Pushing projecteuler ahead

I am busy writing dad’s pdf script. The real troubles begin when you parse 1000 pages using trash like pyparsing (yes, I am back to it). Anyway, let me see how it goes. I did another projecteuler.net sum today ( question 8 ). The question is:

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

Here is what I wrote to solve this problem:

bigno = input('Number:' ) #I am doing this because typing in the number distorts this blog.

divideIntoGroups(bigno):

        bigno_to_str = str(bigno)

        lower_limit = 0

        upper_limit = 5

        fives_list = []

        while upper_limit < len(bigno_to_str):

                lower_limit = lower_limit + 1

                upper_limit = upper_limit + 1

                fives_list.append(bigno_to_str[lower_limit:upper_limit])

        return fives_list

def splitFurther(fives_list):

        another_list = []

        for stringed_number in fives_list:

                commode = []

                for char in stringed_number:

                        commode.append(int(char))

                another_list.append(commode)

        return another_list

def multiplyIndividualNumbers(another_list):

        product_list = []

        for commode in another_list:

                products = reduce(mul, commode)

                product_list.append(products)

        return product_list

def checkGreatest(product_list):

        return max(product_list)

print checkGreatest(multiplyIndividualNumbers(splitFurther(divideIntoGroups(bigno))))

There. It took 0.03 seconds to work. Not bad for a budding computer scientist I suppose :D

Right, I’ll write again later.

One response to “Pushing projecteuler ahead”

Paul McGuire January 20, 2008 at 2:05 am | Permalink

Please don’t hesitate to ask for help if you are struggling with any pyparsing issues. Have you installed just the Windows binary? If so, this omits all of the documentation and examples directories. There is more information if you download the source or doc packages from SourceForge. You can post questions to the pyparsing mailing list, or just add them to the discussion tab on the pyparsing wiki home page.

Your Projecteuler solution *is* good for a beginner, but here are some handy Python idioms that will really get you going:
- treat strings like lists
- use zip to build tuples from other lists
- use iterators
- use generator expressions or list comprehensions

– Paul

number = “”"\
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
…
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450″”"

number = “”.join(number.splitlines())

from operator import mul
def prod(lst):
return reduce(mul, lst)

# create 5 iterators into number string, then position each
# iterator forward ‘n’ places
iters = [iter(number) for i in range(5)]
for i in range(5):
for j in range(i):
iters[i].next()

# use zip(*iters) to create list of tuples
# use prod to multiply the values of each tuple
# use max to give the maximum value
print max(prod(map(int,z)) for z in zip(*iters))

# dump out all 5-digit products
partials = [ number[i:i+5] for i in range(len(number)-5) ]
for p in partials:
print p, prod(map(int,p))

Pushing projecteuler ahead

One response to “Pushing projecteuler ahead”

Leave a Reply Click here to cancel reply.