Parallel calculations
You are encouraged to solve this task according to the task description, using any language you may know.
Many programming languages allow you to specify computations to be run in parallel. While Concurrent computing is focused on concurrency, the purpose of this task is to distribute time-consuming calculations on as many CPUs as possible.
Assume we have a collection of numbers, and want to find the one with the largest minimal prime factor (that is, the one that contains relatively large factors). To speed up the search, the factorization should be done in parallel using separate threads or processes, to take advantage of multi-core CPUs.
Show how this can be formulated in your language. Parallelize the factorization of those numbers, then search the returned list of numbers and factors for the largest minimal factor, and return that number and its prime factors.
For the prime number decomposition you may use the solution of the Prime decomposition task.
Ada
I took the version from Prime decomposition and adjusted it to use tasks.
prime_numbers.ads: <lang Ada>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;
procedure Put (List : Number_List);
task type Calculate_Factors is
entry Start (The_Number : in Number);
entry Get_Size (Size : out Natural);
entry Get_Result (List : out Number_List);
end Calculate_Factors;
end Prime_Numbers;</lang>
prime_numbers.adb: <lang Ada>with Ada.Text_IO; package body Prime_Numbers is
procedure Put (List : Number_List) is
begin
for Index in List'Range loop
Ada.Text_IO.Put (Image (List (Index)));
end loop;
end Put;
task body Calculate_Factors is
Size : Natural := 0;
N : Number;
M : Number;
K : Number := Two;
begin
accept Start (The_Number : in Number) do
N := The_Number;
M := N;
end Start;
-- 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;
Index := Index - 1;
accept Get_Size (Size : out Natural) do
Size := Index;
end Get_Size;
accept Get_Result (List : out Number_List) do
List (1 .. Index) := Result (1 .. Index);
end Get_Result;
end;
end Calculate_Factors;
end Prime_Numbers;</lang>
Example usage:
parallel.adb: <lang Ada>with Ada.Text_IO; with Prime_Numbers; procedure Parallel is
package Integer_Primes is new Prime_Numbers (
Number => Integer, -- use Large_Integer for longer numbers
Zero => 0,
One => 1,
Two => 2,
Image => Integer'Image);
My_List : Integer_Primes.Number_List :=
( 12757923,
12878611,
12757923,
15808973,
15780709,
197622519);
Decomposers : array (My_List'Range) of Integer_Primes.Calculate_Factors; Lengths : array (My_List'Range) of Natural; Max_Length : Natural := 0;
begin
for I in My_List'Range loop
-- starts the tasks
Decomposers (I).Start (My_List (I));
end loop;
for I in My_List'Range loop
-- wait until task has reached Get_Size entry
Decomposers (I).Get_Size (Lengths (I));
if Lengths (I) > Max_Length then
Max_Length := Lengths (I);
end if;
end loop;
declare
Results :
array (My_List'Range) of Integer_Primes.Number_List (1 .. Max_Length);
Largest_Minimal_Factor : Integer := 0;
Winning_Index : Positive;
begin
for I in My_List'Range loop
-- after Get_Result, the tasks terminate
Decomposers (I).Get_Result (Results (I));
if Results (I) (1) > Largest_Minimal_Factor then
Largest_Minimal_Factor := Results (I) (1);
Winning_Index := I;
end if;
end loop;
Ada.Text_IO.Put_Line
("Number" & Integer'Image (My_List (Winning_Index)) &
" has largest minimal factor:");
Integer_Primes.Put (Results (Winning_Index) (1 .. Lengths (Winning_Index)));
Ada.Text_IO.New_Line;
end;
end Parallel;</lang>
Output:
Number 12878611 has largest minimal factor: 47 101 2713
C
C code using OpenMP. Compiled with gcc -Wall -std=c99 -fopenmp, where GCC 4.2 or later is required. Note that the code finds the largest first prime factor, but does not return the factor list: it's just a matter of repeating the prime factor test, which adds clutter but does not make the code any more interesting. For that matter, the code uses the dumbest prime factoring method, and doesn't even test if the numbers can be divided by 2.
