ML anonymous function alternating sum

问题

For my homework assignment in ML I have to use the fold function and an anonymous function to turn a list of integers into the alternating sum. If the list is empty, the result is 0. This is what I have so far. I think what I have is correct, but my biggest problem is I cannot figure out how to write what I have as an anonymous function. Any help would be greatly appreciated.

    fun foldl f y nil = y
    | foldl f y (x::xr) = 
    foldl f(f(x,y))xr;

    val sum = foldl (op -) ~6[1,2,3,4,5,6];

    val sum = foldl (op -) ~4[1,2,3,4];

    val sum = foldl (op -) ~2[1,2];

These are just some examples that I tested to see if what I had worked and I think all three are correct.

回答1:

There are two cases: one when the list length is even and one when the list length is odd. If we have a list [a,b,c,d,e] then the alternating sum is a - b + c - d + e. You can re-write this as

e - (d - (c - (b - a)))

If the list has an even length, for example [a,b,c,d] then we can write its alternating sum as

- (d - (c - (b - a))).

So to address these two cases, we can have our accumulator for fold be a 3-tuple, where the first entry is the correct value if the list is odd, the second entry is the correct value if the list is even, and the third value tells us the number of elements we've looked at, which we can use to know at the end if the answer is the first or second entry.

So an anonymous function like

fn (x,y,n) => (x - #1 y, ~(x + #2 y), n + 1)

will work, and we can use it with foldl with a starting accumulator of (0,0,0), so

fun alternating_sum xs =
  let
    (v1, v2, n) = foldl (fn (x,y,n) => (x - #1 y, ~(x + #2 y), n + 1)) (0,0,0) xs
  in
    if n mod 2 = 0 then v2 else v1
  end

来源：https://stackoverflow.com/questions/26677750/ml-anonymous-function-alternating-sum

标签

function

anonymous