Multiplicative order

From Rosetta Code
Task
Multiplicative order
You are encouraged to solve this task according to the task description, using any language you may know.

The multiplicative order of a relative to m is the least positive integer n such that a^n is 1 (modulo m).


Example

The multiplicative order of 37 relative to 1000 is 100 because 37^100 is 1 (modulo 1000), and no number smaller than 100 would do.


One possible algorithm that is efficient also for large numbers is the following: By the Chinese Remainder Theorem, it's enough to calculate the multiplicative order for each prime exponent p^k of m, and combine the results with the least common multiple operation.

Now the order of a with regard to p^k must divide Φ(p^k). Call this number t, and determine it's factors q^e. Since each multiple of the order will also yield 1 when used as exponent for a, it's enough to find the least d such that (q^d)*(t/(q^e)) yields 1 when used as exponent.


Task

Implement a routine to calculate the multiplicative order along these lines. You may assume that routines to determine the factorization into prime powers are available in some library.


An algorithm for the multiplicative order can be found in Bach & Shallit, Algorithmic Number Theory, Volume I: Efficient Algorithms, The MIT Press, 1996:

Exercise 5.8, page 115:

Suppose you are given a prime p and a complete factorization of p-1.   Show how to compute the order of an element a in (Z/(p))* using O((lg p)4/(lg lg p)) bit operations.

Solution, page 337:

Let the prime factorization of p-1 be q1e1q2e2...qkek . We use the following observation: if x^((p-1)/qifi) = 1 (mod p) , and fi=ei or x^((p-1)/qifi+1) != 1 (mod p) , then qiei-fi||ordp x.   (This follows by combining Exercises 5.1 and 2.10.) Hence it suffices to find, for each i , the exponent fi such that the condition above holds.

This can be done as follows: first compute q1e1, q2e2, ... , qkek . This can be done using O((lg p)2) bit operations. Next, compute y1=(p-1)/q1e1, ... , yk=(p-1)/qkek . This can be done using O((lg p)2) bit operations. Now, using the binary method, compute x1=ay1(mod p), ... , xk=ayk(mod p) . This can be done using O(k(lg p)3) bit operations, and k=O((lg p)/(lg lg p)) by Theorem 8.8.10. Finally, for each i , repeatedly raise xi to the qi-th power (mod p) (as many as ei-1 times), checking to see when 1 is obtained. This can be done using O((lg p)3) steps. The total cost is dominated by O(k(lg p)3) , which is O((lg p)4/(lg lg p)).

Ada[edit]

Instead of assuming a library call to factorize the modulus, we assume the caller of our Find_Order function has already factorized it. The Multiplicative_Order package is specified as follows ("multiplicative_order.ads").

package Multiplicative_Order is
 
type Positive_Array is array (Positive range <>) of Positive;
 
function Find_Order(Element, Modulus: Positive) return Positive;
-- naive algorithm
-- returns the smallest I such that (Element**I) mod Modulus = 1
 
function Find_Order(Element: Positive;
Coprime_Factors: Positive_Array) return Positive;
-- faster algorithm for the same task
-- computes the order of all Coprime_Factors(I)
-- and returns their least common multiple
-- this gives the same result as Find_Order(Element, Modulus)
-- with Modulus being the product of all the Coprime_Factors(I)
--
-- preconditions: (1) 1 = GCD(Coprime_Factors(I), Coprime_Factors(J))
-- for all pairs I, J with I /= J
-- (2) 1 < Coprime_Factors(I) for all I
 
end Multiplicative_Order;

Here is the implementation ("multiplicative_order.adb"):

package body Multiplicative_Order is
 
function Find_Order(Element, Modulus: Positive) return Positive is
 
function Power(Exp, Pow, M: Positive) return Positive is
-- computes Exp**Pow mod M;
-- note that Ada's native integer exponentiation "**" may overflow on
-- computing Exp**Pow before ever computing the "mod M" part
Result: Positive := 1;
E: Positive := Exp;
P: Natural := Pow;
begin
while P > 0 loop
if P mod 2 = 1 then
Result := (Result * E) mod M;
end if;
E := (E * E) mod M;
P := P / 2;
end loop;
return Result;
end Power;
 
begin -- Find_Order(Element, Modulus)
for I in 1 .. Modulus loop
if Power(Element, I, Modulus) = 1 then
return Positive(I);
end if;
end loop;
raise Program_Error with
Positive'Image(Element) &" is not coprime to" &Positive'Image(Modulus);
end Find_Order;
 
function Find_Order(Element: Positive;
Coprime_Factors: Positive_Array) return Positive is
 
function GCD (A, B : Positive) return Integer is
M : Natural := A;
N : Natural := B;
T : Natural;
begin
while N /= 0 loop
T := M;
M := N;
N ;:= T mod N;
end loop;
return M;
end GCD; -- from http://rosettacode.org/wiki/Least_common_multiple#Ada
 
function LCM (A, B : Natural) return Integer is
begin
if A = 0 or B = 0 then
return 0;
end if;
return abs (A * B) / Gcd (A, B);
end LCM; -- from http://rosettacode.org/wiki/Least_common_multiple#Ada
 
Result : Positive := 1;
 
begin -- Find_Order(Element, Coprime_Factors)
for I in Coprime_Factors'Range loop
Result := LCM(Result, Find_Order(Element, Coprime_Factors(I)));
end loop;
return Result;
end Find_Order;
 
end Multiplicative_Order;

This is a sample program using the Multiplicative_Order package:

with Ada.Text_IO, Multiplicative_Order;
 
procedure Main is
package IIO is new Ada.Text_IO.Integer_IO(Integer);
use Multiplicative_Order;
begin
IIO.Put(Find_Order(3,10));
IIO.Put(Find_Order(37,1000));
IIO.Put(Find_Order(37,10_000));
IIO.Put(Find_Order(37, 3343));
IIO.Put(Find_Order(37, 3344));
-- IIO.Put(Find_Order( 2,1000));
--would raise Program_Error, because there is no I with 2**I=1 mod 1000
Ada.Text_IO.New_Line;
IIO.Put(Find_Order(3, (2,5))); -- 3 * 5 = 10
IIO.Put(Find_Order(37, (8, 125))); -- 8 * 125 = 1000
IIO.Put(Find_Order(37, (16, 625))); -- 16 * 625 = 10_000
IIO.Put(Find_Order(37, (1 => 3343))); -- 1-element-array: 3343 is a prime
IIO.Put(Find_Order(37, (11, 19, 16))); -- 11 * 19 * 16 = 3344
 
-- this violates the precondition, because 8 and 2 are not coprime
-- it gives an incorrect result
IIO.Put(Find_Order(37, (11, 19, 8, 2)));
end Main;

The output from the sample program:

          4        100        500       1114         20
          4        100        500       1114         20         10

ALGOL 68[edit]

Translation of: python
Works with: ALGOL 68 version Standard - with preludes manually inserted
Works with: ALGOL 68G version Any - tested with release mk15-0.8b.fc9.i386
MODE LOOPINT = INT;
 
MODE POWMODSTRUCT = LONG INT;
PR READ "prelude/pow_mod.a68" PR;
 
MODE SORTSTRUCT = LONG INT;
PR READ "prelude/sort.a68" PR;
 
MODE GCDSTRUCT = LONG INT;
PR READ "prelude/gcd.a68" PR;
 
PR READ "prelude/iterator.a68" PR;
 
PROC is prime = (LONG INT p)BOOL:
( p > 1 |#ANDF# ALL((YIELDBOOL yield)VOID: factored(p, (LONG INT f, LONG INT e)VOID: yield(f = p))) | FALSE );
 
FLEX[4]LONG INT prime list := (2,3,5,7);
 
OP +:= = (REF FLEX[]LONG INT lhs, LONG INT rhs)VOID: (
[UPB lhs +1] LONG INT next lhs;
next lhs[:UPB lhs] := lhs;
lhs := next lhs;
lhs[UPB lhs] := rhs
);
 