<lang C>#include <stdio.h>
- include <omp.h>
int main() {
int data[] = {12757923, 12878611, 12878893, 12757923, 15808973, 15780709, 197622519};
int largest, largest_factor = 0;
omp_set_num_threads(4);
/* "omp parallel for" turns the for loop multithreaded by making each thread
* iterating only a part of the loop variable, in this case i; variables declared
* as "shared" will be implicitly locked on access
*/
#pragma omp parallel for shared(largest_factor, largest)
for (int i = 0; i < 7; i++) {
int p, n = data[i];
for (p = 3; p * p <= n && n % p; p += 2);
if (p * p > n) p = n;
if (p > largest_factor) {
largest_factor = p;
largest = n;
printf("thread %d: found larger: %d of %d\n",
omp_get_thread_num(), p, n);
} else {
printf("thread %d: not larger: %d of %d\n",
omp_get_thread_num(), p, n);
}
}
printf("Largest factor: %d of %d\n", largest_factor, largest);
return 0;
}</lang>Output (YMMV regarding the order of output):<lang>thread 1: found larger: 47 of 12878893 thread 0: not larger: 3 of 12757923 thread 0: not larger: 47 of 12878611 thread 3: not larger: 3 of 197622519 thread 2: not larger: 29 of 15808973 thread 2: not larger: 7 of 15780709 thread 1: not larger: 3 of 12757923 Largest factor: 47 of 12878893</lang>
C++
This uses C++0x features including lambda functions.
<lang cpp>#include <iostream>
- include <iterator>
- include <vector>
- include <ppl.h> // MSVC++
- include <concurrent_vector.h> // MSVC++
struct Factors {
int number; std::vector<int> primes;
};
const int data[] = {
12757923, 12878611, 12878893, 12757923, 15808973, 15780709, 197622519
};
int main() {
// concurrency-safe container replaces std::vector<> Concurrency::concurrent_vector<Factors> results;
// parallel algorithm replaces std::for_each()
Concurrency::parallel_for_each(std::begin(data), std::end(data), [&](int n)
{
Factors factors;
factors.number = n;
for (int f = 2; n > 1; ++f)
{
while (n % f == 0)
{
factors.primes.push_back(f);
n /= f;
}
}
results.push_back(factors); // add factorization to results
});
// end of parallel calculations
// find largest minimal prime factor in results
auto max = std::max_element(results.begin(), results.end(), [](const Factors &a, const Factors &b)
{
return a.primes.front() < b.primes.front();
});
// print number(s) and factorization
std::for_each(results.begin(), results.end(), [&](const Factors &f)
{
if (f.primes.front() == max->primes.front())
{
std::cout << f.number << " = [ ";
std::copy(f.primes.begin(), f.primes.end(), std::ostream_iterator<int>(std::cout, " "));
std::cout << "]\n";
}
});
}</lang> Output:
12878611 = [ 47 101 2713 ] 12878893 = [ 47 274019 ]
C#
<lang csharp>using System; using System.Collections.Generic; using System.Linq; using System.Threading.Tasks;
private static void Main(string[] args) {
int j = 0, m = 0;
decimal[] n = {12757923, 12878611, 12757923, 15808973, 15780709, 197622519};
var l = new List<int>[n.Length];
Parallel.For(0, n.Length, i => { l[i] = getPrimes(n[i]); });
for (int i = 0; i<n.Length; i++)
if (l[i].Min()>m)
{
m = l[i].Min();
j = i;
}
Console.WriteLine("Number {0} has largest minimal factor:", n[j]);
foreach (int list in l[j])
Console.Write(" "+list);
}</lang>
Number 12878611 has largest minimal factor: 47 101 2713
Clojure
<lang Clojure>(use '[clojure.contrib.lazy-seqs :only [primes]])
(defn lpf [n]
[n (or (last
