Author: Martin Giepmans
Date: 17:26:11 12/19/01
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On December 19, 2001 at 19:05:41, Martin Giepmans wrote: >On December 19, 2001 at 16:57:46, Michel Langeveld wrote: > >>Problem = 1 - ( 16777216! / (16746008! * 16777216 ^ 31208) ) >> >>16777216 = Number of records in my hashtable >>31208 = Number of nodes searched >> >>Windows calculator says BOOM. >> >>If Problem > 0.5 then I've spend 15 hours debugging for nothing :-) > > >problem = 1 - f(b) > >where f(b)=a!/((a-b)!*a^b) > >"15 hours for nothing" if problem>0.5 >this is true if f(b)<=0.5 > >well, f(b+1)=a!/((a-b-1)!*a^(b+1)) = a!/((a-b)!/((a-b)*a^b*a)) >=((a-b)/a)*f(b) ok? > >in your case: a-b/a <0.999 > >f(1)=1 >f(2)=((a-b)/a)*f(1)<0.999*f(1)=0.999 >f(3)=((a-b)/a)*f(2)<0.999^f(2)<0.999^2 >.... >f(b)<0.999^(b-1) > > >f(1000)<0.999^(1000-1) >this is allready <=0.5 and your f(b) is even smaller, so ... > >Martin "in your case: a-b/a<0.999" is not correct It should be: for your a and b>10000 is (a-b)/a < 0.9995 now f(10000)<1 and f(20001)<0.9995^10000*f(10000) (same reasoning as above). this is less then 0.003 and f(b) for your b is even smaller, so the conclusion is correct.
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