# My Weblog

## Project Euler Problem 100

Project Euler problem 100 has nothing to do with probability. At first i chose this problem only for probability sake and it turns out to Pell’s equation. Here is bit analysis. Let number of blue balls are b and red balls are r. We have
2*b*(b-1)=(b+r)*(b+r-1). Solving for b,we have $b= \frac {(1+2r)+\sqrt (1+8{r^2})} 2$ and it turns out $1+8{r^2} = {y^2}$. Find value of r and b for which $(b+r)>=10^{12}$ .I wrote Pell’s solver in Haskell for this problem and solved. 150th problem with Haskell. (pellsolver n) function generates infinite list of solution of equation $x^2-dy^2 = 1$. You may look MathWorld.

import Data.List

contiNuedFraction::Integer->[Integer]
contiNuedFraction n=
let
d=truncate.sqrt.fromIntegral $n lst=iterate helpFun (d,0,1) helpFun (a_0,p_0,q_0)=(a_1,p_1,q_1) where p_1=a_0*q_0-p_0 q_1=div (n-p_1^2) q_0 a_1=div (d+p_1) q_1 in [a|(a,_,_)<-lst] pellsolver::Integer->[(Integer,Integer)] pellsolver n= let d=truncate.sqrt.fromIntegral$ n
lst=takeWhile (/=2*d) $contiNuedFraction n len=length lst r=take (if even len then len else 2*len)$ contiNuedFraction n
p_0=r!!0
p_1=(r!!0)*(r!!1)+1
q_0=1
q_1=r!!1
(pfinal,_)=foldl recFun (p_1,p_0) $drop 2 r (qfinal,_)=foldl recFun (q_1,q_0)$ drop 2 r where
recFun (p_1,p_0) a=(a*p_1+p_0,p_1)
in iterate (helpFun (pfinal,qfinal)) $(pfinal,qfinal) where helpFun (x_1,y_1) (x_k,y_k)=(x_1*x_k+n*y_1*y_k,x_1*y_k+y_1*x_k) ans=let r=snd.head.dropWhile (\(_,r)->(r+(div (1+2*r+(truncate.sqrt.fromIntegral$ (1+8*r^2))) 2))<(10^12)) $pellsolver 8 b=div (1+2*r+(truncate.sqrt.fromIntegral$ (1+8*r^2))) 2
in b