Modular exponentiation
You are encouraged to solve this task according to the task description, using any language you may know.
Find the last 40 decimal digits of , where
A computer is too slow to find the entire value of .
Instead, the program must use a fast algorithm for modular exponentiation: .
The algorithm must work for any integers , where and .
11l
F pow_mod(BigInt =base, BigInt =exponent, BigInt modulus)
BigInt result = 1
L exponent != 0
I exponent % 2 != 0
result = (result * base) % modulus
exponent I/= 2
base = (base * base) % modulus
R result
print(pow_mod(BigInt(‘2988348162058574136915891421498819466320163312926952423791023078876139’),
BigInt(‘2351399303373464486466122544523690094744975233415544072992656881240319’),
BigInt(10) ^ 40))- Output:
1527229998585248450016808958343740453059
Ada
Using the big integer implementation from a cryptographic library [1].
with Ada.Text_IO, Ada.Command_Line, Crypto.Types.Big_Numbers;
procedure Mod_Exp is
A: String :=
"2988348162058574136915891421498819466320163312926952423791023078876139";
B: String :=
"2351399303373464486466122544523690094744975233415544072992656881240319";
D: constant Positive := Positive'Max(Positive'Max(A'Length, B'Length), 40);
-- the number of decimals to store A, B, and result
Bits: constant Positive := (34*D)/10;
-- (slightly more than) the number of bits to store A, B, and result
package LN is new Crypto.Types.Big_Numbers (Bits + (32 - Bits mod 32));
-- the actual number of bits has to be a multiple of 32
use type LN.Big_Unsigned;
function "+"(S: String) return LN.Big_Unsigned
renames LN.Utils.To_Big_Unsigned;
M: LN.Big_Unsigned := (+"10") ** (+"40");
begin
Ada.Text_IO.Put("A**B (mod 10**40) = ");
Ada.Text_IO.Put_Line(LN.Utils.To_String(LN.Mod_Utils.Pow((+A), (+B), M)));
end Mod_Exp;
- Output:
A**B (mod 10**40) = 1527229998585248450016808958343740453059
ALGOL 68
The code below uses Algol 68 Genie which provides arbitrary precision arithmetic for LONG LONG modes.
BEGIN
PR precision=1000 PR
MODE LLI = LONG LONG INT; CO For brevity CO
PROC mod power = (LLI base, exponent, modulus) LLI :
BEGIN
LLI result := 1, b := base, e := exponent;
IF exponent < 0
THEN
put (stand error, (("Negative exponent", exponent, newline)))
ELSE
WHILE e > 0
DO
(ODD e | result := (result * b) MOD modulus);
e OVERAB 2; b := (b * b) MOD modulus
OD
FI;
result
END;
LLI a = 2988348162058574136915891421498819466320163312926952423791023078876139;
LLI b = 2351399303373464486466122544523690094744975233415544072992656881240319;
LLI m = 10000000000000000000000000000000000000000;
printf (($"Last 40 digits = ", 40dl$, mod power (a, b, m)))
END- Output:
Last 40 digits = 1527229998585248450016808958343740453059
Arturo
a: 2988348162058574136915891421498819466320163312926952423791023078876139
b: 2351399303373464486466122544523690094744975233415544072992656881240319
loop [40 80 180 888] 'm ->
print ["(a ^ b) % 10 ^" m "=" powmod a b 10^m]
- Output:
(a ^ b) % 10 ^ 40 = 1527229998585248450016808958343740453059 (a ^ b) % 10 ^ 80 = 53259517041910225328867076245698908287781527229998585248450016808958343740453059 (a ^ b) % 10 ^ 180 = 31857295076204937005344367438778481743660325586328069392203762862423884839076695547212682454523811053259517041910225328867076245698908287781527229998585248450016808958343740453059 (a ^ b) % 10 ^ 888 = 261284964380836515397030706363442226571397237057488951313684545241085642329943676248755716124260447188788530017182951051652748425560733974835944416069466176713156182727448301838517000343485327001656948285381173038339073779331230132340669899896448938858785362771190460312412579875349871655999446205426049662261450633448468931573506876255644749155348923523680730999869785472779116009356696816952771965930728940530517799329942590114178284009260298426735086579254282591289756840358811822151307479352856856983393715348870715239020037962938019847992960978849852850613063177471175191444262586321233906926671000476591123695550566585083205841790404069511972417770392822283604206143472509425391114072344402850867571806031857295076204937005344367438778481743660325586328069392203762862423884839076695547212682454523811053259517041910225328867076245698908287781527229998585248450016808958343740453059
ATS
For its multiple precision support, you will need ats2-xprelude with GNU MP support enabled.
There is GNU MP support that comes with the ATS2 compiler, however, so perhaps someone will write a demonstration using that. Unlike ats2-xprelude (which assumes there is garbage collection), that support represents multi-precision numbers as linear types.
(* You will need
https://sourceforge.net/p/chemoelectric/ats2-xprelude/ *)
#include "share/atspre_staload.hats"
#include "xprelude/HATS/xprelude.hats"
staload "xprelude/SATS/exrat.sats"
staload _ = "xprelude/DATS/exrat.dats"
val a = exrat_make_string_exn "2988348162058574136915891421498819466320163312926952423791023078876139"
val b = exrat_make_string_exn "2351399303373464486466122544523690094744975233415544072992656881240319"
val modulus = exrat_make (10, 1) ** 40
(* xprelude/SATS/exrat.sats includes the "exrat_numerator_modular_pow"
function, based on GMP's mpz_powm. *)
val result1 = exrat_numerator_modular_pow (a, b, modulus)
(* But that was too easy. Here is the right-to-left binary method,
https://en.wikipedia.org/w/index.php?title=Modular_exponentiation&oldid=1136216610#Right-to-left_binary_method
*)
val result2 =
(lam (base : exrat,
exponent : exrat,
modulus : exrat) : exrat =>
let
val zero = exrat_make (0, 1)
and one = exrat_make (1, 1)
and two = exrat_make (2, 1)
macdef divrem = exrat_numerator_euclid_division
macdef rem = exrat_numerator_euclid_remainder
in
if modulus = one then
zero
else
let
fun
loop (result : exrat,
base : exrat,
exponent : exrat) : exrat =
if iseqz exponent then
result
else
let
val @(exponent, remainder) = exponent \divrem two
val result =
if remainder = one then
(result * base) \rem modulus
else
result
val base = (base * base) \rem modulus
in
loop (result, base, exponent)
end
in
loop (one, base \rem modulus, exponent)
end
end) (a, b, modulus)
implement
main0 () =
begin
println! result1;
println! result2
end- Output:
$ patscc -std=gnu2x -g -O2 -DATS_MEMALLOC_GCBDW $(pkg-config --cflags ats2-xprelude) $(pkg-config --variable=PATSCCFLAGS ats2-xprelude) modular-exponentiation.dats $(pkg-config --libs ats2-xprelude) -lgc && ./a.out 1527229998585248450016808958343740453059 1527229998585248450016808958343740453059
AutoHotkey
#NoEnv
#SingleInstance, Force
SetBatchLines, -1
#Include mpl.ahk
MP_SET(base, "2988348162058574136915891421498819466320163312926952423791023078876139")
, MP_SET(exponent, "2351399303373464486466122544523690094744975233415544072992656881240319")
, MP_SET(modulus, "10000000000000000000000000000000000000000")
, NumGet(exponent,0,"Int") = -1 ? return : ""
, MP_SET(result, "1")
, MP_SET(TWO, "2")
while !MP_IS0(exponent)
MP_DIV(q, r, exponent, TWO)
, (MP_DEC(r) = 1
? (MP_MUL(temp, result, base)
, MP_DIV(q, result, temp, modulus))
: "")
, MP_DIV(q, r, exponent, TWO)
, MP_CPY(exponent, q)
, MP_CPY(base1, base)
, MP_MUL(base2, base1, base)
, MP_DIV(q, base, base2, modulus)
msgbox % MP_DEC(result)
Return
- Output:
1527229998585248450016808958343740453059
BBC BASIC
Uses the Huge Integer Math & Encryption library.
INSTALL @lib$+"HIMELIB"
PROC_himeinit("")
PROC_hiputdec(1, "2988348162058574136915891421498819466320163312926952423791023078876139")
PROC_hiputdec(2, "2351399303373464486466122544523690094744975233415544072992656881240319")
PROC_hiputdec(3, "10000000000000000000000000000000000000000")
h1% = 1 : h2% = 2 : h3% = 3 : h4% = 4
SYS `hi_PowMod`, ^h1%, ^h2%, ^h3%, ^h4%
PRINT FN_higetdec(4)
- Output:
1527229998585248450016808958343740453059
bc
define p(n, e, m) {
auto r
for (r = 1; e > 0; e /= 2) {
if (e % 2 == 1) r = n * r % m
n = n * n % m
}
return(r)
}
a = 2988348162058574136915891421498819466320163312926952423791023078876139
b = 2351399303373464486466122544523690094744975233415544072992656881240319
p(a, b, 10 ^ 40)
- Output:
1527229998585248450016808958343740453059
Bracmat
( ( mod-power
= base exponent modulus result
. !arg:(?base,?exponent,?modulus)
& !exponent:~<0
& 1:?result
& whl
' ( !exponent:>0
& ( ( mod$(!exponent.2):1
& mod$(!result*!base.!modulus):?result
& -1
| 0
)
+ !exponent
)
* 1/2
: ?exponent
& mod$(!base^2.!modulus):?base
)
& !result
)
& ( a
= 2988348162058574136915891421498819466320163312926952423791023078876139
)
& ( b
= 2351399303373464486466122544523690094744975233415544072992656881240319
)
& out$("last 40 digits = " mod-power$(!a,!b,10^40))
)- Output:
last 40 digits = 1527229998585248450016808958343740453059
C
Given numbers are too big for even 64 bit integers, so might as well take the lazy route and use GMP:
#include <gmp.h>
int main()
{
mpz_t a, b, m, r;
mpz_init_set_str(a, "2988348162058574136915891421498819466320"
"163312926952423791023078876139", 0);
mpz_init_set_str(b, "2351399303373464486466122544523690094744"
"975233415544072992656881240319", 0);
mpz_init(m);
mpz_ui_pow_ui(m, 10, 40);
mpz_init(r);
mpz_powm(r, a, b, m);
gmp_printf("%Zd\n", r); /* ...16808958343740453059 */
mpz_clear(a);
mpz_clear(b);
mpz_clear(m);
mpz_clear(r);
return 0;
}
Output:
1527229998585248450016808958343740453059
C#
We can use the built-in function from BigInteger.
