*Editor's note: Today 3/14/15 at some point between 9:26:53 and 9:26:54 it was the most π day of them all. Below is a repost from last year. *

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.