Re: Eureqa - create bounded solution?

[Michael Schmidt]

There's a few options here, you could build the range into the model using the target expression:

y = constrain( f(x), -10, 10)

or

y = max(-10, min(-10, f(x) ))

or

y = 20*logistic( f(x) ) - 10

There's probably some interesting things you could do with a custom error metric to reward or penalize for being in the range also:

http://blog.eureqa.com/2010/07/custom-error-metrics-and-special-search.html

Finally you could try having a hard restriction (not recommended for performance reasons):

y = f(x) + 0*require(between( f(x), -10, 10))

Michael

On Sat, May 18, 2013 at 11:43 AM, Henry Albertson <shug...@gmail.com> wrote:

Hi all. I am working to define a formula that takes in a large number of values and returns one based on this one value. I want the output to be within a pre-defined range of integer values.

I have found some solutions for ranges of outputs per set of inputs but this is not quite what I want.

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