Prime decomposition: Difference between revisions
FALSE |
|||
Line 690: | Line 690: | ||
return ans; |
return ans; |
||
}</lang> |
}</lang> |
||
Alternate version, optimised to be faster. Uses java.util.List, java.util.LinkedList and java.math.BigInteger. |
|||
<lang java>public boolean isPrime(BigInteger i);</lang> |
|||
<lang java> |
|||
private static final BigInteger two = BigInteger.valueOf(2); |
|||
public List<BigInteger> primeDecomp(BigInteger a) { |
|||
//Is it even possible |
|||
if(a.compareTo(two)==-1) { |
|||
return null; //impossible for values lower than 2 |
|||
} |
|||
//quickly handle even values |
|||
List<BigInteger> result = new c<BigInteger>(); |
|||
while(a.and(BigInteger.ONE).equals(BigInteger.ONE)) { |
|||
a = a.shiftRight(1); |
|||
result.add(two); |
|||
} |
|||
//left with odd values |
|||
if(!a.equals(BigInteger.ONE)) { |
|||
BigInteger b = BigInteger.valueOf(3); |
|||
while(b.compareTo(a)==-1) { |
|||
if(isPrime(b)) { |
|||
BigInteger[] dr = a.divideAndRemainder(b); |
|||
if(dr[1].equals(BigInteger.ZERO)) { |
|||
result.add(b); |
|||
a = dr[0]; |
|||
} |
|||
} |
|||
b = b.add(two); |
|||
} |
|||
result.add(b); //b will always be prime here... |
|||
} |
|||
return result; |
|||
} |
|||
</lang> |
|||
=={{header|JavaScript}}== |
=={{header|JavaScript}}== |
Revision as of 12:43, 14 September 2009
You are encouraged to solve this task according to the task description, using any language you may know.
The prime decomposition of a number is defined as a list of prime numbers which when all multiplied together, are equal to that number. Example: 12 = 2 * 2 * 3, so its prime decomposition is {2, 2, 3}
Write a function which returns an array or collection which contains the prime decomposition of a given number, n, greater than 1. If your language does not have an isPrime-like function available, you may assume that you have a function which determines whether a number is prime (note its name before your code).
If you would like to test code from this task, you may use code from trial division or the Sieve of Eratosthenes.
Note: The program must not be limited by the word size of your computer or some other artificial limit; it should work for any number regardless of size (ignoring the physical limits of RAM etc).
Ada
<lang ada>with Ada.Text_IO; use Ada.Text_IO;
procedure Test_Prime is
generic type Number is private; Zero : Number; One : Number; Two : Number; with function Image (X : Number) return String is <>; with function "+" (X, Y : Number) return Number is <>; with function "/" (X, Y : Number) return Number is <>; with function "mod" (X, Y : Number) return Number is <>; with function ">=" (X, Y : Number) return Boolean is <>; package Prime_Numbers is type Number_List is array (Positive range <>) of Number; function Decompose (N : Number) return Number_List; procedure Put (List : Number_List); end Prime_Numbers;
package body Prime_Numbers is function Decompose (N : Number) return Number_List is Size : Natural := 0; M : Number := N; K : Number := Two; begin -- Estimation of the result length from above while M >= Two loop M := (M + One) / Two; Size := Size + 1; end loop; M := N; -- Filling the result with prime numbers declare Result : Number_List (1..Size); Index : Positive := 1; begin while N >= K loop -- Divisors loop while Zero = (M mod K) loop -- While divides Result (Index) := K; Index := Index + 1; M := M / K; end loop; K := K + One; end loop; return Result (1..Index - 1); end; end Decompose;
