(* Counting Sort. We sort an array of integers, assuming all elements are in the range 0..k-1 We simply count the elements equal to x, for each x in 0..k-1, and then (re)fill the array with two nested loops. TODO: Implement and prove a *stable* variant of counting sort, as proposed in Introduction to Algorithms Cormen, Leiserson, Rivest The MIT Press (2nd edition) Section 9.2, page 175 *) module Spec use export int.Int use int.NumOf as N use export array.Array use export array.IntArraySorted (* values of the array are in the range 0..k-1 *) constant k: int axiom k_positive: 0 < k predicate k_values (a: array int) = forall i: int. 0 <= i < length a -> 0 <= a[i] < k (* we introduce two predicates: - [numeq a v l u] is the number of values in a[l..u[ equal to v - [numlt a v l u] is the number of values in a[l..u[ less than v *) function numeq (a: array int) (v i j : int) : int = N.numof (\ k. a[k] = v) i j function numlt (a: array int) (v i j : int) : int = N.numof (\ k. a[k] < v) i j (* an ovious lemma relates numeq and numlt *) let rec lemma eqlt (a: array int) (v: int) (l u: int) requires { k_values a } requires { 0 <= v < k } requires { 0 <= l < u <= length a } ensures { numlt a v l u + numeq a v l u = numlt a (v+1) l u } variant { u - l } = if l < u-1 then eqlt a v (l+1) u (* permutation of two arrays is here conveniently defined using [numeq] i.e. as the equality of the two multi-sets *) predicate permut (a b: array int) = length a = length b /\ forall v: int. 0 <= v < k -> numeq a v 0 (length a) = numeq b v 0 (length b) end module CountingSort use import Spec use import ref.Refint (* sorts array a into array b *) let counting_sort (a: array int) (b: array int) requires { k_values a /\ length a = length b } ensures { sorted b /\ permut a b } = let c = make k 0 in for i = 0 to length a - 1 do invariant { forall v: int. 0 <= v < k -> c[v] = numeq a v 0 i } let v = a[i] in c[v] <- c[v] + 1 done; let j = ref 0 in for v = 0 to k-1 do invariant { !j = numlt a v 0 (length a) } invariant { sorted_sub b 0 !j } invariant { forall e: int. 0 <= e < !j -> 0 <= b[e] < v } invariant { forall f: int. 0 <= f < v -> numeq b f 0 !j = numeq a f 0 (length a) } for i = 1 to c[v] do invariant { !j -i+1 = numlt a v 0 (length a) } invariant { sorted_sub b 0 !j } invariant { forall e: int. 0 <= e < !j -> 0 <= b[e] <= v } invariant { forall f: int. 0 <= f < v -> numeq b f 0 !j = numeq a f 0 (length a) } invariant { numeq b v 0 !j = i-1 } b[!j] <- v; incr j done done; assert { !j = length b } end module InPlaceCountingSort use import Spec use import ref.Refint (* sorts array a in place *) let in_place_counting_sort (a: array int) requires { k_values a } ensures { sorted a /\ permut (old a) a } = 'L: let c = make k 0 in for i = 0 to length a - 1 do invariant { forall v: int. 0 <= v < k -> c[v] = numeq a v 0 i } let v = a[i] in c[v] <- c[v] + 1 done; let j = ref 0 in for v = 0 to k-1 do invariant { !j = numlt (at a 'L) v 0 (length a) } invariant { sorted_sub a 0 !j } invariant { forall e: int. 0 <= e < !j -> 0 <= a[e] < v } invariant { forall f: int. 0 <= f < v -> numeq a f 0 !j = numeq (at a 'L) f 0 (length a) } for i = 1 to c[v] do invariant { !j -i+1 = numlt (at a 'L) v 0 (length a) } invariant { sorted_sub a 0 !j } invariant { forall e: int. 0 <= e < !j -> 0 <= a[e] <= v } invariant { forall f: int. 0 <= f < v -> numeq a f 0 !j = numeq (at a 'L) f 0 (length a) } invariant { numeq a v 0 !j = i-1 } a[!j] <- v; incr j; assert {forall f. 0 <= f < v -> numeq a f 0 !j = numeq a f 0 (!j - 1)} done done; assert { !j = length a } end module Harness use import Spec use import InPlaceCountingSort let harness () requires { k = 2 } = (* a is [0;1;0] *) let a = make 3 0 in a[1] <- 1; in_place_counting_sort a; (* b is now [0;0;1] *) assert { numeq a 0 0 3 = 2 }; assert { numeq a 1 0 3 = 1 }; assert { a[0] = 0 }; assert { a[1] = 0 }; assert { a[2] = 1 } end