[[!toc]] Recall back in [[Assignment4]], we asked you to enumerate the "fringe" of a leaf-labeled tree. Both of these trees: . . / \ / \ . 3 1 . / \ / \ 1 2 2 3 have the same fringe: `[1;2;3]`. We also asked you to write a function that determined when two trees have the same fringe. The way you approached that back then was to enumerate each tree's fringe, and then compare the two lists for equality. Today, and then again in a later class, we'll encounter two new ways to approach the problem of determining when two trees have the same fringe. ##List Zippers## Say you've got some moderately-complex function for searching through a list, for example: let find_nth (test : 'a -> bool) (n : int) (lst : 'a list) : (int * 'a) -> let rec helper (position : int) n lst = match lst with | [] -> failwith "not found" | x :: xs when test x -> (if n = 1 then (position, x) else helper (position + 1) (n - 1) xs ) | x :: xs -> helper (position + 1) n xs in helper 0 n lst;; This searches for the `n`th element of a list that satisfies the predicate `test`, and returns a pair containing the position of that element, and the element itself. Good. But now what if you wanted to retrieve a different kind of information, such as the `n`th element matching `test`, together with its preceding and succeeding elements? In a real situation, you'd want to develop some good strategy for reporting when the target element doesn't have a predecessor and successor; but we'll just simplify here and report them as having some default value: let find_nth' (test : 'a -> bool) (n : int) (lst : 'a list) (default : 'a) : ('a * 'a * 'a) -> let rec helper (predecessor : 'a) n lst = match lst with | [] -> failwith "not found" | x :: xs when test x -> (if n = 1 then (predecessor, x, match xs with [] -> default | y::ys -> y) else helper x (n - 1) xs ) | x :: xs -> helper x n xs in helper default n lst;; This duplicates a lot of the structure of `find_nth`; it just has enough different code to retrieve different information when the matching element is found. But now what if you wanted to retrieve yet a different kind of information...? Ideally, there should be some way to factor out the code to find the target element---the `n`th element of the list satisfying the predicate `test`---from the code that retrieves the information you want once the target is found. We might build upon the initial `find_nth` function, since that returns the *position* of the matching element. We could hand that result off to some other function that's designed to retrieve information of a specific sort surrounding that position. But suppose our list has millions of elements, and the target element is at position 600512. The search function will already have traversed 600512 elements of the list looking for the target, then the retrieval function would have to *start again from the beginning* and traverse those same 600512 elements again. It could go a bit faster, since it doesn't have to check each element against `test` as it traverses. It already knows how far it has to travel. But still, this should seem a bit wasteful. Here's an idea. What if we had some way of representing a list as "broken" at a specific point. For example, if our base list is: [10; 20; 30; 40; 50; 60; 70; 80; 90] we might imagine the list "broken" at position 3 like this (positions are numbered starting from 0): 40; 30; 50; 20; 60; [10; 70; 80; 90] Then if we move one step forward in the list, it would be "broken" at position 4: 50; 40; 60; 30; 70; 20; 80; [10; 90] If we had some convenient representation of these "broken" lists, then our search function could hand *that* off to the retrieval function, and the retrieval function could start right at the position where the list was broken, without having to start at the beginning and traverse many elements to get there. The retrieval function would also be able to inspect elements both forwards and backwards from the position where the list was "broken". The kind of data structure we're looking for here is called a **list zipper**. To represent our first broken list, we'd use two lists: (1) containing the elements in the left branch, preceding the target element, *in the order reverse to their appearance in the base list*. (2) containing the target element and the rest of the list, in normal order. So: 40; 30; 50; 20; 60; [10; 70; 80; 90] would be represented as `([30; 20; 10], [40; 50; 60; 70; 80; 90])`. To move forward in the base list, we pop the head element `40` off of the head element of the second list in the zipper, and push it onto the first list, getting `([40; 30; 20; 10], [50; 60; 70; 80; 90])`. To move backwards again, we pop off of the first list, and push it onto the second. To reconstruct the base list, we just "move backwards" until the first list is empty. (This is supposed to evoke the image of zipping up a zipper; hence the data structure's name.) This is a very handy datastructure, and it will become even more handy when we translate it over to more complicated base structures, like trees. To help get a good conceptual grip on how to do that, it's useful to introduce a kind of symbolism for talking about zippers. This is just a metalanguage notation, for us theorists; we don't need our programs to interpret the notation. We'll use a specification like this: [10; 20; 30; *; 50; 60; 70; 80; 90], * filled by 40 to represent a list zipper where the break is at position 3, and the element occupying that position is 40. For a list zipper, this is implemented using the pairs-of-lists structure described above. ##Tree Zippers## Now how could we translate a zipper-like structure over to trees? What we're aiming for is a way to keep track of where we are in a tree, in the same way that the "broken" lists let us keep track of where we are in the base list. For simplicity, we'll only implement this for binary trees---trees where internal nodes always have exactly two subtrees. But to get the guiding idea, it's helpful first to think about trees which permit nodes to have many subtrees. Suppose we had the following tree: 1000 / | \ / | \ / | \ / | \ / | \ 500 920 950 / | \ / | \ / | \ 20 50 80 91 92 93 94 95 96 1 2 3 4 5 6 7 8 9 and we want to represent that we're at the node labeled `50`. We might use the following metalanguage notation to specify this: {parent = ...; siblings = [node 20; *; node 80]}, * filled by node 50 This is modeled on the notation suggested above for list zippers. Here `node 20` refers not to a `int` label attached to that node, but rather to the whole subtree rooted at that node: 20 / | \ 1 2 3 Similarly for `node 50` and `node 80`. We haven't said yet what goes in the `parent = ...