PDA

View Full Version : Python: Project Euler 8

July 8th, 2008, 02:29 PM
Hello. I'm trying out some of the Euler problems in an attempt to learn Python. Here's the problem:

http://projecteuler.net/index.php?section=problems&id=8

And here's my code:

string = # that really long number, edited out for page-width-sanity

num = []

for char in string:
num.append(int(char))

i = 0

while i <= len(num) - 5:
k = i + 1
l = i + 2
m = i + 3
n = i + 4
prod = ( num[i] * num[k] * num[l] * num[m] * num[n])
i += 1

I get the correct answer but the code is clunky and I'm pretty sure there are some built-in python functions that would make this a lot easier to write, more memory efficient and easier on the eye.

What nice functions am I missing out on here?

ghostdog74
July 8th, 2008, 02:49 PM
import sys,textwrap
def mul(x,y): return x*y
string = """731671........"""
num = []
n=textwrap.wrap(string,5)
for i in n: num.append(reduce(mul,map(int,list(i))))
print max(num)

Martin Witte
July 8th, 2008, 05:49 PM
If I use ghostdogs code I get the wrong answer. Building on his code I have

#!/usr/bin/env python
string = '7316717...'

i = 0
max = 0
while i < len(string) - 4:
prod = reduce(lambda x,y: x*y, map(int, string[i:i+5]))
if prod > max:
max = prod
i+= 1
print max

textwrap (see the docs (http://docs.python.org/lib/module-textwrap.html)) chops a string into pieces of equal length, here you need to advance through the string 5 digits each

nanotube
July 8th, 2008, 05:54 PM
import sys,textwrap
def mul(x,y): return x*y
string = """731671........"""
num = []
n=textwrap.wrap(string,5)
for i in n: num.append(reduce(mul,map(int,list(i))))
print max(num)

ghostdog, your solution is not correct, as it doesn't try /all/ consecutive 5-digit numbers, but only non-overlapping sequences thereof. in the case of, eg, a 10-digit string, you have only two products, of digits 1-5, and of digits 6-10. however, the problem asks to check product of digits 1-5, digits 2-6, digits 3-7, etc.

despite that, thanks for introducing me to the textwrap module - i didn't even know that was there. :)

ghostdog74
July 8th, 2008, 06:17 PM

ghostdog74
July 8th, 2008, 06:19 PM
i didn't even know that was there. :)
if you master this (http://docs.python.org/lib/), you are invincible.

Martin Witte
July 8th, 2008, 07:12 PM
if you prefer compact code you can use a list comprehension (http://docs.python.org/tut/node7.html#SECTION007140000000000000000):

big_number = '731671765313....'
print max([ reduce(lambda x,y: x*y, map(int, big_number[i:i+5])) for i in range(0, len(big_number) -4) ])

nanotube
July 8th, 2008, 08:53 PM
if you master this (http://docs.python.org/lib/), you are invincible.

as long as you don't misread problem requirements, that is. ;) :popcorn:

rye_
July 8th, 2008, 09:20 PM
hmmm.... I had a go at this just to try to outsmart you all with the shortest code.... Things didn't go to plan.

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

class Array
def product
return self.inject() {|x,j| j*=x}
end
end

j = 0
a = array.split(/\n/).join("").split(//).collect {|x| x.to_i}
(0..a.length-5).each { |x| j = [a[x,5].product, j].max }
p j