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(************************************************************************)
(* * The Coq Proof Assistant / The Coq Development Team *)
(* v * INRIA, CNRS and contributors - Copyright 1999-2018 *)
(* <O___,, * (see CREDITS file for the list of authors) *)
(* \VV/ **************************************************************)
(* // * This file is distributed under the terms of the *)
(* * GNU Lesser General Public License Version 2.1 *)
(* * (see LICENSE file for the text of the license) *)
(************************************************************************)
(***************************************************)
(* Basic operations on (unbounded) integer numbers *)
(***************************************************)
(* An integer is canonically represented as an array of k-digits blocs,
i.e. in base 10^k.
0 is represented by the empty array and -1 by the singleton [|-1|].
The first bloc is in the range ]0;base[ for positive numbers.
The first bloc is in the range [-base;-1[ for numbers < -1.
All other blocs are numbers in the range [0;base[.
Negative numbers are represented using 2's complementation :
one unit is "borrowed" from the top block for complementing
the other blocs. For instance, with 4-digits blocs,
[|-5;6789|] denotes -43211
since -5.10^4+6789=-((4.10^4)+(10000-6789)) = -43211
The base is a power of 10 in order to facilitate the parsing and printing
of numbers in digital notation.
All functions, to the exception of to_string and of_string should work
with an arbitrary base, even if not a power of 10.
In practice, we set k=4 on 32-bits machines, so that no overflow in ocaml
machine words (i.e. the interval [-2^30;2^30-1]) occur when multiplying two
numbers less than (10^k). On 64-bits machines, k=9.
*)
(* The main parameters *)
let size =
let rec log10 n = if n < 10 then 0 else 1 + log10 (n / 10) in
(log10 max_int) / 2
let format_size =
(* How to parametrize a printf format *)
if Int.equal size 4 then Printf.sprintf "%04d"
else if Int.equal size 9 then Printf.sprintf "%09d"
else fun n ->
let rec aux j l n =
if Int.equal j size then l else aux (j+1) (string_of_int (n mod 10) :: l) (n/10)
in String.concat "" (aux 0 [] n)
(* The base is 10^size *)
let base =
let rec exp10 = function 0 -> 1 | n -> 10 * exp10 (n-1) in exp10 size
(******************************************************************)
(* First, we represent all numbers by int arrays.
Later, we will optimize the particular case of small integers *)
(******************************************************************)
module ArrayInt = struct
(* Basic numbers *)
let zero = [||]
let is_zero = function
| [||] -> true
| _ -> false
(* An array is canonical when
- it is empty
- it is [|-1|]
- its first bloc is in [-base;-1[U]0;base[
and the other blocs are in [0;base[. *)
(*
let canonical n =
let ok x = (0 <= x && x < base) in
let rec ok_tail k = (Int.equal k 0) || (ok n.(k) && ok_tail (k-1)) in
let ok_init x = (-base <= x && x < base && not (Int.equal x (-1)) && not (Int.equal x 0))
in
(is_zero n) || (match n with [|-1|] -> true | _ -> false) ||
(ok_init n.(0) && ok_tail (Array.length n - 1))
*)
(* [normalize_pos] : removing initial blocks of 0 *)
let normalize_pos n =
let k = ref 0 in
while !k < Array.length n && Int.equal n.(!k) 0 do incr k done;
Array.sub n !k (Array.length n - !k)
(* [normalize_neg] : avoid (-1) as first bloc.
input: an array with -1 as first bloc and other blocs in [0;base[
output: a canonical array *)
let normalize_neg n =
let k = ref 1 in
while !k < Array.length n && Int.equal n.(!k) (base - 1) do incr k done;
let n' = Array.sub n !k (Array.length n - !k) in
if Int.equal (Array.length n') 0 then [|-1|] else (n'.(0) <- n'.(0) - base; n')
(* [normalize] : avoid 0 and (-1) as first bloc.
