问题
I'm struggling a little to come up with a notion of membership proof for Data.AVL trees. I would like to be able to pass around a value of type n ∈ m, to mean that n appears as a key in in the AVL tree, so that given such a proof, get n m can always successfully yield a value associated with n. You can assume my AVL trees always contain values drawn from a join semilattice (A, ∨, ⊥) over a setoid (A, ≈), although below the idempotence is left implicit.
module Temp where
open import Algebra.FunctionProperties
open import Algebra.Structures renaming (IsCommutativeMonoid to IsCM)
open import Data.AVL
open import Data.List
open import Data.Nat hiding (_⊔_)
import Data.Nat.Properties as ℕ-Prop
open import Data.Product hiding (_-×-_)
open import Function
open import Level
open import Relation.Binary renaming (IsEquivalence to IsEq)
open import Relation.Binary.PropositionalEquality
module ℕ-AVL {v} (V : Set v)
= Data.AVL (const V) (StrictTotalOrder.isStrictTotalOrder ℕ-Prop.strictTotalOrder)
data ≈-List {a ℓ : Level} {A : Set a} (_≈_ : Rel A ℓ) : Rel (List A) (a ⊔ ℓ) where
[] : ≈-List _≈_ [] []
_∷_ : {x y : A} {xs ys : List A} → (x≈y : x ≈ y) → (xs≈ys : ≈-List _≈_ xs ys) → ≈-List _≈_ (x ∷ xs) (y ∷ ys)
_-×-_ : {a b c d ℓ₁ ℓ₂ : Level} {A : Set a} {B : Set b} {C : Set c} {D : Set d} →
REL A C ℓ₁ → REL B D ℓ₂ → A × B → C × D → Set (ℓ₁ ⊔ ℓ₂)
(R -×- S) (a , b) (c , d) = R a c × S b d
-- Two AVL trees are equal iff they have equal toList images.
≈-AVL : {a ℓ : Level} {A : Set a} → Rel A ℓ → Rel (ℕ-AVL.Tree A) (a ⊔ ℓ)
≈-AVL _≈_ = ≈-List ( _≡_ -×- _≈_ ) on (ℕ-AVL.toList _)
_∈_ : {a ℓ : Level} {A : Set a} {_≈_ : Rel A ℓ} {_∨_ : Op₂ A} {⊥ : A}
{{_ : IsCM _≈_ _∨_ ⊥}} → ℕ → ℕ-AVL.Tree A → Set (a ⊔ ℓ)
n ∈ m = {!!}
get : {a ℓ : Level} {A : Set a} {_≈_ : Rel A ℓ} {_∨_ : Op₂ A} {⊥ : A} →
{{_ : IsCM _≈_ _∨_ ⊥}} → (n : ℕ) → (m : ℕ-AVL.Tree A) → n ∈ m → A
get n m n∈m = {!!}
I feel like this should be easy, but I'm finding it hard. One option would be to use my notion of equivalence for AVL-trees, which says that two trees are equal iff they have the same toList image, and exploit the commutative monoid over A, defining
n ∈ m = ≈-AVL ≈ m (ℕ-AVL.unionWith _ ∨ m (ℕ-AVL.singleton _ n ⊥))
This essentially says that m contains n iff the singleton map (n, ⊥) is "below" m in the partial order induced by the commutative monoid (technically we need the idempotence for this interpretation to make sense). However given such a definition I'm not at all sure how to implement get.
I have also experimented with defining my own inductive ∈ relation but found that hard as I seemed to end up having to know about the complicated internal indices used by Data.AVL.
Finally I also tried using a value of type n ∈? m ≡ true, where ∈? is defined in Data.AVL, but didn't have much success there either. I would appreciate any suggestions.
回答1:
I think your best bet is to define _∈_ as an inductive relation. Yes, this requires you to know the internals of Data.AVL, but I'm fairly sure this will be the most pleasant representation to work with.
The internal structure of Data.AVL is actually quite simple. We have a type Indexed.Tree, which is indexed by three values: the lower bound, the upper bound and the height. Given a tree t : Indexed.Tree lb ub h, all values inside t are within the range (lb, ub).
There's a slight twist to it, though: since we need to have a tree that can contain arbitrary values, we need to artifically extend the _<_ relation given by isStrictTotalOrder with two new values - you can think of those as a negative and a positive infinity. In the Data.AVL module, these are called ⊥⁺ and ⊤⁺. Trees that can contain arbitrary values are then of type Tree ⊥⁺ ⊤⁺ h.
The last piece is the balancing: each node requires heights of its subtrees to be at most one level apart. We don't actually need to touch balancing, but the function signatures might mention it.
