Question

I’m working on solving the Project Euler problem 25:

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

My piece of code works for smaller digits, but when I try a 1000 digits, i get the error:

OverflowError: (34, 'Result too large')

I’m thinking it may be on how I compute the fibonacci numbers, but i’ve tried several different methods, yet i get the same error.

Here’s my code:

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

def fibonacci(n):
    phi = (1 + pow(5, 0.5))/2           #Golden Ratio
    return int((pow(phi, n) - pow(-phi, -n))/pow(5, 0.5))   #Formula: http://bit.ly/qDumIg


n = 0
while len(str(fibonacci(n))) < 1000:
    n += 1
print n

Do you know what may the cause of this problem and how i could alter my code avoid this problem?

Thanks in advance.

Answer 1

The problem here is that only integers in Python have unlimited length, floating point values are still calculated using normal IEEE types which has a maximum precision.

As such, since you’re using an approximation, using floating point calculations, you will get that problem eventually.

Instead, try calculating the Fibonacci sequence the normal way, one number (of the sequence) at a time, until you get to 1000 digits.

ie. calculate 1, 1, 2, 3, 5, 8, 13, 21, 34, etc.

By “normal way” I mean this:

         /  1                        , n < 3
Fib(n) = |
         \  Fib(n-2) + Fib(n-1)      , n >= 3

Note that the “obvious” approach given the above formulas is wrong for this particular problem, so I’ll post the code for the wrong approach just to make sure you don’t waste time on that:

def fib(n):
    if n <= 3:
        return 1
    else:
        return fib(n-2) + fib(n-1)

n = 1
while True:
    f = fib(n)
    if len(str(f)) >= 1000:
        print("#%d: %d" % (n, f))
        exit()
    n += 1

On my machine, the above code starts going really slow at around the 30th fibonacci number, which is still only 6 digits long.

I modified the above recursive approach to output the number of calls to the fib function for each number, and here are some values:

#1: 1
#10: 67
#20: 8361
#30: 1028457
#40: 126491971

I can reveal that the first Fibonacci number with 1000 digits or more is the 4782th number in the sequence (unless I miscalculated), and so the number of calls to the fib function in a recursive approach will be this number:

1322674645678488041058897524122997677251644370815418243017081997189365809170617080397240798694660940801306561333081985620826547131665853835988797427277436460008943552826302292637818371178869541946923675172160637882073812751617637975578859252434733232523159781720738111111789465039097802080315208597093485915332193691618926042255999185137115272769380924184682248184802491822233335279409301171526953109189313629293841597087510083986945111011402314286581478579689377521790151499066261906574161869200410684653808796432685809284286820053164879192557959922333112075826828349513158137604336674826721837135875890203904247933489561158950800113876836884059588285713810502973052057892127879455668391150708346800909439629659013173202984026200937561704281672042219641720514989818775239313026728787980474579564685426847905299010548673623281580547481750413205269166454195584292461766536845931986460985315260676689935535552432994592033224633385680958613360375475217820675316245314150525244440638913595353267694721961

And that is just for the 4782th number. The actual value is the sum of all those values for all the fibonacci numbers from 1 up to 4782. There is no way this will ever complete.

In fact, if we would give the code 1 year of running time (simplified as 365 days), and assuming that the machine could make 10.000.000.000 calls every second, the algorithm would get as far as to the 83rd number, which is still only 18 digits long.