Geek Answers Handbook

Ocaml noobie Q - how to use cumulative parameters?

I am trying to learn Ocaml while working on Problem 18 from Project Euler. I know what I want to do, I just can’t figure out how to do it.

I have three lists:

let list1 = [1;2;3;4;5];;
let list2 = [ 6;7;8;9];;
let line = [9999];;

I want to add the numbers list2 to the maximum neighboring number in list1, IOW I would add 6 + 2, 7 + 3, 8 + 4 and 9 + 5 to get the list [8; 10; 12; 14], List String [] is a dummy variable.

Here is my third attempt:

let rec meld3 l1 l2 accum =
   if List.length l2 = 1 then
     List.append accum [ (hd l2 + max (hd l1) (hd (tl l1)))]
else
    (
     List.append accum [ (hd l2 + max (hd l1) (hd (tl l1)))];
     meld3 (tl l1) (tl l2) accum ;
    )
;;

let fu = meld3 list1 list2 line ;;

List.iter print_int fu;;

After doing this, I would expect line = [9999; 8; 10; 12; 14], but instead, string = [9999]. OTOH, fu is printed as [999914].

When I look at the code, the code executes as I expect, but nothing changes; An asset in an else block never changes.

I just do not get this language. Can anyone advise?

+5
2

, . .

let rec meld3 l1 l2 accum =
   if List.length l2 = 1 then
     List.append accum [ (hd l2 + max (hd l1) (hd (tl l1)))]
else
    (
     List.append accum [ (hd l2 + max (hd l1) (hd (tl l1)))];
     meld3 (tl l1) (tl l2) accum ;
    )

, , , Caml , . hd tl. ; , . , , l2 .

let rec meld3 l1 l2 accum = match l1, l2 with
| x1::x2::xs, [y] ->   (* here the length of l2 is exactly 1 *)
     List.append accum [ y + max x1 x2 ]
| x1::x2::xs, y::ys ->   (* here the length of l2 is at least 1 *)    
    ( List.append accum [ y + max x1 x2 ]
    ; meld3 (x2::xs) ys accum
    )

, . (e1; e2), , e1 ( printf), e1 . , , e1 accum . , e1, ( , ):

let rec meld3 l1 l2 accum = match l1, l2 with
| x1::x2::xs, [y] ->   (* here the length of l2 is exactly 1 *)
     List.append accum [ y + max x1 x2 ]
| x1::x2::xs, y::ys ->   (* here the length of l2 is at least 1 *)    
    ( 
      meld3 (x2::xs) ys (List.append accum [ y + max x1 x2 ])
    )

- , Do not Repeat Yourself, , , l2 :

let rec meld3 l1 l2 accum = match l1, l2 with
| x1::x2::xs, [] ->   (* here the length of l2 is 0 *)
     accum 
| x1::x2::xs, y::ys ->   (* here the length of l2 is at least 1 *)    
    ( 
      meld3 (x2::xs) ys (List.append accum [ y + max x1 x2 ])
    )

:

let rec meld3 l1 l2 accum = match l1, l2 with
| _, [] -> accum 
| x1::x2::xs, y::ys -> meld3 (x2::xs) ys (List.append accum [ y + max x1 x2 ])

, append . : :

let rec meld3 l1 l2 accum' = match l1, l2 with
| _, [] -> List.rev accum' 
| x1::x2::xs, y::ys -> meld3 (x2::xs) ys (y + max x1 x2 :: accum')

accum accum'; . , , - . ( ).

, .

+9

, , : List.append accum .

, . ( ) , - . , :

(* function to merge a list l of length N with a list l' of length N+1,
   such that each element of the merged lists consists of a triple
     (l[i], l'[i], l'[i+1])
 *)

let rec merge_rows l l' = match l, l' with
  | [], [last] -> []   (* correct end of list *)
  | x::xs, y1::y2::ys -> (x, y1, y2) :: merge_rows xs (y2::ys)
  | _ -> raise (Failure "bad length in merge_rows")

let sum_max (cur, left, right) = cur + max left right

let merge_and_sum l l' = List.map sum_max (merge_rows l l')

let list1 = [1;2;3;4;5]
let list2 = [ 6;7;8;9]

let answer = merge_and_sum list2 list1

Euler 18, " ".

+5

More articles:


All Articles