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.

austerity in MCMC land (#2)

After reading the arXiv paper by Korattikara, Chen and Welling, I wondered about the expression of the acceptance step of the Metropolis-Hastings algorithm as a mean of log-likelihoods over the sample. More specifically the long sleepless nights at the hospital led me to ponder the rather silly question of the impact of replacing mean by median. I thus tried running a Metropolis-Hastings algorithm with the substitute and it (of course!) let to a nonsensical answer, as shown by the above graph. The true posterior is the one for a normal model and the histogram indicates a lack of convergence of the Markov chain to this posterior even though it does converge to some posterior. Here is the R code for this tiny experiment:

#data generation
N=100
x=rnorm(N)

#HM steps
T=10^5
theta=rep(0,T)
curlike=dnorm(x,log=TRUE)
for (t in 2:T){

  prop=theta[t-1]+.1*rnorm(1)
  proplike=dnorm(x,mean=prop,log=TRUE)
  u=runif(1)
  bound=log(u)-dnorm(prop,sd=10,log=TRUE)+
         dnorm(theta[t-1],sd=10,log=TRUE)
  if (median(proplike)-median(curlike)>bound/N){
   theta[t]=prop;curlike=proplike
   } else { theta[t]=theta[t-1]}
 }

interesting puzzle

In addition to its weekly mathematics puzzles, Le Monde is now publishing a series of vulgarisation books on mathematics, under the patronage of Cédric Villani. Jean-Michel Marin brought me two from the series, one on the golden number and one on Pythagoras’ theorem. (This is actually a translation of a series published by El Pais last year.) Those books are a bit stretched given the topic, even though I enjoyed the golden number (the second one having a lot of redundancy with the first one.) However, I came upon an interesting question, namely about the maximum size of a cube that could fit through a tunnel drilled through the unit cube. Sadly, I could not find an answer to this problem on the web, even though the book mentions a solution with a side larger than one…