>
> Professor Wolff,
>
> I have a few questions on the homework solutions.
> In question 4, a>=b was done in two parts: take the NOT of the sign bit 
> of a-b OR if A=B.  Won't the NOT of the sign bit of a-b be 1 if a==b
> making the second a=b redundant?

You are correct for 2's complement representation!
 
  a | b | a-b | sign(a-b) | !sign(a-b)| sign(b-a)| !sign(b-a)
  --+---+-----+-----------+-----------+----------+-----------
  5 | 3 |  2  |   0       |   1       |  1       |   0
  3 | 3 |  0  |   0       |   1       |  0       |   1
  3 | 5 | -2  |   1       |   0       |  0       |   1
    |   |     | if (a<b)  | if (a>=b) | if (b<a) | if (b>=a)
    |   |     | if (b>a)  | if (b<=a) | if (a>b) | if (a<=b)

What about 1's complementation representation??? possible extra credit    
question!

>  Likewise, if we wanted just a>b, can't we just take the sign bit of b-a?

correct!

> I don't understand the need for two stages in a comparator function.

>Also, for IF(a), is NOT(a==0) box really necessary?

The C language hides the implied comparison in order to write less code,
which to the minimalist programmer is desirable.

> Isn't this just  another way of writing IF(NOT(a==0))
> which can be simplified to IF(a)?  The box I just used was an IF box.

The IF box you propose is exactly the logic function NOT(a==0) or !(a==0).
Somewhere there has to be hardware has to do it.

--Prof. Wolff.