Kirit,
Forgive me for using Ruby--the scoping won't even be "Haskell correct"
until version 1.9, but I think it will do for this problem. I believe
the answer should be
def peasant(n, power)
case power.divmod(2).collect{|i|i.zero?}
when [true,true] : 1
when [false,true] : peasant(n * n, power/2)
else n * peasant(n * n, power/2)
end
end
puts peasant(3 ,1000).to_s
=>
132207081948080663689045525975214436596542203275214816766492036822682859734
670489954077831385060806196390977769687258235595095458210061891186534272525
795367402762022519832080387801477422896484127439040011758861804112894781562
309443806156617305408667449050617812548034440554705439703889581746536825491
613622083026856377858229022841639830788789691855640408489893760937324217184
635993869551676501894058810906042608967143886410281435038564874716583201061
4366132173102768902855220001
On Oct 7, 11:32 am, "Kirit Sælensminde" <[EMAIL PROTECTED]>
wrote:
> I don't know where we should meet on Thursday, but I'm pleased to say that
> at least the BeerCamp puzzles are back again:
> http://www.kirit.com/Blog:/2008-10-07/Powers
>
> Kirit
>
> 2008/10/6 Dvlong <[EMAIL PROTECTED]>
>
>
>
> > Krua Nai Baan on Langsuan has reasonable food and prices, and a
> > private room or two that would be large enough. It's near Sarasin on
> > the West side of the Road.
>
> > On Oct 5, 3:52 pm, "proteus guy" <[EMAIL PROTECTED]> wrote:
> > > Any suggestions/ideas for a location for the upcoming BeerCamp this week?
>
> > > -- Ben
>
> --http://www.kirit.com/
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