| Total Complexity | 3 |
| Total Lines | 69 |
| Duplicated Lines | 0 % |
| Changes | 0 | ||
| 1 | """ |
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| 2 | Project Euler Problem 25: :math:`1000`-Digit Fibonacci Number |
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| 3 | ============================================================= |
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| 4 | |||
| 5 | .. module:: solutions.problem25 |
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| 6 | :synopsis: My solution to problem #25. |
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| 7 | |||
| 8 | The source code for this problem can be |
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| 9 | `found here <https://bitbucket.org/nekedome/project-euler/src/master/solutions/problem25.py>`_. |
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| 10 | |||
| 11 | Problem Statement |
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| 12 | ################# |
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| 13 | |||
| 14 | The Fibonacci sequence is defined by the recurrence relation: |
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| 15 | |||
| 16 | .. math:: |
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| 17 | |||
| 18 | F_n = F_{n-1} + F_{n-2}, \\mbox{ where } F_1 = 1 \\mbox{ and } F_2 = 1. |
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| 19 | |||
| 20 | Hence the first :math:`12` terms will be: |
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| 21 | |||
| 22 | .. math:: |
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| 23 | |||
| 24 | F_1 &= 1 \\\\ |
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| 25 | F_2 &= 1 \\\\ |
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| 26 | F_3 &= 2 \\\\ |
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| 27 | F_4 &= 3 \\\\ |
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| 28 | F_5 &= 5 \\\\ |
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| 29 | F_6 &= 8 \\\\ |
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| 30 | F_7 &= 13 \\\\ |
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| 31 | F_8 &= 21 \\\\ |
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| 32 | F_9 &= 34 \\\\ |
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| 33 | F_{10} &= 55 \\\\ |
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| 34 | F_{11} &= 89 \\\\ |
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| 35 | F_{12} &= 144 |
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| 36 | |||
| 37 | The :math:`12^{th}` term, :math:`F_{12}`, is the first term to contain three digits. |
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| 38 | |||
| 39 | What is the index of the first term in the Fibonacci sequence to contain :math:`1000` digits? |
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| 40 | |||
| 41 | Solution Discussion |
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| 42 | ################### |
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| 43 | |||
| 44 | Simply iterate over the Fibonacci sequence until an :math:`F_n` is encountered containing :math:`1000` digits. |
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| 45 | |||
| 46 | Solution Implementation |
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| 47 | ####################### |
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| 48 | |||
| 49 | .. literalinclude:: ../../solutions/problem25.py |
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| 50 | :language: python |
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| 51 | :lines: 54- |
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| 52 | """ |
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| 53 | |||
| 54 | from itertools import dropwhile |
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| 55 | |||
| 56 | from lib.digital import num_digits |
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| 57 | from lib.sequence import Fibonaccis |
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| 58 | |||
| 59 | |||
| 60 | def solve(): |
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| 61 | """ Compute the answer to Project Euler's problem #25 """ |
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| 62 | target = 1000 |
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| 63 | fibs = Fibonaccis() |
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| 64 | for n, F_n in dropwhile(lambda elt: num_digits(elt[1]) < target, enumerate(fibs)): |
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| 65 | return n |
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| 66 | |||
| 67 | |||
| 68 | expected_answer = 4782 |
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| 69 |
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