Question about parallel FULL ADDER
- Thread starter NotAlive
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How do you know that it does?View attachment 195462View attachment 195463
Hello, I would like to get help in this question :
a=a2a1a0 , b=b2b1b0, a>b
why s equals to above?
Thank you
What approaches are available to you?
What approach would you like to take in verifying the statement, or alternatively proving it false?
The answer given is correct.
Start with being sure how to combine and dissect the values and their bits.
If a = a2a1a0, then mathematically
a = (2^2)·a2 + (2^1)·a1 + (2^0)·a0
a = 4·a2 + 2·a1 + 1·a0
and similar for the other two.
Now, given 'a', how could you pick off each of its bits using div (integer division) and mod (remainder after integer division) operations?
Once you have that, then you can either piece together the equation they have, or simply brute force the result from the circuit.
Perhaps it might help to group and slightly rewrite the result as
S = [b - (b mod2)] + [a - {[(a div 2) mod 2[ * 2}] + [3 + 1]
Start with being sure how to combine and dissect the values and their bits.
If a = a2a1a0, then mathematically
a = (2^2)·a2 + (2^1)·a1 + (2^0)·a0
a = 4·a2 + 2·a1 + 1·a0
and similar for the other two.
Now, given 'a', how could you pick off each of its bits using div (integer division) and mod (remainder after integer division) operations?
Once you have that, then you can either piece together the equation they have, or simply brute force the result from the circuit.
Perhaps it might help to group and slightly rewrite the result as
S = [b - (b mod2)] + [a - {[(a div 2) mod 2[ * 2}] + [3 + 1]
Of course it is, but checking by any means necessary may lead to an ah-ha moment, and that is really the point. Isn't it??
