**Simply Statistics » R**, and kindly contributed to R-bloggers)

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 Y_{1}, …, Y_{100} 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.

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