Keeping a value during a recursive function

Question

In Haskell, I'm trying to solve a problem where I need to have a function that receives a list of integers and returns the biggest product of two adjacent numbers.

Example: f [5,6,1,3,9] would return 30, that is the product of 5 and 6

The function type would be something like this:

f :: [Int] -> Int

I thought to solve that using recursion to iterate the list getting the 2 head elements with a pattern like this: (x1:x2:xs) The problem is that I don't know how to keep the product value to compare if the current product is bigger than the last product.

Any ideas?

What if one gives the empty list? Or a list with one element? — willeM_ Van Onsem
– willeM_ Van Onsem, Commented Oct 24, 2017 at 22:04
Can you think of a way to create a list with every pair of adjacent numbers, f [5, 6, 1, 3, 9] == [(5, 6), (6, 1), (1, 3), (3, 9)]? — Ry-
– Ry- ♦, Commented Oct 24, 2017 at 22:06
Another possibility would be to make a recursive helper function go, that has an additional argument used as intermediate result (your current maximum). The problem: you have to find an initial value. Therefore consider changing the return type of your function and of the intermediate result to an Maybe Int. — mschmidt
– mschmidt, Commented Oct 24, 2017 at 22:12

Ry- · Accepted Answer · 2017-10-24 22:21:09Z

5

Since Haskell lists are lazy, you can solve this problem using a list-based approach instead of explicitly holding onto a maximum without losing efficiency. Starting with the original list:

> let f x = x
> f [5,6,1,3,9]
[5,6,1,3,9]

get a list of pairs by zipping the entire list with a left-shifted list:

> let f x = zip x (tail x)
> f [5,6,1,3,9]
[(5,6),(6,1),(1,3),(3,9)]

use the related function zipWith to get products instead of pairs:

> let f x = zipWith (*) x (tail x)
> f [5,6,1,3,9]
[30,6,3,27]

and get the maximum from that list:

> let f x = maximum (zipWith (*) x (tail x))
> f [5,6,1,3,9]
30

answered Oct 24, 2017 at 22:21

Ry-♦

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