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Maximum subarray problem


Authors: Jean-Christophe Filliâtre / Andrei Paskevich / Guillaume Melquiond

Topics: Ghost code / Array Data Structure

Tools: Why3

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(* Maximum subarray problem

   Given an array of integers, find the contiguous subarray with the
   largest sum. Subarrays of length 0 are allowed (which means that an
   array with negative values only has a maximal sum of 0).

   Authors: Jean-Christophe FilliĆ¢tre (CNRS)
            Guillaume Melquiond       (Inria)
            Andrei Paskevich          (U-PSUD)
 *)

module Spec
  use import int.Int
  use export array.Array

  use export array.ArraySum
  (* provides [sum a l h] = the sum of a[l..h[ and suitable lemmas *)

  (* s is no smaller than sums of subarrays a[l..h[ with 0 <= l < maxlo *)
  predicate maxsublo (a: array int) (maxlo: int) (s: int) =
    forall l h: int. 0 <= l < maxlo -> l <= h <= length a -> sum a l h <= s

  (* s is no smaller than sums of subarrays of a *)
  predicate maxsub (a: array int) (s: int) =
    forall l h: int. 0 <= l <= h <= length a -> sum a l h <= s

end

(* In all codes below, reference ms stands for the maximal sum found so far
   and ghost references lo and hi hold the bounds for this sum *)

(* Naive solution, in O(N^3) *)
module Algo1

  use import int.Int
  use import ref.Refint
  use import Spec

  let maximum_subarray (a: array int) (ghost lo hi: ref int): int
    ensures { 0 <= !lo <= !hi <= length a && result = sum a !lo !hi }
    ensures { maxsub a result }
  = lo := 0;
    hi := 0;
    let n = length a in
    let ms = ref 0 in
    for l = 0 to n-1 do
      invariant { 0 <= !lo <= l && !lo <= !hi <= n && !ms = sum a !lo !hi }
      invariant { maxsublo a l !ms }
      for h = l to n do
        invariant { 0 <= !lo <= l && !lo <= !hi <= n && !ms = sum a !lo !hi }
        invariant { maxsublo a l !ms }
        invariant { forall h': int. l <= h' < h -> sum a l h' <= !ms }
        (* compute the sum of a[l..h[ *)
        let s = ref 0 in
        for i = l to h-1 do
          invariant { !s = sum a l i }
          invariant { 0 <= !lo <= l && !lo <= !hi <= n && !ms = sum a !lo !hi }
          s += a[i]
        done;
        assert { !s = sum a l h };
        if !s > !ms then begin ms := !s; lo := l; hi := h end
      done
    done;
    !ms

end

(* Slightly less naive solution, in O(N^2)
   Do not recompute the sum, simply update it *)

module Algo2

  use import int.Int
  use import ref.Refint
  use import Spec

  let maximum_subarray (a: array int) (ghost lo hi: ref int): int
    ensures { 0 <= !lo <= !hi <= length a && result = sum a !lo !hi }
    ensures { maxsub a result }
  = lo := 0;
    hi := 0;
    let n = length a in
    let ms = ref 0 in
    for l = 0 to n-1 do
      invariant { 0 <= !lo <= l && !lo <= !hi <= n && 0 <= !ms = sum a !lo !hi }
      invariant { maxsublo a l !ms }
      let s = ref 0 in
      for h = l+1 to n do
        invariant
                { 0 <= !lo <= l && !lo <= !hi <= n && 0 <= !ms = sum a !lo !hi }
        invariant { maxsublo a l !ms }
        invariant { forall h': int. l <= h' < h -> sum a l h' <= !ms }
        invariant { !s = sum a l (h-1) }
        s += a[h-1]; (* update the sum *)
        assert { !s = sum a l h };
        if !s > !ms then begin ms := !s; lo := l; hi := h end
      done
    done;
    !ms

end

(* Divide-and-conqueer solution, in O(N log N) *)

module Algo3

  use import int.Int
  use import ref.Refint
  use import int.ComputerDivision
  use import Spec

