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PONon Brute Force Solution to Project Euler 25
 

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  1. PONon Brute Force Solution to Project Euler 25
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    <p><a href="https://projecteuler.net/problem=25" rel="nofollow noreferrer">Project Euler problem 25</a>:</p> <blockquote> <p>The Fibonacci sequence is defined by the recurrence relation:</p> <p>F<sub>n</sub> = F<sub>n−1</sub> + F<sub>n−2</sub>, where F<sub>1 = 1</sub> and F<sub>2</sub> = 1. Hence the first 12 terms will be F<sub>1</sub> = 1, F<sub>2</sub> = 1, F<sub>3</sub> = 2, F<sub>4</sub> = 3, F<sub>5</sub> = 5, F<sub>6</sub> = 8, F<sub>7</sub> = 13, F<sub>8</sub> = 21, F<sub>9</sub> = 34, F<sub>10</sub> = 55, F<sub>11</sub> = 89, F<sub>12</sub> = 144</p> <p>The 12th term, F<sub>12</sub>, is the first term to contain three digits.</p> <p>What is the first term in the Fibonacci sequence to contain 1000 digits?</p> </blockquote> <p>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?</p> <pre><code>def Fibonacci(NthTerm): if NthTerm == 1 or NthTerm == 2: return 1 # Challenge defines 1st and 2nd term as == 1 else: # recursive definition of Fib term return 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 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." </code></pre>
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    1. COa brute-force solution does not need to take for ever. it is your recursion that is way too slow. just google for a better fibonacci algorithm
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    2. COYour bruteforce complexity is horrible, you got O(2^n). Try to use three arguments: the number of terms to create, and the two previous results. That what you'll get linear time, and it should be fast enough.
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