dependently typed data structures hongwei xi [email protected] presented by james hook...
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Dependently Typed Data Structures
Hongwei Xi
presented byJames Hook ([email protected])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
Examplelet 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.
Typecheckingde Camlsource
Caml type check plus constraint generation
ML-style typeerrors
Constraint Solver
Constraintfailures
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
Example6
7
51
4
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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
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51
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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
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Repairing a Red Red Violation
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Example
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Insert 2
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Example6
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Insert 2
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Okasaki’s Solution
data Color = R | Bdata RedBlackSet a = E | T Color (RedBlackSet a) a (RedBlackSet a)
The Types Capture the Tree Structure:
Okasaki’s SolutionThe balance function does the rotation if there is a red red conflict
yx z
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balance :: Color -> RedBlackSet a -> a -> RedBlackSet a -> RedBlackSet abalance 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 heightthese clauses produce a tree of black heightone greater
data Color = R | Bdata 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 s@(T 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 ainsert :: 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 s@(T B a y b) | x < y = balance B (ins a) y b | x > y = balance B a y (ins b) | True = s ins s@(T 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 treebecause it recolors root
Hongwei’s Solution
data Color = R | Bdata 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 violationsA black node of height bh + 1 with no violationsis constructed from two nodes of arbitrary colorof 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 colorsof 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 Balancewithtype {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 -> twithtype {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)