using System;
using System.Numerics;
class Program
{
static void Main() {
var a = BigInteger.Parse("2988348162058574136915891421498819466320163312926952423791023078876139");
var b = BigInteger.Parse("2351399303373464486466122544523690094744975233415544072992656881240319");
var m = BigInteger.Pow(10, 40);
Console.WriteLine(BigInteger.ModPow(a, b, m));
}
}
- Output:
1527229998585248450016808958343740453059
C++
#include <iostream>
#include <boost/multiprecision/cpp_int.hpp>
#include <boost/multiprecision/integer.hpp>
int main() {
using boost::multiprecision::cpp_int;
using boost::multiprecision::pow;
using boost::multiprecision::powm;
cpp_int a("2988348162058574136915891421498819466320163312926952423791023078876139");
cpp_int b("2351399303373464486466122544523690094744975233415544072992656881240319");
std::cout << powm(a, b, pow(cpp_int(10), 40)) << '\n';
return 0;
}
- Output:
1527229998585248450016808958343740453059
Clojure
(defn powerMod "modular exponentiation" [b e m]
(defn m* [p q] (mod (* p q) m))
(loop [b b, e e, x 1]
(if (zero? e) x
(if (even? e) (recur (m* b b) (/ e 2) x)
(recur (m* b b) (quot e 2) (m* b x))))))
(defn modpow
" b^e mod m (using Java which solves some cases the pure clojure method has to be modified to tackle--i.e. with large b & e and
calculation simplications when gcd(b, m) == 1 and gcd(e, m) == 1) "
[b e m]
(.modPow (biginteger b) (biginteger e) (biginteger m)))
Common Lisp
(defun rosetta-mod-expt (base power divisor)
"Return BASE raised to the POWER, modulo DIVISOR.
This function is faster than (MOD (EXPT BASE POWER) DIVISOR), but
only works when POWER is a non-negative integer."
(setq base (mod base divisor))
;; Multiply product with base until power is zero.
(do ((product 1))
((zerop power) product)
;; Square base, and divide power by 2, until power becomes odd.
(do () ((oddp power))
(setq base (mod (* base base) divisor)
power (ash power -1)))
(setq product (mod (* product base) divisor)
power (1- power))))
(let ((a 2988348162058574136915891421498819466320163312926952423791023078876139)
(b 2351399303373464486466122544523690094744975233415544072992656881240319))
(format t "~A~%" (rosetta-mod-expt a b (expt 10 40))))
;; CLISP provides EXT:MOD-EXPT
(let ((a 2988348162058574136915891421498819466320163312926952423791023078876139)
(b 2351399303373464486466122544523690094744975233415544072992656881240319))
(format t "~A~%" (mod-expt a b (expt 10 40))))
Implementation with LOOP
(defun mod-expt (a n m)
(loop with c = 1 while (plusp n) do
(if (oddp n) (setf c (mod (* a c) m)))
(setf n (ash n -1))
(setf a (mod (* a a) m))
finally (return c)))
Crystal
require "big"
module Integer
module Powmod
# Compute self**e mod m
def powmod(e, m)
r, b = 1, self.to_big_i
while e > 0
r = (b * r) % m if e.odd?
b = (b * b) % m
e >>= 1
end
r
end
end
end
struct Int; include Integer::Powmod end
a = "2988348162058574136915891421498819466320163312926952423791023078876139".to_big_i
b = "2351399303373464486466122544523690094744975233415544072992656881240319".to_big_i
m = 10.to_big_i ** 40
puts a.powmod(b, m)
- Output:
1527229998585248450016808958343740453059
D
Compile this module with -version=modular_exponentiation to see the output.
module modular_exponentiation;
private import std.bigint;
BigInt powMod(BigInt base, BigInt exponent, in BigInt modulus)
pure nothrow /*@safe*/ in {
assert(exponent >= 0);
} body {
BigInt result = 1;
while (exponent) {
if (exponent & 1)
result = (result * base) % modulus;
exponent /= 2;
base = base ^^ 2 % modulus;
}
return result;
}
version (modular_exponentiation)
void main() {
import std.stdio;
powMod(BigInt("29883481620585741369158914214988194" ~
"66320163312926952423791023078876139"),
BigInt("235139930337346448646612254452369009" ~
"4744975233415544072992656881240319"),
BigInt(10) ^^ 40).writeln;
}
- Output:
1527229998585248450016808958343740453059
Dc
2988348162058574136915891421498819466320163312926952423791023078876139 2351399303373464486466122544523690094744975233415544072992656881240319 10 40^|pDelphi
Thanks for Rudy Velthuis, BigIntegers library [2].
program Modular_exponentiation;
{$APPTYPE CONSOLE}
uses
System.SysUtils,
Velthuis.BigIntegers;
var
a, b, m: BigInteger;
begin
a := BigInteger.Parse('2988348162058574136915891421498819466320163312926952423791023078876139');
b := BigInteger.Parse('2351399303373464486466122544523690094744975233415544072992656881240319');
m := BigInteger.Pow(10, 40);
Writeln(BigInteger.ModPow(a, b, m).ToString);
readln;
end.
- Output:
1527229998585248450016808958343740453059
EchoLisp
(lib 'bigint)
(define a 2988348162058574136915891421498819466320163312926952423791023078876139)
(define b 2351399303373464486466122544523690094744975233415544072992656881240319)
(define m 1e40)
(powmod a b m)
→ 1527229998585248450016808958343740453059
;; powmod is a native function
;; it could be defined as follows :
(define (xpowmod base exp mod)
(define result 1)
(while ( !zero? exp)
(when (odd? exp) (set! result (% (* result base) mod)))
(/= exp 2)
(set! base (% (* base base) mod)))
result)
(xpowmod a b m)
→ 1527229998585248450016808958343740453059
Emacs Lisp
(let ((a "2988348162058574136915891421498819466320163312926952423791023078876139")
(b "2351399303373464486466122544523690094744975233415544072992656881240319"))
;; "$ ^ $$ mod (10 ^ 40)" performs modular exponentiation.
;; "unpack(-5, x)_1" unpacks the integer from the modulo form.
(message "%s" (calc-eval "unpack(-5, $ ^ $$ mod (10 ^ 40))_1" nil a b)))
Erlang
%%% For details of the algorithms used see
%%% https://en.wikipedia.org/wiki/Modular_exponentiation
-module modexp_rosetta.
-export [mod_exp/3,binary_exp/2,test/0].
mod_exp(Base,Exp,Mod) when
is_integer(Base),
is_integer(Exp),
is_integer(Mod),
Base > 0,
Exp > 0,
Mod > 0 ->
binary_exp_mod(Base,Exp,Mod).
binary_exp(Base,Exponent) ->
binary_exp(Base,Exponent,1).
binary_exp(_,0,Result) ->
Result;
binary_exp(Base,Exponent,Acc) ->
binary_exp(Base*Base,Exponent bsr 1,Acc * exp_factor(Base,Exponent)).
binary_exp_mod(Base,Exponent,Mod) ->
binary_exp_mod(Base rem Mod,Exponent,Mod,1).
binary_exp_mod(_,0,_,Result) ->
Result;
binary_exp_mod(Base,Exponent,Mod,Acc) ->
binary_exp_mod((Base*Base) rem Mod,
Exponent bsr 1,Mod,(Acc * exp_factor(Base,Exponent))rem Mod).
exp_factor(_,0) ->
1;
exp_factor(Base,1) ->
Base;
exp_factor(Base,Exponent) ->
exp_factor(Base,Exponent band 1).
test() ->
445 = mod_exp(4,13,497),
%% Rosetta code example:
mod_exp(2988348162058574136915891421498819466320163312926952423791023078876139,
2351399303373464486466122544523690094744975233415544072992656881240319,
binary_exp(10,40)).
34> modexp_rosetta:test(). modexp_rosetta:test(). 1527229998585248450016808958343740453059 35>
F#
let expMod a b n =
let rec loop a b c =
if b = 0I then c else
loop (a*a%n) (b>>>1) (if b&&&1I = 0I then c else c*a%n)
loop a b 1I
[<EntryPoint>]
let main argv =
let a = 2988348162058574136915891421498819466320163312926952423791023078876139I
let b = 2351399303373464486466122544523690094744975233415544072992656881240319I
printfn "%A" (expMod a b (10I**40)) // -> 1527229998585248450016808958343740453059
0
Factor
! Built-in
2988348162058574136915891421498819466320163312926952423791023078876139
2351399303373464486466122544523690094744975233415544072992656881240319
10 40 ^
^mod .
- Output:
1527229998585248450016808958343740453059
Fortran
module big_integers ! Big (but not very big) integers.
!
! A very primitive multiple precision module, using the classical
! algorithms (long multiplication, long division) and a mere 8-bit
! "digit" size. Some procedures might assume integers are in a
! two's-complement representation. This module is good enough for us
! to fulfill the task.
!
