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# Help with boolean expression to logic gate circuits watch

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1. Hi everyone iv nearly finished my assignment but I'm really stuck on the last part where i need to convert boolean expressions to logic gate circuits.

1. !A!BC!D + !A!BCD + !AB!C!D + !AB!D + !ABC!D + !ABCD + AB!CD + ABC!D + ABCD!A!BC + !ABC + A!BC + ABC

2. !A!B!C!D + ABC!D + !ABC!D + !ABCD + AB!C!D + AB!CD

3. !A!B!C!D + !A!B!CD + !AB!C!D + !ABC!D + A!B!C!D + A!BC!D + AB!C!D + ABC!D

4. !A!B!CD + !A!BC!D + !ABCD + A!B!C!D + ABC!D + ABCD

5. !A!BCD + !ABC!D + A!B!CD + AB!CD + ABCD

Here are the 5 expressions I have to convert, could anyone help with working one or more of these out and give me some insight on how to work out the rest. Any help would be fantastic.

Thanks!
2. basically where you can, you factorise the common factor (with an AND gate between if there are 2 common factors) and then OR them with the other factors you have found

For example
ABC!D + AB!C!D + !A!B!C!D + ABC!D

Would become NOT D since they all have not D in common
3. (Original post by CompSci in2k17)
basically where you can, you factorise the common factor (with an AND gate between if there are 2 common factors) and then OR them with the other factors you have found

For example
ABC!D + AB!C!D + !A!B!C!D + ABC!D

Would become NOT D since they all have not D in common
Correct me if I'm wrong but are you sure that is right as letters next to each other are put through the AND operation, so you need to evaluate the expression that NOT D has been factored out of as well.
4. (Original post by Faction Paradox)
Correct me if I'm wrong but are you sure that is right as letters next to each other are put through the AND operation, so you need to evaluate the expression that NOT D has been factored out of as well.
That is correct, excuse my poor wording

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Updated: February 9, 2018
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