PROC primes = (PROC (LONG INT)VOID yield)VOID: (
LONG INT p;
FOR p index TO UPB prime list DO
p:= prime list[p index];
yield(p)
OD;
DO
p +:= 2;
WHILE NOT is prime(p) DO
p +:= 2
OD;
prime list +:= p;
yield(p)
OD
);
 
PROC factored = (LONG INT in a, PROC (LONG INT,LONG INT)VOID yield)VOID: (
LONG INT a := in a;
# FOR p IN # primes( # DO #
(LONG INT p)VOID:(
LONG INT j := 0;
WHILE a MOD p = 0 DO
a := a % p;
j +:= 1
OD;
IF j > 0 THEN yield (p,j) FI;
IF a < p*p THEN done FI
)
# ) OD # );
done:
IF a > 1 THEN yield (a,1) FI
);
 
PROC mult0rdr1 = (LONG INT a, p, e)LONG INT: (
LONG INT m := p ** SHORTEN e;
LONG INT t := (p-1)*(p**SHORTEN (e-1)); # = Phi(p**e) where p prime #
LONG INT q;
FLEX[0]LONG INT qs := (1);
# FOR f0,f1 IN # factored(t # DO #,
(LONG INT f0,f1)VOID: (
FLEX[SHORTEN((f1+1)*UPB qs)]LONG INT next qs;
FOR j TO SHORTEN f1 + 1 DO
FOR q index TO UPB qs DO
q := qs[q index];
next qs[(j-1)*UPB qs+q index] := q * f0**(j-1)
OD
OD;
qs := next qs
)
# OD # );
VOID(in place shell sort(qs));
 
FOR q index TO UPB qs DO
q := qs[q index];
IF pow mod(a,q,m)=1 THEN done FI
OD;
done:
q
);
 
PROC reduce = (PROC (LONG INT,LONG INT)LONG INT diadic, FORLONGINT iterator, LONG INT initial value)LONG INT: (
LONG INT out := initial value;
# FOR next IN # iterator( # DO #
(LONG INT next)VOID:
out := diadic(out, next)
# OD # );
out
);
 
PROC mult order = (LONG INT a, LONG INT m)LONG INT: (
PROC mofs = (YIELDLONGINT yield)VOID:(
# FOR p, count IN # factored(m, # DO #
(LONG INT p, LONG INT count)VOID:
yield(mult0rdr1(a,p,count))
)
# OD # );
reduce(lcm, mofs, 1)
);
 
main:(
FORMAT d = $g(-0)$;
printf((d, mult order(37, 1000), $l$)); # 100 #
LONG INT b := LENG 10**20-1;
printf((d, mult order(2, b), $l$)); # 3748806900 #
printf((d, mult order(17,b), $l$)); # 1499522760 #
b := 100001;
printf((d, mult order(54,b), $l$));
printf((d, pow mod( 54, mult order(54,b),b), $l$));
IF ANY( (YIELDBOOL yield)VOID: FOR r FROM 2 TO SHORTEN mult order(54,b)-1 DO yield(1=pow mod(54,r, b)) OD )
THEN
printf(($g$, "Exists a power r < 9090 where pow mod(54,r,b) = 1", $l$))
ELSE
printf(($g$, "Everything checks.", $l$))
FI
)

Output:

100
3748806900
1499522760
9090
1
Everything checks.

C[edit]

Uses prime/factor functions from Factors of an integer#Prime factoring. This implementation is not robust because of integer overflows. To properly deal with even moderately large numbers, an arbitrary precision integer package is a must.

ulong mpow(ulong a, ulong p, ulong m)
{
ulong r = 1;
while (p) {
if ((1 & p)) r = r * a % m;
a = a * a % m;
p >>= 1;
}
return r;
}
 
ulong ipow(ulong a, ulong p) {
ulong r = 1;
while (p) {
if ((1 & p)) r = r * a;
a *= a;
p >>= 1;
}
return r;
}
 
ulong gcd(ulong m, ulong n)
{
ulong t;
while (m) { t = m; m = n % m; n = t; }
return n;
}
 
ulong lcm(ulong m, ulong n)
{
ulong g = gcd(m, n);
return m / g * n;
}
 
ulong multi_order_p(ulong a, ulong p, ulong e)
{
ulong fac[10000];
ulong m = ipow(p, e);
ulong t = m / p * (p - 1);
int i, len = get_factors(t, fac);
for (i = 0; i < len; i++)
if (mpow(a, fac[i], m) == 1)
return fac[i];
return 0;
}
 
ulong multi_order(ulong a, ulong m)
{
prime_factor pf[100];
int i, len = get_prime_factors(m, pf);
ulong res = 1;
for (i = 0; i < len; i++)
res = lcm(res, multi_order_p(a, pf[i].p, pf[i].e));
return res;
}
 
int main()
{
sieve();
printf("%lu\n", multi_order(37, 1000));
printf("%lu\n", multi_order(54, 100001));
return 0;
}

C#[edit]

Translation of: Java
using System;
using System.Collections.Generic;
using System.Numerics;
using System.Threading;
 
namespace MultiplicativeOrder {
// Taken from https://stackoverflow.com/a/33918233
public static class PrimeExtensions {
// Random generator (thread safe)
private static ThreadLocal<Random> s_Gen = new ThreadLocal<Random>(
() => {
return new Random();
}
);
 
// Random generator (thread safe)
private static Random Gen {
get {
return s_Gen.Value;
}
}
 
public static bool IsProbablyPrime(this BigInteger value, int witnesses = 10) {
if (value <= 1)
return false;
 
if (witnesses <= 0)
witnesses = 10;
 
BigInteger d = value - 1;
int s = 0;
 
while (d % 2 == 0) {
d /= 2;
s += 1;
}
 
byte[] bytes = new byte[value.ToByteArray().LongLength];
BigInteger a;
 
for (int i = 0; i < witnesses; i++) {
do {
Gen.NextBytes(bytes);
 
a = new BigInteger(bytes);
}
while (a < 2 || a >= value - 2);
 
BigInteger x = BigInteger.ModPow(a, d, value);
if (x == 1 || x == value - 1)
continue;
 
for (int r = 1; r < s; r++) {
x = BigInteger.ModPow(x, 2, value);
 
if (x == 1)
return false;
if (x == value - 1)
break;
}
 
if (x != value - 1)
return false;
}
 
return true;
}
}
 
static class Helper {
public static BigInteger Sqrt(this BigInteger self) {
BigInteger b = self;
while (true) {
BigInteger a = b;
b = self / a + a >> 1;
if (b >= a) return a;
}
}
 
public static long BitLength(this BigInteger self) {
BigInteger bi = self;
long bitlength = 0;
while (bi != 0) {
bitlength++;
bi >>= 1;
}
return bitlength;
}
 
public static bool BitTest(this BigInteger self, int pos) {
byte[] arr = self.ToByteArray();
int idx = pos / 8;
int mod = pos % 8;
if (idx >= arr.Length) {
return false;
}
return (arr[idx] & (1 << mod)) > 0;
}
}
 
class PExp {
public PExp(BigInteger prime, int exp) {
Prime = prime;
Exp = exp;
}
 
public BigInteger Prime { get; }
 
public int Exp { get; }
}
 
class Program {
static void MoTest(BigInteger a, BigInteger n) {
if (!n.IsProbablyPrime(20)) {
Console.WriteLine("Not computed. Modulus must be prime for this algorithm.");
return;
}
if (a.BitLength() < 100) {
Console.Write("ord({0})", a);
} else {
Console.Write("ord([big])");
}
if (n.BitLength() < 100) {
Console.Write(" mod {0} ", n);
} else {
Console.Write(" mod [big] ");
}
BigInteger mob = MoBachShallit58(a, n, Factor(n - 1));
Console.WriteLine("= {0}", mob);
}
 