! NOTE: I assume that iachar and achar do not alter the most
! significant bit.
use, intrinsic :: iso_fortran_env, only: int16
implicit none
private
public :: big_integer
public :: integer2big
public :: string2big
public :: big2string
public :: big_sgn
public :: big_cmp, big_cmpabs
public :: big_neg, big_abs
public :: big_addabs, big_add
public :: big_subabs, big_sub
public :: big_mul ! One might also include a big_muladd.
public :: big_divrem ! One could also include big_div and big_rem.
public :: big_pow
public :: operator(+)
public :: operator(-)
public :: operator(*)
public :: operator(**)
type :: big_integer
! The representation is sign-magnitude. The radix is 256, which
! is not speed-efficient, but which seemed relatively easy to
! work with if one were writing in standard Fortran (and assuming
! iachar and achar were "8-bit clean").
logical :: sign = .false. ! .false. => +sign, .true. => -sign.
character, allocatable :: bytes(:)
end type big_integer
character, parameter :: zero = achar (0)
character, parameter :: one = achar (1)
! An integer type capable of holding an unsigned 8-bit value.
integer, parameter :: bytekind = int16
interface operator(+)
module procedure big_add
end interface
interface operator(-)
module procedure big_neg
module procedure big_sub
end interface
interface operator(*)
module procedure big_mul
end interface
interface operator(**)
module procedure big_pow
end interface
contains
elemental function logical2byte (bool) result (byte)
logical, intent(in) :: bool
character :: byte
if (bool) then
byte = one
else
byte = zero
end if
end function logical2byte
elemental function logical2i (bool) result (i)
logical, intent(in) :: bool
integer :: i
if (bool) then
i = 1
else
i = 0
end if
end function logical2i
elemental function byte2i (c) result (i)
character, intent(in) :: c
integer :: i
i = iachar (c)
end function byte2i
elemental function i2byte (i) result (c)
integer, intent(in) :: i
character :: c
c = achar (i)
end function i2byte
elemental function byte2bk (c) result (i)
character, intent(in) :: c
integer(bytekind) :: i
i = iachar (c, kind = bytekind)
end function byte2bk
elemental function bk2byte (i) result (c)
integer(bytekind), intent(in) :: i
character :: c
c = achar (i)
end function bk2byte
elemental function bk2i (i) result (j)
integer(bytekind), intent(in) :: i
integer :: j
j = int (i)
end function bk2i
elemental function i2bk (i) result (j)
integer, intent(in) :: i
integer(bytekind) :: j
j = int (iand (i, 255), kind = bytekind)
end function i2bk
! Left shift of the least significant 8 bits of a bytekind integer.
elemental function lshftbk (a, i) result (c)
integer(bytekind), intent(in) :: a
integer, intent(in) :: i
integer(bytekind) :: c
c = ishft (ibits (a, 0, 8 - i), i)
end function lshftbk
! Right shift of the least significant 8 bits of a bytekind integer.
elemental function rshftbk (a, i) result (c)
integer(bytekind), intent(in) :: a
integer, intent(in) :: i
integer(bytekind) :: c
c = ibits (a, i, 8 - i)
end function rshftbk
! Left shift an integer.
elemental function lshfti (a, i) result (c)
integer, intent(in) :: a
integer, intent(in) :: i
integer :: c
c = ishft (a, i)
end function lshfti
! Right shift an integer.
elemental function rshfti (a, i) result (c)
integer, intent(in) :: a
integer, intent(in) :: i
integer :: c
c = ishft (a, -i)
end function rshfti
function integer2big (i) result (a)
integer, intent(in) :: i
type(big_integer), allocatable :: a
!
! To write a more efficient implementation of this procedure is
! left as an exercise for the reader.
!
character(len = 100) :: buffer
write (buffer, '(I0)') i
a = string2big (trim (buffer))
end function integer2big
function string2big (s) result (a)
character(len = *), intent(in) :: s
type(big_integer), allocatable :: a
integer :: n, i, istart, iend
integer :: digit
if ((s(1:1) == '-') .or. s(1:1) == '+') then
istart = 2
else
istart = 1
end if
iend = len (s)
n = (iend - istart + 2) / 2
allocate (a)
allocate (a%bytes(n))
a%bytes = zero
do i = istart, iend
digit = ichar (s(i:i)) - ichar ('0')
if (digit < 0 .or. 9 < digit) error stop
a = short_multiplication (a, 10)
a = short_addition (a, digit)
end do
a%sign = (s(1:1) == '-')
call normalize (a)
end function string2big
function big2string (a) result (s)
type(big_integer), intent(in) :: a
character(len = :), allocatable :: s
type(big_integer), allocatable :: q
integer :: r
integer :: sgn
sgn = big_sgn (a)
if (sgn == 0) then
s = '0'
else
q = a
s = ''
do while (big_sgn (q) /= 0)
call short_division (q, 10, q, r)
s = achar (r + ichar ('0')) // s
end do
if (sgn < 0) s = '-' // s
end if
end function big2string
function big_sgn (a) result (sgn)
type(big_integer), intent(in) :: a
integer :: sgn
integer :: n, i
n = size (a%bytes)
i = 1
sgn = 1234
do while (sgn == 1234)
if (i == n + 1) then
sgn = 0
else if (a%bytes(i) /= zero) then
if (a%sign) then
sgn = -1
else
sgn = 1
end if
else
i = i + 1
end if
end do
end function big_sgn
function big_cmp (a, b) result (cmp)
type(big_integer(*)), intent(in) :: a, b
integer :: cmp
if (a%sign) then
if (b%sign) then
cmp = -big_cmpabs (a, b)
else
cmp = -1
end if
else
if (b%sign) then
cmp = 1
else
cmp = big_cmpabs (a, b)
end if
end if
end function big_cmp
function big_cmpabs (a, b) result (cmp)
type(big_integer(*)), intent(in) :: a, b
integer :: cmp
integer :: n, i
integer :: ia, ib
cmp = 1234
n = max (size (a%bytes), size (b%bytes))
i = n
do while (cmp == 1234)
if (i == 0) then
cmp = 0
else
ia = byteval (a, i)
ib = byteval (b, i)
if (ia < ib) then
cmp = -1
else if (ia > ib) then
cmp = 1
else
i = i - 1
end if
end if
end do
end function big_cmpabs
function big_neg (a) result (c)
type(big_integer), intent(in) :: a
type(big_integer), allocatable :: c
c = a
c%sign = .not. c%sign
end function big_neg
function big_abs (a) result (c)
type(big_integer), intent(in) :: a
type(big_integer), allocatable :: c
c = a
c%sign = .false.
end function big_abs
function big_add (a, b) result (c)
type(big_integer), intent(in) :: a
type(big_integer), intent(in) :: b
type(big_integer), allocatable :: c
logical :: sign
if (a%sign) then
if (b%sign) then ! a <= 0, b <= 0
c = big_addabs (a, b)
sign = .true.
else ! a <= 0, b >= 0
c = big_subabs (a, b)
sign = .not. c%sign
end if
else
if (b%sign) then ! a >= 0, b <= 0
c = big_subabs (a, b)
sign = c%sign
else ! a >= 0, b >= 0
c = big_addabs (a, b)
sign = .false.
end if
end if
c%sign = sign
end function big_add
function big_sub (a, b) result (c)
type(big_integer), intent(in) :: a
type(big_integer), intent(in) :: b
type(big_integer), allocatable :: c
logical :: sign
if (a%sign) then
if (b%sign) then ! a <= 0, b <= 0
c = big_subabs (a, b)
sign = .not. c%sign
else ! a <= 0, b >= 0
c = big_addabs (a, b)
sign = .true.
end if
else
if (b%sign) then ! a >= 0, b <= 0
c = big_addabs (a, b)
sign = .false.
else ! a >= 0, b >= 0
c = big_subabs (a, b)
sign = c%sign
end if
end if
c%sign = sign
end function big_sub
function big_addabs (a, b) result (c)
type(big_integer), intent(in) :: a, b
type(big_integer), allocatable :: c
! Compute abs(a) + abs(b).
integer :: n, nc, i
logical :: carry
type(big_integer), allocatable :: tmp
n = max (size (a%bytes), size (b%bytes))
nc = n + 1
allocate(tmp)
allocate(tmp%bytes(nc))
call add_bytes (get_byte (a, 1), get_byte (b, 1), .false., tmp%bytes(1), carry)
do i = 2, n
call add_bytes (get_byte (a, i), get_byte (b, i), carry, tmp%bytes(i), carry)
end do
tmp%bytes(nc) = logical2byte (carry)
call normalize (tmp)
c = tmp
end function big_addabs
function big_subabs (a, b) result (c)
type(big_integer), intent(in) :: a, b
type(big_integer), allocatable :: c
! Compute abs(a) - abs(b). The result is signed.
integer :: n, i
logical :: carry
type(big_integer), allocatable :: tmp
n = max (size (a%bytes), size (b%bytes))
allocate(tmp)
allocate(tmp%bytes(n))
if (big_cmpabs (a, b) >= 0) then
tmp%sign = .false.
call sub_bytes (get_byte (a, 1), get_byte (b, 1), .false., tmp%bytes(1), carry)
do i = 2, n
call sub_bytes (get_byte (a, i), get_byte (b, i), carry, tmp%bytes(i), carry)
end do
else
tmp%sign = .true.
call sub_bytes (get_byte (b, 1), get_byte (a, 1), .false., tmp%bytes(1), carry)
do i = 2, n
call sub_bytes (get_byte (b, i), get_byte (a, i), carry, tmp%bytes(i), carry)
end do
end if
call normalize (tmp)
c = tmp
end function big_subabs
function big_mul (a, b) result (c)
type(big_integer), intent(in) :: a, b
type(big_integer), allocatable :: c
!
! This is Knuth, Volume 2, Algorithm 4.3.1M.
!
integer :: na, nb, nc
integer :: i, j
integer :: ia, ib, ic
integer :: carry
type(big_integer), allocatable :: tmp
na = size (a%bytes)
nb = size (b%bytes)
nc = na + nb + 1
allocate (tmp)
allocate (tmp%bytes(nc))
tmp%bytes = zero
j = 1
do j = 1, nb
ib = byte2i (b%bytes(j))
if (ib /= 0) then
carry = 0
do i = 1, na
ia = byte2i (a%bytes(i))
ic = byte2i (tmp%bytes(i + j - 1))
ic = (ia * ib) + ic + carry
tmp%bytes(i + j - 1) = i2byte (iand (ic, 255))
carry = ishft (ic, -8)
end do
tmp%bytes(na + j) = i2byte (carry)
end if
end do
tmp%sign = (a%sign .neqv. b%sign)
call normalize (tmp)
c = tmp
end function big_mul
subroutine big_divrem (a, b, q, r)
type(big_integer), intent(in) :: a, b
type(big_integer), allocatable, intent(inout) :: q, r
!
! Division with a remainder that is never negative. Equivalently,
! this is floor division if the divisor is positive, and ceiling
! division if the divisor is negative.
!
! See Raymond T. Boute, "The Euclidean definition of the functions
! div and mod", ACM Transactions on Programming Languages and
! Systems, Volume 14, Issue 2, pp. 127-144.
