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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]

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

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

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

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 * ]) @ (>:@[email protected]>[email protected]%:@((2%3)&*))
Output:
   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)

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.
 

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)

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[^(@i.first: * > $n, :k)]] 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

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