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.

Authors: Jean-Christophe Filliâtre

Topics: Array Data Structure

Tools: Why3

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```(* 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 *)
val constant k: int
ensures { 0 < result }

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 (fun k -> a[k] = v) i j

function numlt (a: array int) (v i j : int) : int =
N.numof (fun 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 Spec
use 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 = Array.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 Spec
use 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 }
= 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 (old a) 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 (old a) f 0 (length a) }
for i = 1 to c[v] do
invariant { !j -i+1 = numlt (old a) 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 (old a) 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 Spec
use 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;
(* a 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
```