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A small puzzle involving a Roberval balance

Auteurs: Jean-Christophe Filliâtre / Léon Gondelman

Catégories: Ghost code

Outils: Why3

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Two puzzles involving a Roberval balance

Note: Ghost code is used to get elegant specifications.

Jean-Christophe Filliâtre (CNRS), December 2013 Léon Gondelman (Université Paris-Sud), April 2014 

module Roberval

  use export int.Int

  type outcome = Left | Equal | Right

the side of the heaviest mass i.e. where the balance leans

  type counter = private { mutable v: int }
  meta coercion function v

  val ghost counter: counter

how many times can we use the balance

  val balance (left right: int) : outcome
    requires { counter > 0 }
    ensures  { match result with
               | Left  -> left > right
               | Equal -> left = right
               | Right -> left < right
               end }
    writes   { counter }
    ensures  { counter = old counter - 1 }

end

You are given 8 balls and a Roberval balance. All balls have the same weight, apart from one, which is lighter. Using the balance at most twice, determine the lighter ball.

Though this problem is not that difficult (though, you may want to think about it before reading any further), it is an interesting exercise in program specification.

The index of the lighter ball is passed as a ghost argument to the program. Thus it cannot be used to compute the answer, but only to write the specification. 

module Puzzle8

  use Roberval
  use array.Array

  predicate spec (balls: array int) (lo hi: int) (lb w: int) =
    0 <= lo <= lb < hi <= length balls &&
    (forall i. lo <= i < hi -> i <> lb -> balls[i] = w) &&
    balls[lb] < w

All values in balls[lo..hi-1] are equal to w, apart from balls[lb] which is smaller.

  let solve3 (balls: array int) (lo: int) (ghost lb: int) (ghost w: int) : int
    requires { counter >= 1 }
    requires { spec balls lo (lo + 3) lb w }
    ensures  { result = lb }
    ensures  { counter = old counter - 1 }
  =
    match balance balls[lo] balls[lo+1] with
    | Right -> lo
    | Left  -> lo+1
    | Equal -> lo+2
    end

Solve the problem for 3 balls, using exactly one weighing. The solution lb is passed as a ghost argument.

  let solve8 (balls: array int) (ghost lb: int) (ghost w: int) : int
    requires { counter = 2 }
    requires { spec balls 0 8 lb w }
    ensures  { result = lb }
  =
    (* first, compare balls 0,1,2 with balls 3,4,5 *)
    match balance (balls[0] + balls[1] + balls[2])
                  (balls[3] + balls[4] + balls[5]) with
    | Right -> solve3 balls 0 lb w
    | Left  -> solve3 balls 3 lb w
    (* 0,1,2 = 3,4,5 thus lb must be 6 or 7 *)
    | Equal -> match balance balls[6] balls[7] with
               | Right -> 6
               | Left  -> 7
               | Equal -> absurd
               end
    end

Solve the problem for 8 balls, using exactly two weighings. The solution lb is passed as a ghost argument.

end

You are given 12 balls, all of the same weight except one (for which you don't know whether it is lighter or heavier)

Given a Roberval balance, you have to find the intruder, and determine whether it is lighter or heavier, using the balance at most three times. 

module Puzzle12

  use Roberval
  use array.Array

 let solve12 (balls: array int) (ghost w j: int) (ghost b: bool) : (int, bool)
    requires { counter = 3 }
    requires { 0 <= j < 12 = length balls }
    requires { forall i. 0 <= i < 12 -> i <> j -> balls[i] = w }
    requires { if b then balls[j] < w else balls[j] > w }
    ensures  { result = (j, b) }
 =
   match balance (balls[0] + balls[1] + balls[2] + balls[3])
                 (balls[4] + balls[5] + balls[6] + balls[7]) with
   | Equal -> (* 0,1,2,3 = 4,5,6,7 *)
      match balance (balls[0] + balls[8]) (balls[9] + balls[10]) with
      | Equal -> (* 0,8 = 9,10 *)
         match balance balls[0] balls[11] with
         | Right -> 11, False | _ -> 11, True end
      | Right -> (* 0,8 < 9,10 *)
         match balance balls[9] balls[10] with
         | Equal ->  8, True
         | Right -> 10, False
         | Left  ->  9, False
         end
      | Left -> (* 0,8 > 9,10 *)
         match balance balls[9] balls[10] with
         | Equal ->  8, False
         | Right ->  9,  True
         | Left  -> 10, True
         end
      end
   | Right -> (* 0,1,2,3 < 4,5,6,7 *)
      match balance (balls[0] + balls[1] + balls[4])
                    (balls[2] + balls[5] + balls[8]) with
      | Equal -> (* 0,1,4 = 2,5,8 *)
         match balance balls[6] balls[7] with
         | Equal -> 3, True
         | Right -> 7, False
         | Left  -> 6, False
         end
      | Right -> (* 0,1,4 < 2,5,8 *)
         match balance balls[0] balls[1] with
         | Equal -> 5, False
         | Right -> 0, True
         | Left  -> 1, True
         end
      | Left -> (* 0,1,4 > 2,5,8 *)
         match balance balls[4] balls[8] with
         | Equal -> 2, True
         | _     -> 4, False
         end
      end
   | Left -> (* 0,1,2,3 > 4,5,6,7 *)
      match balance (balls[0] + balls[1] + balls[4])
                    (balls[2] + balls[5] + balls[8]) with
      | Equal -> (* 0,1,4 = 2,5,8 *)
         match balance balls[6] balls[7] with
         | Equal -> 3, False
         | Right -> 6, True
         | Left  -> 7, True
         end
      | Right -> (* 0,1,4 < 2,5,8 *)
         match balance balls[2] balls[5] with
         | Equal -> 4, True
         | Right -> 5, False
         | Left  -> 2, False
         end
      | Left -> (* 0,1,4 > 2,5,8 *)
         match balance balls[0] balls[1] with
         | Equal -> 5, True
         | Right -> 1, False
         | Left  -> 0, False
         end
      end
    end

