Multiplicative order

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Revision as of 06:50, 22 March 2010 by rosettacode>NevilleDNZ (→‎[[Multiplicative_order#ALGOL 68]]: tidy FOR iterator DO loop OD)
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). 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.

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(k(lg p)4/(lg lg p)) .

ALGOL 68

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

<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;
  1. 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.

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>

Mathematica

In Mathematica this is really easy, as this function is built-in: <lang Mathematica>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.</lang>

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: <lang Mathematica>100 3959 15485862 22801763488 109609547199756140150989321269669269476675495992554276140800 2583496112724752500580158969425549088007844580826869433740066152289289764829816356800</lang>

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>

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

Slavishly

Translation of: Python

Uses struct::list package from

Library: tcllib

<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

}

  2. 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

}

  2. 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

}

  1. python's pow()

proc ::tcl::mathfunc::pypow {x y {z ""}} { 

   expr {$z eq "" ? $x ** $y : ($x ** $y) % $z}

}

  2. prime number generator
  3. 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

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