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(This article was first published on <strong><a href="https://xianblog.wordpress.com/2018/04/18/le-monde-puzzle-1049/"> R – Xi'an's Og</a></strong>, and kindly contributed to <a rel="nofollow" href="https://www.r-bloggers.com/">R-bloggers)</a> </div>
<p><strong><a rel="nofollow" href="https://xianblog.files.wordpress.com/2018/03/sieglemond.jpg" target="_blank"><img data-orig-file="https://xianblog.files.wordpress.com/2018/03/sieglemond.jpg" data-orig-size="660,281" data-medium-file="https://xianblog.files.wordpress.com/2018/03/sieglemond.jpg?w=300" data-attachment-id="35577" data-image-meta="{" data-image-description="" src="https://xianblog.files.wordpress.com/2018/03/sieglemond.jpg?w=450&h=192&fit=450%2C192&resize=450%2C192" data-permalink="https://xianblog.wordpress.com/2018/03/12/le-monde-puzzle-1044/sieglemond/" data-large-file="https://xianblog.files.wordpress.com/2018/03/sieglemond.jpg?w=450&amp;h=192&amp;fit=450%2C192&amp;resize=450%2C192" alt="" data-image-title="sieglemond" sizes="(max-width: 450px) 100vw, 450px" width="450" data-comments-opened="1" class="wp-image-35577" data-recalc-dims="1" height="192"></a>A</strong>n algorithmic <a rel="nofollow" href="https://xianblog.wordpress.com/2011/09/03/le-monde-puzzle-website/" title="Le Monde puzzle [website]" target="_blank">Le Monde mathematical puzzle</a> with a direct</p>
<blockquote>
<p><em>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).</em></p>
</blockquote>
<p>The problem is manageable by a recursive function</p>
<pre>facten=factorial(10)
pick=function(play=1,remz=matrix(0,2,5)){
 if (sum(remz==0)==1){#finale
  remz[remz==0]=(1:10)[!(1:10)%in%remz]
  return(prod(remz[play,]))
  }else{
   gainz=0
   if (min(remz[1,])==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))}}
   if (min(remz[2,])==0){
    for (i in (1:10)[remz==0]){
     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)}}
</pre>
<p>that shows the optimal gain for Alice is 6480=3x5x6x8x9, versus Bob getting 1680=1x2x4x7x10. The moves ensuring the gain are 2-10-…</p>
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