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ocupee=function(M){ ok=rep(0,M) ok[1]=ok[M]=1 ok[trunc((1+M/2))]=1 while (max(diff((1:M)[ok!=0])>2)){ i=order(-diff((1:M)[ok!=0]))[1] ok[(1:M)[ok!=0][i]+trunc((diff((1:M)[ok!=0])[i]/2))]=1 } return(sum(ok>0)) }
with maximal occupation illustrated by the graph below:
Meaning that the efficiency of the positioning scheme is not optimal when following the sequential positioning, requiring $latexN+2^{\lceil log_2(N-1) \rceil}$ urinals. Rather than one out of two, requiring 2N-1 urinals. What is most funny in this simple exercise is the connection exposed in the Riddler with an Xkcd blag written a few years go about the topic.
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