  let rec maximum_subarray_rec (a: array int) (l h: int) (ghost lo hi: ref int)
    : int
    requires { 0 <= l <= h <= length a }
    ensures  { l <= !lo <= !hi <= h && result = sum a !lo !hi }
    ensures  { forall l' h': int. l <= l' <= h' <= h -> sum a l' h' <= result }
    variant  { h - l }
  = if h = l then begin
      (* base case: no element at all *)
      lo := l; hi := h; 0
    end else begin
      (* at least one element *)
      let mid = l + div (h - l) 2 in
      (* first consider all sums that include a[mid] *)
      lo := mid; hi := mid;
      let ms = ref 0 in
      let s  = ref !ms in
      for i = mid-1 downto l do
        invariant { l <= !lo <= mid = !hi && !ms = sum a !lo !hi }
        invariant { forall l': int. i < l' <= mid -> sum a l' mid <= !ms }
        invariant { !s = sum a (i+1) mid }
        s += a[i];
        assert { !s = sum a i mid };
        if !s > !ms then begin ms := !s; lo := i end
      done;
      assert { forall l': int. l <= l' <= mid ->
               sum a l' mid <= sum a !lo mid };
      s := !ms;
      for i = mid to h-1 do
        invariant { l <= !lo <= mid <= !hi <= h && !ms = sum a !lo !hi }
        invariant { forall l' h': int. l <= l' <= mid <= h' <= i ->
                    sum a l' h' <= !ms }
        invariant { !s = sum a !lo i }
        s += a[i];
        assert { !s = sum a !lo (i+1) };
        assert { !s = sum a !lo mid + sum a mid (i+1) };
        if !s > !ms then begin ms := !s; hi := (i+1) end
      done;
      (* then consider sums in a[l..mid[ and a[mid+1..h[, recursively *)
      begin
         let ghost lo' = ref 0 in
         let ghost hi' = ref 0 in
         let left = maximum_subarray_rec a l mid lo' hi' in
         if left > !ms then begin ms := left; lo := !lo'; hi := !hi' end
      end;
      begin
         let ghost lo' = ref 0 in
         let ghost hi' = ref 0 in
         let right = maximum_subarray_rec a (mid+1) h lo' hi' in
         if right > !ms then begin ms := right; lo := !lo'; hi := !hi' end
      end;
      !ms
    end

 let maximum_subarray (a: array int) (ghost lo hi: ref int): int
    ensures { 0 <= !lo <= !hi <= length a && result = sum a !lo !hi }
    ensures { maxsub a result }
  = maximum_subarray_rec a 0 (length a) lo hi

end

(* Optimal solution, in O(N)
   Known as Kadane's algorithm *)

module Algo4

  use import int.Int
  use import ref.Refint
  use import Spec

  let maximum_subarray (a: array int) (ghost lo hi: ref int): int
    ensures { 0 <= !lo <= !hi <= length a && result = sum a !lo !hi }
    ensures { maxsub a result }
  = lo := 0;
    hi := 0;
    let n = length a in
    let ms = ref 0 in
    let ghost l = ref 0 in
    let s = ref 0 in
    for i = 0 to n-1 do
      invariant { 0 <= !lo <= !hi <= i && 0 <= !ms = sum a !lo !hi }
      invariant { forall l' h': int. 0 <= l' <= h' <= i -> sum a l' h' <= !ms }
      invariant { 0 <= !l <= i && !s = sum a !l i }
      invariant { forall l': int. 0 <= l' < i -> sum a l' i <= !s }
      if !s < 0 then begin s := a[i]; l := i end else s += a[i];
      if !s > !ms then begin ms := !s; lo := !l; hi := (i+1) end
    done;
    !ms

end

(* A slightly different variation of Kadane's algorithm *)

module Algo5

  use import int.Int
  use import ref.Refint
  use export array.Array
  use export array.ArraySum

(*
    [| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |]
     ......|###### maxsum #######|..............
     ............................. |## curmax ##
*)

  let maximum_subarray (a: array int): int
    ensures { forall l h: int. 0 <= l <= h <= length a -> sum a l h <= result }
    ensures { exists l h: int. 0 <= l <= h <= length a /\ sum a l h  = result }
  =
    let maxsum = ref 0 in
    let curmax = ref 0 in
    let ghost lo = ref 0 in
    let ghost hi = ref 0 in
    let ghost cl = ref 0 in
    for i = 0 to a.length - 1 do
      invariant { forall l : int. 0 <= l <= i -> sum a l i <= !curmax }
      invariant { 0 <= !cl <= i /\ sum a !cl i  = !curmax }
      invariant { forall l h: int. 0 <= l <= h <= i -> sum a l h <= !maxsum }
      invariant { 0 <= !lo <= !hi <= i /\ sum a !lo !hi = !maxsum }
      curmax += a[i];
      if !curmax < 0 then begin curmax := 0; cl := i+1 end;
      if !curmax > !maxsum then begin maxsum := !curmax; lo := !cl; hi := i+1 end
    done;
    !maxsum

end

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