! https://doi.org/10.1145/128861.128862
!
call nonnegative_division (a, b, .true., .true., q, r)
if (a%sign) then
if (big_sgn (r) /= 0) then
q = short_addition (q, 1)
r = big_sub (big_abs (b), r)
end if
q%sign = .not. b%sign
else
q%sign = b%sign
end if
end subroutine big_divrem
function short_addition (a, b) result (c)
type(big_integer), intent(in) :: a
integer, intent(in) :: b
type(big_integer), allocatable :: c
! Compute abs(a) + b.
integer :: na, nc, i
logical :: carry
type(big_integer), allocatable :: tmp
na = size (a%bytes)
nc = na + 1
allocate(tmp)
allocate(tmp%bytes(nc))
call add_bytes (a%bytes(1), i2byte (b), .false., tmp%bytes(1), carry)
do i = 2, na
call add_bytes (a%bytes(i), zero, carry, tmp%bytes(i), carry)
end do
tmp%bytes(nc) = logical2byte (carry)
call normalize (tmp)
c = tmp
end function short_addition
function short_multiplication (a, b) result (c)
type(big_integer), intent(in) :: a
integer, intent(in) :: b
type(big_integer), allocatable :: c
integer :: i, na, nc
integer :: ia, ic
integer :: carry
type(big_integer), allocatable :: tmp
na = size (a%bytes)
nc = na + 1
allocate (tmp)
allocate (tmp%bytes(nc))
tmp%sign = a%sign
carry = 0
do i = 1, na
ia = byte2i (a%bytes(i))
ic = (ia * b) + carry
tmp%bytes(i) = i2byte (iand (ic, 255))
carry = ishft (ic, -8)
end do
tmp%bytes(nc) = i2byte (carry)
call normalize (tmp)
c = tmp
end function short_multiplication
! Division without regard to signs.
subroutine nonnegative_division (a, b, want_q, want_r, q, r)
type(big_integer), intent(in) :: a, b
logical, intent(in) :: want_q, want_r
type(big_integer), intent(inout), allocatable :: q, r
integer :: na, nb
integer :: remainder
na = size (a%bytes)
nb = size (b%bytes)
! It is an error if b has "significant" zero-bytes or is equal to
! zero.
if (b%bytes(nb) == zero) error stop
if (nb == 1) then
if (want_q) then
call short_division (a, byte2i (b%bytes(1)), q, remainder)
else
block
type(big_integer), allocatable :: bit_bucket
call short_division (a, byte2i (b%bytes(1)), bit_bucket, remainder)
end block
end if
if (want_r) then
if (allocated (r)) deallocate (r)
allocate (r)
allocate (r%bytes(1))
r%bytes(1) = i2byte (remainder)
end if
else
if (na >= nb) then
call long_division (a, b, want_q, want_r, q, r)
else
if (want_q) q = string2big ("0")
if (want_r) r = a
end if
end if
end subroutine nonnegative_division
subroutine short_division (a, b, q, r)
type(big_integer), intent(in) :: a
integer, intent(in) :: b
type(big_integer), intent(inout), allocatable :: q
integer, intent(inout) :: r
!
! This is Knuth, Volume 2, Exercise 4.3.1.16.
!
! The divisor is assumed to be positive.
!
integer :: n, i
integer :: ia, ib, iq
type(big_integer), allocatable :: tmp
ib = b
n = size (a%bytes)
allocate (tmp)
allocate (tmp%bytes(n))
r = 0
do i = n, 1, -1
ia = (256 * r) + byte2i (a%bytes(i))
iq = ia / ib
r = mod (ia, ib)
tmp%bytes(i) = i2byte (iq)
end do
tmp%sign = a%sign
call normalize (tmp)
q = tmp
end subroutine short_division
subroutine long_division (a, b, want_quotient, want_remainder, quotient, remainder)
type(big_integer), intent(in) :: a, b
logical, intent(in) :: want_quotient, want_remainder
type(big_integer), intent(inout), allocatable :: quotient
type(big_integer), intent(inout), allocatable :: remainder
!
! This is Knuth, Volume 2, Algorithm 4.3.1D.
!
! We do not deal here with the signs of the inputs and outputs.
!
! It is assumed size(a%bytes) >= size(b%bytes), and that b has no
! leading zero-bytes and is at least two bytes long. If b is one
! byte long and nonzero, use short division.
!
integer :: na, nb, m, n
integer :: num_lz, num_nonlz
integer :: j
integer :: qhat
logical :: carry
!
! We will NOT be working with VERY large numbers, and so it will
! be safe to put temporary storage on the stack. (Note: your
! Fortran might put this storage in a heap instead of the stack.)
!
! v = b, normalized to put its most significant 1-bit all the
! way left.
!
! u = a, shifted left by the same amount as b.
!
! q = the quotient.
!
! The remainder, although shifted left, will end up in u.
!
integer(bytekind) :: u(0:size (a%bytes) + size (b%bytes))
integer(bytekind) :: v(0:size (b%bytes) - 1)
integer(bytekind) :: q(0:size (a%bytes) - size (b%bytes))
na = size (a%bytes)
nb = size (b%bytes)
n = nb
m = na - nb
! In the most significant byte of the divisor, find the number of
! leading zero bits, and the number of bits after that.
block
integer(bytekind) :: tmp
tmp = byte2bk (b%bytes(n))
num_nonlz = bit_size (tmp) - leadz (tmp)
num_lz = 8 - num_nonlz
end block
call normalize_v (b%bytes) ! Make the most significant bit of v be one.
call normalize_u (a%bytes) ! Shifted by the same amount as v.
! Assure ourselves that the most significant bit of v is a one.
if (.not. btest (v(n - 1), 7)) error stop
do j = m, 0, -1
call calculate_qhat (qhat)
call multiply_and_subtract (carry)
q(j) = i2bk (qhat)
if (carry) call add_back
end do
if (want_quotient) then
if (allocated (quotient)) deallocate (quotient)
allocate (quotient)
allocate (quotient%bytes(m + 1))
quotient%bytes = bk2byte (q)
call normalize (quotient)
end if
if (want_remainder) then
if (allocated (remainder)) deallocate (remainder)
allocate (remainder)
allocate (remainder%bytes(n))
call unnormalize_u (remainder%bytes)
call normalize (remainder)
end if
contains
subroutine normalize_v (b_bytes)
character, intent(in) :: b_bytes(n)
!
! Normalize v so its most significant bit is a one. Any
! normalization factor that achieves this goal will suffice; we
! choose 2**num_lz. (Knuth uses (2**32) div (y[n-1] + 1).)
!
! Strictly for readability, we use linear stack space for an
! intermediate result.
!
integer :: i
integer(bytekind) :: btmp(0:n - 1)
btmp = byte2bk (b_bytes)
v(0) = lshftbk (btmp(0), num_lz)
do i = 1, n - 1
v(i) = ior (lshftbk (btmp(i), num_lz), &
& rshftbk (btmp(i - 1), num_nonlz))
end do
end subroutine normalize_v
subroutine normalize_u (a_bytes)
character, intent(in) :: a_bytes(m + n)
!
! Shift a leftwards to get u. Shift by as much as b was shifted
! to get v.
!
! Strictly for readability, we use linear stack space for an
! intermediate result.
!
integer :: i
integer(bytekind) :: atmp(0:m + n - 1)
atmp = byte2bk (a_bytes)
u(0) = lshftbk (atmp(0), num_lz)
do i = 1, m + n - 1
u(i) = ior (lshftbk (atmp(i), num_lz), &
& rshftbk (atmp(i - 1), num_nonlz))
end do
u(m + n) = rshftbk (atmp(m + n - 1), num_nonlz)
end subroutine normalize_u
subroutine unnormalize_u (r_bytes)
character, intent(out) :: r_bytes(n)
!
! Strictly for readability, we use linear stack space for an
! intermediate result.
!
integer :: i
integer(bytekind) :: rtmp(0:n - 1)
do i = 0, n - 1
rtmp(i) = ior (rshftbk (u(i), num_lz), &
& lshftbk (u(i + 1), num_nonlz))
end do
rtmp(n - 1) = rshftbk (u(n - 1), num_lz)
r_bytes = bk2byte (rtmp)
end subroutine unnormalize_u
subroutine calculate_qhat (qhat)
integer, intent(out) :: qhat
integer :: itmp, rhat
logical :: adjust
itmp = ior (lshfti (bk2i (u(j + n)), 8), &
& bk2i (u(j + n - 1)))
qhat = itmp / bk2i (v(n - 1))
rhat = mod (itmp, bk2i (v(n - 1)))
adjust = .true.
do while (adjust)
if (rshfti (qhat, 8) /= 0) then
continue
else if (qhat * bk2i (v(n - 2)) &
& > ior (lshfti (rhat, 8), &
& bk2i (u(j + n - 2)))) then
continue
else
adjust = .false.
end if
if (adjust) then
qhat = qhat - 1
rhat = rhat + bk2i (v(n - 1))
if (rshfti (rhat, 8) == 0) then
adjust = .false.
end if
end if
end do
end subroutine calculate_qhat
subroutine multiply_and_subtract (carry)
logical, intent(out) :: carry
integer :: i
integer :: qhat_v
integer :: mul_carry, sub_carry
integer :: diff
mul_carry = 0
sub_carry = 0
do i = 0, n
! Multiplication.
qhat_v = mul_carry
if (i /= n) qhat_v = qhat_v + (qhat * bk2i (v(i)))
mul_carry = rshfti (qhat_v, 8)
qhat_v = iand (qhat_v, 255)
! Subtraction.
diff = bk2i (u(j + i)) - qhat_v + sub_carry
sub_carry = -(logical2i (diff < 0)) ! Carry 0 or -1.
u(j + i) = i2bk (diff)
end do
carry = (sub_carry /= 0)
end subroutine multiply_and_subtract
subroutine add_back
integer :: i, carry, sum
q(j) = q(j) - 1_bytekind
carry = 0
do i = 0, n - 1
sum = bk2i (u(j + i)) + bk2i (v(i)) + carry
carry = ishft (sum, -8)
u(j + i) = i2bk (sum)
end do
end subroutine add_back
end subroutine long_division
function big_pow (a, i) result (c)
type(big_integer), intent(in) :: a
integer, intent(in) :: i
type(big_integer), allocatable :: c
type(big_integer), allocatable :: base
integer :: exponent, exponent_halved
integer :: j, last_set
if (i < 0) error stop
if (i == 0) then
c = integer2big (1)
else
last_set = bit_size (i) - leadz (i)
base = a
exponent = i
c = integer2big (1)
do j = 0, last_set - 1
exponent_halved = exponent / 2
if (2 * exponent_halved /= exponent) then
c = c * base
end if
exponent = exponent_halved
base = base * base
end do
end if
end function big_pow
subroutine add_bytes (a, b, carry_in, c, carry_out)
character, intent(in) :: a, b
logical, value :: carry_in
character, intent(inout) :: c
logical, intent(inout) :: carry_out
integer :: ia, ib, ic
ia = byte2i (a)
if (carry_in) ia = ia + 1
ib = byte2i (b)
ic = ia + ib
c = i2byte (iand (ic, 255))
carry_out = (ic >= 256)
end subroutine add_bytes
subroutine sub_bytes (a, b, carry_in, c, carry_out)
character, intent(in) :: a, b
logical, value :: carry_in
character, intent(inout) :: c
logical, intent(inout) :: carry_out
integer :: ia, ib, ic
ia = byte2i (a)
ib = byte2i (b)
if (carry_in) ib = ib + 1
ic = ia - ib
carry_out = (ic < 0)
if (carry_out) ic = ic + 256
c = i2byte (iand (ic, 255))
end subroutine sub_bytes
function get_byte (a, i) result (byte)
type(big_integer), intent(in) :: a
integer, intent(in) :: i
character :: byte
if (size (a%bytes) < i) then
byte = zero
else
byte = a%bytes(i)
end if
end function get_byte
function byteval (a, i) result (v)
type(big_integer), intent(in) :: a
integer, intent(in) :: i
integer :: v
if (size (a%bytes) < i) then
v = 0
else
v = byte2i (a%bytes(i))
end if
end function byteval
subroutine normalize (a)
type(big_integer), intent(inout) :: a
logical :: done
integer :: i
character, allocatable :: fewer_bytes(:)
! Shorten to the minimum number of bytes.