input: an array with first bloc in [-base;base[ and others in [0;base[
output: a canonical array *)
let normalize n =
if Int.equal (Array.length n) 0 then n
else if Int.equal n.(0) (-1) then normalize_neg n
else if Int.equal n.(0) 0 then normalize_pos n
else n
(* Opposite (expects and returns canonical arrays) *)
let neg m =
if is_zero m then zero else
let n = Array.copy m in
let i = ref (Array.length m - 1) in
while !i > 0 && Int.equal n.(!i) 0 do decr i done;
if Int.equal !i 0 then begin
n.(0) <- - n.(0);
(* n.(0) cannot be 0 since m is canonical *)
if Int.equal n.(0) (-1) then normalize_neg n
else if Int.equal n.(0) base then (n.(0) <- 0; Array.append [| 1 |] n)
else n
end else begin
(* here n.(!i) <> 0, hence 0 < base - n.(!i) < base for n canonical *)
n.(!i) <- base - n.(!i); decr i;
while !i > 0 do n.(!i) <- base - 1 - n.(!i); decr i done;
(* since -base <= n.(0) <= base-1, hence -base <= -n.(0)-1 <= base-1 *)
n.(0) <- - n.(0) - 1;
(* since m is canonical, m.(0)<>0 hence n.(0)<>-1,
and m=-1 is already handled above, so here m.(0)<>-1 hence n.(0)<>0 *)
n
end
let push_carry r j =
let j = ref j in
while !j > 0 && r.(!j) < 0 do
r.(!j) <- r.(!j) + base; decr j; r.(!j) <- r.(!j) - 1
done;
while !j > 0 && r.(!j) >= base do
r.(!j) <- r.(!j) - base; decr j; r.(!j) <- r.(!j) + 1
done;
(* here r.(0) could be in [-2*base;2*base-1] *)
if r.(0) >= base then (r.(0) <- r.(0) - base; Array.append [| 1 |] r)
else if r.(0) < -base then (r.(0) <- r.(0) + 2*base; Array.append [| -2 |] r)
else normalize r (* in case r.(0) is 0 or -1 *)
let add_to r a j =
if is_zero a then r else begin
for i = Array.length r - 1 downto j+1 do
r.(i) <- r.(i) + a.(i-j);
if r.(i) >= base then (r.(i) <- r.(i) - base; r.(i-1) <- r.(i-1) + 1)
done;
r.(j) <- r.(j) + a.(0);
push_carry r j
end
let add n m =
let d = Array.length n - Array.length m in
if d > 0 then add_to (Array.copy n) m d else add_to (Array.copy m) n (-d)
let sub_to r a j =
if is_zero a then r else begin
for i = Array.length r - 1 downto j+1 do
r.(i) <- r.(i) - a.(i-j);
if r.(i) < 0 then (r.(i) <- r.(i) + base; r.(i-1) <- r.(i-1) - 1)
done;
r.(j) <- r.(j) - a.(0);
push_carry r j
end
let sub n m =
let d = Array.length n - Array.length m in
if d >= 0 then sub_to (Array.copy n) m d
else let r = neg m in add_to r n (Array.length r - Array.length n)
let mult m n =
if is_zero m || is_zero n then zero else
let l = Array.length m + Array.length n in
let r = Array.make l 0 in
for i = Array.length m - 1 downto 0 do
for j = Array.length n - 1 downto 0 do
let p = m.(i) * n.(j) + r.(i+j+1) in
let (q,s) =
if p < 0
then (p + 1) / base - 1, (p + 1) mod base + base - 1
else p / base, p mod base in
r.(i+j+1) <- s;
if not (Int.equal q 0) then r.(i+j) <- r.(i+j) + q;
done
done;
normalize r
(* Comparisons *)
let is_strictly_neg n = not (is_zero n) && n.(0) < 0
let is_strictly_pos n = not (is_zero n) && n.(0) > 0
let is_neg_or_zero n = is_zero n || n.(0) < 0
let is_pos_or_zero n = is_zero n || n.(0) > 0
(* Is m without its i first blocs less then n without its j first blocs ?
