Le Monde puzzle [#737 re-read]

Posted on August 27, 2011 by xi'an in R bloggers | 0 Comments

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As a coincidence, while I was waiting for the solution to puzzle #737 published this Friday in Le Monde, the delivery (wo)man forgot to include the weekend magazine and I had to buy it this morning with my baguette (as if anyone cares!). The solution is (y₀,z₀,w₀)=(38,40,46) and…it does not work! The value of (x₁,y₁,z₁,w₁) is indeed (19,39,43,8.5). Now, it does not work because the update mechanism used by the writers of the Le Monde column is to move all positions simultaneously, rather than one at a time: the update thus reads as

if uⁿ=(x_n,y_n,z_n,w_n), for i=1,…,4,

$u_i^n=\frac{1}{2}\left(u_i^{n-1} \min_{j;u_j^{n-1}>u_{i}^{n-1}}u_{j}^{n-1}\right)\,\text{mod}\,48$ u_{i}^{n-1}}u_{j}^{n-1}\right)\,\text{mod}\,48′ title=’u_i^n=\frac{1}{2}\left(u_i^{n-1}+\min_{j;u_j^{n-1}>u_{i}^{n-1}}u_{j}^{n-1}\right)\,\text{mod}\,48′ class=’latex’ />

if u_i^n-1 is not the maximum of u^n-1 and

$u_{i}^{n}=\frac{1}{2}\left(u_{i}^{n-1}+48+\min_{j\ne i}u_{j}^{n-1}\right)\,\text{mod}\,48$

otherwise.

(to compare with my earlier proposal). In this new configuration, the core of the R code I use is

while (nodd){

nsuite=suite=start=c(0,sort(2*sample(1:23,3)))
for (t in 1:20){
#some values of start never produce non-integer values
#of suite and never visit suite[1]=0

  nsuite[-4]=(suite[-4]+suite[-1]+48*(suite[-1]-suite[-4]<0))/2
  nsuite[4]=(suite[4]+suite[1]+48*(suite[1]-suite[4]<0))/2
  suite=nsuite%%48

  if ((suite[1]==0)||(max(is.decimal(suite))==1)){
    nodd=(max(is.decimal(suite))==1)
    print(t)
    break()}
  }}

This new version leads to solutions (unsurprisingly!), returning