Anyways, I'm working directly with this raw (indexed) variant. The opaque, unindexed version is just something like:
data Tree : Set ? where
tree : ∀ {h} → Indexed.Tree ⊥⁺ ⊤⁺ h
Some boilerplate first:
open import Data.Empty
open import Data.Product
open import Level
open import Relation.Binary
open import Relation.Binary.PropositionalEquality as P using (_≡_)
open import Relation.Nullary
import Data.AVL
module Membership
{k v ℓ}
{Key : Set k} (Value : Key → Set v)
{_<_ : Rel Key ℓ}
(isStrictTotalOrder : IsStrictTotalOrder _≡_ _<_) where
open Data.AVL Value isStrictTotalOrder public
open Extended-key public
open Height-invariants public
open IsStrictTotalOrder isStrictTotalOrder
Here's the _∈_ as an inductive relation:
data _∈_ {lb ub} (K : Key) :
∀ {n} → Indexed.Tree lb ub n → Set (k ⊔ v ⊔ ℓ) where
here : ∀ {hˡ hʳ} V
{l : Indexed.Tree lb [ K ] hˡ}
{r : Indexed.Tree [ K ] ub hʳ}
{b : hˡ ∼ hʳ} →
K ∈ Indexed.node (K , V) l r b
left : ∀ {hˡ hʳ K′} {V : Value K′}
{l : Indexed.Tree lb [ K′ ] hˡ}
{r : Indexed.Tree [ K′ ] ub hʳ}
{b : hˡ ∼ hʳ} →
K < K′ →
K ∈ l →
K ∈ Indexed.node (K′ , V) l r b
right : ∀ {hˡ hʳ K′} {V : Value K′}
{l : Indexed.Tree lb [ K′ ] hˡ}
{r : Indexed.Tree [ K′ ] ub hʳ}
{b : hˡ ∼ hʳ} →
K′ < K →
K ∈ r →
K ∈ Indexed.node (K′ , V) l r b
This is the sort of inductive definition you'd expect: either the key is in this inner node or it's down in one of the subtrees. We could also do without the less-than, greater-than proofs, but this is more convenient - when you want to show that a tree does not contain a particular element, you only have to track the path lookup would take, instead of searching the whole tree.
How to interpret those l and r implicit arguments? Notice that is makes perfect sense: we have a key K and naturally we require that the values contained in l fall between lb and K (it's actually [ K ], since we are using the extended _<_) and the values in r fall between K and ub. The balancing (b : hˡ ∼ hʳ) is there just so we can construct an actual tree node.
Your get function is then very simple:
get : ∀ {h lb ub n} {m : Indexed.Tree lb ub h} → n ∈ m → Value n
get (here V) = V
get (left _ p) = get p
get (right _ p) = get p
Well, I told you that this representation is fairly convenient to work it and I'm going to prove it. One of the properties we'd like _∈_ to have is decidability: that is, we can construct a program that tells us whether an element is in a tree or not:
find : ∀ {h lb ub} n (m : Indexed.Tree lb ub h) → Dec (n ∈ m)
find will return either yes p, where p is a proof that n is inside m (n ∈ m), or no ¬p, where ¬p is refutation of n ∈ m, n ∈ m → ⊥. We'll need one lemma:
lem : ∀ {lb ub hˡ hʳ K′ n} {V : Value K′}
{l : Indexed.Tree lb [ K′ ] hˡ}
{r : Indexed.Tree [ K′ ] ub hʳ}
{b : hˡ ∼ hʳ} →
n ∈ Indexed.node (K′ , V) l r b →
(n ≯ K′ → n ≢ K′ → n ∈ l) × (n ≮ K′ → n ≢ K′ → n ∈ r)
lem (here V) =
(λ _ eq → ⊥-elim (eq P.refl)) , (λ _ eq → ⊥-elim (eq P.refl))
lem (left x p) = (λ _ _ → p) , (λ ≮ _ → ⊥-elim (≮ x))
lem (right x p) = (λ ≯ _ → ⊥-elim (≯ x)) , (λ _ _ → p)
This tells us that if we know n is contained in t and we know n is less than the key of the root of t, then n must be in the left subtree (and similarly for right subtree).
Here's the implementation of find function:
find : ∀ {h lb ub} n (m : Indexed.Tree lb ub h) → Dec (n ∈ m)
find n (Indexed.leaf _) = no λ ()
find n (Indexed.node (k , v) l r _) with compare n k
find n (Indexed.node (k , v) l r _) | tri< a ¬b ¬c with find n l
... | yes p = yes (left a p)
... | no ¬p = no λ ¬∈l → ¬p (proj₁ (lem ¬∈l) ¬c ¬b)
find n (Indexed.node (k , v) l r _) | tri≈ ¬a b ¬c
rewrite (P.sym b) = yes (here v)
find n (Indexed.node (k , v) l r _) | tri> ¬a ¬b c with find n r
... | yes p = yes (right c p)
... | no ¬p = no λ ¬∈r → ¬p (proj₂ (lem ¬∈r) ¬a ¬b)
The implementation is fairly straightforward, but I would suggest loading it up in Emacs, trying to replace some of the right-hand-sides with holes and see what the types are. And finally, here are some tests:
open import Data.Nat
open import Data.Nat.Properties
open Membership
(λ _ → ℕ)
(StrictTotalOrder.isStrictTotalOrder strictTotalOrder)
un-tree : Tree → ∃ λ h → Indexed.Tree ⊥⁺ ⊤⁺ h
un-tree (tree t) = , t
test : Indexed.Tree _ _ _
test = proj₂ (un-tree
(insert 5 55 (insert 7 77 (insert 4 44 empty))))
Extract : ∀ {p} {P : Set p} → Dec P → Set _
Extract {P = P} (yes _) = P
Extract {P = P} (no _) = ¬ P
extract : ∀ {p} {P : Set p} (d : Dec P) → Extract d
extract (yes p) = p
extract (no ¬p) = ¬p
∈-test₁ : ¬ (2 ∈ test)
∈-test₁ = extract (find 2 test)
∈-test₂ : 4 ∈ test
∈-test₂ = extract (find 4 test)
来源：https://stackoverflow.com/questions/21081583/membership-proofs-for-avl-trees