Partition function P

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


The Partition Function P, often notated P(n) is the number of solutions where n∈ℤ can be expressed as the sum of a set of positive integers.


Example
 P(4) = 5   because   4 = Σ(4) = Σ(3,1) = Σ(2,2) = Σ(2,1,1) = Σ(1,1,1,1)


P(n) can be expressed as the recurrence relation:

 P(n) = P(n-1) +P(n-2) -P(n-5) -P(n-7) +P(n-12) +P(n-15) -P(n-22) -P(n-26) +P(n-35) +P(n-40) ...

The successive numbers in the above equation have the differences:   1, 3, 2, 5, 3, 7, 4, 9, 5, 11, 6, 13, 7, 15, 8 ...

This task may be of popular interest because Mathologer made the video, The hardest "What comes next?" (Euler's pentagonal formula), where he asks the programmers among his viewers to calculate P(666). The video has been viewed more than 100,000 times in the first couple of weeks since its release.

In Wolfram Language, this function has been implemented as PartitionsP.


Task

Write a function which returns the value of PartitionsP(n). Solutions can be iterative or recursive.

Bonus task: show how long it takes to compute PartitionsP(6666).


References


Related tasks




11l[edit]

Translation of: Python: Alternative
F partitions(n)
   V p = [BigInt(1)] [+] [BigInt(0)] * n
   L(i) 1 .. n
      V k = 0
      L
         k++
         V j = (k * (3 * k - 1)) I/ 2
         I j > i
            L.break
         I k [&] 1
            p[i] += p[i - j]
         E
            p[i] -= p[i - j]
         j = (k * (3 * k + 1)) I/ 2
         I j > i
            L.break
         I k [&] 1
            p[i] += p[i - j]
         E
            p[i] -= p[i - j]
   R p[n]

print(‘Partitions: ’(0.<15).map(x -> partitions(x)))

V start = time:perf_counter()
print(partitions(6666))
print(time:perf_counter() - start)
Output:
Partitions: [1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135]
193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
0.598528

C[edit]

Library: GMP
#include <stdint.h>
#include <stdlib.h>
#include <stdio.h>
#include <time.h>
#include <gmp.h>

mpz_t* partition(uint64_t n) {
	mpz_t *pn = (mpz_t *)malloc((n + 2) * sizeof(mpz_t));
	mpz_init_set_ui(pn[0], 1);
	mpz_init_set_ui(pn[1], 1);
	for (uint64_t i = 2; i < n + 2; i ++) {
		mpz_init(pn[i]);
		for (uint64_t k = 1, penta; ; k++) {
			penta = k * (3 * k - 1) >> 1;
			if (penta >= i) break;
			if (k & 1) mpz_add(pn[i], pn[i], pn[i - penta]);
			else mpz_sub(pn[i], pn[i], pn[i - penta]);
			penta += k;
			if (penta >= i) break;
			if (k & 1) mpz_add(pn[i], pn[i], pn[i - penta]);
			else mpz_sub(pn[i], pn[i], pn[i - penta]);
		}
	}
	mpz_t *tmp = &pn[n + 1];
	for (uint64_t i = 0; i < n + 1; i ++) mpz_clear(pn[i]);
	free(pn);
	return tmp;
}

int main(int argc, char const *argv[]) {
	clock_t start = clock();
	mpz_t *p = partition(6666);
	gmp_printf("%Zd\n", p);
	printf("Elapsed time: %.04f seconds\n",
		(double)(clock() - start) / (double)CLOCKS_PER_SEC);
	return 0;
}
Output:
193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
Elapsed time: 0.0136 seconds

C#[edit]

This can also be done using BigIntegers, but will take around five times longer. Since only adding and subtracting are required, some simple routines can be created to handle the tabulations. Also note that detecting odd and even numbers on each loop iteration is avoided by coding four increments per loop.

using System;

class Program {

    const long Lm = (long)1e18;
    const string Fm = "D18";

    // provides 5 x 18 = 90 decimal digits
    struct LI { public long lo, ml, mh, hi, tp; }

    static void inc(ref LI d, LI s) { // d += s
        if ((d.lo += s.lo) >= Lm) { d.ml++; d.lo -= Lm; }
        if ((d.ml += s.ml) >= Lm) { d.mh++; d.ml -= Lm; }
        if ((d.mh += s.mh) >= Lm) { d.hi++; d.mh -= Lm; }
        if ((d.hi += s.hi) >= Lm) { d.tp++; d.hi -= Lm; }
        d.tp += s.tp;
    }
 
    static void dec(ref LI d, LI s) { // d -= s
        if ((d.lo -= s.lo) < 0) { d.ml--; d.lo += Lm; }
        if ((d.ml -= s.ml) < 0) { d.mh--; d.ml += Lm; }
        if ((d.mh -= s.mh) < 0) { d.hi--; d.mh += Lm; }
        if ((d.hi -= s.hi) < 0) { d.tp--; d.hi += Lm; }
        d.tp -= s.tp;
    }

    static LI set(long s) { LI d;
      d.lo = s; d.ml = d.mh = d.hi = d.tp = 0; return d; }

  static string fmt(LI x) { // returns formatted string value of x
    if (x.tp > 0) return x.tp.ToString() + x.hi.ToString(Fm) + x.mh.ToString(Fm) + x.ml.ToString(Fm) + x.lo.ToString(Fm);
    if (x.hi > 0) return x.hi.ToString() + x.mh.ToString(Fm) + x.ml.ToString(Fm) + x.lo.ToString(Fm);
    if (x.mh > 0) return x.mh.ToString() + x.ml.ToString(Fm) + x.lo.ToString(Fm);
    if (x.ml > 0) return x.ml.ToString() + x.lo.ToString(Fm);
    return x.lo.ToString();
  }

  static LI partcount(int n) {
    var P = new LI[n + 1]; P[0] = set(1);
    for (int i = 1; i <= n; i++) {
      int k = 0, d = -2, j = i;
      LI x = set(0);
      while (true) {
        if ((j -= (d += 3) -k) >= 0) inc(ref x, P[j]); else break;
        if ((j -= ++k)         >= 0) inc(ref x, P[j]); else break;
        if ((j -= (d += 3) -k) >= 0) dec(ref x, P[j]); else break;
        if ((j -= ++k)         >= 0) dec(ref x, P[j]); else break;
      }
      P[i] = x;
    }
    return P[n];
  }

  static void Main(string[] args) {
    var sw = System.Diagnostics.Stopwatch.StartNew ();
    var res = partcount(6666); sw.Stop();
    Console.Write("{0}  {1} ms", fmt(res), sw.Elapsed.TotalMilliseconds);
  }
}
Output:
193655306161707661080005073394486091998480950338405932486880600467114423441282418165863  12.9365 ms

Timing from Tio.run.