static BigInteger MoBachShallit58(BigInteger a, BigInteger n, List<PExp> pf) {
BigInteger n1 = n - 1;
BigInteger mo = 1;
foreach (PExp pe in pf) {
BigInteger y = n1 / BigInteger.Pow(pe.Prime, pe.Exp);
int o = 0;
BigInteger x = BigInteger.ModPow(a, y, BigInteger.Abs(n));
while (x > 1) {
x = BigInteger.ModPow(x, pe.Prime, BigInteger.Abs(n));
o++;
}
BigInteger o1 = BigInteger.Pow(pe.Prime, o);
o1 = o1 / BigInteger.GreatestCommonDivisor(mo, o1);
mo = mo * o1;
}
return mo;
}
 
static List<PExp> Factor(BigInteger n) {
List<PExp> pf = new List<PExp>();
BigInteger nn = n;
int e = 0;
while (!nn.BitTest(e)) e++;
if (e > 0) {
nn = nn >> e;
pf.Add(new PExp(2, e));
}
BigInteger s = nn.Sqrt();
BigInteger d = 3;
while (nn > 1) {
if (d > s) d = nn;
e = 0;
while (true) {
BigInteger div = BigInteger.DivRem(nn, d, out BigInteger rem);
if (rem.BitLength() > 0) break;
nn = div;
e++;
}
if (e > 0) {
pf.Add(new PExp(d, e));
s = nn.Sqrt();
}
d = d + 2;
}
 
return pf;
}
 
static void Main(string[] args) {
MoTest(37, 3343);
MoTest(BigInteger.Pow(10, 100) + 1, 7919);
MoTest(BigInteger.Pow(10, 1000) + 1, 15485863);
MoTest(BigInteger.Pow(10, 10000) - 1, 22801763489);
MoTest(1511678068, 7379191741);
MoTest(3047753288, 2257683301);
}
}
}
Output:
ord(37) mod 3343 = 1114
ord([big]) mod 7919 = 3959
ord([big]) mod 15485863 = 15485862
ord([big]) mod 22801763489 = 22801763488
ord(1511678068) mod 7379191741 = 614932645
ord(3047753288) mod 2257683301 = 62713425

D[edit]

Translation of: Java
import std.bigint;
import std.random;
import std.stdio;
 
struct PExp {
BigInt prime;
int exp;
}
 
BigInt gcd(BigInt x, BigInt y) {
if (y == 0) {
return x;
}
return gcd(y, x % y);
}
 
/// https://en.wikipedia.org/wiki/Modular_exponentiation#Right-to-left_binary_method
BigInt modPow(BigInt b, BigInt e, BigInt n) {
if (n == 1) return BigInt(0);
BigInt result = 1;
b = b % n;
while (e > 0) {
if (e % 2 == 1) {
result = (result * b) % n;
}
e >>= 1;
b = (b*b) % n;
}
return result;
}
 
BigInt pow(long b, long e) {
return pow(BigInt(b), BigInt(e));
}
BigInt pow(BigInt b, BigInt e) {
if (e == 0) {
return BigInt(1);
}
 
BigInt result = 1;
while (e > 1) {
if (e % 2 == 0) {
b *= b;
e /= 2;
} else {
result *= b;
b *= b;
e = (e - 1) / 2;
}
}
 
return b * result;
}
 
BigInt sqrt(BigInt self) {
BigInt b = self;
while (true) {
BigInt a = b;
b = self / a + a >> 1;
if (b >= a) return a;
}
}
 
long bitLength(BigInt self) {
BigInt bi = self;
long length;
while (bi != 0) {
length++;
bi >>= 1;
}
return length;
}
 
PExp[] factor(BigInt n) {
PExp[] pf;
BigInt nn = n;
int b = 0;
int e = 1;
while ((nn & e) == 0) {
e <<= 1;
b++;
}
if (b > 0) {
nn = nn >> b;
pf ~= PExp(BigInt(2), b);
}
BigInt s = nn.sqrt();
BigInt d = 3;
while (nn > 1) {
if (d > s) d = nn;
e = 0;
while (true) {
BigInt div, rem;
nn.divMod(d, div, rem);
if (rem.bitLength > 0) break;
nn = div;
e++;
}
if (e > 0) {
pf ~= PExp(d, e);
s = nn.sqrt();
}
d += 2;
}
 
return pf;
}
 
BigInt moBachShallit58(BigInt a, BigInt n, PExp[] pf) {
BigInt n1 = n - 1;
BigInt mo = 1;
foreach(pe; pf) {
BigInt y = n1 / pe.prime.pow(BigInt(pe.exp));
int o = 0;
BigInt x = a.modPow(y, n);
while (x > 1) {
x = x.modPow(pe.prime, n);
o++;
}
BigInt o1 = pe.prime.pow(BigInt(o));
o1 = o1 / gcd(mo, o1);
mo = mo * o1;
}
return mo;
}
 
void moTest(ulong a, ulong n) {
moTest(BigInt(a), n);
}
void moTest(BigInt a, ulong n) {
// Commented out because the implementations tried all failed for the -2 and -3 tests.
// if (!n.isProbablePrime()) {
// writeln("Not computed. Modulus must be prime for this algorithm.");
// return;
// }
if (a.bitLength < 100) {
write("ord(", a, ")");
} else {
write("ord([big])");
}
write(" mod ", n, " ");
BigInt nn = n;
BigInt mob = moBachShallit58(a, nn, factor(nn - 1));
writeln("= ", mob);
}
 
void main() {
moTest(37, 3343);
 
moTest(pow(10, 100) + 1, 7919);
moTest(pow(10, 1000) + 1, 15485863);
moTest(pow(10, 10000) - 1, 22801763489);
 
moTest(1511678068, 7379191741);
moTest(3047753288, 2257683301);
}
Output:
ord(37) mod 3343 = 1114
ord([big]) mod 7919 = 3959
ord([big]) mod 15485863 = 15485862
ord([big]) mod 22801763489 = 22801763488
ord(1511678068) mod 7379191741 = 614932645
ord(3047753288) mod 2257683301 = 62713425

EchoLisp[edit]

 
(require 'bigint)
 
;; factor-exp returns a list ((p k) ..) : a = p1^k1 * p2^k2 ..
(define (factor-exp a)
(map (lambda (g) (list (first g) (length g)))
(group* (prime-factors a))))
 
;; copied from Ruby
(define (_mult_order a p k (x))
(define pk (expt p k))
(define t (* (1- p) (expt p (1- k))))
(define r 1)
(for [((q e) (factor-exp t))]
(set! x (powmod a (/ t (expt q e)) pk))
(while (!= x 1)
(*= r q)
(set! x (powmod x q pk))))
r)
 
(define (order a m)
"multiplicative order : (order a m) → n : a^n = 1 (mod m)"
(assert (= 1 (gcd a m)) "a and m must be coprimes")
(define mopks (for/list [((p k) (factor-exp m))] (_mult_order a p k)))
(for/fold (n 1) ((mopk mopks)) (lcm n mopk)))
 
;; results
order 37 1000)
100
(order (+ (expt 10 100) 1) 7919)
3959
(order (+ (expt 10 1000) 1) 15485863)
15485862
 

Clojure[edit]

Translation of Julie, then revised to be more clojure idiomatic. It references some external modules for factoring and integer exponentiation.

