## April 2014

The bottom line answer for the expected value of the maximum number in the array is, for N=5, 937669227/67108864 (which is approximately 13.97).

Kudus to Christopher Marks who was the first to solve this month's challenge without using programming at all, but rather by merely doing a careful examination of cases.

Among others, we also received the following solutions:

A one-line Mathematica solution from Todd Will:

ave[x_]:=ave[x]=If[Length[x]==5,Max[x],Mean[ave[Prepend[x,#]//.{a_,a_,b___}:>{2a,b}]&/@{2,4}]]

A small Haskel program from John Tromp:

import Data.Ratio

avg :: [Integer] -> Ratio Integer

avg ps | length ps >= 5 = maximum ps % 1

avg ps = (avg' (2:ps) + avg' (4:ps)) / 2

avg' :: [Integer] -> Ratio Integer

avg' = avg . reduce

reduce :: [Integer] -> [Integer]

reduce (a:b:ps) | a == b = reduce (a+b : ps)

reduce ps = ps

main = print $ avg []

And a PARI-GP program from Claus Andersen and David Brink.

Replace(v)=

{

if(#v<2,return(v));

if(v[#v-1]<>v[#v],return(v));

Replace(vector(#v-1,i,if(i<#v-1,v[i],v[i]+v[i+1])))

}

Next()=

{

if(t==0,return());

if(e[t]==2,

e[t]=4;L[t]=Replace(concat(if(t>1,L[t-1],[]),4)),

t--;Next())

}

April(n)=

{

p=0;

e=[2];

L=[[2]];

t=1;

s=0;c=0;j=0;

while(t>0,

if(#L[t]==n,

\\print(L[t]" "vector(t,i,e[i]));

j++;p+=vecmax(L[t])/2^t;Next(),

t++;

if(#e<t,e=concat(e,0));e[t]=2;

if(#L<t,L=concat(L,0));L[t]=Replace(concat(L[t-1],2))

);

);

return(p)

}

April(1)

April(2)

April(3)

April(4)

April(5)

John Snyder sent us a very detailed analysis of the problem. We chose to share his nice graphic representation of the Markov process here:

Graphic representation of the Markov process by John Snyder

Oleg Vlasii computed the answer for N=1,2,...,10 and noted the pattern

A(n) = (F(n)*2^(f(n)-1)+3^f(n))/2^(2^N-N-1)

where f(n) = 2^(n-1)-1

and F(n) are integers; but a closed form formula still eludes us.

*If you have any problems you think we might enjoy, please send them in. All replies should be sent to:* ponder@il.ibm.com