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Logical Effort (G) and Effort (H) in Parallel

Alon123

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Hi Everyone,

I am struggling with this question in which I need to understand what is the route with the least effort (H).
Usually, I'm given a route but it's very simple to calculate, because all the other inputs of the logical gates are regular inputs without any logical gates connected to them, like in this example:

1602939648252.png

The main difference and what I am struggling with, is in the example above the route is clear, and the legs of the logical gates are just regular inputs..

The circuit which I need to analyze is this:

1602939630935.png


And the question is which route has the least effort (H) - A,C, or they both the same.

My hunch is that they're both the same, but I can't figure out how to calculate it.
I think the fan-out (f) is the same for both routes - 4/1=4 (I might be wrong here so please let me know if I do).
But I'm not sure how to calculate the logical effort (G) in the routes.
For example, if I look at route A, so the first logic gate is a NOT (g=1), and the next one is NAND2 (g=4/3), but, the other input of the NAND also has a NOT gate connected to it. In my opinion, this shouldn't matter (because again, g of NOT is 1). But when I look further to this route I have a XOR2 (g=4). so far, G=g1*g2*g3=1*4/3*4=16/3, and H=G*F=64/3.
But I don't think it's right, since the other input of the XOR2 is connected to route C, so it has a NOR and a NOT beforehand.

So finally - my question is how do you calculate the effort of a route, if it has more logical gates from the other input?

Thanks in advance!!
 

ThisIsNotSam

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path A is NOT-NAND2-XOR2
path C is NOT-NOR2-XOR2

what you need to evaluate is the difference between NAND2 and NOR2. this will determine the slowest path. you have to look at paths from input to output, you don't look backwards otherwise you would take loads into account multiple times.
 

Alon123

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path A is NOT-NAND2-XOR2
path C is NOT-NOR2-XOR2

what you need to evaluate is the difference between NAND2 and NOR2. this will determine the slowest path. you have to look at paths from input to output, you don't look backwards otherwise you would take loads into account multiple times.
Great, thanks!
In that case, path A is the least effort since the NAND2 has less logical effort than NOR2.

Thanks again!
 

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