-USING: accessors arrays classes.tuple io kernel locals math math.functions
-    math.ranges prettyprint project-euler.common sequences ;
+USING: accessors arrays classes.tuple io kernel locals math
+math.functions math.ranges prettyprint project-euler.common
+sequences ;
 IN: project-euler.064
+
+! http://projecteuler.net/index.php?section=problems&id=64
+
+! DESCRIPTION
+! -----------
+
+! All square roots are periodic when written as continued
+! fractions and can be written in the form:
+
+! √N=a0+1/(a1+1/(a2+1/a3+...))
+
+! For example, let us consider √23:
+
+! √23=4+√(23)−4=4+1/(1/(√23−4)=4+1/(1+((√23−3)/7)
+
+! If we continue we would get the following expansion:
+
+! √23=4+1/(1+1/(3+1/(1+1/(8+...))))
+
+! The process can be summarised as follows:
+
+! a0=4, 1/(√23−4) = (√23+4)/7 = 1+(√23−3)/7
+! a1=1, 7/(√23−3) = 7*(√23+3)/14 = 3+(√23−3)/2
+! a2=3, 2/(√23−3) = 2*(√23+3)/14 = 1+(√23−4)/7
+! a3=1, 7/(√23−4) = 7*(√23+4)/7 = 8+√23−4
+! a4=8, 1/(√23−4) = (√23+4)/7 = 1+(√23−3)/7
+! a5=1, 7/(√23−3) = 7*(√23+3)/14 = 3+(√23−3)/2
+! a6=3, 2/(√23−3) = 2*(√23+3)/14 = 1+(√23−4)/7
+! a7=1, 7/(√23−4) = 7*(√23+4)/7 = 8+√23−4
+
+! It can be seen that the sequence is repeating. For
+! conciseness, we use the notation √23=[4;(1,3,1,8)], to
+! indicate that the block (1,3,1,8) repeats indefinitely.
+
+! The first ten continued fraction representations of
+! (irrational) square roots are:
+
+! √2=[1;(2)] , period=1
+! √3=[1;(1,2)], period=2
+! √5=[2;(4)], period=1
+! √6=[2;(2,4)], period=2
+! √7=[2;(1,1,1,4)], period=4
+! √8=[2;(1,4)], period=2
+! √10=[3;(6)], period=1
+! √11=[3;(3,6)], period=2
+! √12=[3;(2,6)], period=2
+! √13=[3;(1,1,1,1,6)], period=5
+
+! Exactly four continued fractions, for N <= 13, have an odd period.
+
+! How many continued fractions for N <= 10000 have an odd period?
 
 <PRIVATE
 

    loop
    drop drop drop ;

: try-all ( -- n ) 2 10000 [a,b]
    [ perfect-square? not ] filter
    [ find-period ] map
    [ odd? ] filter

: euler064a ( -- n ) try-all ;

<PRIVATE

! (√n + a)/b
TUPLE: cfrac n a b ;

C: <cfrac> cfrac

: >cfrac< ( fr -- n a b )
    [ n>> ] [ a>> ] [ b>> ] tri ;

! (√n + a) / b = 1 / (k + (√n + a') / b')
!
! b / (√n + a) = b (√n - a) / (n - a^2) = (√n - a) / ((n - a^2) / b)
:: reciprocal ( fr -- fr' )
    fr >cfrac< :> ( n a b )
    n
    a neg
    n a sq - b /
    <cfrac> ;

:: split ( fr -- k fr' )
    fr >cfrac< :> ( n a b )
    n sqrt a + b / >integer
    dup n swap
    b * a swap -
    b
    <cfrac> ;

: pure ( n -- fr )
    0 1 <cfrac> ;

: next ( fr -- fr' )
    reciprocal split nip ;

:: period ( n -- period )
    n sqrt >integer sq n = [ 0 ] [
        n pure split nip :> start
        1 start next
        [ dup start = not ]
        [ next [ 1 + ] dip ]
        while drop
    ] if ;

PRIVATE>

: euler064b ( -- ct )
    10000 [1,b] [ period odd? ] count ;

SOLUTION: euler064b
