why3/examples/binary_search.mlw

(* Binary search

   A classical example. Searches a sorted array for a given value v. *)

module BinarySearch

  use int.Int
  use int.ComputerDivision
  use ref.Ref
  use array.Array

  (* the code and its specification *)

  exception Not_found (* raised to signal a search failure *)

  let binary_search (a: array int) (v: int) : int
    requires { forall i1 i2. 0 <= i1 <= i2 < length a -> a[i1] <= a[i2] }
    ensures  { 0 <= result < length a /\ a[result] = v }
    raises   { Not_found -> forall i. 0 <= i < length a -> a[i] <> v }
  =
    let ref l = 0 in
    let ref u = length a - 1 in
    while l <= u do
      invariant { 0 <= l /\ u < length a }
      invariant {
        forall i. 0 <= i < length a -> a[i] = v -> l <= i <= u }
      variant { u - l }
      let m = l + div (u - l) 2 in
      assert { l <= m <= u };
      if a[m] < v then
        l := m + 1
      else if a[m] > v then
        u := m - 1
      else
        return m
    done;
    raise Not_found

end

(* A generalization: the midpoint is computed by some abstract function.
   The only requirement is that it lies between l and u. *)

module BinarySearchAnyMidPoint

  use int.Int
  use ref.Ref
  use array.Array

  exception Not_found (* raised to signal a search failure *)

  val midpoint (l: int) (u: int) : int
    requires { l <= u } ensures { l <= result <= u }

  let binary_search (a: array int) (v: int) : int
    requires { forall i1 i2. 0 <= i1 <= i2 < length a -> a[i1] <= a[i2] }
    ensures  { 0 <= result < length a /\ a[result] = v }
    raises   { Not_found -> forall i. 0 <= i < length a -> a[i] <> v }
  =
    let ref l = 0 in
    let ref u = length a - 1 in
    while l <= u do
      invariant { 0 <= l /\ u < length a }
      invariant { forall i. 0 <= i < length a -> a[i] = v -> l <= i <= u }
      variant { u - l }
      let m = midpoint l u in
      if a[m] < v then
        l := m + 1
      else if a[m] > v then
        u := m - 1
      else
        return m
    done;
    raise Not_found

end

(* The following version of binary search is faster in practice, by being
   friendlier with the branch predictor of most processors. It also happens
   to be stable, since it always return the highest index. *)

module BinarySearchBranchless

  use int.Int
  use int.ComputerDivision
  use ref.Ref
  use array.Array

  exception Not_found (* raised to signal a search failure *)

  let binary_search (a: array int) (v: int) : int
    requires { forall i1 i2. 0 <= i1 <= i2 < length a -> a[i1] <= a[i2] }
    ensures  { 0 <= result < length a /\ a[result] = v }
    ensures  { forall i. result < i < length a -> a[i] <> v }
    raises   { Not_found -> forall i. 0 <= i < length a -> a[i] <> v }
  =
    let ref l = 0 in
    let ref s = length a in
    if s = 0 then raise Not_found;
    while s > 1 do
      invariant { 0 <= l /\ l + s <= length a /\ s >= 1 }
      invariant {
        forall i. 0 <= i < length a -> a[i] = v -> a[l] <= v /\ i < l + s }
      variant { s }
      let h = div s 2 in
      let m = l + h in
      l := if a[m] > v then l else m;
      s := s - h;
    done;
    if a[l] = v then l
    else raise Not_found

end

(* binary search using 32-bit integers *)

module BinarySearchInt32

  use int.Int
  use mach.int.Int32
  use ref.Ref
  use mach.array.Array32

  exception Not_found   (* raised to signal a search failure *)

  let binary_search (a: array int32) (v: int32) : int32
    requires { forall i1 i2. 0 <= i1 <= i2 < a.length -> a[i1] <= a[i2] }
    ensures  { 0 <= result < a.length /\ a[result] = v }
    raises   { Not_found -> forall i. 0 <= i < a.length -> a[i] <> v }
  =
    let ref l = 0 in
    let ref u = length a - 1 in
    while l <= u do
      invariant { 0 <= l /\ u < a.length }
      invariant { forall i. 0 <= i < a.length -> a[i] = v -> l <= i <= u }
      variant   { u - l }
      let m = l + (u - l) / 2 in
      assert { l <= m <= u };
      if a[m] < v then
        l := m + 1
      else if a[m] > v then
        u := m - 1
      else
        return m
    done;
    raise Not_found

end

(* A particular case with Boolean values (0 or 1) and a sentinel 1 at the end.
   We look for the first position containing a 1. *)

