With a given positive integer list and the addition and the multiplication as operators, I want to compute the highest value.

So if my list is `[2,3,4]`

, it will be : `2 * 3 * 4 = 24`

. If there is at least one `1`

in the list, it is a little bit more difficult. I found by testing on several examples that to compute the highest value, I have to add `1`

to the minimum of the list and etc... until there is no more `1`

in the list and compute the product of all integers. For example : `[1,2,3,4]`

would return `( 2 + 1 ) * 3 * 4 = 36`

; `[1,1,2,3,4]`

would return `( 1 + 1 ) * 2 * 3 * 4 = 48`

.

My first question is concerning the algorithm part : **how to prove properly that my method is correct and will always return the highest value** ?

My second concerns the implementation :

I wrote the following code in OCaml to programm it :

```
(* Return 'list' with the first occurence of 'x' removed *)
let rec remove list x =
match list with
| [] -> []
| hd :: tl -> if hd = x then tl else hd :: remove tl x
;;
(* Return the maximum value which can be computed with the given list*)
let rec test_max list = match list with
| [] -> 0
| hd :: [] -> hd
| hd :: tl ->
(* If the list contains 1 : remove it from the list, then sort the list to get the min value and add 1 to it *)
if List.mem 1 list then begin
let list = List.sort (fun x y -> if (x < y) then 1 else 0) (remove list 1) in
let test = (List.hd list) + 1 in
test_max test::(List.tl list);
end
else
List.fold_left ( * ) 1 list;;
```

And at the instruction

```
let test = (List.hd list) + 1 in
test_max test::(List.tl list);
```

in `test_max`

I get the error :

```
Error: This expression has type 'a list
but an expression was expected of type int
```

Well, I don't really understand w**hy it is expecting an expression of type int in the recursive call** ... ?

Thank you in advance :)

Answer:

Function application has higher precedence than `::`

. So, your code `test_max test::(List.tl list)`

should indeed be `test_max (test::List.tl list)`

.

Also, you shouldn't use a `;`

in this place, as it's used to separate unit instructions. Just put nothing and it'll be fine.

As for some proof hints:

```
n + m > n * m
<=> ( n + m ) / (n * m) > 1
<=> 1/m + 1/n > 1
```

As we use positive integers, and as `/`

is decreasing in its second argument. It's easy to see that this is true iff `m or n = 1`

(equality can be attained when `n and m = 2`

).

Now, how do we use this `+1`

?

```
(n+1) * m > n * (m+1)
<=> n * m + m > n * m + n
<=> m > n
```

So here are your two lemmas. I'll let you search for a full proof.

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