That's what I have trouble with. Basically you need to see which bits cancel out. For instance, if a grouping has one bit that is A'B and another bit that is AB, then the A cancels out leaving just B. Another way to look at it is if there is an odd number of one type of state and an even number of its complement, then that state is part of the final term. If there is an even number of a state and its complement, then that state gets cancelled out.
If the state changes then ignore it but if not count it. Look at the sets of two numbers together to the right of AB and below CD (i.e. 00, 01, 11, 10). For example - Red bit (AB and CD are changing states): AB AB 01 -----> 11 A goes from 0 to 1 (ignore it) B goes from 1 to 1 (keep it) ---> 1 = B CD CD 00 -----> 01 C goes from 0 to 0 (keep it) ---> 0 = C' D goes from 0 to 1 (ignore it) Final expression for red bit - BC' ________________________________________ Green bit (Only CD are changing states): AB AB 01 -----> 01 A goes from 0 to 0 (keep it) ---> 0 = A' B goes from 1 to 1 (keep it) ---> 1 = B CD CD CD CD 00 -----> 01 -----> 11 -----> 10 C goes from 0 to 0 to 1 to 1 (ignore it) D goes from 0 to 1 to 1 to 0 (ignore it) Final expression for green bit - A'B ________________________________________ Magenta bit (Only B is changing states): AB AB 00 -----> 01 A goes from 0 to 0 (keep it) ---> 0 = A' B goes from 0 to 1 (ignore it) CD CD 11 -----> 11 C goes from 1 to 1 (keep it) ---> 1 = C D goes from 1 to 1 (keep it) ---> 1 = D Final expression for magenta bit - A'CD Final Final expression - E = BC' + A'B + A'CD Hope that helps!
It's easier if you write out the letters of each group, I do them on top of each other. Now eliminate the values that differ (ie. a group of four having A and A-Not varying throughout but BCD consistent then becomes just BCD). If you have more than one group, you continue on in the equation by +'ing it (BCD + AD + etc..). Once you're good enough you can map more quickly using 1's and 0's. I still just write it out
Each bit listed is one bit apart in all directions. Even wrapping around from one end to the other. So 00, 01, 11, 10 | 00, 01, 11, 10 | ... Note that you aren't counting. You are just making sure that each category is different by only 1 bit. So even though 01 is 1 and 11 is 3, they only differ by 1 bit. If you so 01 then 10 (2), they would differ by 2 bits (both bits need to swap states).
