Dependently Typed Data Structures Hongwei Xi presented by James Hook Pacific Software Research Center Oregon Graduate Institute

Hongwei’s Program Extend ML-like typechecking with computationally tractable mechanisms modeled on dependent type systems to get more expressive type systems without sacrificing practicality. Dependent ML --- his thesis at CMU de Caml --- a prototype implementation based on Caml

This Talk Apply “Hongwei’s types” to “Chris’ programs” Goal is to express invariants of data structures in the extended type system I will focus on red black trees

The Types ML types indexed by integers and integer expressions datatype ‘a list with nat = nil(0) | {n:nat} cons(n+1) of ‘a * ‘a list(n) Nil is the list of length 0 Cons builds a list of length n+1 from an element and a list of length n {}’s are explicit universal quantifiers for constraints Index expression

Example let rec append = function ([], ys) -> ys | (x :: xs, ys) -> x :: append(xs, ys) withtype {m:nat}{n:nat} ‘a list(m) * ‘a list(n) -> ‘a list(m+n) Given a list of length m and a list of length n append produces a list of length m+n.

Typechecking de Caml source Caml type check plus constraint generation ML-style type errors Constraint Solver Constraint failures

Red Black Trees Balanced, ordered, binary trees, values at nodes Every node is colored red or black Balance Invariant: –No red node has a red node as a child –There is an equal number of black nodes on all root/leaf paths

Example Insert 8 88 Violates equal number of black nodes on all leaf root paths 8 All insertions must be red to maintain the global invariant of equal numbers of black nodes on leaf root paths

Example Insert 2 Now we have a “red red” violation Red red violations are repairable; they are limited to the path to the root

Repairing a Red Red Violation x z y a bc d y xz abcd

y xz abcd x z y a bc d x z y a b c d z y x a b c d z x y a bc d

Example Insert 2 x z y a bc d y xz abcd 2 1 3

Example Insert

Okasaki’s Solution data Color = R | B data RedBlackSet a = E | T Color (RedBlackSet a) a (RedBlackSet a) The Types Capture the Tree Structure:

Okasaki’s Solution The balance function does the rotation if there is a red red conflict y xz abcd x z y a bc d x z y a b c d z y x a b c d z x y a bc d balance :: Color -> RedBlackSet a -> a -> RedBlackSet a -> RedBlackSet a balance B (T R (T R a x b) y c) z d = T R (T B a x b) y (T B c z d) balance B (T R a x (T R b y c)) z d = T R (T B a x b) y (T B c z d) balance B a x (T R (T R b y c) z d) = T R (T B a x b) y (T B c z d) balance B a x (T R b y (T R c z d)) = T R (T B a x b) y (T B c z d) balance color a x b = T color a x b When given two trees of equal black height these clauses produce a tree of black height one greater

data Color = R | B data RedBlackSet a = E | T Color (RedBlackSet a) a (RedBlackSet a) balance :: Color -> RedBlackSet a -> a -> RedBlackSet a -> RedBlackSet a insert :: a -> RedBlackSet a -> RedBlackSet a insert x s = T B a y b where T _ a y b = ins s ins E = T R E x E ins color a y b) | x < y = balance color (ins a) y b | x > y = balance color a y (ins b) | True = s Okasaki’s Solution Where do red red conflicts come from? To answer this question we expand ins to distinguish on color and we recall that balance only rotates trees with black roots

Okasaki’s Solution y a b ins a balance :: Color -> RedBlackSet a -> a -> RedBlackSet a -> RedBlackSet a insert :: Ord a => a -> RedBlackSet a -> RedBlackSet a insert x s = T B a y b where T _ a y b = ins s ins E = T R E x E ins B a y b) | x < y = balance B (ins a) y b | x > y = balance B a y (ins b) | True = s ins R a y b) | x < y = T R (ins a) y b | x > y = T R a y (ins b) | True = s ins on black yields red black tree ins on red may yield one red red violation at the root insert always yields a good tree because it recolors root