C++[edit]

GMP version[edit]

Library: GMP
#include <chrono>
#include <iostream>
#include <vector>
#include <gmpxx.h>

using big_int = mpz_class;

big_int partitions(int n) {
    std::vector<big_int> p(n + 1);
    p[0] = 1;
    for (int i = 1; i <= n; ++i) {
        for (int k = 1;; ++k) {
            int j = (k * (3*k - 1))/2;
            if (j > i)
                break;
            if (k & 1)
                p[i] += p[i - j];
            else
                p[i] -= p[i - j];
            j = (k * (3*k + 1))/2;
            if (j > i)
                break;
            if (k & 1)
                p[i] += p[i - j];
            else
                p[i] -= p[i - j];
        }
    }
    return p[n];
}

int main() {
    auto start = std::chrono::steady_clock::now();
    auto result = partitions(6666);
    auto end = std::chrono::steady_clock::now();
    std::chrono::duration<double, std::milli> ms(end - start);
    std::cout << result << '\n';
    std::cout << "elapsed time: " << ms.count() << " milliseconds\n";
}
Output:
193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
elapsed time: 8.99497 milliseconds

Non GMP version[edit]

Translation of: C#
#include <chrono>
#include <iostream>

using namespace std;
using namespace chrono;

const long long Lm = (long)1e18;
const int Fm = 18;

struct LI { long long lo, ml, mh, hi, tp; };

LI set(long long s) { LI d;
  d.lo = s; d.ml = d.mh = d.hi = d.tp = 0; return d; }

void inc(LI& d, LI s) { // d += s
  if ((d.lo += s.lo) >= Lm) { d.ml++; d.lo -= Lm; }
  if ((d.ml += s.ml) >= Lm) { d.mh++; d.ml -= Lm; }
  if ((d.mh += s.mh) >= Lm) { d.hi++; d.mh -= Lm; }
  if ((d.hi += s.hi) >= Lm) { d.tp++; d.hi -= Lm; }
  d.tp += s.tp;
}

void dec(LI& d, LI s) { // d -= s
  if ((d.lo -= s.lo) < 0) { d.ml--; d.lo += Lm; }
  if ((d.ml -= s.ml) < 0) { d.mh--; d.ml += Lm; }
  if ((d.mh -= s.mh) < 0) { d.hi--; d.mh += Lm; }
  if ((d.hi -= s.hi) < 0) { d.tp--; d.hi += Lm; }
  d.tp -= s.tp;
}

inline string sf(long long n) {
  int len = Fm;
  string result(len--, '0');
  for (long long i = n; len >= 0 && i > 0; --len, i /= 10)
    result[len] = '0' + i % 10;
  return result;
}

string fmt(LI x) { // returns formatted string value of x
  if (x.tp > 0) return to_string(x.tp) + sf(x.hi) + sf(x.mh) + sf(x.ml) + sf(x.lo);
  if (x.hi > 0) return to_string(x.hi) + sf(x.mh) + sf(x.ml) + sf(x.lo);
  if (x.mh > 0) return to_string(x.mh) + sf(x.ml) + sf(x.lo);
  if (x.ml > 0) return to_string(x.ml) + sf(x.lo);
  return to_string(x.lo);
}

LI partcount(int n) { LI p[n + 1]; p[0] = set(1);
  for (int i = 1; i <= n; i++) {
    int k = 0, d = -2, j = i; LI x = set(0); while (true) {
      if ((j -= (d += 3) - k) >= 0) inc(x, p[j]); else break;
      if ((j -= ++k)          >= 0) inc(x, p[j]); else break;
      if ((j -= (d += 3) - k) >= 0) dec(x, p[j]); else break;
      if ((j -= ++k)          >= 0) dec(x, p[j]); else break;
    } p[i] = x; } return p[n]; }

int main() {
  auto start = steady_clock::now();
  auto result = partcount(6666);
  auto end = steady_clock::now();
  duration<double, milli> ms(end - start);
  cout << fmt(result) << "  " << ms.count() << " ms";
}
Output:

Timing from Tio.run, but execution time can't be directly compared to the GMP version, as GMP isn't accessible at Tio.run.

193655306161707661080005073394486091998480950338405932486880600467114423441282418165863  7.32706 ms

Delphi[edit]

Translation of: Go
program Partition_function_P;

{$APPTYPE CONSOLE}

uses
  System.SysUtils,
  Velthuis.BigIntegers,
  System.Diagnostics;

var
  p: TArray<BigInteger>;
  pd: TArray<Integer>;

function PartiDiffDiff(n: Integer): Integer;
begin
  if n and 1 = 1 then
    exit((n + 1) div 2);
  Result := n + 1;
end;

function partDiff(n: Integer): Integer;
begin
  if n < 2 then
    exit(1);

  pd[n] := pd[n - 1] + PartiDiffDiff(n - 1);
  Result := pd[n];
end;

procedure partitionP(n: Integer);
begin
  if n < 2 then
    exit;

  var psum: BigInteger := 0;
  for var i := 1 to n do
  begin
    var pdi := partDiff(i);
    if pdi > n then
      Break;

    var sign: Int64 := -1;

    if (i - 1) mod 4 < 2 then
      sign := 1;

    var t: BigInteger := BigInteger(p[n - pdi]) * BigInteger(sign);
    psum := psum + t;
  end;
  p[n] := psum;
end;

begin
  var stopwatch := TStopwatch.Create;
  const n = 6666;
  SetLength(p, n + 1);
  SetLength(pd, n + 1);
  stopwatch.Start;
  p[0] := 1;
  pd[0] := 1;
  p[1] := 1;
  pd[1] := 1;
  for var i := 2 to n do
    partitionP(i);
  stopwatch.Stop;
  writeln(format('p[%d] = %s', [n, p[n].ToString]));
  writeln('Took ', stopwatch.ElapsedMilliseconds, 'ms');
  Readln;
end.
Output:
p[6666] = 193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
Took 131ms

Elixir[edit]

Loosely based on the Erlang version.

use Bitwise, skip_operators: true

defmodule Partition do
   def init(), do:
      :ets.new :pN, [:set, :named_table, :private]

   def gpentagonals(), do: Stream.unfold {1, 0}, &next/1
   defp next({m, n}) do
      a = case rem m, 2 do
             0 -> div m, 2
             1 -> m
          end
      {n, {m + 1, n + a}}
   end
 
   def p(0), do: 1
   def p(n) do
      case :ets.lookup :pN, n do
         [{^n, val}] -> val
         [] ->
            {val, _} = gpentagonals()
                       |> Stream.drop(1)
                       |> Stream.take_while(fn m -> m <= n end)
                       |> Stream.map(fn g -> p(n - g) end)
                       |> Enum.reduce({0, 0},
                              fn n, {a, sgn} -> {
                                          a + (if sgn < 2, do: n, else: -n),
                                          band(sgn + 1, 3)
                                          }
                              end)
            :ets.insert :pN, {n, val}
            val
      end
   end
end

Partition.init
IO.puts Partition.p 6666
Output:
$ time ./partition6666.ex
193655306161707661080005073394486091998480950338405932486880600467114423441282418165863

real	0m1.106s
user	0m1.191s
sys	0m0.116s

Erlang[edit]

-mode(compile).

main(_) ->
    ets:new(pN, [set, named_table, protected]),
    io:format("~w~n", [p(6666)]).

p(0) -> 1;
p(N) ->
    case ets:lookup(pN, N) of
        [{N, Pn}] -> Pn;
        [] ->
            Terms = [p(N - G) || G <- gpentagonals(N)],
            Pn = sum_partitions(Terms),
            ets:insert(pN, {N, Pn}),
            Pn
    end.

