## the large half now

**T**he little half puzzle proposed a “dumb’ solution in that players play a minimax strategy. There are 34 starting values less than 100 guaranteeing a sure win to dumb players. If instead the players maximise their choice at each step, the R code looks like this:

solveO=function(n){ if (n<3){ solve=(n==2)}else{ solve=(!(solveO(n-1)))||(!solveO(ceiling(n/2)))} solve}

and there are now 66 (=100-34, indeed!) starting values for which the starting player can win.

**I**ncidentally, I typed

> solveO(1113) Error: evaluation nested too deeply: infinite recursion / options(expressions=)?

which shows R cannot handle heavy recursion without further programming. Testing for the upper limit, I found that the largest acceptable value is 555 (which takes forever to return a value, predicted at more than one hour by a linear regression on the run times till 300…).

October 29, 2012 at 5:53 pm

Yup, too much overhead in R to do heavy recursion well. On the other hand, that’s probably not a very efficient way to code it anyways (too many duplicate calculations). Something like this is much speedier (apologies for inconsistent format):

temps <- rep(NA, 10000)

solveO<-function(n){

if (is.na(temps[[n]]))

{ if (n<=3){ solve=(n==2)}else{

solve=(!(solveO(n-1)))||(!solveO(ceiling(n/2)))}

temps[n] <<- solve

solve

} else

{ temps[n]

}

}

sum(mapply(solveO, 1:10000))