# Project Euler Problem 8 in F#

After two easy problems, it's now time for Problem 8.

At first, this problem also seemed to me more complex than it really is. Let's have a look at the question:

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

So, basically what we need to do is:

- consider all chunks of this list of numbers that can be created by taking \(5\) consecutive digits from the beginning till the end (there will be \(n-5\) chunks if \(n\) is the number of digits in the list)
- for each chunk, multiply all the digits together (that's \(n-1\) products per chunk)
- pick the value corresponding to the greatest product.

At first I have considered resolving an easier problem first, like: find the chunk
whose *sum* is bigger and then calculate that product. Assuming that sum is
cheaper to calculate than product (and it is), we would have had \((n-5) \cdot (n-1)\)
sums and just 1 product. However you can easily find examples where the sum of the digits
is the same while the product is different: e.g. \(4 + 5 = 7 + 2 = 9\) but \(4 \cdot 5 = 20 > 7 \cdot 2 = 14\).
Therefore that was a dead end.

A more mathematically sound approach to find the chunk giving the highest product
might have been to first locate the chunk with the highest sum of *logarithms* of
each digit and then calculate the corresponding product. However, a quick test revealed
that logarithm is *way* slower than product.

Finally, a couple of micro-optimizations that occurred to me.

- You can skip all chunks containing at least one 0 (trivially, the product will be zero)
- You don't really have to multiply all the digits for each chunk (that is \((n-5) \cdot (n-1)\) products):
once you have computed the product of the first chunk, the product of the next chunk would be the product of the previous chunk
*times*the sixth digit*divided*by the first digit, and so on for subsequent chunks.

Eventually, I did not try to implement these two things, since brute force, which I am about to show you, is so quick that it's wasn't really worth it:

As you can see, we are generating a sequence of products and then pick the maximum value.
Each product is created by first taking a chunk of the array (using the `Array.sub`

)
and then using `Array.fold`

to compute the product of all elements in the chunk.

Again, I have tried to keep the imperative part to a minimum and I was able to get by without using any mutable variable.

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