sum_partitions(Terms) -> sum_partitions(Terms, 0, 0).
sum_partitions([], _, Sum) -> Sum;
sum_partitions([N|Ns], Sgn, Sum) ->
    Summand = case Sgn < 2 of
        true  -> N;
        false -> -N
    end,
    sum_partitions(Ns, (Sgn+1) band 3, Sum + Summand).

gpentagonals(Max) -> gpentagonals(1, Max, [0]).
gpentagonals(M, Max, Ps = [N|_]) ->
    GP = N + case M rem 2 of
                0 -> M div 2;
                1 -> M
             end,
    if
        GP > Max -> tl(lists:reverse(Ps));
        true -> gpentagonals(M + 1, Max, [GP|Ps])
    end.
Output:
$ time ./partition6666.erl 
193655306161707661080005073394486091998480950338405932486880600467114423441282418165863

real	0m0.480s
user	0m0.490s
sys	0m0.080s

F#[edit]

An implementation of the formula in the task description. P(123456) is included for comparison with the largest value in the related task.

// PartionP: Nigel Galloway. April 12th., 2021
let pP g=let rec fN i g e l=seq{yield(l,e+i);yield(-l,e+i+g);yield! fN(i+1)(g+2)(e+i+g)(-l)}
         let N,G=Array.create(g+1) 1I,seq{yield (1I,1);yield! fN 1 3 1 1I}|>Seq.takeWhile(fun(_,n)->n<=g)|>List.ofSeq
         seq{2..g}|>Seq.iter(fun p->N.[p]<-G|>List.takeWhile(fun(_,n)->n<=p)|>Seq.fold(fun Σ (n,g)->Σ+n*N.[p-g]) 0I); N.[g]
printfn "666->%A\n\n6666->%A\n\n123456->%A" (pP 666) (pP 6666) (pP 123456)
Output:
666->11956824258286445517629485

6666->193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
Real: 00:00:00.096

123456->30817659578536496678545317146533980855296613274507139217608776782063054452191537379312358383342446230621170608408020911309259407611257151683372221925128388387168451943800027128045369650890220060901494540459081545445020808726917371699102825508039173543836338081612528477859613355349851184591540231790254269948278726548570660145691076819912972162262902150886818986555127204165221706149989

Factor[edit]

Works with: Factor version 0.99 2020-08-14
USING: kernel lists lists.lazy math sequences sequences.extras ;

! Compute the nth pentagonal number.
: penta ( n -- m ) [ sq 3 * ] [ - 2/ ] bi ;

! An infinite lazy list of indices to add and subtract in the
! sequence of partitions to find the next P.
: seq ( -- list )
    1 lfrom [ penta 1 - ] <lazy-map> 1 lfrom [ neg penta 1 - ]
    <lazy-map> lmerge ;

! Reduce a sequence by adding two, subtracting two, adding two,
! etc...
: ++-- ( seq -- n ) 0 [ 2/ odd? [ - ] [ + ] if ] reduce-index ;

! Find the next member of the partitions sequence.
: step ( seq pseq -- seq 'pseq )
    dup length [ < ] curry pick swap take-while over <reversed>
    nths ++-- suffix! ;

: partitions ( m -- n )
    [ seq swap [ <= ] curry lwhile list>array ]
    [ V{ 1 } clone swap [ step ] times last nip ] bi ;
Output:
IN: scratchpad [ 6666 partitions ] time .

Running time: 0.084955341 seconds

193655306161707661080005073394486091998480950338405932486880600467114423441282418165863


FreeBASIC[edit]

Unsiged 64bit version[edit]

Translation of: Python
Function PartitionsP(n As UInteger) As ULongInt
    ' if n > 416, the result becomes to large for a unsigned 64bit integer
    Dim As ULongInt p(n)
    Dim As UInteger k, j

    p(0) = 1
    For i As UInteger = 1 To n
        k = 0
        While TRUE
            k += 1
            j = (k * (3*k - 1)) \ 2
            If (j > i) Then Exit While
            If (k And 1) Then
                p(i) += p(i - j)
            Else
                p(i) -= p(i - j)
            End If
            'j = (k * (3*k + 1)) \ 2
            j += k
            If (j > i) Then Exit While
            If (k And 1) Then
                p(i) += p(i - j)
            Else
                p(i) -= p(i - j)
            End If
        Wend
    Next i
    Return p(n)
End Function

Print !"\nPartitionsP: ";
For x As UInteger = 0 To 12
    Print PartitionsP(x);"  ";
Next x

Print !"\n\ndone"
Sleep
Output:
PartitionsP: 1  1  2  3  5  7  11  15  22  30  42  56  77

Big numbers version[edit]

Library: GMP

From the 9_billion_names_of_God_the_integer entry

' version 26-06-2021
' compile with: fbc -s console

#Include Once "gmp.bi"

Sub PartitionsP(max As ULong, p() As MpZ_ptr)
    ' based on Numericana code example
    Dim As ULong a, b, i, k
    Dim As Long j

    Dim As Mpz_ptr s = Allocate(Len(__mpz_struct)) : Mpz_init(s)

    Mpz_set_ui(p(0), 1)

    For i = 1 To max
        j = 1 : k = 1 : b = 2 : a = 5
        While j > 0
            ' j = i - (3*k*k+k) \ 2
            j = i - b : b = b + a : a = a + 3
            If j >= 0 Then
                If k And 1 Then Mpz_add(s, s, p(j)) Else Mpz_sub(s, s, p(j))
            End If
            j = j + k
            If j >= 0 Then
                If k And 1 Then Mpz_add(s, s, p(j)) Else Mpz_sub(s, s, p(j))
            End If
            k = k +1
        Wend
        Mpz_swap(p(i), s)
    Next

    Mpz_clear(s)

End Sub

' ------=< MAIN >=------

#Define max 6666

Dim As UInteger n
Dim As ZString Ptr ans
Dim As Double t = Timer

ReDim big_p(max) As Mpz_ptr
For n = 0 To max
    big_p(n) = Allocate(Len(__mpz_struct)) : Mpz_init(big_p(n))
Next

PartitionsP(max, big_p())
ans = Mpz_get_str (0, 10, big_p(max))
Print "PartitionsP("; Str(max); ") = "; "  "; *ans

For n = 0 To max
    Mpz_clear(big_p(n))
Next

Print Using "time = ###.## ms"; (Timer - t) * 1000

' empty keyboard buffer
While InKey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
Output:
PartitionsP(6666) =   193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
time =  32.97 ms

Frink[edit]

Frink has a built-in function for counting partitions that uses Euler's pentagonal method. It works for arbitrarily-large integers and caches results.

println[partitionCount[6666]]
Output:
193655306161707661080005073394486091998480950338405932486880600467114423441282418165863

Go[edit]

Translation of: Julia

I also tried using Euler's generating function but it was about 20 times slower than this version.

