>We aren't dealing with mathematicians, but programmers.
I am attempting to reconcile this with Colin's rationale in terms of
congruence classes over rings Z/nZ. ;)
b.
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On Fri, Oct 15, 2010 at 5:51 AM, John Baldwin wrote:
> On Thursday, October 14, 2010 11:49:23 pm Garrett Cooper wrote:
>> On Thu, Oct 14, 2010 at 6:37 AM, John Baldwin wrote:
>> > On Thursday, October 14, 2010 7:58:32 am Andriy Gapon wrote:
>> >> on 14/10/2010 00:30 Garrett Cooper said the follow
On Thursday, October 14, 2010 11:49:23 pm Garrett Cooper wrote:
> On Thu, Oct 14, 2010 at 6:37 AM, John Baldwin wrote:
> > On Thursday, October 14, 2010 7:58:32 am Andriy Gapon wrote:
> >> on 14/10/2010 00:30 Garrett Cooper said the following:
> >> > I was talking to someone today about this m
Garrett Cooper wrote:
> I was talking to someone today about this macro, and he noted that
> the algorithm is incorrect -- it fails the base case with ((x) == 0 --
> which makes sense because 2^(x) cannot equal 0 (mathematically
> impossible, unless you consider the limit as x goes to negative
On Thu, Oct 14, 2010 at 6:37 AM, John Baldwin wrote:
> On Thursday, October 14, 2010 7:58:32 am Andriy Gapon wrote:
>> on 14/10/2010 00:30 Garrett Cooper said the following:
>> > I was talking to someone today about this macro, and he noted that
>> > the algorithm is incorrect -- it fails the
On Thursday, October 14, 2010 7:58:32 am Andriy Gapon wrote:
> on 14/10/2010 00:30 Garrett Cooper said the following:
> > I was talking to someone today about this macro, and he noted that
> > the algorithm is incorrect -- it fails the base case with ((x) == 0 --
> > which makes sense because 2
on 14/10/2010 00:30 Garrett Cooper said the following:
> I was talking to someone today about this macro, and he noted that
> the algorithm is incorrect -- it fails the base case with ((x) == 0 --
> which makes sense because 2^(x) cannot equal 0 (mathematically
> impossible, unless you consider
I was talking to someone today about this macro, and he noted that
the algorithm is incorrect -- it fails the base case with ((x) == 0 --
which makes sense because 2^(x) cannot equal 0 (mathematically
impossible, unless you consider the limit as x goes to negative
infinity as log (0) / log(2) i
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