Re: problem with EvenQ
- To: mathgroup at smc.vnet.net
- Subject: [mg105950] Re: [mg105923] problem with EvenQ
- From: "David Park" <djmpark at comcast.net>
- Date: Sun, 27 Dec 2009 02:26:49 -0500 (EST)
- References: <8534260.1261873319290.JavaMail.root@n11>
The trouble is that EvenQ[y] returns False unless y is manifestly an even
integer.
You could obtain a result in the following manner:
FindInstance[2 y < 100 \[And] y == 2 x, {x, y}, Integers, 3]
{{x -> 24, y -> 48}, {x -> -247, y -> -494}, {x -> -245, y -> -490}}
Here is another example. Find some odd integers between 6 and 25.
FindInstance[6 <= y <= 25 \[And] y == 2 x + 1, {x, y}, Integers, 4];
y /. %
{23, 7, 21, 17}
Here is the case using a RandomSeed. This gives different results each time
you evaluate.
FindInstance[6 <= y <= 25 \[And] y == 2 x + 1, {x, y}, Integers, 4,
RandomSeed -> RandomInteger[{0, 100}]];
y /. %
{13, 21, 11, 17} one example
But if we ask for just a single instance then RandomSeed appears not to
work. That looks like a bug to me.
David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/
From: dvholten [mailto:info at dvholten.de]
Hi folks,
i puzzled all afternoon, but couldnt get this one solved:
what is the proper way of using EvenQ[] within FindInstance[] ?
I cant use it like
FindInstance[ y*2 < 100 && EvenQ[y], {y}, Integers ]
or even
FindInstance[ EvenQ[y], {y}, Integers ]
Actually, the expression used in FindInstance is much more complex,
but i condensed the problem to be EvenQ[] - i expect some kind of
special notation to help here.
thanks
dvh