We often need to determine the minimum circuitry for a particular Boolean expression in order to reduce the cost and surface area of the circuit or to improve the performance of our circuit. Karnaugh maps provide an alternative solution to Boolean algebra in determining the minimum circuitry required to implement a logic circuit. Boolean algebra can be very tedious and it can be difficult to determine a consistent solution to a minimisation problem. Karnaugh maps on the other hand provide a formalised method for solving such a problem and, if the the rules are followed correctly, it will result in a consistent solution. Karnaugh maps work well for up to 5/6 variable, but after that they too become unwieldy and a software assisted solution is required.

There is a very good discussion on Karnaugh maps in the notes: Course Notes

The problem that we wish to solve is as follows:

A system has 4 momentary push buttons (A, B, C, D). We wish to light a single LED when the following conditions arise:

• Any 3 buttons are pressed at the same time.
• All 4 buttons are pressed at the same time.
• The outer, or inner pair of buttons are pressed (i.e. AD or BC)

In all other circumstances the LED should be off.

You should produce a minimum logical expression and implement your solution using a mixture of AND/OR gates. Also, you should provide a solution using only NAND gates.

Please try to solve this problem yourself and if you have problems then work through the videos below. The solution to this problem is provided in two parts, which is due to duration constraints with YouTube videos. 

Part 2 Continued (YouTube limits video length).

Hopefully this will give you a good idea of Karnaugh maps and how we can use them to solve logic minimisation problems. Please read the course notes to ensure that you have a good feel for all of the rules for minimisation.