Invariant : |m|-i = |n|-j *)
let rec less_than_same_size m n i j =
i < Array.length m &&
(m.(i) < n.(j) || (Int.equal m.(i) n.(j) && less_than_same_size m n (i+1) (j+1)))
let less_than m n =
if is_strictly_neg m then
is_pos_or_zero n || Array.length m > Array.length n
|| (Int.equal (Array.length m) (Array.length n) && less_than_same_size m n 0 0)
else
is_strictly_pos n && (Array.length m < Array.length n ||
(Int.equal (Array.length m) (Array.length n) && less_than_same_size m n 0 0))
(* For this equality test it is critical that n and m are canonical *)
let rec array_eq len v1 v2 i =
if Int.equal len i then true
else
Int.equal v1.(i) v2.(i) && array_eq len v1 v2 (succ i)
let equal m n =
let lenm = Array.length m in
let lenn = Array.length n in
(Int.equal lenm lenn) && (array_eq lenm m n 0)
(* Is m without its k top blocs less than n ? *)
let less_than_shift_pos k m n =
(Array.length m - k < Array.length n)
|| (Int.equal (Array.length m - k) (Array.length n) && less_than_same_size m n k 0)
let rec can_divide k m d i =
(Int.equal i (Array.length d)) ||
(m.(k+i) > d.(i)) ||
(Int.equal m.(k+i) d.(i) && can_divide k m d (i+1))
(* For two big nums m and d and a small number q,
computes m - d * q * base^(|m|-|d|-k) in-place (in m).
Both m d and q are positive. *)
let sub_mult m d q k =
if not (Int.equal q 0) then
for i = Array.length d - 1 downto 0 do
let v = d.(i) * q in
m.(k+i) <- m.(k+i) - v mod base;
if m.(k+i) < 0 then (m.(k+i) <- m.(k+i) + base; m.(k+i-1) <- m.(k+i-1) -1);
if v >= base then begin
m.(k+i-1) <- m.(k+i-1) - v / base;
let j = ref (i-1) in
while m.(k + !j) < 0 do (* result is positive, hence !j remains >= 0 *)
m.(k + !j) <- m.(k + !j) + base; decr j; m.(k + !j) <- m.(k + !j) -1
done
end
done
(** Euclid division m/d = (q,r), with m = q*d+r and |r|<|q|.
This is the "Trunc" variant (a.k.a "Truncated-Toward-Zero"),
as with ocaml's / (but not as ocaml's Big_int.quomod_big_int).
We have sign r = sign m *)
let euclid m d =
let isnegm, m =
if is_strictly_neg m then (-1),neg m else 1,Array.copy m in
let isnegd, d = if is_strictly_neg d then (-1),neg d else 1,d in
if is_zero d then raise Division_by_zero;
let q,r =
if less_than m d then (zero,m) else
let ql = Array.length m - Array.length d in
let q = Array.make (ql+1) 0 in
let i = ref 0 in
while not (less_than_shift_pos !i m d) do
if Int.equal m.(!i) 0 then incr i else
if can_divide !i m d 0 then begin
let v =
if Array.length d > 1 && not (Int.equal d.(0) m.(!i)) then
(m.(!i) * base + m.(!i+1)) / (d.(0) * base + d.(1) + 1)
else
m.(!i) / d.(0) in
q.(!i) <- q.(!i) + v;
sub_mult m d v !i
end else begin
let v = (m.(!i) * base + m.(!i+1)) / (d.(0) + 1) in
q.(!i) <- q.(!i) + v / base;
sub_mult m d (v / base) !i;
q.(!i+1) <- q.(!i+1) + v mod base;
if q.(!i+1) >= base then
(q.(!i+1) <- q.(!i+1)-base; q.(!i) <- q.(!i)+1);
sub_mult m d (v mod base) (!i+1)
end
done;
(normalize q, normalize m) in
(if Int.equal (isnegd * isnegm) (-1) then neg q else q),
(if Int.equal isnegm (-1) then neg r else r)
(* Parsing/printing ordinary 10-based numbers *)
