Quote from: ChuckOne on February 10, 2014, 11:19:11 AM

Quote from: jkoil on February 10, 2014, 12:10:11 AM

Quote from: ChuckOne on February 09, 2014, 11:57:51 PM

Quote from: gimre on February 09, 2014, 11:32:30 PM

Quote from: ChuckOne on February 09, 2014, 11:01:02 PM

Quote from: gimre on February 09, 2014, 10:45:24 PM

if you're asking if negative input is legal, yes it is.

To simplify it, it's similar to how modulo operates in C.
If you do:

Code:

int x = (1-3) % 5; // -2 % 5

you'll get "-2" as a result, but what you're actually interested in is:

Code:

int x = (1-3 + group_order) % 5; // (-2 + 5) % 5 = 3 % 5 == 5

group_order == 5, and you'll get 3 as a result...

Yes, I know.

So, the problem is that -2 or 3 is processed further and we are not certain if ~~-2 behaves differently than 3 does~~. That is certainly a problem.

EDIT: we are not certain if the 3 resulting from a -2 is process the same way as we would give it another try.

"-2" is passed as input to verify() which was expecting to see "3" not "-2"...

But it is not sure, if -2 in form as a 3 would result in the same outcome as it would be a 'real' 3 from the beginning.

In math -2 mod 5 equals 3. Is that the result, which is wanted? ... in that case C implementation is wrong there and should not use % with negative numbers. (change to positive by adding 5)

I think that's not the point here. Consider this:

Code:

b = f(a)

c = g(b)

d = h(a, c)

Let's assume c results in -2 with a certain b.

Even if c=-2 and c=3 are interchangeable mathematically, h might not work as expected as the pair (a, c) belong together.

So, it could be the case that h(a, -2) != h(a, 3) for the very same a. This must be avoided at any cost.

yes ... and no

No; I didn't mean that c=-2 and c=3 are interchangeable. The results and consequences are likely different, possibly in "many" cases, where the c will be used (in h(a,c) as you stated).
Yes; as you said "this must be avoided at any cost." I agree, when not knowing the background of the original implementation (has it been assumed that no negative input or something else...)