module BinarySearchBoolean

  use int.Int
  use int.ComputerDivision
  use ref.Ref
  use array.Array

  let binary_search (a: array int) : int
    requires { 0 < a.length }
    requires { forall i j. 0 <= i <= j < a.length -> 0 <= a[i] <= a[j] <= 1 }
    requires { a[a.length - 1] = 1 }
    ensures  { 0 <= result < a.length }
    ensures  { a[result] = 1 }
    ensures  { forall i. 0 <= i < result -> a[i] = 0 }
 =
    let ref lo = 0 in
    let ref hi = length a - 1 in
    while lo < hi do
      invariant { 0 <= lo <= hi < a.length }
      invariant { a[hi] = 1 }
      invariant { forall i. 0 <= i < lo -> a[i] = 0 }
      variant   { hi - lo }
      let mid = lo + div (hi - lo) 2 in
      if a[mid] = 1 then
        hi := mid
      else
        lo := mid + 1
    done;
    lo

end

module Complexity

  use int.Int
  use int.ComputerDivision
  use ref.Ref
  use array.Array

  let rec function log2 (n: int) : int
    variant { n }
  = if n <= 1 then 0 else 1 + log2 (div n 2)

  let rec lemma log2_monotone (x y: int)
    requires { x <= y }
    ensures  { log2 x <= log2 y }
    variant  { y }
  = if y > 1 then log2_monotone (div x 2) (div y 2)

  let function f (n: int) : int
  = if n = 0 then 0 else 1 + log2 n

  lemma upper_bound:
    forall n. n >= 2 -> f n <= 2 * log2 n

  val ref time: int

  let binary_search (a: array int) (v: int) : int
    requires { forall i1 i2. 0 <= i1 <= i2 < length a -> a[i1] <= a[i2] }
    requires { time = 0 }
    ensures  { 0 <= result < length a && a[result] = v
            || result = -1 && forall i. 0 <= i < length a -> a[i] <> v }
    ensures  { time - old time <= f (length a) }
  =
    let ref lo = 0 in
    let ref hi = length a in
    while lo < hi do
      invariant { 0 <= lo <= hi <= length a }
      invariant { forall i. 0 <= i < lo || hi <= i < length a -> a[i] <> v }
      invariant { (time - old time) + f (hi - lo) <= f (length a) }
      variant   { hi - lo }
      let mid = lo + div (hi - lo) 2 in
      if a[mid] < v then
        lo <- mid + 1
      else if a[mid] > v then
        hi <- mid
      else
        return mid;
      time <- time + 1
    done;
    -1

end

(* Search in a two-dimensional grid where all rows and columns are
   sorted.  Here is an example:

                            j
     +---+---+---+---+---+-->
     | 1 | 3 | 4 | 4 | 7*|
     +---+---+---+---+---+
     | 1 | 4 | 6 | 6 | 8*|
     +---+---+---+---+---+
     | 2 | 5 | 7 | 9*|11*|
     +---+---+---+---+---+
     | 4 | 8 | 8*|12*|13 |
     +---+---+---+---+---+
     |         *
   i v

  Algorithm: start from the upper right corner and then move left
  (resp. down) when the value is greater (resp. smaller) than the
  value we search for.

  In the example above, the stars depict the elements that are
  examined when we search for the value 10. Here we end up outside of
  the grid and thus the search is unsuccessful.
*)

module TwoDimensional

  use int.Int
  use matrix.Matrix

  let search (m: matrix int) (v: int) : bool
    requires  { [@expl: rows are sorted] forall i. 0 <= i < rows m ->
                forall j1 j2. 0 <= j1 <= j2 < columns m ->
                get m i j1 <= get m i j2  }
    requires  { [@expl: columns are sorted] forall j. 0 <= j < columns m ->
                forall i1 i2. 0 <= i1 <= i2 < rows m ->
                get m i1 j <= get m i2 j  }
    ensures   { result <->
      exists i j. 0 <= i < rows m && 0 <= j < columns m && get m i j = v }
  = let ref i = 0 in
    let ref j = columns m - 1 in
    while i < rows m && 0 <= j do
      invariant {  0 <= i <= rows    m }
      invariant { -1 <= j <  columns m }
      invariant { forall i' j'. 0 <= i' < rows m -> 0 <= j' < columns m ->
                  i' < i || j' > j -> get m i' j' <> v }
      variant   { rows m - i + j }
      let x = get m i j in
      if x = v then return true;
      if x < v then i <- i + 1 else j <- j - 1
    done;
    return false

end