(for [p (take-while #(<= (* % %) n) primes)
:when (zero? (rem n p))]
p))
1)])
(->> (range 2 100000)
(pmap lpf)
(apply max-key second)
println
time)</lang>
Output:
[99847 313] "Elapsed time: 2547.53566 msecs"
D
<lang d>import std.stdio, std.math, std.algorithm, std.typecons; import core.thread, core.stdc.time;
final class MinFactor: Thread {
private ulong num ; private ulong[] fac; private ulong minFac ;
this(ulong n) nothrow {
super(&run);
num = n;
fac = new ulong[](0) ;
}
ulong number() @property { return num ; }
ulong[] factors() @property { return fac.dup ; }
ulong minFactor() @property { return minFac ; }
private void run() {
immutable clock_t begin = clock();
switch(num) {
case 0: fac = [] ; break ;
case 1: fac = [1] ; break ;
default:
uint limit = cast(uint) (1 + sqrt(num)) ;
ulong n = num ;
for (ulong div = 3; div < limit; div += 2) {
if(n == 1) break ;
if((n % div) == 0) {
while((n > 1) && ((n % div) == 0)) {
fac ~= div ;
n /= div ;
}
limit = cast(uint) (1 + sqrt(n)) ;
}
}
if(n > 1)
fac ~= n ;
}
minFac = reduce!min(fac) ;
immutable clock_t end = clock();
writefln("num: %20d --> min. factor: %20d ticks(%7d -> %7d)",
num, minFac, begin, end);
}
}
void main() {
auto numbers = [2UL^^59-1, 2UL^^59-1, 2UL^^59-1, 112272537195293UL,
115284584522153, 115280098190773, 115797840077099,
112582718962171, 112272537095293, 1099726829285419];
auto tgroup = new ThreadGroup;
foreach (n; numbers)
tgroup.add(new MinFactor(n));
writeln("Minimum factors for respective numbers are:");
foreach (t; tgroup)
t.start();
tgroup.joinAll();
Tuple!(ulong, ulong[], ulong) maxmin ;
maxmin = tuple(0UL, [0UL], 0UL) ;
foreach (t; tgroup) {
auto s = cast(MinFactor)t;
if (s !is null && maxmin[2] < s.minFactor)
maxmin = tuple(s.number, s.factors, s.minFactor) ;
}
writefln("Number with largest min. factor is %16d, with factors:\n\t%s",
maxmin[0], maxmin[1]);
}</lang> Output (1 core CPU, edited to fit page width):
Minimum factors for respective numbers are:
num: 576460752303423487 --> min. factor: 179951 ticks( 16 -> 78)
num: 576460752303423487 --> min. factor: 179951 ticks( 78 -> 125)
num: 576460752303423487 --> min. factor: 179951 ticks( 141 -> 203)
num: 112272537195293 --> min. factor: 173 ticks( 203 -> 203)
num: 115284584522153 --> min. factor: 513937 ticks( 203 -> 219)
num: 115280098190773 --> min. factor: 513917 ticks( 219 -> 250)
num: 115797840077099 --> min. factor: 544651 ticks( 250 -> 266)
num: 112582718962171 --> min. factor: 3121 ticks( 266 -> 266)
num: 112272537095293 --> min. factor: 131 ticks( 266 -> 266)
num: 1099726829285419 --> min. factor: 271 ticks( 266 -> 266)
Number with largest min. factor is 115797840077099, with factors:
[544651, 212609249]
F#
<lang fsharp>open System open PrimeDecomp // Has the decompose function from the Prime decomposition task
let data = [112272537195293L; 112582718962171L; 112272537095293L; 115280098190773L; 115797840077099L; 1099726829285419L] let decomp num = decompose num 2L
let largestMinPrimeFactor (numbers: int64 list) =
let decompDetails = Async.Parallel [ for n in numbers -> async { return n, decomp n } ] // Compute the number and its prime decomposition list
|> Async.RunSynchronously // Start and wait for all parallel computations to complete.
|> Array.sortBy (snd >> List.min >> (~-)) // Sort in descending order, based on the min prime decomp number.
