[This article was first published on Simply Statistics » R, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

Happy π day everybody!

I wanted to write some simple code (included below) to the test parallelization capabilities of my  new cluster. So, in honor of  π day, I decided to check for evidence that π is a normal number. A normal number is a real number whose infinite sequence of digits has the property that picking any given random m digit pattern is 10−m. For example, using the Poisson approximation, we can predict that the pattern “123456789” should show up between 0 and 3 times in the first billion digits of π (it actually shows up twice starting, at the 523,551,502-th and  773,349,079-th decimal places).

To test our hypothesis, let Y1, …, Y100 be the number of “00”, “01”, …,”99″ in the first billion digits of  π. If  π is in fact normal then the Ys should be approximately IID binomials with N=1 billon and p=0.01.  In the qq-plot below I show Z-scores (Y – 10,000,000) /  √9,900,000) which appear to follow a normal distribution as predicted by our hypothesis. Further evidence for π being normal is provided by repeating this experiment for 3,4,5,6, and 7 digit patterns (for 5,6 and 7 I sampled 10,000 patterns). Note that we can perform a chi-square test for the uniform distribution as well. For patterns of size 1,2,3 the p-values were 0.84, 0.89, 0.92, and 0.99. Another test we can perform is to divide the 1 billion digits into 100,000 non-overlapping segments of length 10,000. The vector of counts for any given pattern should also be binomial. Below I also include these qq-plots. These observed counts should also be independent, and to explore this we can look at autocorrelation plots: To do this in about an hour and with just a few lines of code (included below), I used the Bioconductor Biostrings package to match strings and the foreach function to parallelize.

library(Biostrings)
library(doParallel)
registerDoParallel(cores = 48)
x=scan("pi-billion.txt",what="c")
x=substr(x,3,nchar(x)) ##remove 3.
x=BString(x)
n<-length(x)
p <- 1/(10^d)
par(mfrow=c(2,3))
for(d in 2:4){
if(d<5){
patterns<-sprintf(paste0("%0",d,"d"),seq(0,10^d-1))
} else{
patterns<-sprintf(paste0("%0",d,"d"),sample(10^d,10^4)-1)
}
res <- foreach(pat=patterns,.combine=c) %dopar% countPattern(pat,x)
z <- (res - n*p ) / sqrt( n*p*(1-p) )
qqnorm(z,xlab="Theoretical quantiles",ylab="Observed z-scores",main=paste(d,"digits"))
abline(0,1)
##correction: original post had length(res)
if(d<5) print(1-pchisq(sum ((res - n*p)^2/(n*p)),length(res)-1))
}
###Now count in segments
d <- 1
m <-10^5

patterns <-sprintf(paste0("%0",d,"d"),seq(0,10^d-1))
res <- foreach(pat=patterns,.combine=cbind) %dopar% {
tmp<-start(matchPattern(pat,x))
tmp2<-floor( (tmp-1)/m)
return(tabulate(tmp2+1,nbins=n/m))
}
##qq-plots
par(mfrow=c(2,5))
p <- 1/(10^d)
for(i in 1:ncol(res)){
z <- (res[,i] - m*p) / sqrt( m*p*(1-p)  )
qqnorm(z,xlab="Theoretical quantiles",ylab="Observed z-scores",main=paste(i-1))
abline(0,1)
}
##ACF plots
par(mfrow=c(2,5))
for(i in 1:ncol(res)) acf(res[,i])

NB: A normal number has the above stated property in any base. The examples above a for base 10.

To leave a comment for the author, please follow the link and comment on their blog: Simply Statistics » R.

R-bloggers.com offers daily e-mail updates about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job.
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

# Never miss an update! Subscribe to R-bloggers to receive e-mails with the latest R posts.(You will not see this message again.)

Click here to close (This popup will not appear again)