Thread Subject: Even/odd

How about even and odd functions built-in:

bool x;

x = even(num);
x = odd(num);

"Edward Seger" <edward.h.seger@saic.com> wrote in message
<fak6ev$t2k$1@fred.mathworks.com>...
> How about even and odd functions built-in:
>
> bool x;
>
> x = even(num);
> x = odd(num);

So what's wrong with <mod> ?

mod(x,2)

0 if even, 1 if odd

Steve Amphlett:
<SNIP down to typical even/odd algorithm...

> mod(x,2)
> 0 if even, 1 if odd

% now, since even/odd should be integers,
% this may fail, eg,
     n=-4:4;
     m=n;
     even=mod(m,2)==0;
     odd=~even;
     [m;even;odd]
% if the user is not careful
     m=pi*n;
     even=mod(m,2)==0;
     odd=~even;
     [m;even;odd]
% a better way may be this
     odd=bitand(abs(n),1);
     even=~odd;
     [n;even;odd]
% ...which would give an error early on
     odd=bitand(abs(m),1);

just a thought
us
     

How about

function [ise] = iseven(x);
%works for sccalar and vector
ise = roundn(x/2,0) == x/2;

In article <fpfhha$86m$1@fred.mathworks.com>,
Amber York <yorksea@gmail.com> wrote:
>How about

>function [ise] = iseven(x);
>%works for sccalar and vector
>ise = roundn(x/2,0) == x/2;

roundn is only in the Mapping toolbox.

Please check the function for a full range of input,
such as iseven([-inf -2 -1 0 1 1.49 1.5 1.51 1.99 2 inf nan])

I do not have roundn to check with, but I suspect you will
find that your iseven function will tell you that -inf and inf
are even.

--
Q: Why did the chicken cross the Mobius strip?

A: There were manifold reasons.

> % a better way may be this
> odd=bitand(abs(n),1);

Is the domain of bitand() limited to 32-bit integers? At
least mod(x,2) covers the range of whole numbers up to 53
bits. Sure, the returned matrix ought to be sparse, omitting
values where it is not certain if the number was even or odd
(e.g. either the fraction is not zero or the value is too
large to be odd).

"Andy Robb"
<SNIP bits and pieces...

> > % a better way may be this
> > =A0 =A0 =A0odd=3Dbitand(abs(n),1);
>
> Is the domain of bitand() limited to 32-bit integers? At
> least mod(x,2) covers the range of whole numbers up to 53
> bits...

so does the bitxxx family of stock functions...
the limit is

     help bitmax;

% the advantage of the BITs...
% they complain if something goes wrong
% and dont let you believe your result is correct as with REM/MOD

     bitand(bitmax,1)
% ans =3D
% these will error out
% bitand(pi,1)
% bitand(bitmax+1,1);
% whereas these don't care to warn you
     b=3Dbitmax+(1:5); % valid doubles
     mod(b,2)
% ans =3D 0 0 0 0 0 %!
     rem(b,2)
% ans =3D 0 0 0 0 0 %!

us