(defn gcd [a b]
(if (zero? b)
a
(recur b (mod a b))))
 
(defn lcm [a b]
(/ (* a b) (gcd a b)))
 
(def NaN (Math/log -1))
 
(defn ord' [a [p e]]
(let [m (imath/expt p e)
t (* (quot m p) (dec p))]
(loop [dv (factor/divisors t)]
(let [d (first dv)]
(if (= (mmath/expm a d m) 1)
d
(recur (next dv)))))))
 
(defn ord [a n]
(if (not= (gcd a n) 1)
NaN
(->>
(factor/factorize n)
(map (partial ord' a))
(reduce lcm))))
 
Output:
user=> (ord 37 1000)
100

Go[edit]

package main
 
import (
"fmt"
"math/big"
)
 
func main() {
moTest(big.NewInt(37), big.NewInt(3343))
b := big.NewInt(100)
moTest(b.Add(b.Exp(ten, b, nil), one), big.NewInt(7919))
moTest(b.Add(b.Exp(ten, b.SetInt64(1000), nil), one), big.NewInt(15485863))
moTest(b.Sub(b.Exp(ten, b.SetInt64(10000), nil), one),
big.NewInt(22801763489))
 
moTest(big.NewInt(1511678068), big.NewInt(7379191741))
moTest(big.NewInt(3047753288), big.NewInt(2257683301))
}
 
func moTest(a, n *big.Int) {
if a.BitLen() < 100 {
fmt.Printf("ord(%v)", a)
} else {
fmt.Print("ord([big])")
}
if n.BitLen() < 100 {
fmt.Printf(" mod %v ", n)
} else {
fmt.Print(" mod [big] ")
}
if !n.ProbablyPrime(20) {
fmt.Println("not computed. modulus must be prime for this algorithm.")
return
}
fmt.Println("=", moBachShallit58(a, n, factor(new(big.Int).Sub(n, one))))
}
 
var one = big.NewInt(1)
var two = big.NewInt(2)
var ten = big.NewInt(10)
 
func moBachShallit58(a, n *big.Int, pf []pExp) *big.Int {
n1 := new(big.Int).Sub(n, one)
var x, y, o1, g big.Int
mo := big.NewInt(1)
for _, pe := range pf {
y.Quo(n1, y.Exp(pe.prime, big.NewInt(pe.exp), nil))
var o int64
for x.Exp(a, &y, n); x.Cmp(one) > 0; o++ {
x.Exp(&x, pe.prime, n)
}
o1.Exp(pe.prime, o1.SetInt64(o), nil)
mo.Mul(mo, o1.Quo(&o1, g.GCD(nil, nil, mo, &o1)))
}
return mo
}
 
type pExp struct {
prime *big.Int
exp int64
}
 
func factor(n *big.Int) (pf []pExp) {
var e int64
for ; n.Bit(int(e)) == 0; e++ {
}
if e > 0 {
n.Rsh(n, uint(e))
pf = []pExp{{big.NewInt(2), e}}
}
s := sqrt(n)
q, r := new(big.Int), new(big.Int)
for d := big.NewInt(3); n.Cmp(one) > 0; d.Add(d, two) {
if d.Cmp(s) > 0 {
d.Set(n)
}
for e = 0; ; e++ {
q.QuoRem(n, d, r)
if r.BitLen() > 0 {
break
}
n.Set(q)
}
if e > 0 {
pf = append(pf, pExp{new(big.Int).Set(d), e})
s = sqrt(n)
}
}
return
}
 
func sqrt(n *big.Int) *big.Int {
a := new(big.Int)
for b := new(big.Int).Set(n); ; {
a.Set(b)
b.Rsh(b.Add(b.Quo(n, a), a), 1)
if b.Cmp(a) >= 0 {
return a
}
}
return a.SetInt64(0)
}
Output:
ord(37) mod 3343 = 1114
ord([big]) mod 7919 = 3959
ord([big]) mod 15485863 = 15485862
ord([big]) mod 22801763489 = 22801763488
ord(1511678068) mod 7379191741 = 614932645
ord(3047753288) mod 2257683301 = 62713425

Haskell[edit]

Assuming a function

powerMod
:: (Integral a, Integral b)
=> a -> a -> b -> a
powerMod m _ 0 = 1
powerMod m x n
| n > 0 = f x_ (n - 1) x_
where
x_ = x `rem` m
f _ 0 y = y
f a d y = g a d
where
g b i
| even i = g (b * b `rem` m) (i `quot` 2)
| otherwise = f b (i - 1) (b * y `rem` m)
powerMod m _ _ = error "powerMod: negative exponent"

to efficiently calculate powers modulo some Integral, we get

import Data.List (foldl1') --'
 
foldl1_ = foldl1' --'
 
multOrder a m
| gcd a m /= 1 = error "Arguments not coprime"
| otherwise = foldl1_ lcm $ map (multOrder_ a) $ primeFacsExp m
 
multOrder_ a (p, k) = r
where
pk = p ^ k
t = (p - 1) * p ^ (k - 1) -- totient \Phi(p^k)
r = product $ map find_qd $ primeFacsExp t
find_qd (q, e) = q ^ d
where
x = powerMod pk a (t `div` (q ^ e))
d = length $ takeWhile (/= 1) $ iterate (\y -> powerMod pk y q) x

J[edit]

The dyadic verb mo converts its arguments to exact numbers a and m, executes mopk on the factorization of m, and combines the result with the least common multiple operation.

mo=: 4 : 0
a=. x: x
m=. x: y
assert. 1=a+.m
*./ a mopk"1 |: __ q: m
)

The dyadic verb mopk expects a pair of prime and exponent in the second argument. It sets up a verb pm to calculate powers module p^k. Then calculates Φ(p^k) as t, factorizes it again into q and e, and calculates a^(t/(q^e)) as x. Now, it finds the least d such that subsequent application of pm yields 1. Finally, it combines the exponents q^d into a product.

mopk=: 4 : 0
a=. x: x
'p k'=. x: y
pm=. (p^k)&|@^
t=. (p-1)*p^k-1 NB. totient
'q e'=. __ q: t
x=. a pm t%q^e
d=. (1<x)+x (pm i. 1:)&> (e-1) */\@$&.> q
*/q^d
)

For example:

   37 mo 1000
100
2 mo _1+10^80x
190174169488577769580266953193403101748804183400400

Java[edit]

Translation of: Kotlin
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.List;
 
public class MultiplicativeOrder {
private static final BigInteger ONE = BigInteger.ONE;
private static final BigInteger TWO = BigInteger.valueOf(2);
private static final BigInteger THREE = BigInteger.valueOf(3);
private static final BigInteger TEN = BigInteger.TEN;
 
private static class PExp {
BigInteger prime;
long exp;
 
PExp(BigInteger prime, long exp) {
this.prime = prime;
this.exp = exp;
}
}
 
private static void moTest(BigInteger a, BigInteger n) {
if (!n.isProbablePrime(20)) {
System.out.println("Not computed. Modulus must be prime for this algorithm.");
return;
}
if (a.bitLength() < 100) System.out.printf("ord(%s)", a);
else System.out.print("ord([big])");
if (n.bitLength() < 100) System.out.printf(" mod %s ", n);
else System.out.print(" mod [big] ");
BigInteger mob = moBachShallit58(a, n, factor(n.subtract(ONE)));
System.out.println("= " + mob);
}
 
private static BigInteger moBachShallit58(BigInteger a, BigInteger n, List<PExp> pf) {
BigInteger n1 = n.subtract(ONE);
BigInteger mo = ONE;
for (PExp pe : pf) {
BigInteger y = n1.divide(pe.prime.pow((int) pe.exp));
long o = 0;
BigInteger x = a.modPow(y, n.abs());
while (x.compareTo(ONE) > 0) {
x = x.modPow(pe.prime, n.abs());
o++;
}
BigInteger o1 = BigInteger.valueOf(o);
o1 = pe.prime.pow(o1.intValue());
o1 = o1.divide(mo.gcd(o1));
mo = mo.multiply(o1);
}
return mo;
}
 
private static List<PExp> factor(BigInteger n) {
List<PExp> pf = new ArrayList<>();
BigInteger nn = n;
Long e = 0L;
while (!nn.testBit(e.intValue())) e++;
if (e > 0L) {
nn = nn.shiftRight(e.intValue());
pf.add(new PExp(TWO, e));
}
BigInteger s = sqrt(nn);
BigInteger d = THREE;
while (nn.compareTo(ONE) > 0) {
if (d.compareTo(s) > 0) d = nn;
e = 0L;
while (true) {
BigInteger[] qr = nn.divideAndRemainder(d);
if (qr[1].bitLength() > 0) break;
nn = qr[0];
e++;
}
if (e > 0L) {
pf.add(new PExp(d, e));
s = sqrt(nn);
}
d = d.add(TWO);
}
return pf;
}
 
private static BigInteger sqrt(BigInteger n) {
BigInteger b = n;
while (true) {
BigInteger a = b;
b = n.divide(a).add(a).shiftRight(1);
if (b.compareTo(a) >= 0) return a;
}
}
 
public static void main(String[] args) {
moTest(BigInteger.valueOf(37), BigInteger.valueOf(3343));
 
