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why3/examples_in_progress/hamming_sequence.mlw
2023-03-08 12:26:34 +00:00

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(*
Hamming sequence is the infinite sequence of integers
1,2,3,4,5,6,8,9,10,12,... divisible by no primes other than 2,3 and 5.
This program computes the n first numbers in this sequence
See Gries, The Science of Programming, page 243
*)
module HammingSequence
use int.Int
use int.MinMax
use number.Divisibility
use number.Prime
use number.Coprime (* for Euclid's lemma *)
predicate is_hamming (n:int) =
forall d:int. prime d /\ divides d n -> d = 2 \/ d = 3 \/ d = 5
lemma is_hamming_times2 :
forall n:int. n >= 1 -> is_hamming n -> is_hamming (2*n)
lemma is_hamming_times3 :
forall n:int. n >= 1 -> is_hamming n -> is_hamming (3*n)
lemma is_hamming_times5 :
forall n:int. n >= 1 -> is_hamming n -> is_hamming (5*n)
use array.Array
use ref.Ref
let hamming (n:int) : array int
requires { n >= 1 }
ensures { forall k:int. 0 <= k < n -> is_hamming result[k] }
ensures { forall k:int. 0 < k < n -> result[k-1] < result[k] }
ensures { forall k m:int. 0 < k < n ->
result[k-1] < m < result[k] -> not (is_hamming m) }
= let t = make n 0 in
t[0] <- 1;
let x2 = ref 2 in let j2 = ref 0 in
let x3 = ref 3 in let j3 = ref 0 in
let x5 = ref 5 in let j5 = ref 0 in
for i=1 to n-1 do
invariant { forall k:int. 0 <= k < i -> t[k] > 0 }
invariant { forall k:int. 0 < k < i -> t[k-1] < t[k] }
invariant { forall k:int. 0 <= k < i -> is_hamming t[k] }
invariant { 0 <= !j2 < i }
invariant { 0 <= !j3 < i }
invariant { 0 <= !j5 < i }
invariant { !x2 = 2*t[!j2] }
invariant { !x3 = 3*t[!j3] }
invariant { !x5 = 5*t[!j5] }
invariant { !x2 > t[i-1] }
invariant { !x3 > t[i-1] }
invariant { !x5 > t[i-1] }
let next = min !x2 (min !x3 !x5) in
t[i] <- next;
while !x2 <= next do
invariant { 0 <= !j2 <= i }
invariant { !x2 = 2*t[!j2] }
variant { next - !x2 }
assert { !j2 < i };
j2 := !j2+1; x2 := 2*t[!j2]
done;
while !x3 <= next do
invariant { 0 <= !j3 <= i }
invariant { !x3 = 3*t[!j3] }
variant { next - !x3 }
assert { !j3 < i };
j3 := !j3+1; x3 := 3*t[!j3]
done;
while !x5 <= next do
invariant { 0 <= !j5 <= i }
invariant { !x5 = 5*t[!j5] }
variant { next - !x5 }
assert { !j5 < i };
j5 := !j5+1; x5 := 5*t[!j5]
done
done;
t
let test () = hamming 30
end