package main

import (
    "fmt"
    "math/big"
    "time"
)

var p []*big.Int
var pd []int

func partDiffDiff(n int) int {
    if n&1 == 1 {
        return (n + 1) / 2
    }
    return n + 1
}

func partDiff(n int) int {
    if n < 2 {
        return 1
    }
    pd[n] = pd[n-1] + partDiffDiff(n-1)
    return pd[n]
}

func partitionsP(n int) {
    if n < 2 {
        return
    }
    psum := new(big.Int)
    for i := 1; i <= n; i++ {
        pdi := partDiff(i)
        if pdi > n {
            break
        }
        sign := int64(-1)
        if (i-1)%4 < 2 {
            sign = 1
        }
        t := new(big.Int).Mul(p[n-pdi], big.NewInt(sign))
        psum.Add(psum, t)
    }
    p[n] = psum
}

func main() {
    start := time.Now()
    const N = 6666
    p = make([]*big.Int, N+1)
    pd = make([]int, N+1)
    p[0], pd[0] = big.NewInt(1), 1
    p[1], pd[1] = big.NewInt(1), 1
    for n := 2; n <= N; n++ {
        partitionsP(n)
    }
    fmt.Printf("p[%d)] = %d\n", N, p[N])
    fmt.Printf("Took %s\n", time.Since(start))
}
Output:
p[6666)] = 193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
Took 54.82683ms

Java[edit]

import java.math.BigInteger;

public class PartitionFunction {
    public static void main(String[] args) {
        long start = System.currentTimeMillis();
        BigInteger result = partitions(6666);
        long end = System.currentTimeMillis();
        System.out.println("P(6666) = " + result);
        System.out.printf("elapsed time: %d milliseconds\n", end - start);
    }

    private static BigInteger partitions(int n) {
        BigInteger[] p = new BigInteger[n + 1];
        p[0] = BigInteger.ONE;
        for (int i = 1; i <= n; ++i) {
            p[i] = BigInteger.ZERO;
            for (int k = 1; ; ++k) {
                int j = (k * (3 * k - 1))/2;
                if (j > i)
                    break;
                if ((k & 1) != 0)
                    p[i] = p[i].add(p[i - j]);
                else
                    p[i] = p[i].subtract(p[i - j]);
                j += k;
                if (j > i)
                    break;
                if ((k & 1) != 0)
                    p[i] = p[i].add(p[i - j]);
                else
                    p[i] = p[i].subtract(p[i - j]);
            }
        }
        return p[n];
    }
}
Output:
P(6666) = 193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
elapsed time: 59 milliseconds

JavaScript[edit]

function p(n){
    var a = new Array(n+1)
    a[0] = 1n
    
    for (let i = 1; i <= n; i++){
        a[i] = 0n
        for (let k = 1, s = 1; s <= i;){
            a[i] += (k & 1 ? a[i-s]:-a[i-s])
            k > 0 ? (s += k, k = -k):(k = -k+1, s = k*(3*k-1)/2)
        }
    }
    
    return a[n]
}

var t = Date.now()
console.log("p(6666) = " + p(6666))
console.log("Computation time in ms: ", Date.now() - t)
Output:
p(6666) = 193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
Computation time in ms: 26

Haskell[edit]

{-# LANGUAGE DeriveFunctor #-}

------------------------------------------------------------
-- memoization utilities

data Memo a = Node a (Memo a) (Memo a)
  deriving (Functor)

memo :: Integral a => Memo p -> a -> p
memo (Node a l r) n
  | n == 0 = a
  | odd n = memo l (n `div` 2)
  | otherwise = memo r (n `div` 2 - 1)

nats :: Memo Int
nats =
  Node
    0
    ((+ 1) . (* 2) <$> nats)
    ((* 2) . (+ 1) <$> nats)

------------------------------------------------------------
-- calculating partitions

partitions :: Memo Integer
partitions = partitionP <$> nats

partitionP :: Int -> Integer
partitionP n
  | n < 2 = 1
  | otherwise = sum $ zipWith (*) signs terms
  where
    terms =
      [ memo partitions (n - i)
        | i <- takeWhile (<= n) ofsets
      ]
    signs = cycle [1, 1, -1, -1]

ofsets :: [Int]
ofsets = scanl1 (+) $ mix [1, 3 ..] [1, 2 ..]
  where
    mix a b = concat $ zipWith (\x y -> [x, y]) a b

main :: IO ()
main = print $ partitionP 6666
*Main> partitionP <$> [1..10]
[1,2,3,5,7,11,15,22,30,42]

*Main> partitionP 100
190569292

*Main> partitionP 1000
24061467864032622473692149727991

*Main> partitionP 6666 
193655306161707661080005073394486091998480950338405932486880600467114423441282418165863

J[edit]

Solution stolen verbatim from the J Wiki. Note the use of memoization (M.) for efficiency:

   pn =: -/@(+/)@:($:"0)@rec ` (x:@(0&=)) @. (0>:]) M.
   rec=: - (-: (*"1) _1 1 +/ 3 * ]) @ (>:@i.@>.@%:@((2%3)&*))
Output:
   pn 6
11
   pn 66
2323520
   pn 666
11956824258286445517629485
   pn 6666
193655306161707661080005073394486091998480950338405932486880600467114423441282418165863

jq[edit]

Translation of: Python:Alternative

def partitions($n):
  def div2: (. - (.%2)) / 2;
  reduce range(1; $n + 1) as $i ( {p: ([1] + [range(0;$n)|0])};
    . + {k: 0, stop: false}
    | until(.stop;
        .k += 1
        | (((.k * (3*.k - 1)) | div2) ) as $j
        | if $j > $i then .stop=true
  	  else if (.k % 2) == 1
  	       then .p[$i] = .p[$i] + .p[$i - $j]
               else .p[$i] = .p[$i] - .p[$i - $j]
               end
          | (((.k * (3*.k + 1)) | div2)) as $j
          | if $j > $i then .stop=true
            elif (.k % 2) == 1
            then .p[$i] = .p[$i] + .p[$i - $j]
            else .p[$i] = .p[$i] - .p[$i - $j]
    	  end
  	end ))
    | .p[$n] ;

[partitions(range(1;15))]
Output:
[1,2,3,5,7,11,15,22,30,42,56,77,101,135]

jq's built-in integer precision is insufficient for computing ``partitions(6666)``, but more as a test of the BigInt.jq library for jq than anything else, here are the results of using it in conjunction with a trivially-modified version of the partitions implementation above. That is, after modifying the lines that refer to "p" (or ".p"), we see that partitions(6666) yields:

   "193655306161707661080005073394486091998480950338405932486880600467114423441282418165863"

The user+sys time is 7m3s as the BigInt.jq library is written in jq.

