## APL Hacking: Project Euler (#26)

Problem #26:

I did not have a clue on the math for this one, but I did look that up, rather than figuring it out myself. The math came down to the biggest prime denominator with D-1 cycle length if D is the denominator. It was rather fast to compute the close version of primes and then to apply a cycle length function to each of the primes to get the cycle length. It's not the fastest solution, because I could actually short circuit the search if I went in reverse, but I figured that it wasn't worth the loss in clarity on this one.

I would have liked to have found a better cycle length function, though.

∇Z←CYCLE∆LENGTH N;A;I;R ⍝ Find the length of the cycle of 1÷N. Z←0 ⋄ I←0 ⋄ A←N⍴0 ⋄ R←1 LP:I←I+1 ⋄ Z←I-A[R] ⋄ →(0≠A[R])/0 A[R]←I ⋄ R←N|R×10 ⋄ →(0=R)/Z←0 →LP ∇∇Z←PETWENTYSIX

⍝ Find the denominator of unit fractions < 1000 that

⍝ has the longest recuring decimal cycle.

Z←1+⌈/CYCLE∆LENGTH¨(2=+⌿0=(⍳1000)∘.|⍳1000)/⍳1000

∇

I did not have a clue on the math for this one, but I did look that up, rather than figuring it out myself. The math came down to the biggest prime denominator with D-1 cycle length if D is the denominator. It was rather fast to compute the close version of primes and then to apply a cycle length function to each of the primes to get the cycle length. It's not the fastest solution, because I could actually short circuit the search if I went in reverse, but I figured that it wasn't worth the loss in clarity on this one.

I would have liked to have found a better cycle length function, though.