I put an example implementation of (nondestructive) binary search trees at
http://ycombinator.com/arc/bst.arc
I'd appreciate it if people could tell me (a) if there are any bugs, (b) if there are bits that could be rewritten to be shorter, and (c) what the equivalent program would look like in other languages. Thanks!
Wouldn't it be better to have f< as a slot in (deftem node ...) ? That way we don't have to keep specifying it, although it does complicate how to specify the null tree.
This code is derived from the bst code I wrote for ACL. I remember when I wrote it thinking people might want to use different ordering functions when operating on the same tree. Now that doesn't seem so likely. But are there conceivable cases when one might want to?
By definition a binary search tree has its l link as < r link. If a different f< suddenly causes the r link to be < l link, then your bst suddently becomes incorrect and weird stuff happens.
Suppose you have a bst that persists for a long time. Suppose the things you're sorting are not mere integers, but a structures that represent something in the world, ranked according to some scoring function. Suppose you're able to figure out a slightly more accurate scoring function. If you're passing f< as an argument on each bst operation, you can just start using the new ranking fn. Whereas if you've been storing the comparison fn in the tree as you've been building it, you're probably going to have to rebuild the tree if the comparison fn changes.
I'm not saying this is going to be common enough to drive the design of a bst library. But it's by no means logically impossible to change f<.
Suppose we have two f<, f<old and f<new, to be used in a binary search tree b.
f<new may safely be used to replace f<old iff:
all x in b: all y in b: (f<old x y) == (f<new x y)
This stems from the binary sort tree invariants:
all x in b!l: (f<old x b!v)
all x in b!r: (f<old b!v x)
And the definition of all x of b:
all x of b = {all x in b!l, b!v, all x in b!r}
all x of nil = {}
This is because if even a single case is wrong, that part of the binary search tree concerned with keeping track of the correct order becomes wrong. In such a case, it is necessary to rebuild the tree.
Of course, note that the above condition is necessary only for all current existing nodes, i.e. if we have a new m:
all x in b: m != x
then it is okay that:
all x in b: (f<old m x) != (f<new m x)
While your style allows such an edge case, it is arguably such an edgy edge case that you might as well rebuild the tree if f<old != f<new; this is largely because you have to check that f<new == f<old for all current members of the tree, and that checking is O(N^2), whereas rebuilding the tree is O(N) taking O(2N) space (and your nondestructive bst takes O(N) space for each operation anyway).
I was thinking about cases where you'd be willing to tolerate a small amount of misordering in the tree (since you were till then using the old scoring function, which presumably misordered things or you wouldn't have improved it). But deletion wouldn't work if you changed the scoring fn, and the number of cases where you both want to change the ordering function and never need to do deletions must be small enough that a general-purpose bst lib doesn't need to support that.
I don't think you quite understand. A misordered binary search tree will fail lookups - (bst-find ...) will fail, even if that object is returned by (bst-elts ...). If you're going to create a general-purpose bst lib with "you can change the sort function without rebuilding the tree!" then you'd better make plenty damn sure there's a big warning saying "...but if you do (bst-find ...) might fail." Personally, I'd rebuild the tree anyway, because if the change is significant enough to change the sort function, it's big enough to completely destroy the meaning of the binary search tree.
I get that. When I think of using a bst I think of what News.YC needs: the top-ranked 100 or so stories out of 100,000, and no need for find or delete. (Though currently News.YC just truncates the list and sorts it, which would stop working at very high submission rates.)
A programming systems product takes about nine times as much effort as the component programs written separately for private use. I estimate that productizing imposes a factor of three; and that designing, integrating, and testing components into a coherent system imposes a factor of three; and that these cost components are essentially independent of each other.
It doesn't look to me like it's obviously shorter.
I haven't tried counting nodes in both either, but in nonwhitespace chars, which should be a reasonable approximation of nodes in two languages so similar, the Arc version is slightly shorter: 755 chars vs 774.
