Project Euler problem 25:

The Fibonacci sequence is defined by the recurrence relation:

F_n = F_n−1 + F_n−2, where F_{1 = 1} and F₂ = 1. Hence the first 12 terms will be F₁ = 1, F₂ = 1, F₃ = 2, F₄ = 3, F₅ = 5, F₆ = 8, F₇ = 13, F₈ = 21, F₉ = 34, F₁₀ = 55, F₁₁ = 89, F₁₂ = 144

The 12th term, F₁₂, is the first term to contain three digits.

What is the first term in the Fibonacci sequence to contain 1000 digits?

I made a brute force solution in Python but it takes forever. Can anyone suggest a non brute force solution?

def Fibonacci(NthTerm):if NthTerm == 1 or NthTerm == 2:return 1 # Challenge defines 1st and 2nd term as == 1else:  # recursive definition of Fib termreturn Fibonacci(NthTerm-1) + Fibonacci(NthTerm-2)FirstTerm = 0 # For scope to include Term in scope of print on line 13
for Term in range(1, 1000): # Arbitrary rangeFibValue = str(Fibonacci(Term)) # Convert integer to string for len()if len(FibValue) == 1000:FirstTerm = Termbreak # Stop thereelse:continue # Go to next number
print "The first term in the\nFibonacci sequence to\ncontain 1000 digits\nis the", FirstTerm, "term."

You can write a fibonacci function that runs in linear time and with constant memory footprint, you don't need a list to keep them. Here's a recursive version (however, if n is big enough, it will just stackoverflow)

def fib(a, b, n):if n == 1:return aelse: return fib(a+b, a, n-1)print fib(1, 0, 10) # prints 55

This function calls itself only once (resulting in around N calls for a parameter N), in contrast with your solution which calls itself twice (around 2^N calls for a parameter N).

Here's a version that won't ever stackoverflow and uses a loop instead of recursion:

def fib(n):a = 1b = 0while n > 1:a, b = a+b, an = n - 1return aprint fib(100000)

And that's fast enough:

$ time python fibo.py 
3364476487643178326662161200510754331030214846068006390656476...real    0m0.869s

But calling fib until you get a result big enough isn't perfect: the first numbers of the serie are calculated multiple times. You can calculate the next fibonacci number and check its size in the same loop:

a = 1
b = 0
n = 1
while len(str(a)) != 1000:a, b = a+b, an = n + 1
print "%d has 1000 digits, n = %d" % (a, n)