# Project Euler: Problem 8

Published: Wed 03 November 2010

In blog.

The website has problem eight stated as such,

 ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22``` ```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

 ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34``` ```#!/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?

 ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15``` ```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 ```