slinky773 - 4 months ago 33
Python Question

# Non Brute Force Solution to Project Euler 25

Project Euler Problem 25:

The Fibonacci sequence is defined by the recurrence relation:

Fn = Fn−1 + Fn−2, where F1 = 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 absolutely forever to calculate the actual solution. 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 == 1
else:
return Fibonacci(NthTerm-1) + Fibonacci(NthTerm-2) # recursive definition of Fib term

FirstTerm = 0 # For scope to include Term in scope of print on line 13
for Term in range(1, 1000): # Arbitrary range
FibValue = str(Fibonacci(Term)) # Convert integer to string for len()
if len(FibValue) == 1000:
FirstTerm = Term
break # Stop there
else: 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 a
else:
return fib(a+b, a, n-1)

print fib(1, 0, 10) # prints 55
``````

Here's a version that won't ever stackoverflow

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

print 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, a
n = n + 1
print "%d has 1000 digits, n = %d" % (a, n)
``````
Source (Stackoverflow)