bean bag win

A quick riddle from The Riddler, where a multiple step game sees a probability of a 3 point increase of .4 and a probability of a 1 point increase of .3 with a first strategy (A), versus a probability of a 3 point increase of .4 and a probability of a 1 point increase of .3 with a second strategy (B), and a sure miss third strategy (C). The goal is to optimise the probability of hitting exactly 3 points after 4 steps.

The optimal strategy is to follow A while the score is zero, C when the score is 3, and B otherwise. The corresponding winning probability is 0.8548, as checked by the following code

win=function(n=1,s=0){
  if(n==4)return((s==3)+.4*(!s)+.8*(s==2))
  else{return(max(c(
    .4*win(n+1,s+3)+.3*win(n+1,s+1)+.3*win(n+1,s),
    .1*win(n+1,s+3)+.8*win(n+1,s+1)+.1*win(n+1,s),
    win(n+1,s))))}}

Le Monde puzzle [#1130]

A two-player game as Le weekly Monde current mathematical puzzle:

Abishag and Caleb fill in alternance a row of N boxes in a row by picking one then two then three &tc. consecutive boxes. When a player is unable to find enough consecutive boxes, the player has lost. Who is winning when N=29? When N=30?

Using a basic recursive search for the optimal strategy, with the status of the row and the number of required boxes as entries,

f<-function(b=!1:N,r=0){
  for(i in 1:(N-r)){
    if(p<-!max(b[j<-i+r:0])){
      q=b;q[j]=1
      if(p<-!f(q,r+1))break}}
  p}

returns Abishag as the winner for N=29 (as well as for N=1,2,7,…,13,19,…,29) and Caleb as the winner for N=30 (as well as for N=3,…,6,14,…,18). I am actually surprised that the recursion operates that deep, even though this means a √N depth for the recursion. While the code took too long to complete, the function operates for N=100. A side phenomenon is the apparent special case of N=47, which makes Abishag the looser, while N=46 and N=48 do not.This is an unusual pattern as otherwise (up to N=59), there are longer and longer stretches of adjacent wins and looses as N increases.

Metropolis in 95 characters

Here is an R function that produces a Metropolis-Hastings sample for the univariate log-target f when the later is defined outside as another function. And when using a Gaussian random walk with scale one as proposal. (Inspired from a X validated question.)

m<-function(T,y=rnorm(1))ifelse(rep(T>1,T),
  c(y*{f({z<-m(T-1)}[1])-f(y+z[1])<rexp(1)}+z[1],z),y)

The function is definitely not optimal, crashes for values of T larger than 580 (unless one modifies the stack size), and operates the most basic version of a Metropolis-Hastings algorithm. But as a codegolf challenge (on a late plane ride), this was a fun exercise.

Le Monde puzzle [#1110]

A low-key sorting problem as Le Monde current mathematical puzzle:

If the numbers from 1 to 67 are randomly permuted and if the sorting algorithm consists in picking a number i with a position higher than its rank i and moving it at the correct i-th position, what is the maximal number of steps to sort this set of integers when the selected integer is chosen optimaly?

As the question does not clearly state what happens to the number j that stood in the i-th position, I made the assumption that the entire sequence from position i to position n is moved one position upwards (rather than having i and j exchanged). In which case my intuition was that moving the smallest moveable number was optimal, resulting in the following R code

sorite<-function(permu){ n=length(permu) p=0 while(max(abs(permu-(1:n)))>0){
    j=min(permu[permu<(1:n)])
    p=p+1
    permu=unique(c(permu[(1:n)<j],j,permu[j:n]))} 
  return(p)}

which takes at most n-1 steps to reorder the sequence. I however checked this insertion sort was indeed the case through a recursive function

resorite<-function(permu){ n=length(permu);p=0 while(max(abs(permu-(1:n)))>0){
    j=cand=permu[permu<(1:n)]
    if (length(cand)==1){
      p=p+1
      permu=unique(c(permu[(1:n)<j],j,permu[j:n]))
    }else{
      sol=n^3
      for (i in cand){
        qermu=unique(c(permu[(1:n)<i],i,permu[i:n]))
        rol=resorite(qermu)
        if (rol<sol)sol=rol}
      p=p+1+sol;break()}} 
  return(p)}

which did confirm my intuition.

Le Monde puzzle [#1049]

An algorithmic Le Monde mathematical puzzle with a direct

Alice and Bob play a game by picking alternatively one of the remaining digits between 1 and 10 and putting it in either one of two available stacks, 1 or 2. Their respective gains are the products of the piles (1 for Alice and 2 for Bob).

The problem is manageable by a recursive function

facten=factorial(10)
pick=function(play=1,remz=matrix(0,2,5)){
 if ((min(remz[1,])>0)||(min(remz[2,])>0)){#finale
  remz[remz==0]=(1:10)[!(1:10)%in%remz]
  return(prod(remz[play,]))
  }else{
   gainz=0
   for (i in (1:10)[!(1:10)%in%remz]){
     propz=rbind(c(remz[1,remz[1,]>0],i,
     rep(0,sum(remz[1,]==0)-1)),remz[2,])
     gainz=max(gainz,facten/pick(3-play,remz=propz))}
   for (i in (1:10)[!(1:10)%in%remz]){
     propz=rbind(remz[1,],c(remz[2,remz[2,]>0],i,
     rep(0,sum(remz[2,]==0)-1)))
     gainz=max(gainz,facten/pick(3-play,remz=propz))}
return(gainz)}}

that shows the optimal gain for Alice is 3360=2x5x6x7x 8, versus Bob getting 1080=1x3x4x9x10. The moves ensuring the gain are 2-10-…