teaching in English

A strange (if very French!) debate is taking place these days in the French main chamber, where some socialist deputies are contesting an incoming change in the regulation of university studies that would allow some courses to be taught in… English! Quelle horreur!!! Since this option has been implemented by many universities, incl. Dauphine, it means that we all are acting outside the law! I do not fear in the least being indicted for teaching R and Bayesian statistics in English… However, I find the action of these deputies missing the point: just like most other Western countries, we need to attract bright students from emerging countries in order to keep our departments open. It is unrealistic to think that those students will accept to learn French in addition to English, just because our universities are that attractive (and they are not!). Plus, our own students are asking for courses in English as they realise that their English level is not that great and that this training is more efficient than regular English courses… This position was better expressed in a Le Monde tribune a few days ago signed by several university professors, incl. Cédric Villani.

Le Monde puzzle [#820]

The current puzzle is… puzzling:

Given the set {1,…,N} with N<61, one iterates the following procedure: take (x,y) within the set and replace the pair with the smallest divider of x+y (bar 1). What are the values of N such that the final value in the set is 61?

I find it puzzling because the way the pairs are selected impacts the final value. Or not, depending upon N. Using the following code (with factors() from the pracma package):

library(pracma)
endof=function(N){
  coll=1:N
  for (t in 1:(N-1)){

    pair=sample(1:length(coll),2)
    dive=min(factors(sum(coll[pair])))
    coll=coll[-pair]
    coll=c(coll,dive)
    }
  print(dive)
  }

I got:

> for (t in 1:10) endof(10)
[1] 5
[1] 3
[1] 3
[1] 5
[1] 7
[1] 5
[1] 5
[1] 7
[1] 3
[1] 3> for (t in 1:10) endof(16)
[1] 2
[1] 2
[1] 2
[1] 2
[1] 2
[1] 2
[1] 2
[1] 2
[1] 2
[1] 2

For N of the form 4k or 4k-1, the final number is always 2 while for N‘s of the form 4k-2 and 4k-3, the final number varies, sometimes producing 61′s. Although I could not find solutions for N less than 17…  Looking more closely into the sequence leading to 61, I could not see a pattern, apart from producing prime numbers as, in, e.g.

61 = 2 + [12 +  (4 + {14 + [13 + 16]})]

for N=17.  (Another puzzle is that 61 plays no particular role: a long run of random calls to endof() return all prime numbers up to 79…)

Udate: Looking at the solution in today’s edition, there exist a solution for N=13 and a solution for N=14. Even though my R code fails to spot it. Of course, an exhaustive search would be feasible in these two cases.  (I had also eliminated values below as not summing up to 61.) The argument for eliminating 4k and 4k-1 is that there must be an odd number of odd numbers in the collection, otherwise, the final number is always 2.

awalé

Following Le Monde puzzle #810, I tried to code an R program (not reproduced here) to optimise an awalé game but the recursion was too rich for R:

Error: evaluation nested too deeply:
infinite recursion / options(expressions=)?

even with a very small number of holes and seeds in the awalé… Searching on the internet, it seems the computer simulation of a winning strategy for an awalé game still is an open problem! Here is a one-step R function that does  not produce sure gains for the first player, far from it, as shown by the histogram below…  I would need a less myopic strategy by iterating  this function at least twice.

onemorestep=function(x,side){
# x current state of the awale,
# side side of the awale (0 vs 1)

M=length(x);N=as.integer(M/2)
rewa=rep(0,M)
newb=matrix(0,ncol=M,nrow=M)

for (i in ((1:N)+N*side)){

 if (x[i]>0){
   y=x
   y[i]=0
   for (t in 0:(x[i]-1))
     y[1+(i+t)%%M]=y[1+(i+t)%%M]+1

   last=1+(i+t)%%M
   if (side){ gain=(last<=N)
    }else{ gain=(last>N)}

   if (gain){# ending up on the right side
     rewa[i]=0
     while (((last>0)&&(side))||((last>N)||(!side)))
     if ((y[last]==2)||(y[last]==3)){
          rewa[i]=rewa[i]+y[last];y[last]=0
          last=last-1
          }else{ break()}
     }
   newb[i,]=y
   }
  }
if (max(rewa)>0){
  sol=order(-rewa)[1]
  }else{ sol=rang=((1:N)+N*side)[x[((1:N)+N*side)]>0]
   if (length(rang)>1) sol=sample(rang,1,prob=x[rang]^3)}

   return(list(reward=max(rewa),board=newb[sol,]))
}

Le Monde puzzle [#818]

The current puzzle is as follows:

Define the symmetric of an integer as the integer obtained by inverting the order of its digits, eg 4321 is the symmetric of 1234. What are the numbers for which the square is equal to the symmetric of the square of the symmetric?

I first consulted stackexchange to find a convenient R function to create the symmetric:

int2digit=function(x){
as.numeric(sapply(sequence(nchar(x)),
  function(y) substr(x, y, y)))}

digit2int=function(a){
as.numeric(paste(a,collapse=""))}

flip=function(x){
  digit2int(rev(int2digit(x)))}

and then found that all integers but the multiples of 10 are symmetric! only some integers involving 1,2,3 and sometimes zero would work:

> for (i in 1:1000){
+   if (i^2==flip(flip(i)^2)) print(i)}
[1] 1
[1] 2
[1] 3
[1] 11
[1] 12
[1] 13
[1] 21
[1] 22
[1] 31
[1] 101
[1] 102
[1] 103
[1] 111
[1] 112
[1] 113
[1] 121
[1] 122
[1] 201
[1] 202
[1] 211
[1] 212
[1] 221
[1] 301
[1] 311

If I am not (again) confused, the symmetric integers would be those (a) not ending with zero and (b) only involving digits whose products are all less than 10.

statistics in Le Monde

In the current weekend edition of Le Monde, science leaflet (soon to disappear from the weekend edition alas!, Pierre Barthélémy wrote his tribune on a (not that recent) PLoS paper on roadkills that seems to use capture-recapture (or should) to evaluate the real number of roadkills from their disappearance rate. And Cédric Villani muses in his carte blanche on the relevance of mathematical models in social sciences, using the recent blunder by Reinhart and Rogoff as an argument: this sounds for the least extreme as there are many counter-examples in political sciences, sociology, psychology, &tc. I think he is missing the point that, while all models are wrong (in the sense physical models can be “right”), there are some models that can prove useful. A last item of interest (?) was the announcement of the new volume in the maths popularisation series, which is dedicated to the fourth dimension. I hope they also deal with higher dimensions, otherwise it could get quickly boring! It reminded me of the textbook I had to teach from the semester I taught basic vector space algebra in Cornell: the chapter on dimension 2 got followed by one on dimension 3, then one on dimension 4…