Recursive[edit]

Translation of: Julia
with memoization
def partDiffDiff($n):
  if ($n % 2) == 1 then ($n+1) / 2 else $n+1 end;

# in:  {n, partDiffMemo} 
# out: object with possibly updated memoization
def partDiff:
  .n as $n
  | if .partDiffMemo[$n] then .
    elif $n<2 then .partDiffMemo[$n]=1
    else ((.n=($n-1)) | partDiff)
    | .partDiffMemo[$n] = .partDiffMemo[$n-1] + partDiffDiff($n-1)
    end;

# in:  {n, memo, partDiffMemo}
#      where `.memo[i]` memoizes partitions(i)
#      and   `.partDiffMemo[i]` memoizes partDiff(i) 
# out: object with possibly updated memoization
def partitionsM:
  .n as $n
  | if .memo[$n] then .
    elif $n<2 then .memo[$n] = 1
    else label $out
    | foreach range(1; $n+2) as $i (.emit = false | .psum = 0;
        if $i > $n then .emit = true
        else ((.n = $i) | partDiff)
	| .partDiffMemo[$i] as $pd
        | if $pd > $n then .emit=true, break $out
          else {psum, emit} as $local  # for restoring relevant state
	  |  ((.n = ($n-$pd)) | partitionsM)
	  | .memo[$n-$pd] as $increment
	  | . + $local                 # restore
          | if (($i-1)%4)<2
            then .psum += $increment
            else .psum -= $increment
            end
	  end
        end;
        select(.emit) )
    | .memo[$n] = .psum
    end ;

def partitionsP:
  . as $n
  | {n: $n, memo:[], partDiffMemo:[]}
  | partitionsM
  | .memo[$n];

# Stretch goal:
6666 | partitionsP
Using gojq, the above program takes 41.35 seconds (u+s) on a 3GHz Mac Mini to produce:
193655306161707661080005073394486091998480950338405932486880600467114423441282418165863

Julia[edit]

Recursive[edit]

using Memoize

function partDiffDiff(n::Int)::Int
    isodd(n) ? (n+1)÷2 : n+1
end

@memoize function partDiff(n::Int)::Int
    n<2 ? 1 : partDiff(n-1)+partDiffDiff(n-1)
end

@memoize function partitionsP(n::Int)
    T=BigInt
    if n<2
        one(T)
    else
        psum = zero(T)
        for i  1:n
            pd = partDiff(i)
            if pd>n
                break
            end
            if ((i-1)%4)<2
                psum += partitionsP(n-pd)
            else
                psum -= partitionsP(n-pd)
            end
        end
        psum
    end
end

n=6666
@time println("p($n) = ", partitionsP(n))
Output:
p(6666) = 193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
  0.260310 seconds (3.58 M allocations: 77.974 MiB, 8.54% gc time)

Lingo[edit]

Lingo natively only supports 32 bit integers, so P(6666) would be way too big.

-- returns number of partitions of n
on partitions(n, res_table)
    if n < 2 then return 1
    if voidP(res_table) then
        res_table = []
        res_table[n] = 0
    else if res_table[n] then
        return res_table[n]
    end if
    res = 0
    i = 0
    param = 1
    repeat while param <= n
        if i mod 4 < 2 then
            res = res + partitions(n - param, res_table)
        else
            res = res - partitions(n - param, res_table)
        end if
        if i mod 2 then
            param = param + i + 2
        else
            param = param + i / 2 + 1
        end if
        i = i + 1
    end repeat
    res_table[n] = res
    return res
end
Output:
ms = _system.milliseconds
put "P(121):", partitions(121)
put "Time (ms):", _system.milliseconds - ms
-- P(121):  2056148051
-- Time (ms):  3

Maple[edit]

p:=proc(n)
  option remember;
  local k,s:=0,m;
  for k from 1 while (m:=iquo(k*(3*k-1),2))<=n do
    s-=(-1)^k*p(n-m);
  od;
  for k from 1 while (m:=iquo(k*(3*k+1),2))<=n do
    s-=(-1)^k*p(n-m);
  od;
  s
end:
p(0):=1:

time(p(6666));
# 0.796

time(combinat[numbpart](6666));
# 0.406

p~([$1..20]);
# [1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627]

combinat[numbpart]~([$1..20]);
# [1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627]

p(1000)
# 24061467864032622473692149727991

combinat[numbpart](1000);
# 24061467864032622473692149727991

Mathematica / Wolfram Language[edit]

PartitionsP /@ Range[15]
PartitionsP[666]
PartitionsP[6666]
Output:
{1,2,3,5,7,11,15,22,30,42,56,77,101,135,176}
11956824258286445517629485
193655306161707661080005073394486091998480950338405932486880600467114423441282418165863

Nim[edit]

Translation of: C++
Library: bignum
import sequtils, strformat, times
import bignum

func partitions(n: int): Int =
  var p = newSeqWith(n + 1, newInt())
  p[0] = newInt(1)
  for i in 1..n:
    var k = 1
    while true:
      var j = k * (3 * k - 1) div 2
      if j > i: break
      if (k and 1) != 0:
        inc p[i], p[i - j]
      else:
        dec p[i], p[i - j]
      j = k * (3 * k + 1) div 2
      if j > i: break
      if (k and 1) != 0:
        inc p[i], p[i - j]
      else:
        dec p[i], p[i - j]
      inc k
  result = p[n]

let t0 = cpuTime()
echo partitions(6666)
echo &"Elapsed time: {(cpuTime() - t0) * 1000:.2f} ms"
Output:
193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
Elapsed time: 9.73 ms

Perl[edit]

use strict;
use warnings;
no warnings qw(recursion);
use Math::AnyNum qw(:overload);
use Memoize;

memoize('partitionsP');
memoize('partDiff');

sub partDiffDiff { my($n) = @_; $n%2 != 0 ? ($n+1)/2 : $n+1 }

sub partDiff { my($n) = @_; $n<2 ? 1 : partDiff($n-1) + partDiffDiff($n-1) }

sub partitionsP {
    my($n) = @_;
    return 1 if $n < 2;

    my $psum = 0;
    for my $i (1..$n) {
        my $pd = partDiff($i);
        last if $pd > $n;
        if ( (($i-1)%4) < 2 ) { $psum += partitionsP($n-$pd) }
        else                  { $psum -= partitionsP($n-$pd) }
    }
    return $psum
}

print partitionsP($_) . ' ' for 0..25; print "\n";
print partitionsP(6666) . "\n";
Output:
1 1 2 3 5 7 11 15 22 30 42 56 77 101 135 176 231 297 385 490 627 792 1002 1255 1575 1958
193655306161707661080005073394486091998480950338405932486880600467114423441282418165863

Phix[edit]

Library: Phix/mpfr

Not exactly super-fast, but this sort of stuff is not really what Phix does best.

with javascript_semantics
function partDiffDiff(integer n)
    return (n+1)/(1+and_bits(n,1))
end function
 
sequence pd = {1}
function partDiff(integer n)
    while n>length(pd) do
        pd &= pd[$] + partDiffDiff(length(pd))
    end while
    return pd[max(1,n)]
end function
 
include mpfr.e
 
sequence pn = {mpz_init(1)}
function partitionsP(integer n)
    mpz res = mpz_init(1)
    while n>length(pn) do
        integer nn = length(pn)+1
        mpz psum = mpz_init(0)
        for i=1 to nn do
            integer pd = partDiff(i)
            if pd>nn then exit end if
            integer sgn = iff(remainder(i-1,4)<2 ? 1 : -1)
            mpz pnmpd = pn[max(1,nn-pd)]
            if sgn=-1 then
                mpz_sub(psum,psum,pnmpd)
            else
                mpz_add(psum,psum,pnmpd)
            end if
        end for
        pn = append(pn,psum)
    end while
    return pn[max(1,n)]
end function
 
atom t0 = time()
integer n=6666
printf(1,"p(%d) = %s (%s)\n",{n,mpz_get_str(partitionsP(n)),elapsed(time()-t0)})
Output:
p(6666) = 193655306161707661080005073394486091998480950338405932486880600467114423441282418165863 (0.8s)

