Re: Some data on where HP users are Message #1 Posted by Rupert (European Union) on 17 Sept 2000, 1:28 a.m.
> Yes, I also found 5743, and the number of CD's dave > Ordered seemed more realistic. But I was astonnished too > of the
> coincidence of same 7 shipments...
It seems I'm missing something.
I tried to compute that number
with the following assumption:
let Ni=X*Pi,
where X = unknown total CDs number,
Pi=(percentual of CDs sent to the country i)/100,
Ni=number of CDs sent to the country i,
then X should satisfy the condition
abs(round(Ni)  Ni) <= epsilon
for every i, because every calculated Ni should be close to
an integer value.
That way, with epsilon=1e3 I get:
X = 851, already found by other posters, and
X = 1,702.
No other results have been found allowing the algorithm to run
up to 40,000+.
The result 5,743 has not been found by the algorithm.
With an higher value for epsilon, 1e2, I get:
X = 851
X = 1,702
X = 2,553
X = 3,404
X = 4,255
X = 5,106
X = 5,957
X = 6,808
etc.
but the 5,743 value is still not found.
It's also possible to impose a tolerance on the sum
of the abs(round(Ni)  Ni) terms so that X should satisfy the
following condition:
sigma(abs(round(Ni)  Ni)) <= epsilon, where Ni=X*Pi as above.
That way with epsilon=.01 I get only X = 851.
With epsilon=.1 the results are again
X = 851
X = 1,702
X = 2,553
X = 3,404
etc.
but the 5,743 value is still not found.
Infact sigma(5,743) > 11, so that seems not to be
a valid solution at all.
Well, I gave it a try because this problem seems to be a bit tricky, or maybe I'm just out of exercise...