decompDetails.[0]
let showLargestMinPrimeFactor numbers =
let number, primeList = largestMinPrimeFactor numbers printf "Number %d has largest minimal factor:\n " number List.iter (printf "%d ") primeList
showLargestMinPrimeFactor data</lang>
Output:
Number 115797840077099 has largest minimal factor:
544651 212609249
Go
<lang go>package main
import (
"big" "fmt"
)
// collection of numbers. A slice is used for the collection. // The elements are big integers, since that's what the function Primes // uses (as was specified by the Prime decomposition task.) var numbers = []*big.Int{
big.NewInt(12757923), big.NewInt(12878611), big.NewInt(12878893), big.NewInt(12757923), big.NewInt(15808973), big.NewInt(15780709),
}
// main just calls the function specified by the task description and // prints results. note it allows for multiple numbers with the largest // minimal factor. the task didn't specify to handle this, but obviously // it's possible. func main() {
rs := lmf(numbers)
fmt.Println("largest minimal factor:", rs[0].decomp[0])
for _, r := range rs {
fmt.Println(r.number, "->", r.decomp)
}
}
// this type associates a number with it's prime decomposition. // the type is neccessary so that they can be sent together over // a Go channel, but it turns out to be convenient as well for // the return type of lmf. type result struct {
number *big.Int decomp []*big.Int
}
// the function specified by the task description, "largest minimal factor." func lmf([]*big.Int) []result {
// construct result channel and start a goroutine to decompose each number.
// goroutines run in parallel as CPU cores are available.
rCh := make(chan result)
for _, n := range numbers {
go decomp(n, rCh)
}
// collect results. <-rCh returns a single result from the result channel.
// we know how many results to expect so code here collects exactly that
// many results, and accumulates a list of those with the largest
// minimal factor.
rs := []result{<-rCh}
for i := 1; i < len(numbers); i++ {
switch r := <-rCh; r.decomp[0].Cmp(rs[0].decomp[0]) {
case 1:
rs = rs[:1]
rs[0] = r
case 0:
rs = append(rs, r)
}
}
return rs
}
// decomp is the function run as a goroutine. multiple instances of this // function will run concurrently, one for each number being decomposed. // it acts as a driver for Primes, calling Primes as needed, packaging // the result, and sending the packaged result on the channel. // "as needed" turns out to mean sending Primes a copy of n, as Primes // as written is destructive on its argument. func decomp(n *big.Int, rCh chan result) {
rCh <- result{n, Primes(new(big.Int).Set(n))}
}
// code below copied from Prime decomposition task var (
ZERO = big.NewInt(0) ONE = big.NewInt(1)
)
func Primes(n *big.Int) []*big.Int {
res := []*big.Int{}
mod, div := new(big.Int), new(big.Int)
for i := big.NewInt(2); i.Cmp(n) != 1; {
div.DivMod(n, i, mod)
for mod.Cmp(ZERO) == 0 {
res = append(res, new(big.Int).Set(i))
n.Set(div)
div.DivMod(n, i, mod)
}
i.Add(i, ONE)
}
return res
}</lang> Output:
largest minimal factor: 47 12878611 -> [47 101 2713] 12878893 -> [47 274019]
JavaScript
This code demonstrates Web Workers. This should work on current versions of Firefox, Safari, Chrome and Opera.
