[Haskell-cafe] Project Euler: request for comments
KC
kc1956 at gmail.com
Tue Aug 30 19:41:42 CEST 2011
You might like this zipping & folding version.
Explicit recursion has the disadvantage that one has to read the
entire function in order to figure out what's going on; whereas using
the higher order functions makes things much easier to grasp.
listof4tuples xs = (zip4 xs (tail xs) (tail (tail xs)) (tail (tail (tail xs))))
prods xs = map prods4 (listof4tuples xs)
prods4 (t,u,v,w) = t*u*v*w
maxprods4 xs = maximum $ prods xs
On Mon, Aug 29, 2011 at 9:40 AM, Oscar Picasso <oscarpicasso at gmail.com> wrote:
> Got it.
>
> f :: [Int] -> Int
> f (t:u:v:xs) = helper t u v xs
>
> helper :: Int -> Int -> Int -> [Int] -> Int
> helper t u v (w:ws)
> | ws == [] = t*u*v*w
> | otherwise = max (t*u*v*w) (f (u:v:w:ws))
>
> I tend to overlook mutual recursion in my toolbox.
>
> Thanks for the nnlightenment.
>
> On Sun, Aug 28, 2011 at 4:54 PM, KC <kc1956 at gmail.com> wrote:
>> Try something like the following:
>>
>> -- Project Euler 11
>>
>> -- In the 20×20 grid below, four numbers along a diagonal line have
>> been marked in red.
>>
>> -- <snip>
>>
>> -- The product of these numbers is 26 × 63 × 78 × 14 = 1788696.
>>
>> -- What is the greatest product of four adjacent numbers in any
>> direction (up, down, left, right, or diagonally) in the 20×20 grid?
>>
>>
>> import Data.List
>>
>> -- Doing the one dimensional case.
>> f011 :: [Int] -> Int
>> f011 (t:u:v:xs) = f011helper t u v xs
>>
>> f011helper :: Int -> Int -> Int -> [Int] -> Int
>> f011helper t u v (w:ws)
>> | ws == [] = t*u*v*w
>> | otherwise = yada nada mada
>>
>> -- What are yada nada mada?
>>
>> -- The 20x20 grid case will become:
>> f0112D :: [[Int]] -> Int
>> -- where [[Int]] is a list of lists of rows, columns, major diagonals,
>> & minor diagonals.
>>
>>
>>
>>
>> On Sun, Aug 28, 2011 at 5:10 AM, Oscar Picasso <oscarpicasso at gmail.com> wrote:
>>> No. The answer I posted is not good.
>>> It worked, by chance, on a couple of small examples I tried but it
>>> could end up comparing sequence of 4 numbers that where not initially
>>> adjacent.
>>>
>>> On Sun, Aug 28, 2011 at 12:32 AM, Oscar Picasso <oscarpicasso at gmail.com> wrote:
>>>> Maybe this?
>>>>
>>>> f x@(a:b:c:d:[]) = x
>>>> f (a:b:c:d:e:ys) = if e >= a
>>>> then f (b:c:d:e:ys)
>>>> else f (a:b:c:d:ys)
>>>>
>>>> On Sat, Aug 27, 2011 at 8:26 PM, KC <kc1956 at gmail.com> wrote:
>>>>> Think of the simplest version of the problem that isn't totally trivial.
>>>>>
>>>>> e.g. A one dimensional list of numbers.
>>>>>
>>>>> What would you do?
>>>>>
>>>>> Note: you only want to touch each element once.
>>>>>
>>>>> The 2 dimensional case could be handled by putting into lists: rows,
>>>>> columns, major diagonals, and minor diagonals.
>>>>>
>>>>> This isn't the fastest way of doing the problem but it has the
>>>>> advantage of avoiding "indexitis".
>>>>>
>>>>>
>>>>> On Fri, Aug 26, 2011 at 6:15 PM, Oscar Picasso <oscarpicasso at gmail.com> wrote:
>>>>>> Like:
>>>>>> 20*19*21*18
>>>>>> is bigger than
>>>>>> 100*100*3*2
>>>>>> ?
>>>>>>
>>>>>> If so I need to think about how to formalize it.
>>>>>>
>>>>>> Thanks for the hint.
>>>>>>
>>>>>> On Fri, Aug 26, 2011 at 8:55 PM, KC <kc1956 at gmail.com> wrote:
>>>>>>> Is Problem 11 the 4 consecutive #'s problem?
>>>>>>>
>>>>>>> If so what must be true for 4 #'s to have a large product?
>>>>>>>
>>>>>>> Hint: x * y * z * 2 is that going to be larger?
>>>>>>>
>>>>>>> --
>>>>>>> --
>>>>>>> Regards,
>>>>>>> KC
>>>>>>>
>>>>>>
>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> --
>>>>> Regards,
>>>>> KC
>>>>>
>>>>
>>>
>>
>>
>>
>> --
>> --
>> Regards,
>> KC
>>
>
--
--
Regards,
KC
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