procedure Put (List : Number_List) is begin for Index in List'Range loop Put (Image (List (Index))); end loop; end Put; end Prime_Numbers; package Integer_Numbers is new Prime_Numbers (Natural, 0, 1, 2, Positive'Image); use Integer_Numbers;
begin
Put (Decompose (12));
end Test_Prime;</lang> The solution is generic. The package is instantiated by a type that supports necessary operations +, /, mod, >=. The constants 0, 1, 2 are parameters too, because the type might have no literals. The package also provides a procedure to output an array of prime numbers and a function to convert a number to string (as a parameter). The function Decompose first estimates the maximal result length as log2 of the argument. Then it allocates the result and starts to enumerate divisors. It does not care to check if the divisors are prime, because non-prime divisors will be automatically excluded. In the example provided, the package is instantiated with plain integer type. Sample output:
2 2 3
ALGOL 68
<lang algol>MODE LINT = LONG INT;
MODE YIELDLINT = PROC(LINT)VOID; MODE FORLINT = PROC(YIELDLINT)VOID;
PROC (LINT, YIELDLINT)VOID for decompose;
INT upb cache = bits width;
BITS cache := 2r0; BITS cached := 2r0;
PROC is prime = (LINT n)BOOL: (
BOOL has factor := FALSE, out := TRUE; # FOR factor IN decompose(n) DO # for decompose(n, (LINT factor)VOID:( IF has factor THEN out := FALSE; GO TO done FI; has factor := TRUE )); done: out
);
PROC is prime cached := (LINT n)BOOL: (
LINT half n = n OVER LONG 2 - LONG 1; IF half n <= LENG upb cache THEN BOOL out=IF SHORTEN half n ELEM cached THEN SHORTEN half n ELEM cache ELSE BOOL out := is prime(n); INT shift = upb cache - SHORTEN half n; cache := cache OR (BIN ABS out SHL shift); cached := cached OR ( 2r1 SHL shift); out FI; out ELSE is prime(n) # above useful cache limit # FI
);
PROC for primes := (YIELDLINT yield)VOID:(
yield(LONG 2); LINT n := LONG 3; WHILE n < long maxint - LONG 2 DO yield(n); n +:= LONG 2; WHILE n < long maxint - LONG 2 AND NOT is prime cached(n) DO n +:= LONG 2 OD OD
);
- PROC # for decompose := (LINT in n, YIELDLINT yield)VOID: (
LINT n := in n; # FOR p IN primes() DO # for primes((LINT p)VOID: IF p*p > n THEN GO TO done ELSE WHILE n MOD p = LONG 0 DO yield(p); n := n OVER p OD FI ); done: IF n > LONG 1 THEN yield(n) FI
);
main:(
- FOR m IN primes() DO #
for primes((LINT m)VOID:( LINT p = LONG 2 ** SHORTEN m - LONG 1; print(("2**",whole(m,0),"-1 = ",whole(p,0),", with factors:")); # FOR factor IN decompose(p) DO # for decompose(p, (LINT factor)VOID: print((" ",whole(factor,0))) ); print(new line); IF m >= LONG 59 THEN GO TO done FI )); done: EMPTY
)</lang> Output:
2**2-1 = 3, with factors: 3 2**3-1 = 7, with factors: 7 2**5-1 = 31, with factors: 31 2**7-1 = 127, with factors: 127 2**11-1 = 2047, with factors: 23 89 2**13-1 = 8191, with factors: 8191 2**17-1 = 131071, with factors: 131071 2**19-1 = 524287, with factors: 524287 2**23-1 = 8388607, with factors: 47 178481 2**29-1 = 536870911, with factors: 233 1103 2089 2**31-1 = 2147483647, with factors: 2147483647 2**37-1 = 137438953471, with factors: 223 616318177 2**41-1 = 2199023255551, with factors: 13367 164511353 2**43-1 = 8796093022207, with factors: 431 9719 2099863 2**47-1 = 140737488355327, with factors: 2351 4513 13264529 2**53-1 = 9007199254740991, with factors: 6361 69431 20394401 2**59-1 = 576460752303423487, with factors: 179951 3203431780337
Note: ALGOL 68G took 49,109,599 BogoMI and ELLA ALGOL 68RS took 1,127,634 BogoMI to complete the example.
