Subject:
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Re: On a scale of 0 to 10?
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Newsgroups:
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lugnet.off-topic.geek
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Date:
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Wed, 26 Mar 2003 18:08:15 GMT
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Viewed:
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333 times
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In lugnet.off-topic.geek, Adrian Drake writes:
> In lugnet.off-topic.geek, David Eaton writes:
> > I've got a function that represents a quality rating, based on 2 parameters,
> > which are on a scale of 0 to 1:
> >
> > (d-k)*k => quality (q)
> >
> > Notice that this yields a sort of wacky result, insofar as the quality can
> > vary between -1 and 0.25, not -1 and 1 or -1 and 0 or 0 and 1. Icky. Anyway,
> > what I'd *like* to do is translate the resulting quality rating (q) onto a
> > scale (s) of 0 to 10. In other words, for q = -1, you get a 0, and for a q
> > of .25, you get a 10.
> >
> > Problem. Assuming a linear distribution for 0.25 => 10, and -1 => 0, we get:
> >
> > s = 8q + 8
> >
> > Well, that's all well and good, except for when q = 0, we get s = 8, and I'd
> > rather that a quality of 0 translated to 5 rather than 8, since an 8 implies
> > that such a quality is really good, and in reality, a quality of 0
> > represents 'middle of the road'.
> >
> > So. I figured why not assume a quadratic solution? IE: s = Aq^2 + Bq + C,
> > with known points (-1,0), (0,5), and (0.25,10). However. This seems to yield:
> >
> > s = 12q^2 + 17q + 5
> >
> > So, what's wrong with that? Well, for some values of q (like q = -0.72), s
> > results in a *negative* value (in this case -1.0192). ACK! That means that
> > *some* things that have better quality ratings actually have *LOWER* 0-to-10
> > rankings!
> >
> > I assume I want a logarithmic scale function instead... How do I set that up
> > & solve for it? Or do I want something totally different?
> >
> > DaveE
>
> How about a double-linear solution? from -1 to 0, translate linearly from 0 >to 5. From 0 to .25, translate linearly from 5 to 10. You won't have a
> single equation, but that seems to accomplish what you want, right?
Thought about that... If necessary I guess, but I'm pretty sure there's a
way to do it. I think what I need is:
s = Ae^(q + B) + C
And at the moment, I've solved:
C = -Ae^(B - 1)
A = 10 / (e^(B + .25) - e^(B - 1))
And I'm scratching my head like crazy to solve for B:
e^(B + .25) + e^(B - 1) = 2e^B
It's been too long since algebra! Anyone know how to solve for B here?
DaveE
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Message has 1 Reply:
 | | Re: On a scale of 0 to 10?
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| (...) *snip* (...) summed exponents like that can be re-written in this way: e^B * e^.25 + e^B * e^-1 = 2e^B (e^.25 + e^-1) * e^B = 2e^B Uh oh. Divide both sides by e^B and we've got (e^.25 + e^-1) = 2 1.652 = 2 Which ain't right. I think your math (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
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Message is in Reply To:
| | Re: On a scale of 0 to 10?
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| (...) How about a double-linear solution? from -1 to 0, translate linearly from 0 to 5. From 0 to .25, translate linearly from 5 to 10. You won't have a single equation, but that seems to accomplish what you want, right? Adrian (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
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