Multiplicative order
The multiplicative order of a relative to m is the least positive integer n such that a^n is 1 (modulo m). For 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.
You are encouraged to solve this task according to the task description, using any language you may know.
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 wrt. 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.
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
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"). <lang Ada>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;</lang>
Here is the implementation ("multiplicative_order.adb"):
<lang Ada>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;</lang>
This is a sample program using the Multiplicative_Order package: <lang Ada>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;</lang>
The output from the sample program:
4 100 500 1114 20 4 100 500 1114 20 10
ALGOL 68
<lang algol68>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
)</lang> Output:
100 3748806900 1499522760 9090 1 Everything checks.
C
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. <lang c>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; }</lang>
Go
<lang go>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 = []pExpTemplate: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)
}</lang>
- 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
Assuming a function
<lang haskell>primeFacsExp :: Integer -> [(Integer, Int)]</lang>
to calculate prime power factors, and another function
<lang haskell>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"</lang>
to efficiently calculate powers modulo some Integral
, we get
<lang haskell>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</lang>
J
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.
<lang j>mo=: 4 : 0
a=. x: x m=. x: y assert. 1=a+.m *./ a mopk"1 |: __ q: m
)</lang>
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.
<lang j>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) */\@$&.> q */q^d
)</lang>
For example:
<lang j> 37 mo 1000 100
2 mo _1+10^80x
190174169488577769580266953193403101748804183400400</lang>
Julia
(Uses the factors
function from Factors of an integer#Julia.)
<lang julia>function factors(n)
f = [one(n)] for (p,e) in factor(n) f = reduce(vcat, f, [f*p^j for j in 1:e]) end return length(f) == 1 ? [one(n), n] : sort!(f)
end
function multorder(a, m)
gcd(a,m) == 1 || error("$a and $m are not coprime") res = one(m) for (p,e) in factor(m) m = p^e t = div(m, p) * (p-1) for f in factors(t) if powermod(a, f, m) == 1 res = lcm(res, f) break end end end res
end</lang>
Example output (using big
to employ arbitrary-precision arithmetic where needed):
julia> multorder(37, 1000) 100 julia> multorder(big(10)^100 + 1, 7919) 3959 julia> multorder(big(10)^1000 + 1, 15485863) 15485862 julia> multorder(big(10)^10000 - 1, 22801763489) 22801763488
Mathematica
In Mathematica this is really easy, as this function is built-in:
MultiplicativeOrder[k,n] gives the multiplicative order of k modulo n, defined as the smallest integer m such that k^m == 1 mod n.
MultiplicativeOrder[k,n,{r1,r2,...}] gives the generalized multiplicative order of k modulo n, defined as the smallest integer m such that k^m==ri mod n for some i.
Examples:
<lang Mathematica>MultiplicativeOrder[37, 1000]
MultiplicativeOrder[10^100 + 1, 7919] (*10^3th prime number Prime[1000]*)
MultiplicativeOrder[10^1000 + 1, 15485863] (*10^6th prime number*)
MultiplicativeOrder[10^10000 - 1, 22801763489] (*10^9th prime number*)
MultiplicativeOrder[13, 1 + 10^80]
MultiplicativeOrder[11, 1 + 10^100]</lang>
gives back:
100 3959 15485862 22801763488 109609547199756140150989321269669269476675495992554276140800 2583496112724752500580158969425549088007844580826869433740066152289289764829816356800
Maxima
<lang maxima>zn_order(37, 1000); /* 100 */
zn_order(10^100 + 1, 7919); /* 3959 */
zn_order(10^1000 + 1, 15485863); /* 15485862 */
zn_order(10^10000 - 1, 22801763489); /* 22801763488 */
zn_order(13, 1 + 10^80); /* 109609547199756140150989321269669269476675495992554276140800 */
zn_order(11, 1 + 10^100); /* 2583496112724752500580158969425549088007844580826869433740066152289289764829816356800 */</lang>
PARI/GP
<lang parigp>znorder(Mod(a,n))</lang>
Perl 6
<lang perl6>my @primes := 2, grep *.is-prime, (3,5,7...*);
sub factor($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 = @qs.map(-> $q { (0..$f.value).map(-> $j { $q * $f.key ** $j }) }); } @qs.sort(); @qs.first(-> $q { expmod( $a, $q, $m ) == 1});
}
sub mo($a, $m) {
$a gcd $m == 1 || die "$a and $m are not relatively prime"; [lcm] 1, factor($m).map(-> $r { mo-prime($a, $r.key, $r.value) });
}
sub MAIN("test") {
use Test; for (10, 21, 25, 150, 1231, 123141, 34131) -> $n { # say factor($n).perl; # say factor($n).map(-> $pair { $pair.key ** $pair.value }).perl; is ([*] factor($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';
}</lang>
- 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
Python
<lang python>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.'</lang>
Racket
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>
- 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)))
</lang>
Output:
<lang racket>
100
3959
15485862
22801763488
109609547199756140150989321269669269476675495992554276140800
</lang>
Ruby
<lang ruby>require 'rational' # for lcm require 'mathn' # for prime_division
def powerMod(b, p, m)
result = 1 bits = p.to_s(2) for bit in bits.split() result = (result * result) % m if bit == '1' result = (result * b) % m end end result
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) {|result, f| result.lcm(multOrder_(a, *f)) }
end
puts multOrder(37, 1000) # 100 b = 10**20-1 puts multOrder(2, b) # 3748806900 puts multOrder(17,b) # 1499522760 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</lang>
Tcl
<lang tcl>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"</lang>