AutoHotkey
<lang AutoHotkey>MsgBox % factor(8388607) ; 47 * 178481
factor(n) {
If (n = 1) Return f = 2 While (f <= n) { If (Mod(n, f) = 0) { next := factor(n / f) factors = %f%`n%next% Return factors } f++ }
}</lang>
AWK
As the examples show, pretty large numbers can be factored in tolerable time: <lang awk>$ awk 'func pfac(n){r="";f=2;while(f<=n){while(!(n%f)){n=n/f;r=r" "f};f=f+2-(f==2)};return r}{print pfac($1)}' 36
2 2 3 3
77
7 11
536870911
233 1103 2089
8796093022207
431 9719 2099863</lang>
C
primedecompose.h
<lang c>#ifndef _PRIMEDECOMPOSE_H_
- define _PRIMEDECOMPOSE_H_
- include <gmp.h>
int decompose(mpz_t n, mpz_t *o);
- endif</lang>
primedecompose.c
<lang c>#include "primedecompose.h"
int decompose(mpz_t n, mpz_t *o) {
int i; mpz_t tmp, d; i = 0; mpz_init(tmp); mpz_init(d); while(mpz_cmp_si(n, 1)) { mpz_set_ui(d, 1); do { mpz_add_ui(tmp, d, 1); mpz_swap(tmp, d); } while(!mpz_divisible_p(n, d)); mpz_divexact(tmp, n, d); mpz_swap(tmp, n); mpz_init(o[i]); mpz_set(o[i], d); i++; } return i;
}</lang>
Testing
<lang c>#include <stdio.h>
- include <stdlib.h>
- include <gmp.h>
- include "primedecompose.h"
mpz_t dest[100]; /* must be big enough to hold all the factors! */
int main(int argc, char **argv) {
mpz_t n; int i, l; if(argc != 2) { puts("Pass a parameter"); return EXIT_SUCCESS; } mpz_init_set_str(n, argv[1], 10); l = decompose(n, dest); for(i=0; i < l; i++) { gmp_printf("%s%Zd", i?" * ":"", dest[i]); mpz_clear(dest[i]); } printf("\n"); return EXIT_SUCCESS;
}</lang>
Using GNU Compiler Collection gcc extensions
Note: The following code sample is experimental as it implements python style iterators for (potentially) infinite sequences. C is not normally written this way, and in the case of this sample it requires the GCC "nested procedure" extension to the C language. <lang c>#include <limits.h>
- include <stdio.h>
- include <math.h>
typedef enum{false=0, true=1}bool; const int max_lint = LONG_MAX;
typedef long long int lint;
- assert sizeof_long_long_int (LONG_MAX>=8) /* XXX */
- ifdef NEED_GOTO
- include <setjmp.h>
/* declare label otherwise it is not visible in sub-scope */
- define LABEL(label) jmp_buf label; if(setjmp(label))goto label;
- define GOTO(label) longjmp(label, true)
- endif
/* the following line is the only time I have ever required "auto" */
- define FOR(i,iterator) auto bool lambda(i); yield_init = (void *)λ iterator; bool lambda(i)
- define DO {
- define YIELD(x) if(!yield(x))return
- define BREAK return false
- define CONTINUE return true
- define OD CONTINUE; }
/* Warning: _Most_ FOR(,){ } loops _must_ have a CONTINUE as the last statement.
* Otherwise the lambda will return random value from stack, and may terminate early */
typedef void iterator, lint_iterator; /* hint at procedure purpose */ static volatile void *yield_init; /* not thread safe */
- define YIELDS(type) bool (*yield)(type) = yield_init
typedef unsigned int bits;
- define ELEM(shift, bits) ( (bits >> shift) & 0b1 )
bits cache = 0b0, cached = 0b0; const lint upb_cache = 8 * sizeof(cache);
lint_iterator decompose(lint); /* forward declaration */
bool is_prime(lint n){
bool has_factor = false, out = true;
/* for factor in decompose(n) do */
FOR(lint factor, decompose(n)){ if( has_factor ){ out = false; BREAK; } has_factor = true; CONTINUE; } return out;
}
bool is_prime_cached (lint n){
lint half_n = n / 2 - 2; if( half_n <= upb_cache){ /* dont cache the initial four, nor the even numbers */ if (ELEM(half_n,cached)){ return ELEM(half_n,cache); } else { bool out = is_prime(n); cache = cache | out << half_n; cached = cached | 0b1 << half_n; return out; } } else { return is_prime(n); }
}
lint_iterator primes (){
YIELDS(lint); YIELD(2); lint n = 3; while( n < max_lint - 2 ){ YIELD(n); n += 2; while( n < max_lint - 2 && ! is_prime_cached(n) ){ n += 2; } }
}
lint_iterator decompose (lint in_n){
YIELDS(lint); lint n = in_n; /* for p in primes do */ FOR(lint p, primes()){ if( p*p > n ){ BREAK; } else { while( n % p == 0 ){ YIELD(p); n = n / p; } } CONTINUE; } if( n > 1 ){ YIELD(n); }
}
main(){
FOR(lint m, primes()){ lint p = powl(2, m) - 1; printf("2**%lld-1 = %lld, with factors:",m,p); FOR(lint factor, decompose(p)){ printf(" %lld",factor); fflush(stdout); CONTINUE; } printf("\n",m); if( m >= 59 )BREAK; CONTINUE; }
}</lang> Output:
2**2-1 = 3, with factors: 3 2**3-1 = 7, with factors: 7 2**5-1 = 31, with factors: 31 2**7-1 = 127, with factors: 127 2**11-1 = 2047, with factors: 23 89 2**13-1 = 8191, with factors: 8191 2**17-1 = 131071, with factors: 131071 2**19-1 = 524287, with factors: 524287 2**23-1 = 8388607, with factors: 47 178481 2**29-1 = 536870911, with factors: 233 1103 2089 2**31-1 = 2147483647, with factors: 2147483647 2**37-1 = 137438953471, with factors: 223 616318177 2**41-1 = 2199023255551, with factors: 13367 164511353 2**43-1 = 8796093022207, with factors: 431 9719 2099863 2**47-1 = 140737488355327, with factors: 2351 4513 13264529 2**53-1 = 9007199254740991, with factors: 6361 69431 20394401 2**59-1 = 576460752303423487, with factors: 179951 3203431780337
Note: gcc took 487,719 BogoMI to complete the example.