i = size (a%bytes)
done = .false.
do while (.not. done)
if (i == 1) then
done = .true.
else if (a%bytes(i) /= zero) then
done = .true.
else
i = i - 1
end if
end do
if (i /= size (a%bytes)) then
allocate (fewer_bytes (i))
fewer_bytes = a%bytes(1:i)
call move_alloc (fewer_bytes, a%bytes)
end if
! If the magnitude is zero, then clear the sign bit.
if (size (a%bytes) == 1) then
if (a%bytes(1) == zero) then
a%sign = .false.
end if
end if
end subroutine normalize
end module big_integers
program modular_exponentiation_task
use, non_intrinsic :: big_integers
implicit none
type(big_integer), allocatable :: zero, one, two
type(big_integer), allocatable :: a, b, modulus
zero = integer2big (0)
one = integer2big (1)
two = integer2big (2)
a = string2big ("2988348162058574136915891421498819466320163312926952423791023078876139")
b = string2big ("2351399303373464486466122544523690094744975233415544072992656881240319")
modulus = string2big ("10") ** 40
write (*,*) big2string (modular_pow (a, b, modulus))
contains
! The right-to-left binary method,
! https://en.wikipedia.org/w/index.php?title=Modular_exponentiation&oldid=1136216610#Right-to-left_binary_method
function modular_pow (base, exponent, modulus) result (retval)
type(big_integer), intent(in) :: base, exponent, modulus
type(big_integer), allocatable :: retval
type(big_integer), allocatable :: bas, expnt, remainder, bit_bucket
if (big_sgn (modulus - one) == 0) then
retval = zero
else
retval = one
bas = base
expnt = exponent
do while (big_sgn (expnt) /= 0)
call big_divrem (expnt, two, expnt, remainder)
if (big_sgn (remainder) /= 0) then
call big_divrem (retval * bas, modulus, bit_bucket, retval)
end if
call big_divrem (bas * bas, modulus, bit_bucket, bas)
end do
end if
end function modular_pow
end program modular_exponentiation_task
- Output:
$ gfortran -g -fbounds-check -Wall -Wextra modular_exponentiation.f90 && ./a.out 1527229998585248450016808958343740453059
FreeBASIC
'From first principles (No external library)
Function _divide(n1 As String,n2 As String,decimal_places As Integer=10,dpflag As String="s") As String
Dim As String number=n1,divisor=n2
dpflag=Lcase(dpflag)
'For MOD
Dim As Integer modstop
If dpflag="mod" Then
If Len(n1)<Len(n2) Then Return n1
If Len(n1)=Len(n2) Then
If n1<n2 Then Return n1
End If
modstop=Len(n1)-Len(n2)+1
End If
If dpflag<>"mod" Then
If dpflag<>"s" Then dpflag="raw"
End If
Dim runcount As Integer
'_______ LOOK UP TABLES ______________
Dim Qmod(0 To 19) As Ubyte
Dim bool(0 To 19) As Ubyte
For z As Integer=0 To 19
Qmod(z)=(z Mod 10+48)
bool(z)=(-(10>z))
Next z
Dim answer As String 'THE ANSWER STRING
'_______ SET THE DECIMAL WHERE IT SHOULD BE AT _______
Dim As String part1,part2
#macro set(decimal)
#macro insert(s,char,position)
If position > 0 And position <=Len(s) Then
part1=Mid(s,1,position-1)
part2=Mid(s,position)
s=part1+char+part2
End If
#endmacro
insert(answer,".",decpos)
answer=thepoint+zeros+answer
If dpflag="raw" Then
answer=Mid(answer,1,decimal_places)
End If
#endmacro
'______________________________________________
'__________ SPLIT A STRING ABOUT A CHARACTRR __________
Dim As String var1,var2
Dim pst As Integer
#macro split(stri,char,var1,var2)
pst=Instr(stri,char)
var1="":var2=""
If pst<>0 Then
var1=Rtrim(Mid(stri,1,pst),".")
var2=Ltrim(Mid(stri,pst),".")
Else
var1=stri
End If
#endmacro
#macro Removepoint(s)
split(s,".",var1,var2)
#endmacro
'__________ GET THE SIGN AND CLEAR THE -ve __________________
Dim sign As String
If Left(number,1)="-" Xor Left (divisor,1)="-" Then sign="-"
If Left(number,1)="-" Then number=Ltrim(number,"-")
If Left (divisor,1)="-" Then divisor=Ltrim(divisor,"-")
'DETERMINE THE DECIMAL POSITION BEFORE THE DIVISION
Dim As Integer lennint,lenddec,lend,lenn,difflen
split(number,".",var1,var2)
lennint=Len(var1)
split(divisor,".",var1,var2)
lenddec=Len(var2)
If Instr(number,".") Then
Removepoint(number)
number=var1+var2
End If
If Instr(divisor,".") Then
Removepoint(divisor)
divisor=var1+var2
End If
Dim As Integer numzeros
numzeros=Len(number)
number=Ltrim(number,"0"):divisor=Ltrim (divisor,"0")
numzeros=numzeros-Len(number)
lend=Len(divisor):lenn=Len(number)
If lend>lenn Then difflen=lend-lenn
Dim decpos As Integer=lenddec+lennint-lend+2-numzeros 'THE POSITION INDICATOR
Dim _sgn As Byte=-Sgn(decpos)
If _sgn=0 Then _sgn=1
Dim As String thepoint=String(_sgn,".") 'DECIMAL AT START (IF)
Dim As String zeros=String(-decpos+1,"0")'ZEROS AT START (IF) e.g. .0009
If dpflag<>"mod" Then
If Len(zeros) =0 Then dpflag="s"
End If
Dim As Integer runlength
If Len(zeros) Then
runlength=decimal_places
answer=String(Len(zeros)+runlength+10,"0")
If dpflag="raw" Then
runlength=1
answer=String(Len(zeros)+runlength+10,"0")
If decimal_places>Len(zeros) Then
runlength=runlength+(decimal_places-Len(zeros))
answer=String(Len(zeros)+runlength+10,"0")
End If
End If
Else
decimal_places=decimal_places+decpos
runlength=decimal_places
answer=String(Len(zeros)+runlength+10,"0")
End If
'___________DECIMAL POSITION DETERMINED _____________
'SET UP THE VARIABLES AND START UP CONDITIONS
number=number+String(difflen+decimal_places,"0")
Dim count As Integer
Dim temp As String
Dim copytemp As String
Dim topstring As String
Dim copytopstring As String
Dim As Integer lenf,lens
Dim As Ubyte takeaway,subtractcarry
Dim As Integer n3,diff
If Ltrim(divisor,"0")="" Then Return "Error :division by zero"
lens=Len(divisor)
topstring=Left(number,lend)
copytopstring=topstring
Do
count=0
Do
count=count+1
copytemp=temp
Do
'___________________ QUICK SUBTRACTION loop _________________
lenf=Len(topstring)
If lens<lenf=0 Then 'not
If Lens>lenf Then
temp= "done"
Exit Do
End If
If divisor>topstring Then
temp= "done"
Exit Do
End If
End If
diff=lenf-lens
temp=topstring
subtractcarry=0
For n3=lenf-1 To diff Step -1
takeaway= topstring[n3]-divisor[n3-diff]+10-subtractcarry
temp[n3]=Qmod(takeaway)
subtractcarry=bool(takeaway)
Next n3
If subtractcarry=0 Then Exit Do
If n3=-1 Then Exit Do
For n3=n3 To 0 Step -1
takeaway= topstring[n3]-38-subtractcarry
temp[n3]=Qmod(takeaway)
subtractcarry=bool(takeaway)
If subtractcarry=0 Then Exit Do
Next n3
Exit Do
Loop 'single run
temp=Ltrim(temp,"0")
If temp="" Then temp= "0"
topstring=temp
Loop Until temp="done"
' INDIVIDUAL CHARACTERS CARVED OFF ________________
runcount=runcount+1
If count=1 Then
topstring=copytopstring+Mid(number,lend+runcount,1)
Else
topstring=copytemp+Mid(number,lend+runcount,1)
End If
copytopstring=topstring
topstring=Ltrim(topstring,"0")
If dpflag="mod" Then
If runcount=modstop Then
If topstring="" Then Return "0"
Return Mid(topstring,1,Len(topstring)-1)
End If
End If
answer[runcount-1]=count+47
If topstring="" And runcount>Len(n1)+1 Then
Exit Do
End If
Loop Until runcount=runlength+1
' END OF RUN TO REQUIRED DECIMAL PLACES
set(decimal) 'PUT IN THE DECIMAL POINT
'THERE IS ALWAYS A DECIMAL POINT SOMEWHERE IN THE ANSWER
'NOW GET RID OF IT IF IT IS REDUNDANT
answer=Rtrim(answer,"0")
answer=Rtrim(answer,".")
