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- Thread starter ibcoding
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- #2

cepheid

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The diagram is confusing a bit, because one of the a lines is very close to one of the b lines. However it looks correct to me.

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- #4

cepheid

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I think you can simplify this alot using DeMorgan's and other things. For instance, take the first input to the OR gate:

(ab)' * (b'c)'

Using DeMorgan's theorem (xy)' = x' + y', you can expand each thing in parentheses:

= (a' + b') * (b + c') [eq. 1]

and now using the distributive property of AND operator (which I believe is present for Boolean algebra?)

= a'b + a'c' + b'b + b'c'

and using the associative property:

a'(b + c') + 0 + b'c'

Anything OR'd with 0 is just itself, so this becomes:

a'(b + c') + b'c' [eq. 2]

I'm not sure that this is any simpler in this case, but the point is that you could, in principle, try to minimize the number of gates. I wrote a little program to generate the truth tables for eq. 1 and eq. 2 and they came out the same.

(ab)' * (b'c)'

Using DeMorgan's theorem (xy)' = x' + y', you can expand each thing in parentheses:

= (a' + b') * (b + c') [eq. 1]

and now using the distributive property of AND operator (which I believe is present for Boolean algebra?)

= a'b + a'c' + b'b + b'c'

and using the associative property:

a'(b + c') + 0 + b'c'

Anything OR'd with 0 is just itself, so this becomes:

a'(b + c') + b'c' [eq. 2]

I'm not sure that this is any simpler in this case, but the point is that you could, in principle, try to minimize the number of gates. I wrote a little program to generate the truth tables for eq. 1 and eq. 2 and they came out the same.

Last edited:

- #5

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1

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- #6

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I think you can simplify this alot using DeMorgan's and other things. For instance, take the first input to the OR gate:

(ab)' * (b'c)'

Using DeMorgan's theorem (xy)' = x' + y', you can expand each thing in parentheses:

= (a' + b') * (b + c') [eq. 1]

and now using the distributive property of AND operator (which I believe is present for Boolean algebra?)

= a'b + a'c' + b'b + b'c'

and using the associative property:

a'(b + c') + 0 + b'c'

Anything OR'd with 0 is just itself, so this becomes:

a'(b + c') + b'c' [eq. 2]

I'm not sure that this is any simpler in this case, but the point is that you could, in principle, try to minimize the number of gates. I wrote a little program to generate the truth tables for eq. 1 and eq. 2 and they came out the same.

Ahh, the a' a' are adjacent so it is absorbed, got ya

- #7

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So the equation should be a'(b + c') + b'pc'

- #8

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which gives me three and gates and two or gates

- #9

cepheid

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Alright, variable p is missingfrom that equation, but I get the idea.

Well, I was ONLY considering the (ab)'(b'c)' in the original equation, as an example. I wasn't trying to simplify the whole thing.

On that note, is this the correct truth table outputs:

1

1

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0

1

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1

1

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0

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0

Here's what I get:

Code:

```
a | b | c | p | f
0 | 0 | 0 | 0 | 1
0 | 0 | 0 | 1 | 1
0 | 0 | 1 | 0 | 0
0 | 0 | 1 | 1 | 0
0 | 1 | 0 | 0 | 1
0 | 1 | 0 | 1 | 1
0 | 1 | 1 | 0 | 1
0 | 1 | 1 | 1 | 1
1 | 0 | 0 | 0 | 1
1 | 0 | 0 | 1 | 1
1 | 0 | 1 | 0 | 0
1 | 0 | 1 | 1 | 0
1 | 1 | 0 | 0 | 0
1 | 1 | 0 | 1 | 0
1 | 1 | 1 | 0 | 0
1 | 1 | 1 | 1 | 0
```

- #10

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Sorry, I wasn't complaining. :) Alright, that's the same as what I have.

- #11

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The equation breaks down to a'b + b'c' + a'bp, I believe.

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