Project Euler: Problem 8

The website has problem eight stated as such,

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Find the greatest product of five consecutive digits in the 1000-digit number.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

Yeah, looks pretty intimadating to me too. ;) Lucky enough for both of us this really isn't all that hard to figure out.

Python

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#!/usr/bin/python3
"""
code solution for Euler Project #8
"""
import operator
from functools import reduce

if __name__ == "__main__":
  index = 0
  high_score = 0
  investigate_this = 7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450


  digit_list = [int(x) for x in str(investigate_this)]
  digit_list_length = len(digit_list)

  while index < digit_list_length:
    if index == 0:
      temp_product = reduce(operator.mul, digit_list[index: index +5])
    elif index == 1:
      temp_product = reduce(operator.mul, digit_list[index - 1: index + 4])
    elif index == digit_list_length - 1:
      temp_product = reduce(operator.mul, digit_list[index - 4: index + 1])
    elif index == digit_list_length - 2:
      temp_product = reduce(operator.mul, digit_list[index - 3: index + 2])
    else:
      temp_product = reduce(operator.mul, digit_list[index -2 : index + 3])

    if temp_product > high_score:
        high_score = temp_product

    index += 1

  print(high_score)

For my first attempt I took an approach of having a travelling index and checking both two digits before and two digits after the index. This worked out pretty well except that I needed to add four special use cases (the first two index points and the last two). Some people may say that it's not “pythonic” to use reduce, and in some ways I can agree with it. However, in this case my reduce fuctions really aren't very complicated and I think they look cleaner than using for loops in this instance. Does anyone out there have any opinions?

Haskell

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module Main where

import Data.Char

investigate_this = map digitToInt $ show 7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450

investigate_length = length investigate_this

productof5 index
  | index >= investigate_length - 4 = []
  | otherwise  = prod5 : productof5 (index + 1)
  where prod5 = product [ investigate_this !! y | y <- [index..(index + 4)]]

main :: IO()
main = print . maximum $ productof5 0

In my next solution, I followed the same idea of using a traveling index, but this time I only paid attention to the index and the next four digits. The advantage to this was that I only needed to worry about one special case, and that was if the number of multipliers were less than five total. This did seem to simplify things.

Sorry to all my Perl & Java readers. Life is kind of chaotic at the moment and I only really had the energy to do these two languages. One day I hope to revisit this problem in those languages, and when I do I'll update this post with the new code additions.

I realize in the last few times I've posted project euler solution I haven't said anything so just to nail this thought home amongst the people reading this. I look forward to hearing any and all constructive comments regarding algorythms and code. Oh, and one more thing, I apologize in advance for the formatting of the rather large number.

Comments !

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