This first portion should be placed in a file called "parallel_worker.js". This file contains the logic used by every worker created. <lang javascript> var onmessage = function(event) {
postMessage({"n" : event.data.n,
"factors" : factor(event.data.n),
"id" : event.data.id});
};
function factor(n) {
var factors = [];
for(p = 2; p <= n; p++) {
if((n % p) == 0) {
factors[factors.length] = p;
n /= p;
}
}
return factors;
} </lang>
This portion appends each of the numbers with their factors as the results arrive into an html tag with the id "result". Once the final worker completes its task (worker_count is reduced to 0), the reduce function is called to determine which number is the answer. <lang javascript> var numbers = [12757923, 12878611, 12757923, 15808973, 15780709, 197622519]; var workers = []; var worker_count = 0;
var results = [];
for(var i = 0; i < numbers.length; i++) {
worker_count++;
workers[i] = new Worker("parallel_worker.js");
workers[i].onmessage = accumulate;
workers[i].postMessage({n: numbers[i], id: i});
}
function accumulate(event) {
n = event.data.n;
factors = event.data.factors;
id = event.data.id;
result = document.getElementById("result");
ne = document.createElement("p");
ne.appendChild(document.createTextNode(n + " : " + factors));
result.appendChild(ne);
results[id] = {n:n, factors:factors};
// Cleanup - kill the worker and countdown until all work is done
workers[id].terminate();
worker_count--;
if(worker_count == 0)
reduce();
}
function reduce() {
answer = 0;
for(i = 1; i < results.length; i++) {
min = results[i].factors[0];
largest_min = results[answer].factors[0];
if(min > largest_min)
answer = i;
}
n = results[answer].n;
factors = results[answer].factors;
result = document.getElementById("result");
ne = document.createElement("p");
ne.appendChild(document.createTextNode("The number with the relatively largest factors is: " + n + " : " + factors));
result.appendChild(ne);
results[id] = {n:n, factors:factors};
}
</lang>
Perl 6
Assuming that factors is defined exactly as in the prime decomposition task: <lang perl6>my @nums = 12757923, 12878611, 123456789, 15808973, 15780709, 197622519;
my @factories; @factories[$_] := factors(@nums[$_]) for ^@nums; my $gmf = ([max] @factories»[0] »=>« @nums).value; </lang> The line with the for loop is just setting up a bunch of lazy lists, one for each number to be factored, but doesn't actually do any of the work of factoring. Most of the parallelizing work is done by the hyperoperators that demand the first value from each of the factories' lists, then builds (again in parallel) the pairs associating each first value with its original value. The [max] reduction finds the pair with the largest key, from which we can easily extract the greatest minimum factor candidate, and then refactor it completely.
The rakudo system does not actually do hypers in parallel yet, but when it does, this can automatically parallelize. (Hypers do parallelize in pugs, but it doesn't do some of the other things we rely on here.) It will be up to each individual compiler to determine how many cores to use for any given hyperoperator; the construct merely promises the compiler that it can be parallelized, but does not require that it must be.
There is also some pipelining that can happen within the factors routine itself, which uses a gather/take construct, which the compiler may implement using either coroutines or threads as it sees fit. Threading pipelines can make more sense on, say, a cell architecture.
PicoLisp
The 'later' function is used in PicoLisp to start parallel computations. The following solution calls 'later' on the 'factor' function from Prime decomposition#PicoLisp, and then 'wait's until all results are available: <lang PicoLisp>(let Lst
(mapcan
'((N)
(later (cons) # When done,
(cons N (factor N)) ) ) # return the number and its factors
(quote
188573867500151328137405845301 # Process a collection of 12 numbers
3326500147448018653351160281
979950537738920439376739947
2297143294659738998811251
136725986940237175592672413
3922278474227311428906119
839038954347805828784081
42834604813424961061749793
2651919914968647665159621
967022047408233232418982157
2532817738450130259664889
122811709478644363796375689 ) )
(wait NIL (full Lst)) # Wait until all computations are done
(maxi '((L) (apply min L)) Lst) ) # Result: Number in CAR, factors in CDR</lang>
Output:
-> (2532817738450130259664889 6531761 146889539 2639871491)
Prolog
This piece needs prime_decomp definition from the Prime decomposition#Prolog example, it worked on my swipl, but I don't know how other Dialects thread.
<lang Prolog>threaded_decomp(Number,ID):- thread_create( (prime_decomp(Number,Y), thread_exit((Number,Y))) ,ID,[]).
threaded_decomp_list(List,Erg):- maplist(threaded_decomp,List,IDs), maplist(thread_join,IDs,Results), maplist(pack_exit_out,Results,Smallest_Factors_List), largest_min_factor(Smallest_Factors_List,Erg).
pack_exit_out(exited(X),X). %Note that here some error handling should happen.
largest_min_factor([(N,Facs)|A],(N2,Fs2)):- min_list(Facs,MF), largest_min_factor(A,(N,MF,Facs),(N2,_,Fs2)).
largest_min_factor([],Acc,Acc). largest_min_factor([(N1,Facs1)|Rest],(N2,MF2,Facs2),Goal):- min_list(Facs1, MF1), (MF1 > MF2-> largest_min_factor(Rest,(N1,MF1,Facs1),Goal); largest_min_factor(Rest,(N2,MF2,Facs2),Goal)).
format_it(List):-
threaded_decomp_list(List,(Number,Factors)),
format('Number with largest minimal Factor is ~w\nFactors are ~w\n',
[Number,Factors]).