answer=Ltrim(answer,"0")
If answer="" Then Return "0"
Return sign+answer
End Function
Dim Shared As Integer _Mod(0 To 99),_Div(0 To 99)
For z As Integer=0 To 99:_Mod(z)=(z Mod 10+48):_Div(z)=z\10:Next
Function Qmult(a As String,b As String) As String
Var flag=0,la = Len(a),lb = Len(b)
If Len(b)>Len(a) Then flag=1:Swap a,b:Swap la,lb
Dim As Ubyte n,carry,ai
Var c =String(la+lb,"0")
For i As Integer =la-1 To 0 Step -1
carry=0:ai=a[i]-48
For j As Integer =lb-1 To 0 Step -1
n = ai * (b[j]-48) + (c[i+j+1]-48) + carry
carry =_Div(n):c[i+j+1]=_Mod(n)
Next j
c[i]+=carry
Next i
If flag Then Swap a,b
Return Ltrim(c,"0")
End Function
'=======================================================================
#define mod_(a,b) _divide((a),(b),,"mod")
#define div_(a,b) iif(len((a))<len((b)),"0",_divide((a),(b),-2))
Function Modular_Exponentiation(base_num As String, exponent As String, modulus As String) As String
Dim b1 As String = base_num
Dim e1 As String = exponent
Dim As String result="1"
b1 = mod_(b1,modulus)
Do While e1 <> "0"
Var L=Len(e1)-1
If e1[L] And 1 Then
result=mod_(Qmult(result,b1),modulus)
End If
b1=mod_(qmult(b1,b1),modulus)
e1=div_(e1,"2")
Loop
Return result
End Function
var base_num="2988348162058574136915891421498819466320163312926952423791023078876139"
var exponent="2351399303373464486466122544523690094744975233415544072992656881240319"
var modulus="10000000000000000000000000000000000000000"
dim as double t=timer
var ans=Modular_Exponentiation(base_num,exponent,modulus)
print "Result:"
Print ans
print "time taken ";(timer-t)*1000;" milliseconds"
Print "Done"
Sleep
Result: 1527229998585248450016808958343740453059 time taken 93.09767815284431 milliseconds Done
Frink
a = 2988348162058574136915891421498819466320163312926952423791023078876139
b = 2351399303373464486466122544523690094744975233415544072992656881240319
println[modPow[a,b,10^40]]- Output:
1527229998585248450016808958343740453059
GAP
# Built-in
a := 2988348162058574136915891421498819466320163312926952423791023078876139;
b := 2351399303373464486466122544523690094744975233415544072992656881240319;
PowerModInt(a, b, 10^40);
1527229998585248450016808958343740453059
# Implementation
PowerModAlt := function(a, n, m)
local r;
r := 1;
while n > 0 do
if IsOddInt(n) then
r := RemInt(r*a, m);
fi;
n := QuoInt(n, 2);
a := RemInt(a*a, m);
od;
return r;
end;
PowerModAlt(a, b, 10^40);
gnuplot
_powm(b, e, m, r) = (e == 0 ? r : (e % 2 == 1 ? _powm(b * b % m, e / 2, m, r * b % m) : _powm(b * b % m, e / 2, m, r)))
powm(b, e, m) = _powm(b, e, m, 1)
# Usage
print powm(2, 3453, 131)
# Where b is the base, e is the exponent, m is the modulus, i.e.: b^e mod m
Go
package main
import (
"fmt"
"math/big"
)
func main() {
a, _ := new(big.Int).SetString(
"2988348162058574136915891421498819466320163312926952423791023078876139", 10)
b, _ := new(big.Int).SetString(
"2351399303373464486466122544523690094744975233415544072992656881240319", 10)
m := big.NewInt(10)
r := big.NewInt(40)
m.Exp(m, r, nil)
r.Exp(a, b, m)
fmt.Println(r)
}
- Output:
1527229998585248450016808958343740453059
Groovy
println 2988348162058574136915891421498819466320163312926952423791023078876139.modPow(
2351399303373464486466122544523690094744975233415544072992656881240319,
10000000000000000000000000000000000000000)
Ouput:
1527229998585248450016808958343740453059
Haskell
modPow :: Integer -> Integer -> Integer -> Integer -> Integer
modPow b e 1 r = 0
modPow b 0 m r = r
modPow b e m r
| e `mod` 2 == 1 = modPow b' e' m (r * b `mod` m)
| otherwise = modPow b' e' m r
where
b' = b * b `mod` m
e' = e `div` 2
main = do
print (modPow 2988348162058574136915891421498819466320163312926952423791023078876139
2351399303373464486466122544523690094744975233415544072992656881240319
(10 ^ 40)
1)
- Output:
1527229998585248450016808958343740453059
or in terms of until:
powerMod :: Integer -> Integer -> Integer -> Integer
powerMod b e m = x
where
(_, _, x) =
until
(\(_, e, _) -> e <= 0)
(\(b, e, x) ->
( mod (b * b) m
, div e 2
, if 0 /= mod e 2
then mod (b * x) m
else x))
(b, e, 1)
main :: IO ()
main =
print $
powerMod
2988348162058574136915891421498819466320163312926952423791023078876139
2351399303373464486466122544523690094744975233415544072992656881240319
(10 ^ 40)
- Output:
1527229998585248450016808958343740453059
Icon and Unicon
This uses the exponentiation procedure from RSA Code an example of the right to left binary method.
- Output:
last 40 digits = 1527229998585248450016808958343740453059
J
Solution:
m&|@^
Example:
a =: 2988348162058574136915891421498819466320163312926952423791023078876139x
b =: 2351399303373464486466122544523690094744975233415544072992656881240319x
m =: 10^40x
a m&|@^ b
1527229998585248450016808958343740453059
Discussion: The phrase a m&|@^ b is the natural expression of a^b mod m in J, and is recognized by the interpreter as an opportunity for optimization, by avoiding the exponentiation.
Java
java.math.BigInteger.modPow solves this task. Inside OpenJDK, BigInteger.java implements BigInteger.modPow with a fast algorithm from Colin Plumb's bnlib. This "window algorithm" caches odd powers of the base, to decrease the number of squares and multiplications. It also exploits both the Chinese remainder theorem and the Montgomery reduction.
import java.math.BigInteger;
public class PowMod {
public static void main(String[] args){
BigInteger a = new BigInteger(
"2988348162058574136915891421498819466320163312926952423791023078876139");
BigInteger b = new BigInteger(
"2351399303373464486466122544523690094744975233415544072992656881240319");
BigInteger m = new BigInteger("10000000000000000000000000000000000000000");
System.out.println(a.modPow(b, m));
}
}
- Output:
1527229998585248450016808958343740453059
jq
Also works with gojq, the Go implementation of jq, and with fq.
Adapted from Wren
# To take advantage of gojq's arbitrary-precision integer arithmetic:
def power($b): . as $in | reduce range(0;$b) as $i (1; . * $in);
# Returns (. ^ $exp) % $mod
# where $exp >= 0, $mod != 0, and the input are integers.
def modPow($exp; $mod):
if $mod == 0 then "Cannot take modPow with modulus 0." | error
elif $exp < 0 then "modPow with exp < 0 is not supported." | error
else . as $x
| {r: 1, base: ($x % $mod), exp: $exp}
| until( .exp <= 0 or .emit;
if .base == 0 then .emit = 0
else if .exp%2 == 1
then .r = (.r * .base) % $mod
| .exp |= (. - 1) / 2
else .exp /= 2
end
| .base |= (. * .) % $mod
end )
| if .emit then .emit else .r end
end;
def task:
2988348162058574136915891421498819466320163312926952423791023078876139 as $a
| 2351399303373464486466122544523690094744975233415544072992656881240319 as $b
| (10|power(40)) as $m
| $a | modPow($b; $m) ;
task- Output:
1527229998585248450016808958343740453059
Julia
We can use the built-in powermod function with the built-in BigInt type (based on GMP):
a = 2988348162058574136915891421498819466320163312926952423791023078876139
b = 2351399303373464486466122544523690094744975233415544072992656881240319
m = big(10) ^ 40
@show powermod(a, b, m)
- Output:
powermod(a, b, m) = 1527229998585248450016808958343740453059
Kotlin
// version 1.0.6
import java.math.BigInteger
fun main(args: Array<String>) {
val a = BigInteger("2988348162058574136915891421498819466320163312926952423791023078876139")
val b = BigInteger("2351399303373464486466122544523690094744975233415544072992656881240319")
val m = BigInteger.TEN.pow(40)
println(a.modPow(b, m))
}
- Output:
1527229998585248450016808958343740453059
Lambdatalk
Following scheme
We will call the lib_BN library for big numbers:
{require lib_BN}
In this library {BN.compare a b} returns 1 if a > b, 0 if a = b and -1 if a < b.
For a better readability we define three small functions
{def BN.= {lambda {:a :b} {= {BN.compare :a :b} 0}}}
-> BN.=
{def BN.even? {lambda {:n} {= {BN.compare {BN.% :n 2} 0} 0}}}
-> BN.even?
{def BN.square {lambda {:n} {BN.* :n :n}}}
-> BN.square
{def mod-exp
{lambda {:a :n :mod}
{if {BN.= :n 0}
then 1
else {if {BN.even? :n}
then {BN.% {BN.square {mod-exp :a {BN./ :n 2} :mod}} :mod}
else {BN.% {BN.* :a {mod-exp :a {BN.- :n 1} :mod}} :mod}}}}}
-> mod-exp
{mod-exp
2988348162058574136915891421498819466320163312926952423791023078876139
2351399303373464486466122544523690094744975233415544072992656881240319
{BN.pow 10 40}}
-> 1527229998585248450016808958343740453059 // 3300ms
Maple
a := 2988348162058574136915891421498819466320163312926952423791023078876139:
b := 2351399303373464486466122544523690094744975233415544072992656881240319:
a &^ b mod 10^40;- Output:
1527229998585248450016808958343740453059
Mathematica /Wolfram Language
a = 2988348162058574136915891421498819466320163312926952423791023078876139;
b = 2351399303373464486466122544523690094744975233415544072992656881240319;
m = 10^40;
PowerMod[a, b, m]
-> 1527229998585248450016808958343740453059
Maxima
a: 2988348162058574136915891421498819466320163312926952423791023078876139$
b: 2351399303373464486466122544523690094744975233415544072992656881240319$
power_mod(a, b, 10^40);
/* 1527229998585248450016808958343740453059 */
Nim
import bigints
proc powmod(b, e, m: BigInt): BigInt =
assert e >= 0
var e = e
var b = b
result = initBigInt(1)
while e > 0:
if e mod 2 == 1:
result = (result * b) mod m
e = e div 2
b = (b.pow 2) mod m
var
a = initBigInt("2988348162058574136915891421498819466320163312926952423791023078876139")
b = initBigInt("2351399303373464486466122544523690094744975233415544072992656881240319")
echo powmod(a, b, 10.pow 40)
- Output:
1527229998585248450016808958343740453059
ObjectIcon
# -*- ObjectIcon -*-
#
# This program is close to being an exact copy of the Icon.
#
import io # <-- Object Icon requires this for I/O.