let of_string s =
let len = String.length s in
let isneg = len > 1 && s.[0] == '-' in
let d = ref (if isneg then 1 else 0) in
while !d < len && s.[!d] == '0' do incr d done;
if Int.equal !d len then zero else
let r = (len - !d) mod size in
let h = String.sub s (!d) r in
let e = match h with "" -> 0 | _ -> 1 in
let l = (len - !d) / size in
let a = Array.make (l + e) 0 in
if Int.equal e 1 then a.(0) <- int_of_string h;
for i = 1 to l do
a.(i+e-1) <- int_of_string (String.sub s ((i-1)*size + !d + r) size)
done;
if isneg then neg a else a
let to_string_pos sgn n =
if Int.equal (Array.length n) 0 then "0" else
sgn ^
String.concat ""
(string_of_int n.(0) :: List.map format_size (List.tl (Array.to_list n)))
let to_string n =
if is_strictly_neg n then to_string_pos "-" (neg n)
else to_string_pos "" n
end
(******************************************************************)
(* Optimized operations on (unbounded) integer numbers *)
(* integers smaller than base are represented as machine integers *)
(******************************************************************)
open ArrayInt
type bigint = Obj.t
(* Since base is the largest power of 10 such that base*base <= max_int,
we have max_int < 100*base*base : any int can be represented
by at most three blocs *)
let small n = (-base <= n) && (n < base)
let mkarray n =
(* n isn't small, this case is handled separately below *)
let lo = n mod base
and hi = n / base in
let t = if small hi then [|hi;lo|] else [|hi/base;hi mod base;lo|]
in
for i = Array.length t -1 downto 1 do
if t.(i) < 0 then (t.(i) <- t.(i) + base; t.(i-1) <- t.(i-1) -1)
done;
t
let ints_of_int n =
if Int.equal n 0 then [| |]
else if small n then [| n |]
else mkarray n
let of_int n =
if small n then Obj.repr n else Obj.repr (mkarray n)
let of_ints n =
let n = normalize n in (* TODO: using normalize here seems redundant now *)
if is_zero n then Obj.repr 0 else
if Int.equal (Array.length n) 1 then Obj.repr n.(0) else
Obj.repr n
let coerce_to_int = (Obj.magic : Obj.t -> int)
let coerce_to_ints = (Obj.magic : Obj.t -> int array)
let to_ints n =
if Obj.is_int n then ints_of_int (coerce_to_int n)
else coerce_to_ints n
let int_of_ints =
let maxi = mkarray max_int and mini = mkarray min_int in
fun t ->
let l = Array.length t in
if (l > 3) || (Int.equal l 3 && (less_than maxi t || less_than t mini))
then failwith "Bigint.to_int: too large";
let sum = ref 0 in
let pow = ref 1 in
for i = l-1 downto 0 do
sum := !sum + t.(i) * !pow;
pow := !pow*base;
done;
!sum
let to_int n =
if Obj.is_int n then coerce_to_int n
else int_of_ints (coerce_to_ints n)
let app_pair f (m, n) =
(f m, f n)
let add m n =
if Obj.is_int m && Obj.is_int n
then of_int (coerce_to_int m + coerce_to_int n)
else of_ints (add (to_ints m) (to_ints n))
let sub m n =
if Obj.is_int m && Obj.is_int n
then of_int (coerce_to_int m - coerce_to_int n)
else of_ints (sub (to_ints m) (to_ints n))
let mult m n =
if Obj.is_int m && Obj.is_int n
then of_int (coerce_to_int m * coerce_to_int n)
else of_ints (mult (to_ints m) (to_ints n))
let euclid m n =
