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847Re: Excercise 18.14, page 677

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  • daashmashty
    Oct 23, 2008
    • 0 Attachment
      The attachment was stripped as suspected, so here's the readable version...

      http://ai.autonomy.net.au/chrome/site/AIMA-18.4.png

      Nima

      --- In aima-talk@yahoogroups.com, "Nima Talebi" <nima.talebi@...> wrote:
      >
      > Okay, I can already see a flaw in this, here's my revised version...
      >
      > I've attached my formula as a png, hope it gets through, in case it
      > doesn't...
      >
      > *Text:*
      > E=|_sum_{{i=0};{i=M-m};{|_pmatrix1x2_{{M};{m+i}}ε^{m+i}}}
      >
      > *LaTeX:
      > *\mbox{E}=\sum_{i=0}^{i=M-m}{\left(\begin{array}{c} M \\ m+i
      > \end{array}\right)\epsilon ^{m+i}}
      >
      > ...where m is the *bigger half* - that is the `half' that has just enough
      > power to out-weight the other half, for example where M = 20, m = 11, where
      > M = 5, m = 3.
      >
      > Again, I don't think I have the right, or at least most elegant solution to
      > the problem, I feel it has to be much simpler - otherwise the question
      > wouldn't seriously expect us to calculate the above formula for M = 20 as
      > that sum would expand into 9 terms!
      >
      > Nima
      >
      >
      > On Fri, Oct 24, 2008 at 12:37 AM, Nima Talebi <nima.talebi@...> wrote:
      >
      > > Hi,
      > >
      > > I'm interested to see if my solution is even remotely correct, the question
      > > is to derive the error formula given the number of experts/hypotheses (M)
      > > and also the error of each hypotheses (e) - assuming all to be the same, and
      > > that all are independent of one another...
      > >
      > > I treated this as a probability problem (and unsure if I should have done
      > > so), and hence derived a formula which yields the following results...
      > >
      > > M:5, e:0.10, 0.0100
      > > M:5, e:0.20, 0.0800
      > > M:5, e:0.40, 0.6400
      > > M:10, e:0.10, 0.0002
      > > M:10, e:0.20, 0.0134
      > > M:10, e:0.40, 0.8602
      > > M:20, e:0.10, 0.0000
      > > M:20, e:0.20, 0.0034
      > > M:20, e:0.40, 7.0448
      > >
      > > The third column is the total error for the network I calculated; it
      > > certainly looks wrong from looking at the last value, but I'd like to know
      > > how to go about solving this as it's rather intriguing.
      > >
      > > My `formula' in Python form is as follows...
      > >
      > > def fact(k):
      > > return reduce(lambda i, j : i*j, range(1, k+1))
      > >
      > > for M in 5, 10, 20:
      > > for e in 0.1, 0.2, 0.4:
      > > m = M/2+1
      > > _m = M-m
      > >
      > > print "M:%d, e:%0.2f, %0.4f"%(M, e, (fact(M) * pow(e, m))/(fact(_m) *
      > > fact(m)))
      > >
      > > Thanks to anyone who has a crack at solving this =)
      > >
      > > Nima
      > >
      > >
      >
      >
      > --
      > Nima Talebiw: http://ai.autonomy.net.au/
      > p: +61-4-0667-7607 m: nima@...
      >
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