The index j of the intruder, its status b (whether it is lighter or heavier), and the weight w of the other balls are all passed as ghost arguments.

end

The solutions above are not perfect, as the code could cheat by simply looking for the smallest value in array balls. Even if the code is only using the balance to compare weights, nothing prevents it from weighing several times the same balls e.g. balance (3*balls[0] + 5*balls[1]) ....

Below is a better approach where the weights of the balls are now in a ghost array and where we indicate which subsets of the balls are put on the balance, using Boolean maps (type subset).

Suggested by Martin Clochard (Adacore).

module NoCheating

  use int.Int
  use array.Array

  type outcome = Left | Equal | Right

the side of the heaviest mass i.e. where the balance leans

  type counter = private { mutable v: int }
  meta coercion function v

  val ghost counter: counter

how many times can we use the balance

  type subset = int -> bool

  function sum (balls: array int) (s: subset) : int =
    (if s 0 then balls[0] else 0) +
    (if s 1 then balls[1] else 0) +
    (if s 2 then balls[2] else 0) +
    (if s 3 then balls[3] else 0) +
    (if s 4 then balls[4] else 0) +
    (if s 5 then balls[5] else 0) +
    (if s 6 then balls[6] else 0) +
    (if s 7 then balls[7] else 0)

  val balance (ghost balls: array int) (left right: subset) : (o: outcome)
    requires { counter > 0 }
    requires { forall i. 0 <= i < 8 -> not (left i) \/ not (right i) }
    ensures  { let left  = sum balls left  in
               let right = sum balls right in
               match o with
               | Left  -> left > right
               | Equal -> left = right
               | Right -> left < right
               end }
    writes   { counter }
    ensures  { counter = old counter - 1 }

  predicate spec (balls: array int) (lo hi: int) (lb w: int) =
    0 <= lo <= lb < hi <= length balls = 8 &&
    (forall i. lo <= i < hi -> i <> lb -> balls[i] = w) &&
    balls[lb] < w

All values in balls[lo..hi-1] are equal to w, apart from balls[lb] which is smaller.

  let solve3 (ghost balls: array int) (lo: int) (ghost lb w: int) : int
    requires { counter >= 1 }
    requires { spec balls lo (lo + 3) lb w }
    ensures  { result = lb }
    ensures  { counter = old counter - 1 }
  =
    match balance balls (fun i -> i=lo) (fun i -> i=lo+1) with
    | Right -> lo
    | Left  -> lo+1
    | Equal -> lo+2
    end

Solve the problem for 3 balls, using exactly one weighing. The solution lb is passed as a ghost argument.

  let solve8 (ghost balls: array int) (ghost lb w: int) : int
    requires { counter = 2 }
    requires { spec balls 0 8 lb w }
    ensures  { result = lb }
  =
    (* first, compare balls 0,1,2 with balls 3,4,5 *)
    match balance balls (fun i -> i=0 || i=1 || i=2)
                        (fun i -> i=3 || i=4 || i=5) with
    | Right -> solve3 balls 0 lb w
    | Left  -> solve3 balls 3 lb w
    (* 0,1,2 = 3,4,5 thus lb must be 6 or 7 *)
    | Equal -> match balance balls (fun i -> i=6) (fun i -> i=7) with
               | Right -> 6
               | Left  -> 7
               | Equal -> absurd
               end
    end

Solve the problem for 8 balls, using exactly two weighings. The solution lb is passed as a ghost argument.

end
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