Project Euler problem 8

Source code available on GitHub.

Ah, an interesting-looking one!

  The four adjacent digits in the 1000-digit number that have the greatest product are $9 \times 9 \times 8 \times 9 = 5832$.

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

  Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?

So, immediately we want to get this big number and open it in a text file. In Python, while you could just open the file, use it and then close it again, there’s a handy construct called withas which is perfect for dealing with text files. We copy and paste that big number into a file called input.txt and then do:

with open('input.txt') as f:

We do what we want to the text f and as soon as we leave the with statement, Python closes the file down and cleans up after us.

We want to read the file with f.read() and get rid of all the new line characters (\n). This does it:

with open('input.txt') as f:
    data = f.read().replace('\n','')

Now data contains a big string of our number and nothing else. We happen to know the number is 1000 digits long and we want 13 adjacent digits at a time, but these numbers could be anything so we’ll be general. The following for loop will iterate through all the possible 13 adjacent digits, find their product, and store it if it’s the biggest yet.

for i in range(len(data)-digitsLength):
    digits = [int(j) for j in data[i:i+digitsLength]]
    product = prod(digits)
    if product > maxProduct:
        maxProduct = product

You may notice that prod() isn’t a built-in function: we’ll define that. It just finds the products of all the elements in an iterable variable like a list, tuple or set. It’s a simple definition:

def prod(iterable):
    product = 1
    for i in iterable:
        product *= i
    return product

Fine, that’s all done. Our program is as follows:

def prod(iterable):
    product = 1
    for i in iterable:
        product *= i
    return product

with open('input.txt') as f:
    data = f.read().replace('\n','')

digitsLength = 13 # The number of adjacent digits we want to find the product of
maxProduct = 0

for i in range(len(data)-digitsLength):
    digits = [int(j) for j in data[i:i+digitsLength]]
    product = prod(digits)
    if product > maxProduct:
        maxProduct = product

print(maxProduct)

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