BigInteger b = TEN.pow(100).add(ONE);
moTest(b, BigInteger.valueOf(7919));
 
b = TEN.pow(1000).add(ONE);
moTest(b, BigInteger.valueOf(15485863));
 
b = TEN.pow(10000).subtract(ONE);
moTest(b, BigInteger.valueOf(22801763489L));
 
moTest(BigInteger.valueOf(1511678068), BigInteger.valueOf(7379191741L));
moTest(BigInteger.valueOf(3047753288L), BigInteger.valueOf(2257683301L));
}
}
Output:
ord(37) mod 3343 = 1114
ord([big]) mod 7919 = 3959
ord([big]) mod 15485863 = 15485862
ord([big]) mod 22801763489 = 22801763488
ord(1511678068) mod 7379191741 = 614932645
ord(3047753288) mod 2257683301 = 62713425

Julia[edit]

(Uses the factors function from Factors of an integer#Julia.)

using Primes
 
function factors(n)
f = [one(n)]
for (p,e) in factor(n)
f = reduce(vcat, [f*p^j for j in 1:e], init=f)
end
return length(f) == 1 ? [one(n), n] : sort!(f)
end
 
function multorder(a, m)
gcd(a,m) == 1 || error("$a and $m are not coprime")
res = one(m)
for (p,e) in factor(m)
m = p^e
t = div(m, p) * (p-1)
for f in factors(t)
if powermod(a, f, m) == 1
res = lcm(res, f)
break
end
end
end
res
end

Example output (using big to employ arbitrary-precision arithmetic where needed):

julia> multorder(37, 1000)
100

julia> multorder(big(10)^100 + 1, 7919)
3959

julia> multorder(big(10)^1000 + 1, 15485863)
15485862

julia> multorder(big(10)^10000 - 1, 22801763489)
22801763488

Kotlin[edit]

Translation of: Go
// version 1.2.10
 
import java.math.BigInteger
 
val bigOne = BigInteger.ONE
val bigTwo = 2.toBigInteger()
val bigThree = 3.toBigInteger()
val bigTen = BigInteger.TEN
 
class PExp(val prime: BigInteger, val exp: Long)
 
fun moTest(a: BigInteger, n: BigInteger) {
if (!n.isProbablePrime(20)) {
println("Not computed. Modulus must be prime for this algorithm.")
return
}
if (a.bitLength() < 100) print("ord($a)") else print("ord([big])")
if (n.bitLength() < 100) print(" mod $n ") else print(" mod [big] ")
val mob = moBachShallit58(a, n, factor(n - bigOne))
println("= $mob")
}
 
fun moBachShallit58(a: BigInteger, n: BigInteger, pf: List<PExp>): BigInteger {
val n1 = n - bigOne
var mo = bigOne
for (pe in pf) {
val y = n1 / pe.prime.pow(pe.exp.toInt())
var o = 0L
var x = a.modPow(y, n.abs())
while (x > bigOne) {
x = x.modPow(pe.prime, n.abs())
o++
}
var o1 = o.toBigInteger()
o1 = pe.prime.pow(o1.toInt())
o1 /= mo.gcd(o1)
mo *= o1
}
return mo
}
 
fun factor(n: BigInteger): List<PExp> {
val pf = mutableListOf<PExp>()
var nn = n
var e = 0L
while (!nn.testBit(e.toInt())) e++
if (e > 0L) {
nn = nn shr e.toInt()
pf.add(PExp(bigTwo, e))
}
var s = bigSqrt(nn)
var d = bigThree
while (nn > bigOne) {
if (d > s) d = nn
e = 0L
while (true) {
val (q, r) = nn.divideAndRemainder(d)
if (r.bitLength() > 0) break
nn = q
e++
}
if (e > 0L) {
pf.add(PExp(d, e))
s = bigSqrt(nn)
}
d += bigTwo
}
return pf
}
 
fun bigSqrt(n: BigInteger): BigInteger {
var b = n
while (true) {
val a = b
b = (n / a + a) shr 1
if (b >= a) return a
}
}
 
fun main(args: Array<String>) {
moTest(37.toBigInteger(), 3343.toBigInteger())
 
var b = bigTen.pow(100) + bigOne
moTest(b, 7919.toBigInteger())
 
b = bigTen.pow(1000) + bigOne
moTest(b, BigInteger("15485863"))
 
b = bigTen.pow(10000) - bigOne
moTest(b, BigInteger("22801763489"))
 
moTest(BigInteger("1511678068"), BigInteger("7379191741"))
moTest(BigInteger("3047753288"), BigInteger("2257683301"))
}
Output:
ord(37) mod 3343 = 1114
ord([big]) mod 7919 = 3959
ord([big]) mod 15485863 = 15485862
ord([big]) mod 22801763489 = 22801763488
ord(1511678068) mod 7379191741 = 614932645
ord(3047753288) mod 2257683301 = 62713425

Maple[edit]

numtheory:-order( a, n )

For example,

> numtheory:-order( 37, 1000 );
100

Mathematica[edit]

In Mathematica this is really easy, as this function is built-in: MultiplicativeOrder[k,n] gives the multiplicative order of k modulo n, defined as the smallest integer m such that k^m == 1 mod n.
MultiplicativeOrder[k,n,{r1,r2,...}] gives the generalized multiplicative order of k modulo n, defined as the smallest integer m such that k^m==ri mod n for some i.
Examples:

MultiplicativeOrder[37, 1000]
MultiplicativeOrder[10^100 + 1, 7919] (*10^3th prime number Prime[1000]*)
MultiplicativeOrder[10^1000 + 1, 15485863] (*10^6th prime number*)
MultiplicativeOrder[10^10000 - 1, 22801763489] (*10^9th prime number*)
MultiplicativeOrder[13, 1 + 10^80]
MultiplicativeOrder[11, 1 + 10^100]

gives back:

100
3959
15485862
22801763488
109609547199756140150989321269669269476675495992554276140800
2583496112724752500580158969425549088007844580826869433740066152289289764829816356800

Maxima[edit]

zn_order(37, 1000);
/* 100 */
 
zn_order(10^100 + 1, 7919);
/* 3959 */
 
zn_order(10^1000 + 1, 15485863);
/* 15485862 */
 
zn_order(10^10000 - 1, 22801763489);
/* 22801763488 */
 
zn_order(13, 1 + 10^80);
/* 109609547199756140150989321269669269476675495992554276140800 */
 
zn_order(11, 1 + 10^100);
/* 2583496112724752500580158969425549088007844580826869433740066152289289764829816356800 */

PARI/GP[edit]

znorder(Mod(a,n))

Perl[edit]

Using modules:

Library: ntheory
use ntheory qw/znorder/;
say znorder(54, 100001);
use bigint; say znorder(11, 1 + 10**100);

or

use Math::Pari qw/znorder Mod/;
say znorder(Mod(54, 100001));
say znorder(Mod(11, 1 + Math::Pari::PARI(10)**100));

