(** {1 Minimum excludant, aka mex} Author: Jean-Christophe FilliĆ¢tre (CNRS) Given a finite set of integers, find the smallest nonnegative integer that does not belong to this set. In the following, the set is given as an array A. If N is the size of this array, it is clear that we have 0 <= mex <= N for we cannot have the N+1 first natural numbers in the N cells of array A (pigeon hole principle). *) (** A simple algorithm is thus to mark values that belong to [0..N[ in some external Boolean array of length N, ignoring any value that is negative or greater or equal than N. Then a second loop scans the marks until we find some unused value. If don't find any, then it means that A contains exactly the integers 0,...,N-1 and the answer is N. The very last step in this reasoning requires to invoke the pigeon hole principle (imported from the standard library). Time O(N) and space O(N). *) module MexArray use int.Int use map.Map use array.Array use ref.Refint use pigeon.Pigeonhole predicate mem (x: int) (a: array int) = exists i. 0 <= i < length a && a[i] = x let mex (a: array int) : int ensures { 0 <= result <= length a } ensures { not (mem result a) } ensures { forall x. 0 <= x < result -> mem x a } = let n = length a in let used = make n false in let ghost idx = ref (fun i -> i) in (* the position of each marked value *) for i = 0 to n - 1 do invariant { forall x. 0 <= x < n -> used[x] -> mem x a && 0 <= !idx x < n && a[!idx x] = x } invariant { forall j. 0 <= j < i -> 0 <= a[j] < n -> used[a[j]] && 0 <= !idx a[j] < n && a[!idx a[j]] = a[j] } let x = a[i] in if 0 <= x && x < n then begin used[x] <- true; idx := set !idx x i end done; let r = ref 0 in let ghost posn = ref (-1) in while !r < n && used[!r] do invariant { 0 <= !r <= n } invariant { forall j. 0 <= j < !r -> used[j] && 0 <= !idx j < n } invariant { if !posn >= 0 then 0 <= !posn < n && a[!posn] = n else forall j. 0 <= j < !r -> a[j] <> n } variant { n - !r } if a[!r] = n then posn := !r; incr r done; (* we cannot have !r=n (all values marked) and !posn>=0 at the same time *) if !r = n && !posn >= 0 then pigeonhole (n+1) n (set !idx n !posn); !r end (** In this second implementation, we assume we are free to mutate array A. The idea is then to scan the array from left to right, while swapping elements to put any value in 0..N-1 at its place in the array. When we are done, a second loop looks for the mex, advancing as long as a[i]=i holds. Since we perform only swaps, it is obvious that the mex of the final array is equal to the mex of the original array. Time O(N) and space O(1). The argument for a linear time complexity is as follows: whenever we do not advance, we swap the element to its place, which is further and did not contain that element; so we can do this only N times. Surprinsingly, proving that N is the answer whenever array A contains a permutation of 0..N-1 is now easy (no need for a pigeon hole principle or any kind of proof by induction). *) module MexArrayInPlace use int.Int use int.NumOf use array.Array use array.ArraySwap use ref.Refint predicate mem (x: int) (a: array int) = exists i. 0 <= i < length a && a[i] = x function placed (a: array int) : int -> bool = fun i -> a[i] = i let mex (a: array int) : int ensures { 0 <= result <= length a } ensures { not (mem result (old a)) } ensures { forall x. 0 <= x < result -> mem x (old a) } = let n = length a in let i = ref 0 in while !i < n do invariant { 0 <= !i <= n } invariant { forall x. mem x a <-> mem x (old a) } invariant { forall j. 0 <= j < !i -> 0 <= a[j] < n -> a[a[j]] = a[j] } variant { n - !i, n - numof (placed a) 0 n } let x = a[!i] in if x < 0 || x >= n then incr i else if x < !i then begin swap a !i x; incr i end else if a[x] = x then incr i else swap a !i x done; assert { forall j. 0 <= j < n -> let x = (old a)[j] in 0 <= x < n -> a[x] = x }; for i = 0 to n - 1 do invariant { forall j. 0 <= j < i -> a[j] = j } if a[i] <> i then return i done; n end