x+y=y+x | xy=yx |
| Theorem 4 | x+(y+z)=(x+y)+z | (xy)z=x(yz) |
| Theorem 5 | (x+y)' = x'y' | (xy)' = x' + y' |
| Theorem 6 | x+xy = x | x(x+y) = x |
Next day
We'll look at Assignment 1, talk about what's expected and what's important to know.
Probably also canonical forms...
CPSC 359: Tutorial 2
By Joshua Horacsek