Perl 6[edit]

my @primes := 2, |grep *.is-prime, (3,5,7...*);
 
sub factor($a is copy) {
gather {
for @primes -> $p {
my $j = 0;
while $a %% $p {
$a div= $p;
$j++;
}
take $p => $j if $j > 0;
last if $a < $p * $p;
}
 
take $a => 1 if $a > 1;
}
}
 
sub mo-prime($a, $p, $e) {
my $m = $p ** $e;
my $t = ($p - 1) * ($p ** ($e - 1)); # = Phi($p**$e) where $p prime
my @qs = 1;
for factor($t) -> $f {
@qs = flat @qs.map(-> $q { (0..$f.value).map(-> $j { $q * $f.key ** $j }) });
}
 
@qs.sort.first(-> $q { expmod( $a, $q, $m ) == 1});
}
 
sub mo($a, $m) {
$a gcd $m == 1 || die "$a and $m are not relatively prime";
[lcm] flat 1, factor($m).map(-> $r { mo-prime($a, $r.key, $r.value) });
}
 
sub MAIN("test") {
use Test;
 
for (10, 21, 25, 150, 1231, 123141, 34131) -> $n {
is ([*] factor($n).map(-> $pair { $pair.key ** $pair.value })), $n, "$n factors correctly";
}
 
is mo(37, 1000), 100, 'mo(37,1000) == 100';
my $b = 10**20-1;
is mo(2, $b), 3748806900, 'mo(2,10**20-1) == 3748806900';
is mo(17, $b), 1499522760, 'mo(17,10**20-1) == 1499522760';
$b = 100001;
is mo(54, $b), 9090, 'mo(54,100001) == 9090';
}
Output:
ok 1 - 10 factors correctly
ok 2 - 21 factors correctly
ok 3 - 25 factors correctly
ok 4 - 150 factors correctly
ok 5 - 1231 factors correctly
ok 6 - 123141 factors correctly
ok 7 - 34131 factors correctly
ok 8 - mo(37,1000) == 100
ok 9 - mo(2,10**20-1) == 3748806900
ok 10 - mo(17,10**20-1) == 1499522760
ok 11 - mo(54,100001) == 9090

Phix[edit]

Translation of: Ruby

Note this is considerably slower than Ruby, which completes in a fraction of a second, mainly I suspect due to the fact that bigatom.e operates on a digit-by-digit basis.

Also, ba_mod_exp is fairly likely to get added to bigatom.e in a future release, likewise the other ba_xxx routines below have been earmarked for possible inclusion.

include bigatom.e
 
function ba_mod_exp(object base, exponent, modulus)
-- base/exponent/modulus can be integer/string/bigatom
-- returns mod(power(base,exponent),modulus) [aka b^e%m], but in bigatoms and faster.
bigatom res = BA_ONE
base = ba_mod(base,modulus)
while ba_compare(exponent,0)!=0 do
if ba_mod(exponent,2)=BA_ONE then
res = ba_mod(ba_multiply(res,base),modulus)
end if
base = ba_mod(ba_multiply(base,base),modulus)
exponent = ba_idivide(exponent,2)
end while
return res
end function
 
function ba_factor(object n)
-- eg ba_factor(1000) -> {{2,3},{5,3}}, ie power(2,3)*power(5,3) == 8*125 == 1000.
-- (note that each res[i] is {bigatom,integer})
if ba_compare(n,BA_ZERO)=0 then return {} end if
sequence pf = {}
integer e = 0
while ba_mod(n,2)=BA_ZERO do
n = ba_idivide(n,2)
e += 1
end while
if e>0 then
pf = {{2,e}}
end if
bigatom s = ba_sqrt(n),
d = ba_new(3)
while ba_compare(n,BA_ONE)>0 do
if ba_compare(d,s)>0 then
d = ba_new(n)
end if
e = 0
while true do
bigatom r = ba_mod(n,d)
if r!=BA_ZERO then exit end if
n = ba_idivide(n,d)
e += 1
end while
if e>0 then
pf = append(pf,{d,e})
s = ba_sqrt(n)
end if
d = ba_add(d,2)
end while
return pf
end function
 
function ba_gcd(object m, n)
while ba_compare(n,BA_ZERO)!=0 do
{m,n} = {n,ba_mod(m,n)}
end while
return m
end function
 
function ba_lcm(object m, n)
return ba_mul(ba_idivide(m,ba_gcd(m,n)),n)
end function
 
function multi_order(object a, sequence p_and_k)
{object p, integer k} = p_and_k
bigatom pk = ba_power(p,k),
t = ba_mul(ba_sub(p,1),ba_power(p,ba_sub(k,1))),
r = BA_ONE
sequence pf = ba_factor(t)
for i=1 to length(pf) do
{object q, integer e} = pf[i]
bigatom x = ba_mod_exp(a,ba_idiv(t,ba_power(q,e)),pk)
--
-- previous attempts at this task resulted in an infinite loop,
-- so I added a guard; feel free to increase limit as needed.
--
integer guard = 0
while x!=BA_ONE do
r = ba_mul(r,q)
x = ba_mod_exp(x,q,pk)
guard += 1
if guard>100 then ?9/0 end if
end while
end for
return r
end function
 
function multiplicative_order(object a, m)
if ba_gcd(a,m)!=BA_ONE then return "(a,m) not coprime" end if
sequence pf = ba_factor(m)
bigatom res = BA_ONE
for i=1 to length(pf) do
res = ba_lcm(res,multi_order(a,pf[i]))
end for
return res
end function
 
constant b100 = ba_power(2,100)
 
function shorta(object n)
return iff(ba_compare(n,b100)>0?"[big]":ba_sprint(n))
end function
 
procedure mo_test(object a, n)
string res = ba_sprint(multiplicative_order(a, n))
printf(1,"ord(%s) mod %s = %s\n",{shorta(a),shorta(n),res})
end procedure
 
atom t = time()
mo_test(37, 1000)
mo_test(37, 3343)
mo_test(ba_add(ba_power(10,100),1), 7919)
mo_test(ba_add(ba_power(10,1000),1), 15485863)
mo_test(ba_sub(ba_power(10,10000),1), 22801763489)
mo_test(1511678068, 7379191741)
mo_test(3047753288, 2257683301)
?"==="
bigatom b = ba_sub(ba_power(10,20),1)
mo_test(2, b)
mo_test(17,b)
mo_test(54,100001)
--/* -- this all works fine, but doubles the runtime...
bigatom b9090 = multiplicative_order(54,100001)
ba_printf(1,"%B\n",b9090)
if ba_compare(b9090,9090)!=0 then ?9/0 end if
ba_printf(1,"%B\n",ba_mod_exp(54,b9090,100001))
bool error = false
atom t1 = time()+1
for r=1 to 9090-1 do
if ba_compare(ba_mod_exp(54,r,100001),1)=0 then
printf(1,"ba_mod_exp(54,%d,100001) gives 1!\n",r)
error = true
exit
end if
if time()>t1 then
printf(1,"checking %d...\r",r)
t1 = time()+1
end if
end for
if not error then puts(1,"Everything checks.\n") end if
--*/
?elapsed(time()-t)
Output:
ord(37) mod 1000 = 100
ord(37) mod 3343 = 1114
ord([big]) mod 7919 = 3959
ord([big]) mod 15485863 = 15485862
ord([big]) mod 22801763489 = 22801763488
ord(1511678068) mod 7379191741 = 614932645
ord(3047753288) mod 2257683301 = 62713425
"==="
ord(2) mod 99999999999999999999 = 3748806900
ord(17) mod 99999999999999999999 = 1499522760
ord(54) mod 100001 = 9090
"1 minute and 11s"

Python[edit]

def gcd(a, b):
while b != 0:
a, b = b, a % b
return a
 
def lcm(a, b):
return (a*b) / gcd(a, b)
 
def isPrime(p):
return (p > 1) and all(f == p for f,e in factored(p))
 
primeList = [2,3,5,7]
def primes():
for p in primeList:
yield p
while 1:
p += 2
while not isPrime(p):
p += 2
primeList.append(p)
yield p
 
def factored( a):
for p in primes():
j = 0
while a%p == 0:
a /= p
j += 1
if j > 0:
yield (p,j)
if a < p*p: break
if a > 1:
yield (a,1)
 
 
def multOrdr1(a,(p,e) ):
m = p**e
t = (p-1)*(p**(e-1)) # = Phi(p**e) where p prime
qs = [1,]
for f in factored(t):
qs = [ q * f[0]**j for j in range(1+f[1]) for q in qs ]
qs.sort()
 
for q in qs:
if pow( a, q, m )==1: break
return q
 
 
def multOrder(a,m):
assert gcd(a,m) == 1
mofs = (multOrdr1(a,r) for r in factored(m))
return reduce(lcm, mofs, 1)
 
 
if __name__ == "__main__":
print multOrder(37, 1000) # 100
b = 10**20-1
print multOrder(2, b) # 3748806900
print multOrder(17,b) # 1499522760
b = 100001
print multOrder(54,b)
print pow( 54, multOrder(54,b),b)
if any( (1==pow(54,r, b)) for r in range(1,multOrder(54,b))):
print 'Exists a power r < 9090 where pow(54,r,b)==1'
else:
print 'Everything checks.'