C++
<lang cpp>#include <iostream>
- include <gmpxx.h>
// This function template works for any type representing integers or // nonnegative integers, and has the standard operator overloads for // arithmetic and comparison operators, as well as explicit conversion // from int. // // OutputIterator must be an output iterator with value_type Integer. // It receives the prime factors. template<typename Integer, typename OutputIterator>
void decompose(Integer n, OutputIterator out)
{
Integer i(2);
while (n != 1) { while (n % i == Integer(0)) { *out++ = i; n /= i; } ++i; }
}
// this is an output iterator similar to std::ostream_iterator, except // that it outputs the separation string *before* the value, but not // before the first value (i.e. it produces an infix notation). template<typename T> class infix_ostream_iterator:
public std::iterator<T, std::output_iterator_tag>
{
class Proxy; friend class Proxy; class Proxy { public: Proxy(infix_ostream_iterator& iter): iterator(iter) {} Proxy& operator=(T const& value) { if (!iterator.first) { iterator.stream << iterator.infix; } iterator.stream << value; } private: infix_ostream_iterator& iterator; };
public:
infix_ostream_iterator(std::ostream& os, char const* inf): stream(os), first(true), infix(inf) { } infix_ostream_iterator& operator++() { first = false; return *this; } infix_ostream_iterator operator++(int) { infix_ostream_iterator prev(*this); ++*this; return prev; } Proxy operator*() { return Proxy(*this); }
private:
std::ostream& stream; bool first; char const* infix;
};
int main() {
std::cout << "please enter a positive number: "; mpz_class number; std::cin >> number; if (number <= 0) std::cout << "this number is not positive!\n;"; else { std::cout << "decomposition: "; decompose(number, infix_ostream_iterator<mpz_class>(std::cout, " * ")); std::cout << "\n"; }
}</lang>
Common Lisp
<lang Lisp>;;; Recursive algorithm (defun factor (n)
"Return a list of factors of N." (when (> n 1) (loop with max-d = (isqrt n)
for d = 2 then (if (evenp d) (+ d 1) (+ d 2)) do (cond ((> d max-d) (return (list n))) ; n is prime ((zerop (rem n d)) (return (cons d (factor (truncate n d)))))))))</lang>
E
This example assumes a function isPrime
and was tested with this one. It could use a self-referential implementation such as the Python task, but the original author of this example did not like the ordering dependency involved.
<lang e>def primes := {
var primesCache := [2] /** A collection of all prime numbers. */ def primes { to iterate(f) { primesCache.iterate(f) for x in (int > primesCache.last()) { if (isPrime(x)) { f(primesCache.size(), x) primesCache with= x } } } }
}
def primeDecomposition(var x :(int > 0)) {
var factors := [] for p in primes { while (x % p <=> 0) { factors with= p x //= p } if (x <=> 1) { break } } return factors
}</lang>
Factor
Factor already offers this functionality in its standard library. Example use: <lang factor>USING: math.primes.factors ; 12 factors . { 2 2 3 } 576460752303423487 factors . { 179951 3203431780337 }</lang>
FALSE
<lang false> [2[\$@$$*@>~][\$@$@$@$@\/*=$[%$." "$@\/\0~]?~[1+1|]?]#%.]d: 27720d;! {2 2 2 3 3 5 7 11} </lang>
Forth
: decomp ( n -- ) 2 begin 2dup dup * >= while 2dup /mod swap if drop 1+ 1 or \ next odd number else -rot nip dup . then repeat drop . ;
Fortran
<lang fortran>module PrimeDecompose
implicit none
integer, parameter :: huge = selected_int_kind(18) ! => integer(8) ... more fails on my 32 bit machine with gfortran(gcc) 4.3.2
contains
subroutine find_factors(n, d) integer(huge), intent(in) :: n integer, dimension(:), intent(out) :: d
integer(huge) :: div, next, rest integer :: i
i = 1 div = 2; next = 3; rest = n do while ( rest /= 1 ) do while ( mod(rest, div) == 0 ) d(i) = div i = i + 1 rest = rest / div end do div = next next = next + 2 end do
