Karnaugh Maps - Helpsheet
Example Problem 1
Logic Gate Diagram:
In Boolean algebra this is represented by:
A∨B
A / A / AB / 0 / 1
B / 0 / 1
B / 1 / 1 / 1
Karnaugh Map:
Solution: A∨B (already simplified as much as it can be!)
Example Problem 2
Logic Gate Diagram:
In Boolean algebra this is represented by:
A^B∨ A^¬B (Remember the order of precedence – NOT then AND then OR)
Karnaugh Map:
First of all do A^B, then do A^¬B (we can ignore the ∨ as it has no overall effect on the result)
A / A / AB / 0 / 1
B / 0 / 1
B / 1 / 1
Solution: Using the map, we can see if A is True, then the expression A^B∨ A^¬B is always True, therefore A^B∨ A^¬B can be simplified to simply A
A^B∨ A^¬B = A
Example Problem 3
Logic Gate Diagram:
In Boolean algebra, this is represented by:
¬C^B∨ A^B∨ C
Karnaugh Map:
First of all look at ¬C^B and place the results on the map, then place the results of A^B on the map then look at C (remember we can ignore any ∨ as it has no overall effect on the result)
AB / AB / AB / AB / ABC / 00 / 01 / 11 / 10
C / 0 / 1 / 1
C / 1 / 1 / 1 / 1 / 1
Solution: Using the map, we can see that if C is True then the expression ¬C^B∨ A^B∨ C is always True. We can also see that if B is True then the expression ¬C^B∨ A^B∨ C is always True. Therefore if C or B are True, then the expression ¬C^B∨ A^B∨ C is always True. So…
¬C^B∨ A^B∨ C = C∨B
Don’t believe me? Try creating the logic diagram in logic.ly to prove it works..