Hongwei’s Solution data Color = R | B data RedBlackSet a = E | T Color (RedBlackSet a) a (RedBlackSet a) sort color == {a:int | 0 <= a <= 1};; datatype tree with (color, nat, nat) = (* color, black height, violation *) E(0, 0, 0) | {cl:color}{cr:color}{bh:nat} B(0, bh+1, 0) of tree(cl, bh, 0) * key * tree(cr, bh, 0) | {cl:color}{cr:color}{bh:nat}{sl:nat}{sr:nat} R(1, bh, cl+cr) of tree(cl, bh, 0) * key * tree(cr, bh, 0) ;; General Approach: Index the datatype to record “black height” and to detect “red red violations” Detecting red red violations in the type indexs requires having an arithmetic encoding of color in the type index set as well as the value distinction in the program

Reading the Datatype sort color == {a:int | 0 <= a <= 1};; datatype tree with (color, nat, nat) = (* color, black height, violation *) E(0, 0, 0) | {cl:color}{cr:color}{bh:nat} B(0, bh+1, 0) of tree(cl, bh, 0) * key * tree(cr, bh, 0) | {cl:color}{cr:color}{bh:nat} R(1, bh, cl+cr) of tree(cl, bh, 0) * key * tree(cr, bh, 0) ;; Convention: 0 = black, 1 = red (permits detecting red red with +) The empty tree is black, has height 0, and no violations A black node of height bh + 1 with no violations is constructed from two nodes of arbitrary color of height bh A red node of height bh + 1 is constructed from two nodes of arbitrary color of height bh. The violation value of the node is the sum of the colors of its children, I.e. non-zero if either child is red. The children contain no violations.

Hongwei’s Solution let balance = function (R(R(a, x, b), y, c), z, d) -> R(B(a, x, b), y, B(c, z, d)) | (R(a, x, R(b, y, c)), z, d) -> R(B(a, x, b), y, B(c, z, d)) | (a, x, R(R(b, y, c), z, d)) -> R(B(a, x, b), y, B(c, z, d)) | (a, x, R(b, y, R(c, z, d))) -> R(B(a, x, b), y, B(c, z, d)) | (a, x, b) -> B(a, x, b) withtype {cl:color}{cr:color}{bh:nat}{vl: nat}{vr:nat | vl+vr <= 1} tree(cl, bh, vl) * key * tree(cr, bh, vr) -> [c:color] tree(c, bh+1, 0) ;; balance :: Color -> RedBlackSet a -> a -> RedBlackSet a -> RedBlackSet a Recall from Okasaki’s solution: As in red red analysis, Hongwei only calls balance on black nodes, hence Color argument is eliminated

Type of Balance withtype {cl:color}{cr:color}{bh:nat}{vl: nat}{vr:nat | vl+vr <= 1} tree(cl, bh, vl) * key * tree(cr, bh, vr) -> [c:color] tree(c, bh+1, 0) Give two trees of equal height bh, arbitrary color, and at most one red red violation, balance yields a tree with unspecified color of height bh+1 containing no violations

Hongwei’s Solution let rec ins = function E -> R(E, x, E) | B(a, y, b) -> if x < y then balance(ins a, y, b) else if y < x then balance(a, y, ins b) else raise Item_already_exists | R(a, y, b) -> if x < y then R(ins a, y, b) else if y < x then R(a, y, ins b) else raise Item_already_exists withtype {c:color}{bh:nat} tree(c, bh, 0) -> [c':color][v:nat | v <= c] tree(c', bh, v) ins is essentially as before The type of ins is now dramatically more expressive! ins produces a tree of unspecified color with height equal to its input. If the root of the argument was red the tree may contain a violation. If it was black it contains no violations.

Hongwei’s Solution let insert x t = let rec ins =... withtype {c:color}{bh:nat} tree(c, bh, 0) -> [c':color][v:nat | v <= c] tree(c', bh, v) in match ins t with R(a, y, b) -> B(a, y, b) | t -> t withtype {c:color}{bh:nat} key -> tree(c, bh, 0) -> [bh’:nat] tree(0, bh’, 0) ;; Insert is also essentially unchanged The type of insert now shows that both invariants are maintained by the operation. In particular, given a key and a red black tree of any height containing no violations, insert produces a tree with black root of some height containing no violations.

The Paper Braun Trees –The type of size guarantees it computes the size Random-Access Lists Binomial Heaps

Limitations Sometimes the programmer knows more than de Caml can figure out Not all integer constraints are decidable

Related Work  Refinement types (Freeman, Davies, Pfenning)  Indexed types (Zenger)  Sized types (Hughes, Pareto, Sabry)  Nested datatypes (Bird & Meertens, Okasaki, Hinze, etc)

Contacting Hongwei http// Tel