flea circus

Posted in Books, Kids, pictures, R, Statistics with tags , , , , , , , , , , , on December 8, 2016 by xi'an

An old riddle found on X validated asking for Monte Carlo resolution  but originally given on Project Euler:

A 30×30 grid of squares contains 30² fleas, initially one flea per square. When a bell is rung, each flea jumps to an adjacent square at random. What is the expected number of unoccupied squares after 50 bell rings, up to six decimal places?

The debate on X validated is whether or not a Monte Carlo resolution is feasible. Up to six decimals, certainly not. But with some lower precision, certainly. Here is a rather basic R code where the 50 steps are operated on the 900 squares, rather than the 900 fleas. This saves some time by avoiding empty squares.

xprmt=function(n=10,T=50){

mean=0
for (t in 1:n){

board=rep(1,900)
for (v in 1:T){

beard=rep(0,900)
if (board[1]>0){
poz=c(0,1,0,30)
ne=rmultinom(1,board[1],prob=(poz!=0))
beard[1+poz]=beard[1+poz]+ne}
#
for (i in (2:899)[board[-1][-899]>0]){
u=(i-1)%%30+1;v=(i-1)%/%30+1
poz=c(-(u>1),(u<30),-30*(v>1),30*(v<30))
ne=rmultinom(1,board[i],prob=(poz!=0))
beard[i+poz]=beard[i+poz]+ne}
#
if (board[900]>0){
poz=c(-1,0,-30,0)
ne=rmultinom(1,board[900],prob=(poz!=0))
beard[900+poz]=beard[900+poz]+ne}
board=beard}
mean=mean+sum(board==0)}
return(mean/n)}


The function returns an empirical average over n replications. With a presumably awkward approach to the borderline squares, since it involves adding zeros to keep the structure the same… Nonetheless, it produces an approximation that is rather close to the approximate expected value, in about 3mn on my laptop.

> exprmt(n=1e3)
[1] 331.082
> 900/exp(1)
[1] 331.0915


Further gains follow from considering only half of the squares, as there are two independent processes acting in parallel. I looked at an alternative and much faster approach using the stationary distribution, with the stationary being the Multinomial (450,(2/1740,3/1740…,4/1740,…,2/1740)) with probabilities proportional to 2 in the corner, 3 on the sides, and 4 in the inside. (The process, strictly speaking, has no stationary distribution, since it is periodic. But one can consider instead the subprocess indexed by even times.) This seems to be the case, though, when looking at the occupancy frequencies, after defining the stationary as:

inva=function(B=30){
return(c(2,rep(3,B-2),2,rep(c(3,
rep(4,B-2),3),B-2),2,rep(3,B-2),2))}


namely

> mn=0;n=1e8 #14 clock hours!
> proz=rep(c(rep(c(0,1),15),rep(c(1,0),15)),15)*inva(30)
> for (t in 1:n)
+ mn=mn+table(rmultinom(1,450,prob=rep(1,450)))[1:4]
> mn=mn/n
> mn[1]=mn[1]-450
> mn
0      1      2     3
166.11 164.92  82.56 27.71
> exprmt(n=1e6) #55 clock hours!!
[1] 165.36 165.69 82.92 27.57

my original confusion being that the Poisson approximation had not yet taken over… (Of course, computing the first frequency for the stationary distribution does not require any simulation, since it is the sum of the complement probabilities to the power 450, i.e., 166.1069.)

sampling by exhaustion

Posted in Books, Kids, R, Statistics with tags , , , , on November 25, 2016 by xi'an

The riddle set by The Riddler of last week sums up as follows:

Within a population of size N, each individual in the population independently selects another individual. All individuals selected at least once are removed and the process iterates until one or zero individual is left. What is the probability that there is zero individual left?

While I cannot see a clean analytical solution to this problem, it reminds me of an enveloppe-versus-letter (matches) problem I saw in graduate school. Indeed, the expected number of removed (or selected) individuals is given by

$N\left\{1-\frac{N-2}{N-1}\right\}^{N-1}$

which is equivalent to (1-e⁻¹)N for N large, meaning that the population decreases by an average factor of e⁻¹ at each round. And that it takes on average approximately log(N) iterations to reach a single individual. A simulation of the first probabilities of ending with one individual led me to the above curve, which wiggles in an almost periodic way around the probability ½, equal to the average of those probabilities. Using the R code

rad=function(N){#next population size
ut=sample(rep(2:N,2),1)
for (i in 2:N)#sampling
ut=c(ut,sample(rep((1:N)[-i],2),1))
return(N-length(unique(ut))}
sal=rep(0,m);sal[1]=1
for (N in 3:M){
prop=0;
for (t in 1:T){#one single step
if (i>0) prop=prop+sal[i]}
sal[N]=prop/T}


which exploits the previously computed probabilities. The variability is most interesting if unexpected, but looking back at Feller‘s sections and exercises on the classical occupancy problem, I could not find a connection with this problem. If it exists. Still, if N is large enough, the exclusion of one index from the selection becomes negligible and the probability of moving from n to m individuals should be approximately [Feller, eqn (2.4), p.102]

$p_n(m)={n\choose m}\sum_{v=}^{n-m} (-1)^v {n-m\choose v} \left(1-\frac{m+v}{n}\right)^n$

This formula approximates quite well the exact probability, but as in a previous post about the birthday problem, it proves quite delicate to compute. As already noticed by Feller.

occupancy rules

Posted in Kids, R, Statistics with tags , , , , , , , on May 23, 2016 by xi'an

While the last riddle on The Riddler was rather anticlimactic, namely to find the mean of the number Y of empty bins in a uniform multinomial with n bins and m draws, with solution

$\mathbb{E}[Y]=n(1-\frac{1}{n})^m,$

[which still has a link with e in that the fraction of empty bins converges to e⁻¹ when n=m], this led me to some more involved investigation on the distribution of Y. While it can be shown directly that the probability that k bins are non-empty is

${n \choose k}\sum_{i=1}^k (-1)^{k-i}{k \choose i}(i/n)^m$

with an R representation by

miss<-function(n,m){
p=rep(0,n)
for (k in 1:n)
p[k]=choose(n,k)*sum((-1)^((k-1):0)*choose(k,1:k)*(1:k)^m)
return(rev(p)/n^m)}


I wanted to take advantage of the moments of Y, since it writes as a sum of n indicators, counting the number of empty cells. However, the higher moments of Y are not as straightforward as its expectation and I struggled with the representation until I came upon this formula

$\mathbb{E}[Y^k]=\sum_{i=1}^k {k \choose i} i! S(k,i) \left( 1-\frac{i}{n}\right)^m$

where S(k,i) denotes the Stirling number of the second kind… Or i!S(n,i) is the number of surjections from a set of size n to a set of size i. Which leads to the distribution of Y by inverting the moment equations, as in the following R code:

diss<-function(n,m){
A=matrix(0,n,n)
mome=rep(0,n)
A[n,]=rep(1,n)
mome[n]=1
for (k in 1:(n-1)){
A[k,]=(0:(n-1))^k
for (i in 1:k)
mome[k]=mome[k]+factorial(i)*as.integer(Stirling2(n,i))*
(1-(i+1)/n)^m*factorial(k)/factorial(k-i-1)}
return(solve(A,mome))}


that I still checked by raw simulations from the multinomial

zample<-function(n,m,T=1e4){
x=matrix(sample(1:n,m*T,rep=TRUE),nrow=T)
x=sapply(apply(x,1,unique),length)
return(n-x)}