The line deriving(Eq,Show) means that you can now do this:
Prelude BST> Tree 'a' Empty Empty
Tree 'a' Empty Empty
Prelude BST> let a = Tree 'a' Empty Empty
Prelude BST> a
Tree 'a' Empty Empty
Prelude BST> let b = Tree 'b' Empty Empty
Prelude BST> b
Tree 'b' Empty Empty
Prelude BST> a == b
False
Prelude BST> let c = Tree 'c' a b
Prelude BST> let d = Tree 'c' a b
Prelude BST> c == d
True
All this is guaranteed at compile time. Moreover, the haskell version is much more readable. The use of modules removes unnecessary function prefixing, and there is no doubt about what the arguments to functions are when they are pattern matched.
Curiously, though, the arc version actually has less tokens by my measurement
but here is the interesting thing- now lets count just the unique tokens
bst.arc 41
bst.hs 47
This is in part due to haskell's pattern matching semantics where the same function definition is repeated multiple times, and the '|' character is repeatedly used. Also, there are more functions in the arc version, and in both pieces of code, variable names are usually one character and function names have multiple characters.
A more balanced tree will speed up 'find' and 'insert'. If those actions are performed far more often than 'remove', it would indeed be a good idea. Without knowing the use case, I cannot say which I would prefer.
It's fairly easy to do by hand. Just count the number of tokens-- in the sense of things whose first character a parser would not treat as merely another character in the name it was previously reading-- that are not closing delimiters of a pair.
If by that you mean counting the nodes, you need to parse the source first. You can either do it by hand (if you know the grammar) or tweak the compiler/interpreter, if it doesn't have any statistics option.
no need for bst-trav since there is an elements function. You can just map over the elements.
in the data declaration 'a' is a generic type, but Haskell will infer that the type must be an instance or class Ord (i.e. implement <,<=,>=,>,max,min,==). (If the type is not an instance of Ord it will tell you so at compile time)
Actually yes, in Haskell it is the same. Lazy evaluation means you could take the 4th element of a map over an infinite list. For example: "map (*2) [1..] !! 4"
I would argue that relying on overloaded comparison functions is a bad idea. Not all things have a 'natural ordering,' and even for those that do you might want to override it at some point.
I think reasonable people can differ on this. Passing comparators as arguments is an approach more typical of dynamically-typed languages (e.g. CL and Scheme 'sort'). In a statically-typed language, it is more common to use a comparison operator that specializes on type (e.g. Haskell Prelude 'sort'). If a set of objects don't have a 'natural ordering', they might not belong in a Binary Search Tree in the first place.
You have just seen a bunch of code that gets the job done, whereas there is still an argument about whether the lisp code should be passed a comparison function. If you are going to make this argument you should show some actual code, because it is obvious from looking at this example which way is better.
apply copies its argument list (guaranteed by Scheme spec). So this line lets the apply copy x, then passes it to a function that just returns the copied list. It may be faster than traversing the list in arc, for apply may be able to take advantage of underlying Scheme.
To me, this code is illustrating that "." and "!" should have their meanings reversed. "b!l" and "b!r" in arc would map to "b.l" or "b.r" in python, while "b.side" in arc, is something more akin to "b.getattr(side)"
Ofcourse, swapping the meanings of ! and . puts a cramp in syntax like "b.0"...
Update: I've rewritten my solution above to be as close as possible to the original Arc code. All the function interfaces should be the same. Equality is determined with (=), which is the closest thing OCaml has to Arc's is.
I've also created a separate version that's closer to how I would prefer to implement it:
1. The comparison function is specified using a functor. This allows the user to specify it only once instead of in every call to insert, find, and so on. (I'd like to see something like ML's functors in Lisp someday.)
2. The comparison function handles equality as well as order.
3. edge is replaced by min_elt and max_elt.
4. rem-edge is gone.
5. The output of to_list (aka elts) isn't backwards.
Maybe I'm not understanding your code (it's kind of cryptic...) but it does seem the same to me. When bst-rem removes a node with one child, it just grabs the other child. This is the same thing bst-rem-edge does.
Your 'bubble' function calls bst-edge to find the left/rightmost node and then calls bst-rem-edge to remove it. You could just as well call bst-edge and then bst-rem on the result; either way you traverse the tree twice. I can't imagine a situation where you'd just want bst-rem-edge by itself.
Come to think of it, I don't know why you need bst-edge either. It's simpler to just have min and max functions that return an element, as in the Haskell solution.