end subroutine find_factors
end module PrimeDecompose</lang>
<lang fortran>program Primes
use PrimeDecompose implicit none
integer, dimension(100) :: outprimes integer i
outprimes = 0
call find_factors(12345649494449_huge, outprimes)
do i = 1, 100 if ( outprimes(i) == 0 ) exit print *, outprimes(i) end do
end program Primes</lang>
Haskell
primes = sieve [2..] where sieve (p:xs) = p : sieve [x|x <- xs, x `mod` p > 0] factorize n pps@(p:ps) = case n `divMod` p of (0,1) -> [] (remainder,0) -> p : factorize remainder pps _ -> factorize n ps
J
q:
Example use:
q: 3684 2 2 3 307 _1+2^128x 340282366920938463463374607431768211455 q: _1+2^128x 3 5 17 257 641 65537 274177 6700417 67280421310721 */ q: _1+2^128x 340282366920938463463374607431768211455
Java
This is a version for arbitrary-precision integers which assumes the existence of a function with the signature: <lang java>public boolean prime(BigInteger i);</lang> You will need to import java.util.List, java.util.LinkedList, and java.math.BigInteger. <lang java>public static List<BigInteger> primeFactorBig(BigInteger a){
List<BigInteger> ans = new LinkedList<BigInteger>(); //loop until we test the number itself or the number is 1 for (BigInteger i = BigInteger.valueOf(2); i.compareTo(a) <= 0 && !a.equals(BigInteger.ONE); i = i.add(BigInteger.ONE)){ while (a.remainder(i).equals(BigInteger.ZERO) && prime(i)) { //if we have a prime factor ans.add(i); //put it in the list a = a.divide(i); //factor it out of the number } } return ans;
}</lang>
Alternate version, optimised to be faster. Uses java.util.List, java.util.LinkedList and java.math.BigInteger. <lang java>public boolean isPrime(BigInteger i);</lang> <lang java>
private static final BigInteger two = BigInteger.valueOf(2); public List<BigInteger> primeDecomp(BigInteger a) { //Is it even possible if(a.compareTo(two)==-1) { return null; //impossible for values lower than 2 } //quickly handle even values List<BigInteger> result = new c<BigInteger>(); while(a.and(BigInteger.ONE).equals(BigInteger.ONE)) { a = a.shiftRight(1); result.add(two); } //left with odd values if(!a.equals(BigInteger.ONE)) { BigInteger b = BigInteger.valueOf(3); while(b.compareTo(a)==-1) { if(isPrime(b)) { BigInteger[] dr = a.divideAndRemainder(b); if(dr[1].equals(BigInteger.ZERO)) { result.add(b); a = dr[0]; } } b = b.add(two); } result.add(b); //b will always be prime here... } return result; }
</lang>
JavaScript
This code uses the BigInteger Library jsbn and jsbn2
http://xenon.stanford.edu/~tjw/jsbn/jsbn.js
http://xenon.stanford.edu/~tjw/jsbn/jsbn2.js
function run_factorize(input, output) { var n = new BigInteger(input.value, 10); var prod = false; var two = new BigInteger("2", 10); var divisor = new BigInteger("3", 10); if(n.compareTo(two) < 0) { return; } output.value = ""; while(true) { var qr = n.divideAndRemainder(two); if(qr[1].compareTo(BigInteger.ZERO) == 0) { if(prod) { output.value += "*"; } else { prod = true; } output.value += "2"; n = qr[0]; } else { break; } } while(n.compareTo(BigInteger.ONE) != 0) { var qr = n.divideAndRemainder(divisor); if(qr[1].compareTo(BigInteger.ZERO) == 0) { if(prod) { output.value += "*"; } else { prod = true; } output.value += divisor; n = qr[0]; } else { divisor = divisor.add(two); } } }
Logo
<lang logo>to decompose :n [:p 2]
if :p*:p > :n [output (list :n)] if less? 0 modulo :n :p [output (decompose :n bitor 1 :p+1)] output fput :p (decompose :n/:p :p)
end</lang>
Mathematica
Bare built-in function does: <lang Mathematica> FactorInteger[2016] => {{2, 5}, {3, 2}, {7, 1}}</lang>
Read as: 2 to the power 5 times 3 squared times 7 (to the power 1). To show them nicely we could use the following functions: <lang Mathematica> supscript[x_,y_]:=If[y==1,x,Superscript[x,y]]
