Re: [gpsim-devel] overflow flag

[Scott D. <sc...@da...>]

On Fri, 11 Jun 2004, Eric Smith wrote:

> Scott wrote:
>> A perhaps somewhat simpler expression is:
>>    v = ( ((sum ^ ~b) & (a ^ ~b)) & 0x80) ? 1 : 0;
>
> Does not seem to give the same results.  An error in simplification?

I obviously didn't test it. Instead, I (mis)read the boolean equation from 
a karnaugh map. I should've written:

  v = ( ((sum ^ a) & (a ^ ~b)) & 0x80) ? 1 : 0;

Or you can write:

  v = ( ((sum ^ a) & (~a ^ b)) & 0x80) ? 1 : 0;

The karnaugh map for addition looks like:

                a,b
            pp  pn  nn  np
           +---+---+---+---+
         p |   |   | v |   |
    sum    +---+---+---+---+
         n | v |   |   |   |
           +---+---+---+---+

the 'p' and 'n refer to the sign (bit 7). The 'v' in the map corresponds 
to the conditions where there is an overflow. To read this, you can see v 
is true for two conditions. The first is when the sum is p and the two 
inputs are n. There are numerous ways to extract the boolean equation from 
the map. The most direct case is the first one I wrote. The others 
involving xor's and and's.


                a,b
            pp  pn  nn  np
           +---+---+---+---+
         p | 1 |   | 1 |   |
    sum    +---+---+---+---+
         n | 1 |   | 1 |   |
           +---+---+---+---+

This map is:  a ^ ~b = ~a ^ b = ~(a ^ b)

                a,b
            pp  pn  nn  np
           +---+---+---+---+
         p |   |   | 1 | 1 |
    sum    +---+---+---+---+
         n | 1 | 1 |   |   |
           +---+---+---+---+

This map is:  sum ^ a = ~(sum ^ ~a) = ~(~sum ^ a)

Anding these two maps yields the overflow map. So there are at least 9 
different ways of writing this with the xor's and and's.

Scott

PS. Do they still teach Karnaugh maps in logic design?