procedure main()
local a, b # <-- Object Icon forces you to declare your variables.
a := 2988348162058574136915891421498819466320163312926952423791023078876139
b := 2351399303373464486466122544523690094744975233415544072992656881240319
# You could leave out the "io." in the call to "write" below,
# because there is some "compatibility with regular Icon" support in
# the io package.
io.write("last 40 digits = ", mod_power(a,b,(10^40)))
end
procedure mod_power(base, exponent, modulus)
local result
result := 1
while exponent > 0 do
{
if exponent % 2 = 1 then
result := (result * base) % modulus
exponent /:= 2
base := base ^ 2 % modulus
}
return result
end$ oiscript modular-exponentiation-OI.icn last 40 digits = 1527229998585248450016808958343740453059
OCaml
Using the zarith library:
let a = Z.of_string "2988348162058574136915891421498819466320163312926952423791023078876139" in
let b = Z.of_string "2351399303373464486466122544523690094744975233415544072992656881240319" in
let m = Z.pow (Z.of_int 10) 40 in
Z.powm a b m
|> Z.to_string
|> print_endline
- Output:
1527229998585248450016808958343740453059
Oforth
Usage : a b mod powmod
: powmod(base, exponent, modulus)
1 exponent dup ifZero: [ return ]
while ( dup 0 > ) [
dup isEven ifFalse: [ swap base * modulus mod swap ]
2 / base sq modulus mod ->base
] drop ;- Output:
>2988348162058574136915891421498819466320163312926952423791023078876139 ok >2351399303373464486466122544523690094744975233415544072992656881240319 ok >10 40 pow ok >powmod println 1527229998585248450016808958343740453059 ok
PARI/GP
a=2988348162058574136915891421498819466320163312926952423791023078876139;
b=2351399303373464486466122544523690094744975233415544072992656881240319;
lift(Mod(a,10^40)^b)Pascal
A port of the C example using gmp.
Program ModularExponentiation(output);
uses
gmp;
var
a, b, m, r: mpz_t;
fmt: pchar;
begin
mpz_init_set_str(a, '2988348162058574136915891421498819466320163312926952423791023078876139', 10);
mpz_init_set_str(b, '2351399303373464486466122544523690094744975233415544072992656881240319', 10);
mpz_init(m);
mpz_ui_pow_ui(m, 10, 40);
mpz_init(r);
mpz_powm(r, a, b, m);
fmt := '%Zd' + chr(13) + chr(10);
mp_printf(fmt, @r); (* ...16808958343740453059 *)
mpz_clear(a);
mpz_clear(b);
mpz_clear(m);
mpz_clear(r);
end.
- Output:
% ./ModularExponentiation 1527229998585248450016808958343740453059
PascalABC.NET
// Modular exponentiation. Nigel Galloway:August 22nd., 2024
type I=System.Numerics.BigInteger;
var bi:string->I:=n->I.Parse(n);
function expMod(a:I;b:I;n:I):I;
function fN(a:I;b:I;c:I):I;
begin
result:=if b=0 then c else fN(a*(a mod n),b/bi('2'),(if (b mod 2)=0 then c else c*a mod n));
end;
begin
result:=fN(a,b,I.One)
end;
begin
var a := bi('2988348162058574136915891421498819466320163312926952423791023078876139');
var b := bi('2351399303373464486466122544523690094744975233415544072992656881240319');
println(expMod(a,b,bi('10')**40).ToString());
end.
- Output:
1527229998585248450016808958343740453059
Perl
Calculating the result both with an explicit algorithm (borrowed from Raku example) and with a built-in available when the use bigint pragma is invoked. Note that bmodpow modifies the base value (here $a) while expmod does not.
use bigint;
sub expmod {
my($a, $b, $n) = @_;
my $c = 1;
do {
($c *= $a) %= $n if $b % 2;
($a *= $a) %= $n;
} while ($b = int $b/2);
$c;
}
my $a = 2988348162058574136915891421498819466320163312926952423791023078876139;
my $b = 2351399303373464486466122544523690094744975233415544072992656881240319;
my $m = 10 ** 40;
print expmod($a, $b, $m), "\n";
print $a->bmodpow($b, $m), "\n";
- Output:
1527229998585248450016808958343740453059 1527229998585248450016808958343740453059
Phix
Already builtin as mpz_powm, but here is an explicit version
with javascript_semantics include mpfr.e procedure mpz_mod_exp(mpz base, exponent, modulus, result) if mpz_cmp_si(exponent,1)=0 then mpz_set(result,base) else mpz _exp = mpz_init_set(exponent) -- (use a copy) bool odd = mpz_odd(_exp) if odd then mpz_sub_ui(_exp,_exp,1) end if mpz_fdiv_q_2exp(_exp,_exp,1) mpz_mod_exp(base,_exp,modulus,result) _exp = mpz_free(_exp) mpz_mul(result,result,result) if odd then mpz_mul(result,result,base) end if end if mpz_mod(result,result,modulus) end procedure mpz base = mpz_init("2988348162058574136915891421498819466320163312926952423791023078876139"), exponent = mpz_init("2351399303373464486466122544523690094744975233415544072992656881240319"), modulus = mpz_init("1"&repeat('0',40)), result = mpz_init() mpz_mod_exp(base,exponent,modulus,result) ?mpz_get_str(result) -- check against the builtin: mpz_powm(result,base,exponent,modulus) ?mpz_get_str(result)
- Output:
"1527229998585248450016808958343740453059" "1527229998585248450016808958343740453059"
PHP
<?php
$a = '2988348162058574136915891421498819466320163312926952423791023078876139';
$b = '2351399303373464486466122544523690094744975233415544072992656881240319';
$m = '1' . str_repeat('0', 40);
echo bcpowmod($a, $b, $m), "\n";
- Output:
1527229998585248450016808958343740453059
PicoLisp
The following function is taken from "lib/rsa.l":
(de **Mod (X Y N)
(let M 1
(loop
(when (bit? 1 Y)
(setq M (% (* M X) N)) )
(T (=0 (setq Y (>> 1 Y)))
M )
(setq X (% (* X X) N)) ) ) )Test:
: (**Mod
2988348162058574136915891421498819466320163312926952423791023078876139
2351399303373464486466122544523690094744975233415544072992656881240319
10000000000000000000000000000000000000000 )
-> 1527229998585248450016808958343740453059Powershell
Function Invoke-ModuloExponentiation ([BigInt]$Base, [BigInt]$Exponent, $Modulo) {
$Result = 1
$Base = $Base % $Modulo
If ($Base -eq 0) {return 0}
While ($Exponent -gt 0) {
If (($Exponent -band 1) -eq 1) {$Result = ($Result * $Base) % $Modulo}
$Exponent = $Exponent -shr 1
$Base = ($Base * $Base) % $Modulo
}
return ($Result % $Modulo)
}
$a = [BigInt]::Parse('2988348162058574136915891421498819466320163312926952423791023078876139')
$b = [BigInt]::Parse('2351399303373464486466122544523690094744975233415544072992656881240319')
$m = [BigInt]::Pow(10, 40)
Invoke-ModuloExponentiation -Base $a -Exponent $b -Modulo $m
- Output:
1527229998585248450016808958343740453059
Prolog
SWI Prolog has a built-in function named powm for this purpose.
main:-
A = 2988348162058574136915891421498819466320163312926952423791023078876139,
B = 2351399303373464486466122544523690094744975233415544072992656881240319,
M is 10 ** 40,
P is powm(A, B, M),
writeln(P).
- Output:
1527229998585248450016808958343740453059
Python
a = 2988348162058574136915891421498819466320163312926952423791023078876139
b = 2351399303373464486466122544523690094744975233415544072992656881240319
m = 10 ** 40
print(pow(a, b, m))
- Output:
1527229998585248450016808958343740453059
def power_mod(b, e, m):
" Without using builtin function "
x = 1
while e > 0:
b, e, x = (
b * b % m,
e // 2,
b * x % m if e % 2 else x
)
return x
a = 2988348162058574136915891421498819466320163312926952423791023078876139
b = 2351399303373464486466122544523690094744975233415544072992656881240319
m = 10 ** 40
print(power_mod(a, b, m))
- Output:
1527229998585248450016808958343740453059
Quackery
[ temp put 1 unrot
[ dup while
dup 1 & if
[ unrot tuck *
temp share mod
swap rot ]
1 >>
swap dup *
temp share mod
swap again ]
2drop temp release ] is **mod ( n n n --> n )
2988348162058574136915891421498819466320163312926952423791023078876139
2351399303373464486466122544523690094744975233415544072992656881240319
10 40 ** **mod echo- Output:
1527229998585248450016808958343740453059
Racket
#lang racket
(require math)
(define a 2988348162058574136915891421498819466320163312926952423791023078876139)
(define b 2351399303373464486466122544523690094744975233415544072992656881240319)
(define m (expt 10 40))
(modular-expt a b m)
- Output:
1527229998585248450016808958343740453059
Raku
(formerly Perl 6) This is specced as a built-in, but here's an explicit version:
sub expmod(Int $a is copy, Int $b is copy, $n) {
my $c = 1;
repeat while $b div= 2 {
($c *= $a) %= $n if $b % 2;
($a *= $a) %= $n;
}
$c;
}
say expmod
2988348162058574136915891421498819466320163312926952423791023078876139,
2351399303373464486466122544523690094744975233415544072992656881240319,
10**40;
- Output:
1527229998585248450016808958343740453059
REXX
/* REXX Modular exponentiation */
say powerMod(,
2988348162058574136915891421498819466320163312926952423791023078876139,,
2351399303373464486466122544523690094744975233415544072992656881240319,,
1e40)
return
powerMod: procedure
parse arg base, exponent, modulus
/* we need a numeric precision of twice the modulus size, */
/* the exponent size, or the base size, whichever is largest */
numeric digits max(2 * length(format(modulus, , , 0)),,
length(format(exponent, , , 0)), length(format(base, , , 0)))
result = 1
base = base // modulus
do while exponent > 0
if exponent // 2 = 1 then
result = result * base // modulus
base = base * base // modulus
exponent = exponent % 2
end
return result
- Output:
1527229998585248450016808958343740453059
RPL
10 40 ^ MODSTO 2988348162058574136915891421498819466320163312926952423791023078876139 2351399303373464486466122544523690094744975233415544072992656881240319 POWMOD
- Output:
1: 1527229998585248450016808958343740453059
Ruby
Built in since version 2.5.