ShowPrimeDecomposition[input_Integer]:=Print@@{input," = ",Sequence@@Riffle[supscript@@@FactorInteger[input]," "]}</lang>
Example for small prime: <lang Mathematica> ShowPrimeDecomposition[1337]</lang> gives: <lang Mathematica> 1337 = 7 191</lang> Examples for large primes: <lang Mathematica> Table[AbsoluteTiming[ShowPrimeDecomposition[2^a-1]]//Print[#1," sec"]&,{a,50,150,10}];</lang> gives back: <lang Mathematica>1125899906842623 = 3 11 31 251 601 1801 4051 0.000231 sec 1152921504606846975 = 3^2 5^2 7 11 13 31 41 61 151 331 1321 0.000146 sec 1180591620717411303423 = 3 11 31 43 71 127 281 86171 122921 0.001008 sec 1208925819614629174706175 = 3 5^2 11 17 31 41 257 61681 4278255361 0.000340 sec 1237940039285380274899124223 = 3^3 7 11 19 31 73 151 331 631 23311 18837001 0.000192 sec 1267650600228229401496703205375 = 3 5^3 11 31 41 101 251 601 1801 4051 8101 268501 0.000156 sec 1298074214633706907132624082305023 = 3 11^2 23 31 89 683 881 2971 3191 201961 48912491 0.001389 sec 1329227995784915872903807060280344575 = 3^2 5^2 7 11 13 17 31 41 61 151 241 331 1321 61681 4562284561 0.000374 sec 1361129467683753853853498429727072845823 = 3 11 31 131 2731 8191 409891 7623851 145295143558111 0.024249 sec 1393796574908163946345982392040522594123775 = 3 5^2 11 29 31 41 43 71 113 127 281 86171 122921 7416361 47392381 0.009419 sec 1427247692705959881058285969449495136382746623 = 3^2 7 11 31 151 251 331 601 1801 4051 100801 10567201 1133836730401 0.007705 sec</lang>
MATLAB
<lang Matlab>function [outputPrimeDecomposition] = primedecomposition(inputValue)
outputPrimeDecomposition = factor(inputValue);</lang>
OCaml
<lang ocaml>open Big_int;;
let prime_decomposition x =
let rec inner c p = if lt_big_int p (square_big_int c) then [p] else if eq_big_int (mod_big_int p c) zero_big_int then c :: inner c (div_big_int p c) else inner (succ_big_int c) p in inner (succ_big_int (succ_big_int zero_big_int)) x;;</lang>
Octave
<lang octave>r = factor(120202039393); disp(r);</lang>
Python
<lang python>import sys, time
def is_prime(n):
return zip((True, False), decompose(n))[-1][0]
class IsPrimeCached(dict):
def __missing__(self, n): r = is_prime(n) self[n] = r return r
is_prime_cached = IsPrimeCached()
def primes():
yield 2 n = 3 while n < sys.maxint - 2: yield n n += 2 while n < sys.maxint - 2 and not is_prime_cached[n]: n += 2
def decompose(n):
for p in primes(): if p*p > n: break while n % p == 0: yield p n /=p if n > 1: yield n
- Example: calculate factors of Mersenne numbers to M59 #
for m in primes():
p = 2 ** m - 1 print "2**%d-1 = %d, with factors:"%(m, p), start = time.time() for factor in decompose(p): print factor, sys.stdout.flush() print "=> %.2fs"%(time.time()-start) if m >= 59: break</lang>
Output:
2**2-1 = 3, with factors: 3 => 0.00s 2**3-1 = 7, with factors: 7 => 0.00s 2**5-1 = 31, with factors: 31 => 0.00s 2**7-1 = 127, with factors: 127 => 0.00s 2**11-1 = 2047, with factors: 23 89 => 0.00s 2**13-1 = 8191, with factors: 8191 => 0.00s 2**17-1 = 131071, with factors: 131071 => 0.00s 2**19-1 = 524287, with factors: 524287 => 0.01s 2**23-1 = 8388607, with factors: 47 178481 => 0.00s 2**29-1 = 536870911, with factors: 233 1103 2089 => 0.01s 2**31-1 = 2147483647, with factors: 2147483647 => 1.67s 2**37-1 = 137438953471, with factors: 223 616318177 => 0.02s 2**41-1 = 2199023255551, with factors: 13367 164511353 => 0.01s 2**43-1 = 8796093022207, with factors: 431 9719 2099863 => 0.01s 2**47-1 = 140737488355327, with factors: 2351 4513 13264529 => 0.00s 2**53-1 = 9007199254740991, with factors: 6361 69431 20394401 => 1.17s 2**59-1 = 576460752303423487, with factors: 179951 3203431780337 => 211.07s
Note: Python took 740,238 BogoMI to complete the example.