Picat[edit]

/* Picat 3.0#5 */
/* Author: Hakan Kjellerstrand */
table
partition1(0) = 1.
partition1(N) = P =>
  S = 0,
  K = 1,
  M = (K*(3*K-1)) // 2,  
  while (M <= N)
     S := S - ((-1)**K)*partition1(N-M),
     K := K + 1, 
     M := (K*(3*K-1)) // 2 
  end,
  K := 1,
  M := (K*(3*K+1)) // 2,
  while (M <= N)
     S := S - ((-1)**K)*partition1(N-M),
     K := K + 1,
     M := (K*(3*K+1)) // 2  
  end,
  P = S.

Picat> time(println('p(6666)'=partition1(6666))) 
p(6666) = 193655306161707661080005073394486091998480950338405932486880600467114423441282418165863

CPU time 0.206 seconds.

PicoLisp[edit]

Based on the Erlang implementation.

(de gpentagonals (Max)
   (make
      (let (N 0  M 1)
         (loop
            (inc 'N (if (=0 (& M 1)) (>> 1 M) M))
            (T (> N Max))
            (link N)
            (inc 'M)))))

(de p (N)
   (cache '(NIL) N
      (if (=0 N)
         1
         (let (Sum 0  Sgn 0)
            (for G (gpentagonals N)
               ((if (< Sgn 2) 'inc 'dec) 'Sum (p (- N G)))
               (setq Sgn (& 3 (inc Sgn))))
            Sum))))
Output:
: (bench (p 6666))
0.959 sec
-> 193655306161707661080005073394486091998480950338405932486880600467114423441282418165863

Prolog[edit]

/* SWI-Prolog 8.3.21 */
/* Author: Jan Burse */
:- table p/2.
p(0, 1) :- !.
p(N, X) :-
    aggregate_all(sum(Z), (between(1,inf,K), M is K*(3*K-1)//2,
           (M>N, !, fail; L is N-M, p(L,Y), Z is (-1)^K*Y)), A),
    aggregate_all(sum(Z), (between(1,inf,K), M is K*(3*K+1)//2,
           (M>N, !, fail; L is N-M, p(L,Y), Z is (-1)^K*Y)), B),
    X is -A-B.

?- time(p(6666,X)).
% 13,962,294 inferences, 2.610 CPU in 2.743 seconds (95% CPU, 5350059 Lips)
X = 1936553061617076610800050733944860919984809503384
05932486880600467114423441282418165863.

Python[edit]

Python: Mathloger[edit]

This follows the algorithm from the Mathloger video closely

from itertools import islice

def posd():
    "diff between position numbers. 1, 2, 3... interleaved with  3, 5, 7..."
    count, odd = 1, 3
    while True:
        yield count
        yield odd
        count, odd = count + 1, odd + 2

def pos_gen():
    "position numbers. 1 3 2 5 7 4 9 ..."
    val = 1
    diff = posd()
    while True:
        yield val
        val += next(diff)
                
def plus_minus():
    "yield (list_offset, sign) or zero for Partition calc"
    n, sign = 0, [1, 1]
    p_gen = pos_gen()
    out_on = next(p_gen)
    while True:
        n += 1
        if n == out_on:
            next_sign = sign.pop(0)
            if not sign:
                sign = [-next_sign] * 2
            yield -n, next_sign
            out_on = next(p_gen)
        else:
            yield 0
            
def part(n):
    "Partition numbers"
    p = [1]
    p_m = plus_minus()
    mods = []
    for _ in range(n):
        next_plus_minus = next(p_m)
        if next_plus_minus:
            mods.append(next_plus_minus)
        p.append(sum(p[offset] * sign for offset, sign in mods))
    return p[-1]
        
print("(Intermediaries):")
print("    posd:", list(islice(posd(), 10)))
print("    pos_gen:", list(islice(pos_gen(), 10)))
print("    plus_minus:", list(islice(plus_minus(), 15)))
print("\nPartitions:", [part(x) for x in range(15)])
Output:
(Intermediaries):
    posd: [1, 3, 2, 5, 3, 7, 4, 9, 5, 11]
    pos_gen: [1, 2, 5, 7, 12, 15, 22, 26, 35, 40]
    plus_minus: [(-1, 1), (-2, 1), 0, 0, (-5, -1), 0, (-7, -1), 0, 0, 0, 0, (-12, 1), 0, 0, (-15, 1)]

Partitions: [1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135]
Stretch goal

From command line after running the above

In [1]: part(6666)
Out[1]: 193655306161707661080005073394486091998480950338405932486880600467114423441282418165863

In [2]: %timeit part(6666)
103 ms ± 477 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

Python: Mathloger video prime generator[edit]

Add the following code after that above

def par_primes():
    "Prime number generator from the partition machine"
    p = [1]
    p_m = plus_minus()
    mods = []
    n = 0
    while True:
        n += 1
        next_plus_minus = next(p_m)
        if next_plus_minus:
            mods.append(next_plus_minus)
        p.append(sum(p[offset] * sign for offset, sign in mods))
        if p[0] + 1 == p[-1]:
            yield p[0]
        p[0] += 1
    yield p

print("\nPrimes:", list(islice(par_primes(), 15)))
Output:
Primes: [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]

Python: Alternative[edit]

Translation of: C++
from typing import List


def partitions(n: int) -> int:
    """Count partitions."""
    p: List[int] = [1] + [0] * n
    for i in range(1, n + 1):
        k: int = 0
        while True:
            k += 1
            j: int = (k * (3*k - 1)) // 2
            if (j > i):
                break
            if (k & 1):
                p[i] += p[i - j]
            else:
                p[i] -= p[i - j]
            j = (k * (3*k + 1)) // 2
            if (j > i):
                break
            if (k & 1):
                p[i] += p[i - j]
            else:
                p[i] -= p[i - j]

    return p[n]


if __name__ == '__main__':
    print("\nPartitions:", [partitions(x) for x in range(15)])
Output:
Partitions: [1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135]
Stretch goal

From command line after running the above

In [3]: partitions(6666)
Out[3]: 193655306161707661080005073394486091998480950338405932486880600467114423441282418165863

In [4]: %timeit partitions(6666)
215 ms ± 1.84 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Racket[edit]

Backwards range was used to get responsive feedback for progress.

