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(** {1 Pancake sorting}
See {h <a href="https://en.wikipedia.org/wiki/Pancake_sorting">Pancake
sorting</a>}.
Author: Jean-Christophe Filliâtre (CNRS)
*)
use mach.int.Int
use ref.Ref
use array.Array
use array.ArraySwap
use array.ArrayPermut
(** We choose to have the bottom of the stack of pancakes at `a[0]`.
So it means we sort the array in reverse order. *)
predicate sorted (a: array int) (hi: int) =
forall j1 j2. 0 <= j1 <= j2 < hi -> a[j1] >= a[j2]
(** Insert the spatula at index `i` and flip the pancakes *)
let flip (a: array int) (i: int)
requires { 0 <= i < length a }
ensures { forall j. 0 <= j < i -> a[j] = (old a)[j] }
ensures { forall j. i <= j < length a -> a[j] = (old a)[length a -1-(j-i)] }
ensures { permut_all (old a) a }
= let n = length a in
for k = 0 to (n - i) / 2 - 1 do
invariant { forall j. 0 <= j < i -> a[j] = (old a)[j] }
invariant { forall j. i <= j < i+k -> a[j] = (old a)[n-1-(j-i)] }
invariant { forall j. i+k <= j < n-k -> a[j] = (old a)[j] }
invariant { forall j. n-k <= j < n -> a[j] = (old a)[n-1-(j-i)] }
invariant { permut_all (old a) a }
swap a (i + k) (n - 1 - k)
done
let pancake_sort (a: array int)
ensures { sorted a (length a) }
ensures { permut_all (old a) a }
= for i = 0 to length a - 2 do
invariant { sorted a i }
invariant { forall j1 j2. 0 <= j1 < i <= j2 < length a -> a[j1] >= a[j2] }
invariant { permut_all (old a) a }
(* 1. look for the maximum of a[i..] *)
let m = ref i in
for k = i + 1 to length a - 1 do
invariant { i <= !m < length a }
invariant { forall j. i <= j < k -> a[j] <= a[!m] }
if a[k] > a[!m] then m := k
done;
(* 2. then flip the pancakes to put it at index i *)
if !m = i then continue;
if !m < length a - 1 then flip a !m;
flip a i
done
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