## 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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