#lang racket

(require math/number-theory)

(define σ
  (let ((memo (make-hash)))
    (λ (z)
      (hash-ref! memo z
                 (λ () (apply + (divisors z)))))))

(define p
  (let ((memo (make-hash '((0 . 1)))))
    (λ (n)
      (hash-ref!
       memo n
       (λ ()
         (let ((r (if (zero? n) 1
             (/ (for/sum ((k (in-range (sub1 n) -1 -1)))
                  (* (σ (- n k))
                     (p k)))
                n))))
           (when (zero? (modulo n 1000)) (displayln (cons n r) (current-error-port)))
           r))))))

(map p (range 1 30))
(p 666)
(p 1000)
(p 10000)
Output:
'(1 2 3 5 7 11 15 22 30 42 56 77 101 135 176 231 297 385 490 627 792 1002 1255 1575 1958 2436 3010 3718 4565)
11956824258286445517629485
(1000 . 24061467864032622473692149727991)
24061467864032622473692149727991
(2000 . 4720819175619413888601432406799959512200344166)
(3000 . 496025142797537184410324879054927095334462742231683423624)
(4000 . 1024150064776551375119256307915896842122498030313150910234889093895)
(5000 . 169820168825442121851975101689306431361757683049829233322203824652329144349)
(6000 . 4671727531970209092971024643973690643364629153270037033856605528925072405349246129)
(7000 . 32856930803440615786280925635924166861950151574532240659699032157432236394374450791229199)
(8000 . 78360264351568349490593145013364599719010769352985864331118600209417827764524450990388402844164)
(9000 . 77133638117808884907320791427403134961639798322072034262647713694605367979684296948790335590435626459)
(10000 . 36167251325636293988820471890953695495016030339315650422081868605887952568754066420592310556052906916435144)
36167251325636293988820471890953695495016030339315650422081868605887952568754066420592310556052906916435144

Raku[edit]

Works with: Rakudo version 2020.09

Not the fastest, but it gets the job done.

my @P = 1, { p(++$) } … *;
my @i = lazy [\+] flat 1, ( 1..* Z (1..*).map: * × 2 + 1 );
sub p ($n) { sum @P[$n X- @i] Z× (flat (1, 1, -1, -1) xx *) }

put @P[^26];
put @P[6666];
Output:
1 1 2 3 5 7 11 15 22 30 42 56 77 101 135 176 231 297 385 490 627 792 1002 1255 1575 1958
193655306161707661080005073394486091998480950338405932486880600467114423441282418165863

REXX[edit]

These three REXX versions are recursive.

version 1[edit]

/*REXX program calculates and displays a specific value (or a range of)  partitionsP(N).*/
numeric digits 1000                              /*able to handle some ginormous numbers*/
parse arg lo hi .                                /*obtain optional arguments from the CL*/
if lo=='' | lo==","  then lo=  0                 /*Not specified?  Then use the default.*/
if hi=='' | hi==","  then hi= lo                 /* "      "         "   "   "     "    */
@.= 0;    @.0= 1; @.1= 1; @.2= 2; @.3= 3; @.4= 5 /*assign default value and low values. */
!.= @.;   !.1= 1; !.3= 1; !.5= 1; !.7= 1; !.9= 1 /*assign default value and even digits.*/
w= length( commas(hi) )                          /*W:  is used for aligning the index.  */

       do j=lo  to hi                            /*compute a range of  partitionsP.     */
       say right( commas(j), w)    ' '     commas( partP(j) )
       end   /*j*/
exit 0                                           /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: parse arg ?;  do jc=length(?)-3  to 1  by -3; ?=insert(',', ?, jc); end;  return ?
/*──────────────────────────────────────────────────────────────────────────────────────*/
partP:  procedure expose @. !.;  parse arg n     /*obtain number (index) for computation*/
        if @.n\==0  then return @.n              /*Is it already computed?   Return it. */
        #= 0                                               /*initialize part  P  number.*/
               do k=1  for n;  z= n  -  (k*3 - 1) * k % 2  /*compute the partition P num*/
               if z<0  then leave                          /*Is Z negative?  Then leave.*/
               if @.z==0  then x= partP(z)                 /*use recursion if not known.*/
                          else x= @.z                      /*use the pre─computed number*/
               z= z - k                                    /*subtract index (K) from Z. */
               if z<0     then y= 0                        /*Is Z negative? Then set Y=0*/
                          else if @.z==0  then y= partP(z) /*use recursion if not known.*/
                                          else y= @.z      /*use the pre─computed number*/
               if k//2    then #= # +  x + y               /*Odd? Then   sum    X and Y.*/
                          else #= # - (x + y)              /*Even?  "  subtract "  "  " */
               end   /*k*/
        @.n= #;                   return #       /*define and return partitionsP of  N. */
output   when using the input of:     6666
6,666   193,655,306,161,707,661,080,005,073,394,486,091,998,480,950,338,405,932,486,880,600,467,114,423,441,282,418,165,863
output   when using the input of:     66666
66,666   931,283,431,869,095,717,238,416,063,534,148,471,363,928,685,651,267,074,563,360,050,977,549,251,436,058,076,515,262,033,789,845,457,356,081,278,451,246,751,375,974,038,318,319,810,923,102,769,109,446,980,055,567,090,089,060,954,748,065,131,666,952,830,623,286,286,824,837,240,058,805,177,319,865,230,913,417,587,625,830,803,675,380,262,765,598,632,742,903,192,085,254,824,621

version 2[edit]

This REXX version is about   25%   faster than REXX version 1.

The biggest part of the improvement was using the expression     k+k+k     instead of     k*3.

/*REXX program calculates and displays a specific value (or a range of)  partitionsP(N).*/
numeric digits 1000                              /*able to handle some ginormous numbers*/
parse arg lo hi .                                /*obtain optional arguments from the CL*/
if lo=='' | lo==","  then lo=  0                 /*Not specified?  Then use the default.*/
if hi=='' | hi==","  then hi= lo                 /* "      "         "   "   "     "    */
@.= 0;   @.0= 1; @.1= 1; @.2= 2; @.3= 3; @.4= 5  /*default values for some low numbers. */
!.= @.;  !.1= 1; !.3= 1; !.5= 1; !.7= 1; !.9= 1  /*   "       "    "  all the 1─digit #s*/
w= length( commas(hi) )                          /*W:  is used for aligning the index.  */

       do j=lo  to hi                            /*compute a range of  partitionsP.     */
       say right( commas(j), w)    ' '     commas( partP(j) )
       end   /*j*/
exit 0                                           /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: parse arg ?;  do jc=length(?)-3  to 1  by -3; ?=insert(',', ?, jc); end;  return ?
/*──────────────────────────────────────────────────────────────────────────────────────*/
partP:  procedure expose @. !.;  parse arg n     /*obtain number (index) for computation*/
        if @.n\==0  then return @.n              /*Is it already computed?   Return it. */
        #= 0                                               /*initialize part  P  number.*/
               do k=1  for n;  z= n  - (k+k+k - 1) * k % 2 /*compute the partition P num*/
               if z<0  then leave                          /*Is Z negative?  Then leave.*/
               if @.z==0  then x= partP(z)                 /*use recursion if not known.*/
                          else x= @.z                      /*use the pre─computed number*/
               z= z - k                                    /*subtract index (K) from Z. */
               if z<0     then y= 0                        /*Is Z negative? Then set Y=0*/
                          else if @.z==0  then y= partP(z) /*use recursion if not known.*/
                                          else y= @.z      /*use the pre─computed number*/
               parse var   k   ''  -1  _                   /*obtain K's last decimal dig*/
               if !._     then #= # +  x + y               /*Odd? Then   sum    X and Y.*/
                          else #= # - (x + y)              /*Even?  "  subtract "  "  " */
               end   /*k*/
        @.n= #;                   return #       /*define and return partitionsP of  N. */
output   is identical to the 1st REXX version.

version 3[edit]

This REXX version is about   90%   faster than REXX version 1.