R
<lang R>findfactors <- function(n) {
d <- c() div <- 2; nxt <- 3; rest <- n while( rest != 1 ) { while( rest%%div == 0 ) { d <- c(d, div) rest <- floor(rest / div) } div <- nxt nxt <- nxt + 2 } d
}
print(findfactors(1005025))</lang>
Ruby
<lang ruby># get prime decomposition of integer i
- this routine is terribly inefficient, but elegance rules :-)
def prime_factors(i)
v = (2..i-1).detect{|j| i % j == 0} v ? ([v] + prime_factors(i/v)) : [i]
end
- example: decompose all possible Mersenne primes up to 2**31-1
(2..31).each do |i|
puts "prime_factors(#{2**i-1}): #{prime_factors(2**i-1).join(' ')}"
end</lang>
A more efficient version, and quite similar to the Integer#prime_division method added by the
package in the Ruby stdlib:
<lang ruby>require 'mathn' def prime_factors(n)
factors = [] prime_number_generator = Prime.new p = prime_number_generator.next while p <= n q, r = n.divmod(p) if r == 0 factors << p n = q elsif p**2 >= n break else p = prime_number_generator.next end end factors << n if n > 1 factors
end
- example: decompose all possible Mersenne primes up to 2**31-1
results = [] png = Prime.new
require 'benchmark' Benchmark.bm(7) do |x|
begin i = png.next n = 2**i- 1 f = prime_factors(n) results << "%2d : %-20d : %s\n" % [i, n, f.inspect] x.report("new-#{i}") {prime_factors(n)} x.report("ruby-#{i}") {n.prime_division} end while i < 53
end puts results</lang>
user system total real new-2 0.000000 0.000000 0.000000 ( 0.000000) ruby-2 0.000000 0.000000 0.000000 ( 0.000000) new-3 0.000000 0.000000 0.000000 ( 0.000000) ruby-3 0.000000 0.000000 0.000000 ( 0.000000) new-5 0.000000 0.000000 0.000000 ( 0.000000) ruby-5 0.000000 0.000000 0.000000 ( 0.000000) new-7 0.000000 0.000000 0.000000 ( 0.000000) ruby-7 0.000000 0.000000 0.000000 ( 0.000000) new-11 0.000000 0.000000 0.000000 ( 0.000000) ruby-11 0.000000 0.000000 0.000000 ( 0.000000) new-13 0.000000 0.000000 0.000000 ( 0.000000) ruby-13 0.000000 0.000000 0.000000 ( 0.002000) new-17 0.000000 0.000000 0.000000 ( 0.005000) ruby-17 0.015000 0.000000 0.015000 ( 0.007000) new-19 0.016000 0.000000 0.016000 ( 0.013000) ruby-19 0.015000 0.000000 0.015000 ( 0.013000) new-23 0.000000 0.000000 0.000000 ( 0.006000) ruby-23 0.000000 0.000000 0.000000 ( 0.006000) new-29 0.047000 0.000000 0.047000 ( 0.024000) ruby-29 0.031000 0.000000 0.031000 ( 0.025000) new-31 13.016000 0.000000 13.016000 ( 13.390000) ruby-31 13.437000 0.000000 13.437000 ( 13.578000) new-37 5.031000 0.000000 5.031000 ( 4.639000) ruby-37 4.750000 0.000000 4.750000 ( 4.648000) new-41 1.594000 0.000000 1.594000 ( 1.668000) ruby-41 1.546000 0.000000 1.546000 ( 1.568000) new-43 0.844000 0.000000 0.844000 ( 0.879000) ruby-43 0.906000 0.000000 0.906000 ( 0.914000) new-47 0.265000 0.000000 0.265000 ( 0.256000) ruby-47 0.266000 0.000000 0.266000 ( 0.240000) new-53 27.938000 0.000000 27.938000 ( 28.369000) ruby-53 28.562000 0.000000 28.562000 ( 28.227000) 2 : 3 : [3] 3 : 7 : [7] 5 : 31 : [31] 7 : 127 : [127] 11 : 2047 : [23, 89] 13 : 8191 : [8191] 17 : 131071 : [131071] 19 : 524287 : [524287] 23 : 8388607 : [47, 178481] 29 : 536870911 : [233, 1103, 2089] 31 : 2147483647 : [2147483647] 37 : 137438953471 : [223, 616318177] 41 : 2199023255551 : [13367, 164511353] 43 : 8796093022207 : [431, 9719, 2099863] 47 : 140737488355327 : [2351, 4513, 13264529] 53 : 9007199254740991 : [6361, 69431, 20394401]
Slate
Admittedly, this is just based on the Smalltalk entry below: <lang slate>n@(Integer traits) primesDo: block "Decomposes the Integer into primes, applying the block to each (in increasing order)." [| div next remaining |
div: 2. next: 3. remaining: n. [[(remaining \\ div) isZero] whileTrue: [block applyTo: {div}.