` slot. Well, the parent of a subtree targetted on `node 50` should intuitively be a tree targetted on `node 500`: {parent = ...; siblings = [*; node 920; node 950]}, * filled by node 500 And the parent of that targetted subtree should intuitively be a tree targetted on `node 1000`: {parent = None; siblings = [*]}, * filled by node 1000 This tree has no parents because it's the root of the base tree. Fully spelled out, then, our tree targetted on `node 50` would be: { parent = { parent = { parent = None; siblings = [*] }, * filled by node 1000; siblings = [*; node 920; node 950] }, * filled by node 500; siblings = [node 20; *; node 80] }, * filled by node 50 In fact, there's some redundancy in this structure, at the points where we have `* filled by node 1000` and `* filled by node 500`. Most of `node 1000`---with the exception of any label attached to node `1000` itself---is determined by the rest of this structure; and so too with `node 500`. So we could really work with: { parent = { parent = { parent = None; siblings = [*] }, label for * position (at node 1000); siblings = [*; node 920; node 950] }, label for * position (at node 500); siblings = [node 20; *; node 80] }, * filled by node 50 Or, if we only had labels on the leafs of our tree: { parent = { parent = { parent = None; siblings = [*] }, siblings = [*; node 920; node 950] }, siblings = [node 20; *; node 80] }, * filled by node 50 We're understanding the `20` here in `node 20` to just be a metalanguage tag to help us theorists keep track of which node we're referring to. We're supposing the tree structure itself doesn't associate any informative labelling information with those nodes. It only associates informative labels with the tree leafs. (We haven't represented any such labels in our diagreams.) We still do need to keep track of what fills the outermost targetted position---`* filled by node 50`---because that contain a subtree of arbitrary complexity, that is not determined by the rest of this data structure. For simplicity, I'll continue to use the abbreviated form: {parent = ...; siblings = [node 20; *; node 80]}, * filled by node 50 But that should be understood as standing for the more fully-spelled-out structure. Structures of this sort are called **tree zippers**, for a reason that will emerge. They should already seem intuitively similar to list zippers, though, at least in what we're using them to represent. I think it may initially be more helpful to call these **targetted trees**, though, and so will be switching back and forth between this different terms. Moving left in our targetted tree that's targetted on `node 50` would be a matter of shifting the `*` leftwards: {parent = ...; siblings = [*; node 50; node 80]}, * filled by node 20 and similarly for moving right. If the sibling list is implemented as a list zipper, you should already know how to do that. If one were designing a tree zipper for a more restricted kind of tree, however, such as a binary tree, one would probably not represented siblings with a list zipper, but with something more special-purpose and economical. Moving downward in the tree would be a matter of constructing a tree targetted on some child of `node 20`, with the first part of the targetted tree above as its parent: { parent = {parent = ...; siblings = [*; node 50; node 80]}; siblings = [*; leaf 2; leaf 3] }, * filled by leaf 1 How would we move upward in a tree? Well, we'd build a regular, untargetted tree with a root node---let's call it `20`---and whose children are given by the outermost sibling list in the targetted tree above, after inserting the targetted subtree into the `*` position: node 20 / | \ / | \ lead 1 leaf 2 leaf 3 We'll call this new untargetted tree `node 20`. The result of moving upward from our tree targetted on `leaf 1` would be the outermost `parent` element of that targetted tree, with `node 20` being the subtree that occupies that parent's `*` position: { parent = ...; siblings = [*; node 50; node 80] }, * filled by node 20 Or, spelling that structure out fully: { parent = { parent = { parent = None; siblings = [*] }, siblings = [*; node 920; node 950] }, siblings = [*; node 50; node 80] }, * filled by node 20 Moving upwards yet again would get us: { parent = { parent = None; siblings = [*] }, siblings = [*; node 920; node 950] }, * filled by node 500 where `node 500` refers to a tree built from a root node whose children are given by the list `[*; node 50; node 80]`, with `node 20` inserted into the `*` position. Moving upwards yet again would get us: { parent = None; siblings = [*] }, * filled by node 1000 where the targetted element is the root of our base tree. Like the "moving backward" operation for the list zipper, this "moving upward" operation is supposed to be reminiscent of closing a zipper, and that's why these data structures are called zippers. We haven't given you a real implementation of the tree zipper, but only a suggestive notation. We have however told you enough that you should be able to implement it yourself. Or if you're lazy, you can read: * [[!wikipedia Zipper (data structure)]] * Huet, Gerard. ["Functional Pearl: The Zipper"](http://www.st.cs.uni-sb.de/edu/seminare/2005/advanced-fp/docs/huet-zipper.pdf) Journal of Functional Programming 7 (5): 549-554, September 1997. * As always, [Oleg](http://okmij.org/ftp/continuations/Continuations.html#zipper) takes this a few steps deeper. [[Same-fringe using a tree zipper]] Supposing you did work out an implementation of the tree zipper, then one way to determine whether two trees have the same fringe would be: go downwards (and leftwards) in each tree as far as possible. Compare the targetted leaves. If they're different, stop because the trees have different fringes. If they're the same, then for each tree, move rightward if possible; if it's not (because you're at the rightmost position in a sibling list), more upwards then try again to move rightwards. Repeat until you are able to move rightwards. Once you do move rightwards, go downwards (and leftwards) as far as possible. Then you'll be targetted on the next leaf in the tree's fringe. The operations it takes to get to "the next leaf" may be different for the two trees. For example, in these trees: . . / \ / \ . 3 1 . / \ / \ 1 2 2 3 you won't move upwards at the same steps. Keep comparing "the next leafs" until they are different, or you exhaust the leafs of only one of the trees (then again the trees have different fringes), or you exhaust the leafs of both trees at the same time, without having found leafs with different labels. In this last case, the trees have the same fringe.