Racket[edit]

The Racket function unit-group-order from racket/math computes the multiplicative order of an element a in Zn. An implementation of the algorithm in the tast description is shown below.

 
#lang racket
(require math)
 
(define (order a n)
(unless (coprime? a n) (error 'order "arguments must be coprime"))
(for/fold ([o 1]) ([r (factorize n)])
(lcm o (order1 a r))))
 
(define (order1 a p&e)
(match-define (list p e) p&e)
(define m (expt p e))
(define t (* (- p 1) (expt p (- e 1))))
(define qs
(for/fold ([qs '(1)]) ([f (factorize t)])
(match f [(list f0 f1)
(for*/list ([q qs] [j (in-range (+ 1 f1))])
(* q (expt f0 j)))])))
(for/or ([q (sort qs <)] #:when (= (modular-expt a q m) 1)) q))
 
 
(order 37 1000)
(order (+ (expt 10 100) 1) 7919)
(order (+ (expt 10 1000) 1) 15485863)
(order (- (expt 10 10000) 1) 22801763489)
(order 13 (+ 1 (expt 10 80)))
 

Output:

 
100
3959
15485862
22801763488
109609547199756140150989321269669269476675495992554276140800
 

REXX[edit]

/*REXX pgm computes multiplicative order of a minimum integer  N  such that  a^n mod m≡1*/
wa= 0; wm= 0 /* ═a═ ══m══ */ /*maximum widths of the A and M values.*/
@.=.; @.1= 3 10
@.2= 37 1000
@.3= 37 10000
@.4= 37 3343
@.5= 37 3344
@.6= 2 1000
pad= left('', 9)
d= 500 /*use 500 decimal digits for a starter.*/
do w=1 for 2 /*when W≡1, find max widths of A and M.*/
do j=1 while @.j\==.; parse var @.j a . 1 r m , n
if w==1 then do; wa= max(wa, length(a) ); wm= max(wm, length(m) ); iterate
end
if m//a==0 then n= ' [solution not possible]' /*test co─prime for A and B. */
numeric digits d /*start with 100 decimal digits. */
if n=='' then do n= 2; p= r * a /*compute product──may have an exponent*/
parse var p 'E' _ /*try to extract the exponent from P. */
if _\=='' then do; numeric digits _+d /*bump the decimal digs.*/
p=r*a /*recalculate integer P.*/
end
if p//m==1 then leave /*now, perform the nitty─gritty modulo.*/
r= p /*assign product to R for next multiply*/
end /*n*/ /* [↑] // is really ÷ remainder.*/
say pad 'a=' right(a,wa) pad "m=" right(m,wm) pad 'multiplicative order:' n
end /*j*/
end /*w*/ /*stick a fork in it, we're all done. */
output:
          a=  3           m=    10           multiplicative order: 4
          a= 37           m=  1000           multiplicative order: 100
          a= 37           m= 10000           multiplicative order: 500
          a= 37           m=  3343           multiplicative order: 1114
          a= 37           m=  3344           multiplicative order: 20
          a=  2           m=  1000           multiplicative order:  [solution not possible]

Ruby[edit]

require 'prime'
 
def powerMod(b, p, m)
p.to_s(2).each_char.inject(1) do |result, bit|
result = (result * result) % m
bit=='1' ? (result * b) % m : result
end
end
 
def multOrder_(a, p, k)
pk = p ** k
t = (p - 1) * p ** (k - 1)
r = 1
for q, e in t.prime_division
x = powerMod(a, t / q**e, pk)
while x != 1
r *= q
x = powerMod(x, q, pk)
end
end
r
end
 
def multOrder(a, m)
m.prime_division.inject(1) do |result, f|
result.lcm(multOrder_(a, *f))
end
end
 
puts multOrder(37, 1000)
b = 10**20-1
puts multOrder(2, b)
puts multOrder(17,b)
b = 100001
puts multOrder(54,b)
puts powerMod(54, multOrder(54,b), b)
if (1...multOrder(54,b)).any? {|r| powerMod(54, r, b) == 1}
puts 'Exists a power r < 9090 where powerMod(54,r,b)==1'
else
puts 'Everything checks.'
end
Output:
100
3748806900
1499522760
9090
1
Everything checks.

Seed7[edit]

$ include "seed7_05.s7i";
include "bigint.s7i";
 
const type: oneFactor is new struct
var bigInteger: prime is 0_;
var integer: exp is 0;
end struct;
 
const func oneFactor: oneFactor (in bigInteger: prime, in integer: exp) is func
result
var oneFactor: aFactor is oneFactor.value;
begin
aFactor.prime := prime;
aFactor.exp := exp;
end func;
 
const func array oneFactor: factor (in var bigInteger: n) is func
result
var array oneFactor: pf is 0 times oneFactor.value;
local
var integer: e is 0;
var bigInteger: d is 0_;
var bigInteger: s is 0_;
begin
e := lowestSetBit(n);
if e > 0 then
n >>:= e;
pf := [] (oneFactor(2_, e));
end if;
s := sqrt(n);
d := 3_;
while n > 1_ do
if d > s then
d := n;
end if;
e := 0;
while n rem d = 0_ do
n := n div d;
incr(e);
end while;
if e > 0 then
pf &:= oneFactor(d, e);
s := sqrt(n);
end if;
d +:= 2_;
end while;
end func;
 
const func bigInteger: moBachShallit58(in bigInteger: a, in bigInteger: n, in array oneFactor: pf) is func
result
var bigInteger: mo is 0_;
local
var bigInteger: n1 is 0_;
var oneFactor: pe is oneFactor.value;
var bigInteger: x is 0_;
var bigInteger: y is 0_;
var integer: o is 0;
var bigInteger: o1 is 0_;
begin
n1 := n - 1_;
mo := 1_;
for pe range pf do
y := n1 div pe.prime ** pe.exp;
x := modPow(a, y, n);
o := 0;
while x > 1_ do
x := modPow(x, pe.prime, n);
incr(o);
end while;
o1 := pe.prime ** o;
mo *:= o1 div gcd(mo, o1);
end for;
end func;
 
const func boolean: isProbablyPrime (in bigInteger: primeCandidate, in var integer: count) is func
result
var boolean: isProbablyPrime is TRUE;
local
var bigInteger: aRandomNumber is 0_;
begin
while isProbablyPrime and count > 0 do
aRandomNumber := rand(1_, pred(primeCandidate));
isProbablyPrime := modPow(aRandomNumber, pred(primeCandidate), primeCandidate) = 1_;
decr(count);
end while;
# writeln(count);
end func;
 
const proc: moTest (in bigInteger: a, in bigInteger: n) is func
begin
if bitLength(a) < 100 then
write("ord(" <& a <& ")");
else
write("ord([big])");
end if;
if bitLength(n) < 100 then
write(" mod " <& n <& " ");
else
write(" mod [big] ");
end if;
if not isProbablyPrime(n, 20) then
writeln("not computed. modulus must be prime for this algorithm.")
else
writeln("= " <& moBachShallit58(a, n, factor(n - 1_)));
end if;
end func;
 
const proc: main is func
local
var bigInteger: b is 100_;
begin
moTest(37_, 3343_);
moTest(10_ ** 100 + 1_, 7919_);
moTest(10_ ** 1000 + 1_, 15485863_);
moTest(10_ ** 10000 - 1_, 22801763489_);
moTest(1511678068_, 7379191741_);
moTest(3047753288_, 2257683301_);
end func;
Output:
ord(37) mod 3343 = 1114
ord([big]) mod 7919 = 3959
ord([big]) mod 15485863 = 15485862
ord([big]) mod 22801763489 = 22801763488
ord(1511678068) mod 7379191741 = 614932645
ord(3047753288) mod 2257683301 = 62713425