</lang>
Example (Numbers Same as in Ada Example):
?- ['prime_decomp.prolog', parallel].
% prime_decomp.prolog compiled 0.00 sec, 3,392 bytes
% parallel compiled 0.00 sec, 4,672 bytes
true.
format_it([12757923,
12878611,
12757923,
15808973,
15780709,
197622519]).
Number with largest minimal factor is 12878611
Factors are [2713, 101, 47]
true.
PureBasic
<lang PureBasic>Structure IO_block
ThreadID.i StartSeamaphore.i Value.q MinimumFactor.i List Factors.i()
EndStructure
- \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
Declare Factorize(*IO.IO_block) Declare main()
- \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
Main() End
- \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
Procedure Main()
Protected AvailableCpu, MainSemaphore
Protected i, j, qData.q, Title$, Message$
NewList T.IO_block()
;
AvailableCpu = Val(GetEnvironmentVariable("NUMBER_OF_PROCESSORS"))
If AvailableCpu<1: AvailableCpu=1: EndIf
MainSemaphore = CreateSemaphore(AvailableCpu)
;
Restore Start_of_data
For i=1 To (?end_of_data-?Start_of_data) / SizeOf(Quad)
; Start all threads at ones, they will then be let to
; self-oganize according to the availiable Cores.
AddElement(T())
Read.q qData
T()\Value = qData
T()\StartSeamaphore = MainSemaphore
T()\ThreadID = CreateThread(@Factorize(), @T())
Next
;
ForEach T()
; Wait for all threads to complete their work and
; find the smallest factor from eact task.
WaitThread(T()\ThreadID)
Next
;
i = OffsetOf(IO_block\MinimumFactor)
SortStructuredList(T(), #PB_Sort_Integer, i, #PB_Sort_Descending)
FirstElement(T())
Title$="Info"
Message$="Number "+Str(T()\Value)+" has largest minimal factor:"+#CRLF$
ForEach T()\Factors()
Message$ + Str(T()\Factors())+" "
Next
MessageRequester(Title$, Message$)
EndProcedure
ProcedureDLL Factorize(*IO.IO_block) ; Fill list Factors() with the factor parts of Number
;Based on http://rosettacode.org/wiki/Prime_decomposition#PureBasic With *IO Protected Value.q=\Value WaitSemaphore(\StartSeamaphore) Protected I = 3 ClearList(\Factors()) While Value % 2 = 0 AddElement(\Factors()) \Factors() = 2 Value / 2 Wend Protected Max = Value While I <= Max And Value > 1 While Value % I = 0 AddElement(\Factors()) \Factors() = I Value / I Wend I + 2 Wend SortList(\Factors(), #PB_Sort_Ascending) FirstElement(\Factors()) \MinimumFactor=\Factors() SignalSemaphore(\StartSeamaphore) EndWith ;*IO
EndProcedure
DataSection
Start_of_data: ; Same numbers as Ada Data.q 12757923, 12878611, 12757923, 15808973, 15780709, 197622519 end_of_data:
EndDataSection
</lang>
Python
Python 3.2 has a new concurrent.futures module that allows for the easy specification of thread-parallel or process-parallel processes. The following is modified from their example and will run, by default, with as many processes as there are available cores on your machine.
Note that there is no need to calculate all prime factors of all NUMBERS when only the prime factors of the number with the lowest overall prime factor is needed.
