Yoriyuki Yamagata <yoriyuki.y@gmail.com> writes:

> Hi, list.
>
> When I read the balancing function of stdlib's Set/Map several years ago, I
> thought I have understand how it works.  But now, I read it again and I'm less
> confident now.  Could someone answer my questions?  Here is the snippet of the
> code.
>
>  let bal l v r =
>       let hl = match l with Empty -> 0 | Node(_,_,_,h) -> h in
>       let hr = match r with Empty -> 0 | Node(_,_,_,h) -> h in
>       if hl > hr + 2 then begin
>         match l with
>
>           Empty -> invalid_arg "Set.bal"
>         | Node(ll, lv, lr, _) ->
>             if height ll >= height lr then
>               create ll lv (create lr v r)
>             else begin
>               match lr with
>
>                 Empty -> invalid_arg "Set.bal"
>               | Node(lrl, lrv, lrr, _)->
>                   create (create ll lv lrl) lrv (create lrr v r)
>             end
>       end else if hr > hl + 2 then begin
>
>         match r with
>           Empty -> invalid_arg "Set.bal"
>         | Node(rl, rv, rr, _) ->
>             if height rr >= height rl then
>               create (create l v rl) rv rr
>             else begin
>
>               match rl with
>                 Empty -> invalid_arg "Set.bal"
>               | Node(rll, rlv, rlr, _) ->
>                   create (create l v rll) rlv (create rlr rv rr)
>             end
>
>       end else
>         Node(l, v, r, (if hl >= hr then hl + 1 else hr + 1))
>
> I have two question.
>
>         | Node(ll, lv, lr, _) ->
>             if height ll >= height lr then
>               create ll lv (create lr v r)
>
>             else begin
>
> Is this code right?  If r is Empty and lr and ll are huge trees,
> doesn't it create a massively unbalanced tree?

If r is empty then lr and ll can not be huge. Otherwise the tree was
massively unbalance beforehand. The balancing prevents this from
hapening.

> Another question is that why OCaml implementation allows
> a balancing factor up to *2*, which is usually allowed only up to 1?

Probably avoids having to do 2 balancings in a single operation. Or
weighs the number of balancing done on average against a slightly
unbalanced tree, i.e. turns out to be faster to be more unbalanced in
practice.

> Maybe my question is naive one, but I would appreciate if your could comment it.
>
> Regards,

MfG
        Goswin