The biggest part of the improvement was using memoization of the expressions     (k+k+k - 1) * k % 2     for all values of (positive)   k   up to   hi.

/*REXX program calculates and displays a specific value (or a range of)  partitionsP(N).*/
numeric digits 1000                              /*able to handle some ginormous numbers*/
parse arg lo hi .                                /*obtain optional arguments from the CL*/
if lo=='' | lo==","  then lo=  0                 /*Not specified?  Then use the default.*/
if hi=='' | hi==","  then hi= lo                 /* "      "         "   "   "     "    */
@.= 0;   @.0= 1; @.1= 1; @.2= 2; @.3= 3; @.4= 5  /*default values for some low numbers. */
!.= @.;  !.1= 1; !.3= 1; !.5= 1; !.7= 1; !.9= 1  /*   "       "    "  all the 1─digit #s*/
w= length( commas(hi) )                          /*W:  is used for aligning the index.  */
       do i=1  for hi;  a.i= (i+i+i - 1) * i % 2 /*calculate HI expressions (for partP).*/
       end   /*i*/

       do j=lo  to hi                            /*compute a range of  partitionsP.     */
       say right( commas(j), w)    ' '     commas( partP(j) )
       end   /*j*/
exit 0                                           /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: parse arg ?;  do jc=length(?)-3  to 1  by -3; ?=insert(',', ?, jc); end;  return ?
/*──────────────────────────────────────────────────────────────────────────────────────*/
partP:  procedure expose @. !. a.;  parse arg n  /*obtain number (index) for computation*/
        if @.n\==0  then return @.n              /*Is it already computed?   Return it. */
        #= 0                                               /*initialize part  P  number.*/
               do k=1  for n;  z= n - a.k                  /*compute the partition P num*/
               if z<0  then leave                          /*Is Z negative?  Then leave.*/
               if @.z==0  then x= partP(z)                 /*use recursion if not known.*/
                          else x= @.z                      /*use the pre─computed number*/
               z= z - k                                    /*subtract index (K) from Z. */
               if z<0     then y= 0                        /*Is Z negative? Then set Y=0*/
                          else if @.z==0  then y= partP(z) /*use recursion if not known.*/
                                          else y= @.z      /*use the pre─computed number*/
               parse var   k   ''  -1  _                   /*obtain K's last decimal dig*/
               if !._     then #= # +  x + y               /*Odd? Then   sum    X and Y.*/
                          else #= # - (x + y)              /*Even?  "  subtract "  "  " */
               end   /*k*/
        @.n= #;                   return #       /*define and return partitionsP of  N. */
output   is identical to the 1st REXX version.


Rust[edit]

// [dependencies]
// rug = "1.11"

use rug::Integer;

fn partitions(n: usize) -> Integer {
    let mut p = Vec::with_capacity(n + 1);
    p.push(Integer::from(1));
    for i in 1..=n {
        let mut num = Integer::from(0);
        let mut k = 1;
        loop {
            let mut j = (k * (3 * k - 1)) / 2;
            if j > i {
                break;
            }
            if (k & 1) == 1 {
                num += &p[i - j];
            } else {
                num -= &p[i - j];
            }
            j += k;
            if j > i {
                break;
            }
            if (k & 1) == 1 {
                num += &p[i - j];
            } else {
                num -= &p[i - j];
            }
            k += 1;
        }
        p.push(num);
    }
    p[n].clone()
}

fn main() {
    use std::time::Instant;
    let n = 6666;
    let now = Instant::now();
    let result = partitions(n);
    let time = now.elapsed();
    println!("P({}) = {}", n, result);
    println!("elapsed time: {} microseconds", time.as_micros());
}
Output:
P(6666) = 193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
elapsed time: 8912 microseconds

Sidef[edit]

Built-in:

say partitions(6666)   # very fast

User-defined:

func partitionsP(n) {
    func (n) is cached {

        n <= 1 && return n

        var a = sum(1..floor((sqrt(24*n + 1) + 1)/6), {|k|
            (-1)**(k-1) * __FUNC__(n - ((k*(3*k - 1)) >> 1))
        })

        var b = sum(1..ceil((sqrt(24*n + 1) - 7)/6), {|k|
            (-1)**(k-1) * __FUNC__(n - ((k*(3*k + 1)) >> 1))
        })

        a + b
    }(n+1)
}

var t = Time.micro

say partitionsP.map(0..25).join(' ')
say partitionsP(6666)

say ("Took %.4f seconds" % Time.micro-t)
Output:
1 1 2 3 5 7 11 15 22 30 42 56 77 101 135 176 231 297 385 490 627 792 1002 1255 1575 1958
193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
Took 24.5225 seconds

Swift[edit]

Translation of: Rust


Using AttaSwift's BigInt library.

import BigInt 

func partitions(n: Int) -> BigInt {
  var p = [BigInt(1)]

  for i in 1...n {
    var num = BigInt(0)
    var k = 1

    while true {
      var j = (k * (3 * k - 1)) / 2

      if j > i {
        break
      }

      if k & 1 == 1 {
        num += p[i - j]
      } else {
        num -= p[i - j]
      }

      j += k

      if j > i {
        break
      }

      if k & 1 == 1 {
        num += p[i - j]
      } else {
        num -= p[i - j]
      }

      k += 1
    }

    p.append(num)
  }

  return p[n]
}

print("partitions(6666) = \(partitions(n: 6666))")
Output:
partitions(6666) = 193655306161707661080005073394486091998480950338405932486880600467114423441282418165863

Wren[edit]

Translation of: Julia
Library: Wren-big

Although it may not look like it, this is actually a decent time for Wren which is interpreted and the above module is written entirely in Wren itself.

import "/big" for BigInt

var p = []
var pd = []

var partDiffDiff = Fn.new { |n| (n&1 == 1) ? (n + 1)/2 : n + 1 }

var partDiff = Fn.new { |n|
    if (n < 2) return 1
    pd[n] = pd[n-1] + partDiffDiff.call(n-1)
    return pd[n]
}

var partitionsP = Fn.new { |n|
    if (n < 2) return
    var psum = BigInt.zero
    for (i in 1..n) {
        var pdi = partDiff.call(i)
        if (pdi > n) break
        var sign = (i-1)%4 < 2 ? 1 : -1
        psum = psum + p[n-pdi] * sign
    }
    p[n] = psum
}

var start = System.clock
var N = 6666
p = List.filled(N+1, null)
pd = List.filled(N+1, 0)
p[0] = BigInt.one
p[1] = BigInt.one
pd[0] = 1
pd[1] = 1
for (n in 2..N) partitionsP.call(n)
System.print("p[%(N)] = %(p[N])")
System.print("Took %(System.clock - start) seconds")
Output:
p[6666] = 193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
Took 1.428762 seconds