## Archive for bug

## golden bug [jatp]

Posted in pictures, Travel with tags bug, garden, jatp, scarab beetle, summer on August 11, 2021 by xi'an## genuine Latin~~square~~ rectangle

Posted in Books, pictures, Statistics, Travel, University life with tags art restoration, bioweapon, bug, Choi Seok-jeong, Firenze, Florence, Italia, latin square, Leonhard Euler, Medicis, NYT, sacrophagus, Serratia ficaria SH7 on June 9, 2021 by xi'an
## the strange occurrence of the one bump

Posted in Books, Kids, R, Statistics with tags accept-reject algorithm, bug, code golf, cross validated, debugging, Gamma generator, mixture of distributions, R, sample on June 8, 2020 by xi'an**W**hen answering an X validated question on running an accept-reject algorithm for the Gamma distribution by using a mixture of Beta and drifted (bt 1) Exponential distributions, I came across the above glitch in the fit of my 10⁷ simulated sample to the target, apparently displaying a wrong proportion of simulations above (or below) one.

a=.9 g<-function(T){ x=rexp(T) v=rt(T,1)<0 x=c(1+x[v],exp(-x/a)[!v]) x[runif(T)<x^a/x/exp(x)/((x>1)*exp(1-x)+a*(x<1)*x^a/x)*a]}

It took me a while to spot the issue, namely that the output of

z=g(T) while(sum(!!z)<T)z=c(z,g(T)) z[1:T]

was favouring simulations from the drifted exponential by truncating. Permuting the elements of z before returning solved the issue (as shown below for a=½)!

## weird bug

Posted in Kids, pictures with tags beetle, bug, garden, ladybug, Nezera viridula, raspberries on August 2, 2015 by xi'an## Le Monde puzzle [#6]

Posted in R, Statistics with tags bug, Le Monde, mathematical puzzle, prime numbers, schoolmath on February 18, 2011 by xi'an**A** simple challenge in ** Le Monde** this week: find the group of four primes such that any sum of three terms in the group is prime and the overall sum is minimised. Here is a quick exploration by simulation, using the schoolmath package (with its imperfections):

A=primes(start=1,end=53)[-1] lengthA=length(A) res=4*53 for (t in 1:10^4){ B=sample(A,4,prob=1/(1:lengthA)) sto=is.prim(sum(B[-1])) for (j in 2:4) sto=sto*is.prim(sum(B[-j])) if ((sto)&(sum(B)<res)){ res=sum(B) sol=B} } }

providing the solution 5 7 17 19.

**A** subsidiary question in the same puzzle is whether or not it is possible to find a group of five primes such that any sum of three terms is still prime. Running the above program with the proper substitutions of 4 by 5 does not produce any solution, even when increasing the upper boundary in A. So it is most likely that the answer is no.