a = 2988348162058574136915891421498819466320163312926952423791023078876139
b = 2351399303373464486466122544523690094744975233415544072992656881240319
m = 10**40
puts a.pow(b, m)
Using OpenSSL standard library
require 'openssl'
a = 2988348162058574136915891421498819466320163312926952423791023078876139
b = 2351399303373464486466122544523690094744975233415544072992656881240319
m = 10 ** 40
puts a.to_bn.mod_exp(b, m)
Written in Ruby
def mod_exp(n, e, mod)
fail ArgumentError, 'negative exponent' if e < 0
prod = 1
base = n % mod
until e.zero?
prod = (prod * base) % mod if e.odd?
e >>= 1
base = (base * base) % mod
end
prod
end
Rust
/* Add this line to the [dependencies] section of your Cargo.toml file:
num = "0.2.0"
*/
use num::bigint::BigInt;
use num::bigint::ToBigInt;
// The modular_exponentiation() function takes three identical types
// (which get cast to BigInt), and returns a BigInt:
fn modular_exponentiation<T: ToBigInt>(n: &T, e: &T, m: &T) -> BigInt {
// Convert n, e, and m to BigInt:
let n = n.to_bigint().unwrap();
let e = e.to_bigint().unwrap();
let m = m.to_bigint().unwrap();
// Sanity check: Verify that the exponent is not negative:
assert!(e >= Zero::zero());
use num::traits::{Zero, One};
// As most modular exponentiations do, return 1 if the exponent is 0:
if e == Zero::zero() {
return One::one()
}
// Now do the modular exponentiation algorithm:
let mut result: BigInt = One::one();
let mut base = n % &m;
let mut exp = e;
// Loop until we can return out result:
loop {
if &exp % 2 == One::one() {
result *= &base;
result %= &m;
}
if exp == One::one() {
return result
}
exp /= 2;
base *= base.clone();
base %= &m;
}
}
Test code:
fn main() {
let (a, b, num_digits) = (
"2988348162058574136915891421498819466320163312926952423791023078876139",
"2351399303373464486466122544523690094744975233415544072992656881240319",
"40",
);
// Covert a, b, and num_digits to numbers:
let a = BigInt::parse_bytes(a.as_bytes(), 10).unwrap();
let b = BigInt::parse_bytes(b.as_bytes(), 10).unwrap();
let num_digits = num_digits.parse().unwrap();
// Calculate m from num_digits:
let m = num::pow::pow(10.to_bigint().unwrap(), num_digits);
// Get the result and print it:
let result = modular_exponentiation(&a, &b, &m);
println!("The last {} digits of\n{}\nto the power of\n{}\nare:\n{}",
num_digits, a, b, result);
}
- Output:
The last 40 digits of 2988348162058574136915891421498819466320163312926952423791023078876139 to the power of 2351399303373464486466122544523690094744975233415544072992656881240319 are: 1527229998585248450016808958343740453059
Scala
import scala.math.BigInt
val a = BigInt(
"2988348162058574136915891421498819466320163312926952423791023078876139")
val b = BigInt(
"2351399303373464486466122544523690094744975233415544072992656881240319")
println(a.modPow(b, BigInt(10).pow(40)))
Scheme
(define (square n)
(* n n))
(define (mod-exp a n mod)
(cond ((= n 0) 1)
((even? n)
(remainder (square (mod-exp a (/ n 2) mod))
mod))
(else (remainder (* a (mod-exp a (- n 1) mod))
mod))))
(define result
(mod-exp 2988348162058574136915891421498819466320163312926952423791023078876139
2351399303373464486466122544523690094744975233415544072992656881240319
(expt 10 40)))
- Output:
> result 1527229998585248450016808958343740453059
Seed7
The library bigint.s7i defines the function modPow, which does modular exponentiation.
$ include "seed7_05.s7i";
include "bigint.s7i";
const proc: main is func
begin
writeln(modPow(2988348162058574136915891421498819466320163312926952423791023078876139_,
2351399303373464486466122544523690094744975233415544072992656881240319_,
10_ ** 40));
end func;- Output:
1527229998585248450016808958343740453059
The library bigint.s7i defines modPow with:
const func bigInteger: modPow (in var bigInteger: base,
in var bigInteger: exponent, in bigInteger: modulus) is func
result
var bigInteger: power is 1_;
begin
if exponent < 0_ or modulus < 0_ then
raise RANGE_ERROR;
else
while exponent > 0_ do
if odd(exponent) then
power := (power * base) mod modulus;
end if;
exponent >>:= 1;
base := base ** 2 mod modulus;
end while;
end if;
end func;Original source: [3]
Sidef
Built-in:
say expmod(
2988348162058574136915891421498819466320163312926952423791023078876139,
2351399303373464486466122544523690094744975233415544072992656881240319,
10**40)
User-defined:
func expmod(a, b, n) {
var c = 1
do {
(c *= a) %= n if b.is_odd
(a *= a) %= n
} while (b //= 2)
c
}
- Output:
1527229998585248450016808958343740453059
Swift
AttaSwift's BigInt has a built-in modPow method, but here we define a generic modPow.
import BigInt
func modPow<T: BinaryInteger>(n: T, e: T, m: T) -> T {
guard e != 0 else {
return 1
}
var res = T(1)
var base = n % m
var exp = e
while true {
if exp & 1 == 1 {
res *= base
res %= m
}
if exp == 1 {
return res
}
exp /= 2
base *= base
base %= m
}
}
let a = BigInt("2988348162058574136915891421498819466320163312926952423791023078876139")
let b = BigInt("2351399303373464486466122544523690094744975233415544072992656881240319")
print(modPow(n: a, e: b, m: BigInt(10).power(40)))
- Output:
1527229998585248450016808958343740453059
Tcl
While Tcl does have arbitrary-precision arithmetic (from 8.5 onwards), it doesn't expose a modular exponentiation function. Thus we implement one ourselves.
Recursive
package require Tcl 8.5
# Algorithm from http://introcs.cs.princeton.edu/java/78crypto/ModExp.java.html
# but Tcl has arbitrary-width integers and an exponentiation operator, which
# helps simplify the code.
proc tcl::mathfunc::modexp {a b n} {
if {$b == 0} {return 1}
set c [expr {modexp($a, $b / 2, $n)**2 % $n}]
if {$b & 1} {
set c [expr {($c * $a) % $n}]
}
return $c
}
Demonstrating:
set a 2988348162058574136915891421498819466320163312926952423791023078876139
set b 2351399303373464486466122544523690094744975233415544072992656881240319
set n [expr {10**40}]
puts [expr {modexp($a,$b,$n)}]
- Output:
1527229998585248450016808958343740453059
Iterative
package require Tcl 8.5
proc modexp {a b n} {
for {set c 1} {$b} {set a [expr {$a*$a % $n}]} {
if {$b & 1} {
set c [expr {$c*$a % $n}]
}
set b [expr {$b >> 1}]
}
return $c
}
Demonstrating:
set a 2988348162058574136915891421498819466320163312926952423791023078876139
set b 2351399303373464486466122544523690094744975233415544072992656881240319
set n [expr {10**40}]
puts [modexp $a $b $n]
- Output:
1527229998585248450016808958343740453059
TXR
From your system prompt:
$ txr -p '(exptmod 2988348162058574136915891421498819466320163312926952423791023078876139
2351399303373464486466122544523690094744975233415544072992656881240319
(expt 10 40)))'
1527229998585248450016808958343740453059
Visual Basic .NET
' Modular exponentiation - VB.Net - 21/01/2019
Imports System.Numerics
Private Sub Main()
Dim a, b, m, x As BigInteger
a = BigInteger.Parse("2988348162058574136915891421498819466320163312926952423791023078876139")
b = BigInteger.Parse("2351399303373464486466122544523690094744975233415544072992656881240319")
m = BigInteger.Pow(10, 40) '=10^40
x = ModPowBig(a, b, m)
Debug.Print("x=" & x.ToString)
End Sub 'Main
Function ModPowBig(ByVal base As BigInteger, ByVal exponent As BigInteger, ByVal modulus As BigInteger) As BigInteger
Dim result As BigInteger
result = 1
Do While exponent > 0
If (exponent Mod 2) = 1 Then
result = (result * base) Mod modulus
End If
exponent = exponent / 2
base = (base * base) Mod modulus
Loop
Return result
End Function 'ModPowBig
- Output:
x=1527229998585248450016808958343740453059
Wren
import "./big" for BigInt
var a = BigInt.new("2988348162058574136915891421498819466320163312926952423791023078876139")
var b = BigInt.new("2351399303373464486466122544523690094744975233415544072992656881240319")
var m = BigInt.ten.pow(40)
System.print(a.modPow(b, m))
- Output:
1527229998585248450016808958343740453059
zkl
Using the GMP big num library:
var BN=Import("zklBigNum");
a:=BN("2988348162058574136915891421498819466320163312926952423791023078876139");
b:=BN("2351399303373464486466122544523690094744975233415544072992656881240319");
m:=BN(10).pow(40);
a.powm(b,m).println();
a.powm(b,m) : "%,d".fmt(_).println();- Output:
1527229998585248450016808958343740453059 1,527,229,998,585,248,450,016,808,958,343,740,453,059
zsh
#!/bin/zsh
#
# I use expr from GNU coreutils.
EXPR=`which expr`
#
# One could use GNU bc or any other such program that is capable of
# handling the large integers.
#
unset PATH # Besides shell commands, the script uses ONLY
# the expression calculator.
mod_power()
{
local base="${1}"
local exponent="${2}"
local modulus="${3}"
local result=1
while [[ `${EXPR} ${exponent} '>' 0` != 0 ]] ; do
if [[ `${EXPR} '(' ${exponent} '%' 2 ')' '=' 1` != 0 ]] ; then
result=`${EXPR} '(' ${result} '*' ${base} ')' '%' ${modulus}`
fi
exponent=`${EXPR} ${exponent} / 2`
base=`${EXPR} '(' ${base} '*' ${base} ')' '%' ${modulus}`
done
echo ${result}
}
a=2988348162058574136915891421498819466320163312926952423791023078876139
b=2351399303373464486466122544523690094744975233415544072992656881240319
# One followed by 40 zeros.
modulus=10000000000000000000000000000000000000000
# Or the following will work, as long as the modulus is at least
# 10**40. :)))))
#modulus=$(mod_power 10 40 1000000000000000000000000000000000000000000000000000000000000)
echo $(mod_power ${a} ${b} ${modulus})
exit 0
- Output:
$ zsh ./modular-exponentiation.zsh 1527229998585248450016808958343740453059
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