if Obj.is_int m && Obj.is_int n
then app_pair of_int
(coerce_to_int m / coerce_to_int n, coerce_to_int m mod coerce_to_int n)
else app_pair of_ints (euclid (to_ints m) (to_ints n))
let less_than m n =
if Obj.is_int m && Obj.is_int n
then coerce_to_int m < coerce_to_int n
else less_than (to_ints m) (to_ints n)
let neg n =
if Obj.is_int n then of_int (- (coerce_to_int n))
else of_ints (neg (to_ints n))
let of_string m = of_ints (of_string m)
let to_string m = to_string (to_ints m)
let zero = of_int 0
let one = of_int 1
let two = of_int 2
let sub_1 n = sub n one
let add_1 n = add n one
let mult_2 n = add n n
let div2_with_rest n =
let (q,b) = euclid n two in
(q, b == one)
let is_strictly_neg n = is_strictly_neg (to_ints n)
let is_strictly_pos n = is_strictly_pos (to_ints n)
let is_neg_or_zero n = is_neg_or_zero (to_ints n)
let is_pos_or_zero n = is_pos_or_zero (to_ints n)
let equal m n =
if Obj.is_block m && Obj.is_block n then
ArrayInt.equal (Obj.obj m) (Obj.obj n)
else m == n
(* spiwack: computes n^m *)
(* The basic idea of the algorithm is that n^(2m) = (n^2)^m *)
(* In practice the algorithm performs :
k*n^0 = k
k*n^(2m) = k*(n*n)^m
k*n^(2m+1) = (n*k)*(n*n)^m *)
let pow =
let rec pow_aux odd_rest n m = (* odd_rest is the k from above *)
if m<=0 then
odd_rest
else
let quo = m lsr 1 (* i.e. m/2 *)
and odd = not (Int.equal (m land 1) 0) in
pow_aux
(if odd then mult n odd_rest else odd_rest)
(mult n n)
quo
in
pow_aux one
(** Testing suite w.r.t. OCaml's Big_int *)
(*
module B = struct
open Big_int
let zero = zero_big_int
let to_string = string_of_big_int
let of_string = big_int_of_string
let add = add_big_int
let opp = minus_big_int
let sub = sub_big_int
let mul = mult_big_int
let abs = abs_big_int
let sign = sign_big_int
let euclid n m =
let n' = abs n and m' = abs m in
let q',r' = quomod_big_int n' m' in
(if sign (mul n m) < 0 && sign q' <> 0 then opp q' else q'),
(if sign n < 0 then opp r' else r')
end
let check () =
let roots = [ 1; 100; base; 100*base; base*base ] in
let rands = [ 1234; 5678; 12345678; 987654321 ] in
let nums = (List.flatten (List.map (fun x -> [x-1;x;x+1]) roots)) @ rands in
let numbers =
List.map string_of_int nums @
List.map (fun n -> string_of_int (-n)) nums
in
let i = ref 0 in
let compare op x y n n' =
incr i;
let s = Printf.sprintf "%30s" (to_string n) in
let s' = Printf.sprintf "%30s" (B.to_string n') in
if s <> s' then Printf.printf "%s%s%s: %s <> %s\n" x op y s s' in
let test x y =
let n = of_string x and m = of_string y in
let n' = B.of_string x and m' = B.of_string y in
let a = add n m and a' = B.add n' m' in
let s = sub n m and s' = B.sub n' m' in
let p = mult n m and p' = B.mul n' m' in
let q,r = try euclid n m with Division_by_zero -> zero,zero
and q',r' = try B.euclid n' m' with Division_by_zero -> B.zero, B.zero
in
compare "+" x y a a';
compare "-" x y s s';
compare "*" x y p p';
compare "/" x y q q';
compare "%" x y r r'
in
List.iter (fun a -> List.iter (test a) numbers) numbers;
Printf.printf "%i tests done\n" !i
*)
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