Goats do room

The riddle of the week is about 10 goats sequentially moving to their room, which they have chosen at random and independently (among ten rooms), unless another goat already occupies the room, in which case they move to the first free room with a higher number or fail. What is the probability that all goats end up in a room?

Coding the experiment is straightforward:

g=sample(1:N,N,rep=TRUE)
o=0*g
for(i in 1:N){
    if(min(o[g[i]:N])){f=f+1;break()
    }else{
      o[min(which(!o[g[i]:N]))+g[i]-1]=1
    }}}

returning an estimated probability of approximately 0.764.

As I had some free time during the early mornings at ISBA 2022, I tried to reformulate the question as a continuous event on uniform order statistics, turning to be at most one uniform larger than (N-1)/N, at most two larger than (N-2)/N, and so on… Asking the question on math.stackexchange quickly produced an answer that reversed engineered my formulation back to the goats (or parking lot), with a generic probability of

which of course coincides with the Monte Carlo approximation!

As an aside, I once drank South-African wines named Goats-do-Roam and Goat-Roti at my friends Jim and Maria’s place,  and they were quite enjoyable!

conditioning on zero probability events

An interesting question on X validated as to how come a statistic T(X) can be sufficient when its support depends on the parameter θ behind the distribution of X. The reasoning there being that the distribution of X given T(X)=t does depend on θ since it is not defined for some values of θ … Which is not correct in that the conditional distribution of X depends on the realisation of T, meaning that if this realisation is impossible, then the conditional is arbitrary and of no relevance. Which also led me to tangentially notice and bemoan that most (Stack) exchanges on conditioning on zero probability events are pretty unsatisfactory in that they insist on interpreting P(X=x) [equal to zero] in a literal sense when it is merely a notation in the continuous case. And undefined when X has a discrete support. (Conditional probability is always a sore point for my students!)

shortened iterations [code golf]

A codegolf lazy morning exercise towards finding the sequence of integers that starts with an arbitrary value n and gets updated by blocks of four as

until the last term is not an integer. While the update can be easily implemented with the appropriate stopping rule, a simple congruence analysis shows that, depending on n, the sequence is 4, 8 or 12 values long when

respectively. But sadly the more interesting fixed length solution

`~`=rep #redefine function
b=(scan()-1)*c(32~4,8,40~4,1,9~3)/32+c(1,1,3,0~3,6,-c(8,1,9,-71,17)/8)
b[!b%%1] #keep integers only

ends up being longer than the more basic one:

a=scan()
while(!a[T]%%1)a=c(a,d<-a[T]*T,d+T+1,e<-d-1,e/((T<-T+4)-1))
a[-T]

where Robin’s suggestion of using T rather than length is very cool as T has double meaning, first TRUE (and 1) then the length of a…

natural LaTeX

Nature must have been out of inspiration in the past weeks for running a two-page article on L^aT_eX [as the toolbox column] and how it compares with… Word! Which is not so obvious since most articles in Nature are not involving equations (much) and are from fields where Word prevails. Besides the long-running whine that LaTeX is not he selling argument for this article seemed to be the increasing facility to use (basic) LaTeX commands in forums (like Stack Exchange) and blogs (like WordPress) via MathJax. But the author also pushes for the lighter (R?)Markdown as, “in LaTeX, there is a greater risk that contributors will make changes that prevent the code compiling into a PDF” (which does not make sense to me). This tribune also got me to find out that there is a blog dedicated to the “LaTeX fetish”, which sounds to me like an perfect illustration of Internet vigilantism, especially with arguments like “free and open source software has a strong tendency towards being difficult to install and get up and running”.

(x=scan())%in%(2*4^(n=0:x)-2^n-1)

One challenge on code golf is to find the shortest possible code to identify whether or not an integer belongs to the binary cyclops numbers which binary expansion is 0, 101, 11011, 1110111, 111101111, &tc. The n-th such number being

this leads to the above solution in R (26 bits). The same length as the C solution [which I do not get]

f(n){n=~n==(n^=-~n)*~n/2;}

And with shorter versions in many esoteric languages I had never heard of, like the 8 bits Brachylog code

ḃD↔Dḍ×ᵐ≠

or the 7 bits Jelly

B¬ŒḂ⁼SƊ

As a side remark, since this was not the purpose of the game, the R code is most inefficient in creating a set of size (x+1), with most terms being Inf.