remaining: remaining // div].
remaining = 1] whileFalse: [div: next. next: next + 2] "Just look at the next odd integer."
].</lang>
Smalltalk
<lang smalltalk> Integer extend [
primesDo: aBlock [ | div next rest | div := 2. next := 3. rest := self. [ [ rest \\ div == 0 ] whileTrue: [ aBlock value: div. rest := rest // div ]. rest = 1] whileFalse: [ div := next. next := next + 2 ] ] ] 123456 primesDo: [ :each | each printNl ]</lang>
Scheme
<lang scheme> (define (prime-decomposition x)
(let inner ((c 2) (p x)) (if (> (* c c) p) (list p) (if (zero? (remainder p c)) (cons c (inner c (quotient p c))) (inner (+ c 1) p)))))</lang>
Tcl
<lang tcl>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 $list] > 0} { set current_index 0 }
}
proc primes::is_prime {candidate} {
variable list
if {$candidate in $list} {return true} 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 $candidate
}
- return the prime factors of a number in a dictionary.
- The keys will be the factors, the value will be the number
- of times the factor divides the given number
- example: 120 = 2**3 * 3 * 5, so
- [primes::factors 120] returns 2 3 3 1 5 1
- so: set prod 1
- dict for {p e} [primes::factors 120] {
- set prod [expr {$prod * $p**$e}]
- }
- expr {$prod == 120} ;# ==> true
proc primes::factors {num} {
restart set factors [dict create] for {set i [get_next_prime]} {$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 [get_next_prime] } } return $factors
}</lang> Testing <lang tcl>primes::reset foreach m {2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59} {
set n [expr {2**$m - 1}] catch {time {set f [dict create {*}[primes::factors $n]]} 1} tm set primes [list] dict for {p e} $f {lappend primes {*}[lrepeat $e $p]} puts [format "2**%02d-1 = %-18s = %-22s => %s" $m $n [join $primes *] $tm]
}</lang> Outputs
2**02-1 = 3 = 3 => 20 microseconds per iteration 2**03-1 = 7 = 7 => 16 microseconds per iteration 2**05-1 = 31 = 31 => 27 microseconds per iteration 2**07-1 = 127 = 127 => 33 microseconds per iteration 2**11-1 = 2047 = 23*89 => 43 microseconds per iteration 2**13-1 = 8191 = 8191 => 159 microseconds per iteration 2**17-1 = 131071 = 131071 => 535 microseconds per iteration 2**19-1 = 524287 = 524287 => 911 microseconds per iteration 2**23-1 = 8388607 = 47*178481 => 162 microseconds per iteration 2**29-1 = 536870911 = 233*1103*2089 => 982 microseconds per iteration 2**31-1 = 2147483647 = 2147483647 => 138831 microseconds per iteration 2**37-1 = 137438953471 = 223*616318177 => 5154 microseconds per iteration 2**41-1 = 2199023255551 = 13367*164511353 => 2901 microseconds per iteration 2**43-1 = 8796093022207 = 431*9719*2099863 => 2141 microseconds per iteration 2**47-1 = 140737488355327 = 2351*4513*13264529 => 1102 microseconds per iteration 2**53-1 = 9007199254740991 = 6361*69431*20394401 => 97472 microseconds per iteration 2**59-1 = 576460752303423487 = 179951*3203431780337 => 12664437 microseconds per iteration
V
like in scheme (using variables)
[prime-decomposition [inner [c p] let [c c * p >] [p unit] [ [p c % zero?] [c c p c / inner cons] [c 1 + p inner] ifte] ifte]. 2 swap inner].
(mostly) the same thing using stack (with out variables)
[prime-decomposition [inner [dup * <] [pop unit] [ [% zero?] [ [p c : [c p c / c]] view i inner cons] [succ inner] ifte] ifte]. 2 inner].
Using it
|1221 prime-decomposition puts =[3 11 37]