<lang python>from concurrent import futures
from math import floor, sqrt
NUMBERS = [
112272537195293, 112582718962171, 112272537095293, 115280098190773, 115797840077099, 1099726829285419]
- NUMBERS = [33, 44, 55, 275]
def lowest_factor(n, _start=3):
if n % 2 == 0:
return 2
search_max = int(floor(sqrt(n))) + 1
for i in range(_start, search_max, 2):
if n % i == 0:
return i
return n
def prime_factors(n, lowest):
pf = []
while n > 1:
pf.append(lowest)
n //= lowest
lowest = lowest_factor(n, max(lowest, 3))
return pf
def prime_factors_of_number_with_lowest_prime_factor(NUMBERS):
with futures.ProcessPoolExecutor() as executor:
low_factor, number = max( (l, f) for l, f in zip(executor.map(lowest_factor, NUMBERS), NUMBERS) )
all_factors = prime_factors(number, low_factor)
return number, all_factors
def main():
print( 'For these numbers:\n ' + '\n '.join(str(p) for p in NUMBERS) )
number, all_factors = prime_factors_of_number_with_lowest_prime_factor(NUMBERS)
print(' The one with the largest minimum prime factor is %i:' % number)
print(' All its prime factors in order are: %s' % all_factors)
if __name__ == '__main__':
main()</lang>
Sample Output
For these numbers:
112272537195293
112582718962171
112272537095293
115280098190773
115797840077099
1099726829285419
The one with the largest minimum prime factor is 115797840077099:
All its prime factors in order are: [544651, 212609249]
Tcl
With Tcl, it is necessary to explicitly perform computations in other threads because each thread is strongly isolated from the others (except for inter-thread messaging). However, it is entirely practical to wrap up the communications so that only a small part of the code needs to know very much about it, and in fact most of the complexity is managed by a thread pool; each value to process becomes a work item to be handled. It is easier to transfer the results by direct messaging instead of collecting the thread pool results, since we can leverage Tcl's vwait command nicely.
<lang tcl>package require Tcl 8.6 package require Thread
- Pooled computation engine; runs event loop internally
namespace eval pooled {
variable poolSize 3; # Needs to be tuned to system size
proc computation {computationDefinition entryPoint values} {
variable result variable poolSize # Add communication shim append computationDefinition [subst -nocommands { proc poolcompute {value target} { set outcome [$entryPoint \$value] set msg [list set ::pooled::result(\$value) \$outcome] thread::send -async \$target \$msg } }]
# Set up the pool set pool [tpool::create -initcmd $computationDefinition \ -maxworkers $poolSize]
# Prepare to receive results unset -nocomplain result array set result {}
# Dispatch the computations foreach value $values { tpool::post $pool [list poolcompute $value [thread::id]] }
# Wait for results while {[array size result] < [llength $values]} {vwait pooled::result}
# Dispose of the pool tpool::release $pool
# Return the results return [array get result]
}
}</lang> This is the definition of the prime factorization engine (a somewhat stripped-down version of the Tcl Prime decomposition solution: <lang tcl># Code for computing the prime factors of a number set computationCode {
namespace eval prime {
variable primes [list 2 3 5 7 11] proc restart {} { variable index -1 variable primes variable current [lindex $primes end] }
proc get_next_prime {} { variable primes variable index if {$index < [llength $primes]-1} { return [lindex $primes [incr index]] } variable current while 1 { incr current 2 set p 1 foreach prime $primes { if {$current % $prime} {} else { set p 0 break } } if {$p} { return [lindex [lappend primes $current] [incr index]] } } }
proc factors {num} { restart set factors [dict create] for {set i [get_next_prime]} {$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 }
}
}
- The values to be factored
set values {
188573867500151328137405845301 3326500147448018653351160281 979950537738920439376739947 2297143294659738998811251 136725986940237175592672413 3922278474227311428906119 839038954347805828784081 42834604813424961061749793 2651919914968647665159621 967022047408233232418982157 2532817738450130259664889 122811709478644363796375689
}</lang> Putting everything together: <lang tcl># Do the computation, getting back a dictionary that maps
- values to its results (itself an ordered dictionary)
set results [pooled::computation $computationCode prime::factors $values]
- Find the maximum minimum factor with sorting magic
set best [lindex [lsort -integer -stride 2 -index {1 0} $results] end-1]
- Print in human-readable form
proc renderFactors {factorDict} {
dict for {factor times} $factorDict {
lappend v {*}[lrepeat $times $factor]
} return [join $v "*"]
} puts "$best = [renderFactors [dict get $results $best]]"</lang>