Sidef[edit]

Translation of: Perl 6
func mo_prime(a, p, e) {
var m = p**e
var t = (p-1)*(p**(e-1))
var qs = [1]
 
for f in (t.factor_exp) {
qs.map! {|q|
0..f[1] -> map {|j| q * f[0]**j }...
}
}
 
qs.sort.first_by {|q| powmod(a, q, m) == 1 }
}
 
func mo(a, m) {
gcd(a, m) == 1 || die "#{a} and #{m} are not relatively prime"
Math.lcm(1, m.factor_exp.map {|r| mo_prime(a, r...) }...)
}
 
say mo(37, 1000)
say mo(54, 100001)
 
with (10**20 - 1) {|b|
say mo(2, b)
say mo(17, b)
}
Output:
100
9090
3748806900
1499522760

Tcl[edit]

Translation of: Python
Library: Tcllib (Package: struct::list)
package require Tcl 8.5
package require struct::list
 
proc multOrder {a m} {
assert {[gcd $a $m] == 1}
set mofs [list]
dict for {p e} [factor_num $m] {
lappend mofs [multOrdr1 $a $p $e]
}
return [struct::list fold $mofs 1 lcm]
}
 
proc multOrdr1 {a p e} {
set m [expr {$p ** $e}]
set t [expr {($p - 1) * ($p ** ($e - 1))}]
set qs [dict create 1 ""]
 
dict for {f0 f1} [factor_num $t] {
dict for {q -} $qs {
foreach j [range [expr {1 + $f1}]] {
dict set qs [expr {$q * $f0 ** $j}] ""
}
}
}
 
dict for {q -} $qs {
if {pypow($a, $q, $m) == 1} break
}
return $q
}
 
####################################################
# utility procs
proc assert {condition {message "Assertion failed!"}} {
if { ! [uplevel 1 [list expr $condition]]} {
return -code error $message
}
}
 
proc gcd {a b} {
while {$b != 0} {
lassign [list $b [expr {$a % $b}]] a b
}
return $a
}
 
proc lcm {a b} {
expr {$a * $b / [gcd $a $b]}
}
 
proc factor_num {num} {
primes::restart
set factors [dict create]
for {set i [primes::get_next_prime]} {$i <= $num} {} {
if {$num % $i == 0} {
dict incr factors $i
set num [expr {$num / $i}]
continue
} elseif {$i*$i > $num} {
dict incr factors $num
break
} else {
set i [primes::get_next_prime]
}
}
return $factors
}
 
####################################################
# a range command akin to Python's
proc range args {
foreach {start stop step} [switch -exact -- [llength $args] {
1 {concat 0 $args 1}
2 {concat $args 1}
3 {concat $args }
default {error {wrong # of args: should be "range ?start? stop ?step?"}}
}] break
if {$step == 0} {error "cannot create a range when step == 0"}
set range [list]
while {$step > 0 ? $start < $stop : $stop < $start} {
lappend range $start
incr start $step
}
return $range
}
 
# python's pow()
proc ::tcl::mathfunc::pypow {x y {z ""}} {
expr {$z eq "" ? $x ** $y : ($x ** $y) % $z}
}
 
####################################################
# prime number generator
# ref http://wiki.tcl.tk/5996
####################################################
namespace eval primes {}
 
proc primes::reset {} {
variable list [list]
variable current_index end
}
 
namespace eval primes {reset}
 
proc primes::restart {} {
variable list
variable current_index
if {[llength $list] > 0} {
set current_index 0
}
}
 
proc primes::is_prime {candidate} {
variable list
 
foreach prime $list {
if {$candidate % $prime == 0} {
return false
}
if {$prime * $prime > $candidate} {
return true
}
}
while true {
set largest [get_next_prime]
if {$largest * $largest >= $candidate} {
return [is_prime $candidate]
}
}
}
 
proc primes::get_next_prime {} {
variable list
variable current_index
 
if {$current_index ne "end"} {
set p [lindex $list $current_index]
if {[incr current_index] == [llength $list]} {
set current_index end
}
return $p
}
 
switch -exact -- [llength $list] {
0 {set candidate 2}
1 {set candidate 3}
default {
set candidate [lindex $list end]
while true {
incr candidate 2
if {[is_prime $candidate]} break
}
}
}
lappend list $candidate
return $candidate
}
 
####################################################
puts [multOrder 37 1000] ;# 100
 
set b [expr {10**20 - 1}]
puts [multOrder 2 $b] ;# 3748806900
puts [multOrder 17 $b] ;# 1499522760
 
set a 54
set m 100001
puts [set n [multOrder $a $m]] ;# 9090
puts [expr {pypow($a, $n, $m)}] ;# 1
 
set lambda {{a n m} {expr {pypow($a, $n, $m) == 1}}}
foreach r [lreverse [range 1 $n]] {
if {[apply $lambda $a $r $m]} {
error "Oops, $n is not the smallest: {$a $r $m} satisfies $lambda"
}
if {$r % 1000 == 0} {puts "$r ..."}
}
puts "OK, $n is the smallest n such that {$a $n $m} satisfies $lambda"

zkl[edit]

Translation of: Python

Using Extensible prime generator#zkl and the GMP library for lcm (least common multiple), pow and powm ((n^e)%m)

It would probably be nice to memoize the prime numbers but that isn't necessary for this task.

var BN   =Import("zklBigNum");
var Sieve=Import("sieve");
 
// factor n into powers of primes
// eg 9090 == 2^1 * 3^2 * 5^1 * 101^1
fcn factor2PP(n){ // lazy factors using lazy primes --> (prime,power) ...
Utils.Generator(fcn(a){
primes:=Utils.Generator(Sieve.postponed_sieve);
foreach p in (primes){
e:=0; while(a%p == 0){ a /= p; e+=1; }
if (e) vm.yield(p,e);
if (a<p*p) break;
}
if (a>1) vm.yield(a,1);
},n)
}
 
fcn _multOrdr1(a,p,e){
m:=p.pow(e);
t:=m/p*(p - 1);
qs:=L(BN(1));
foreach p2,e2 in (factor2PP(t)){
qs=[[(e,q); [0..e2]; qs; '{ q*BN(p2).pow(e) }]];
}
qs.filter1('wrap(q){ a.powm(q,m)==1 });
}
 
fcn multiOrder(a,m){
if (m.gcd(a)!=1) throw(Exception.ValueError("Not co-prime"));
res:=BN(1);
foreach p,e in (factor2PP(m)){
res = res.lcm(_multOrdr1(BN(a),BN(p),e));
}
return(res);
}
multiOrder(37,1000).println();
b:=BN(10).pow(20)-1;
multiOrder(2,b).println();
multiOrder(17,b).println();
 
b=0d10_0001;
[BN(1)..multiOrder(54,b)-1].filter1('wrap(r,b54){b54.powm(r,b)==1},BN(54)) :
if (_) println("Exists a power r < 9090 where (54^r)%b)==1");
else println("Everything checks.");
Output:
100
3748806900
1499522760
Everything checks.