Partition function P
(Redirected from 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 is the function P(n), where n∈ℤ, defined as the number of distinct ways in which n can be expressed as the sum of non-increasing 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 was viewed more than 100,000 times in the first couple of weeks after 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
- The hardest "What comes next?" (Euler's pentagonal formula) The explanatory video by Mathologer that makes this task a popular interest.
- Partition Function P Mathworld entry for the Partition function.
- Partition function (number theory) Wikipedia entry for the Partition function.
- Related tasks
11l
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
#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#
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++
GMP version
#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
#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
Dart
BigInt partitions(n) {
var p = List.filled(n + 1, BigInt.zero);
p[0] = BigInt.one;
for (var i = 1; i < n + 1; i++) {
var k = 0;
while (true) {
k += 1;
var j = (k * (3 * k - 1)) ~/ 2;
if (j > i) break;
var t = p[i - j];
p[i] = (p[i] + ((k % 2 == 1) ? t : -t)) ;
j = (k * (3 * k + 1)) ~/ 2;
if (j > i) break;
t = p[i - j];
p[i] = (p[i] + ((k % 2 == 1) ? t : -t));
}
}
return p[n];
}
main() => print(partitions(6666));
- Output:
193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
Sys. time: 0.197 s on TIO
Delphi
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
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
-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#
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
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
Unsiged 64bit version
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
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
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
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
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
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
{-# 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 <$> [1..10] [1,2,3,5,7,11,15,22,30,42] *Main> partitionP 100 190569292 *Main> partitionP 1000 24061467864032622473692149727991 *Main> partitionP 6666 193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
J
Solution stolen verbatim from the J Wiki. Note the use of memoization (M.) for efficiency:
pn =: -/@(+/)@:($:"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
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]
Using gojq 0.12.11, `partitions(6666)` yields (in about 12 minutes (u+s) on a 3GHz machine):
193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
The integer precision of the C implementation of jq is insufficient for computing ``partitions(6666)``, but as a test of the BigInt.jq library for jq, 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"
Curiously, the u+s time is 7m3s, which is significantly less than the above-mentioned gojq time, even though the BigInt library is written in jq.
Recursive
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 | partitionsPUsing gojq, the above program takes 41.35 seconds (u+s) on a 3GHz Mac Mini to produce:
193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
Julia
Recursive
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
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
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);
# 24061467864032622473692149727991Mathematica / Wolfram Language
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
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
PascalABC.NET
const
n = 6666;
var
p: array of biginteger;
pd: array of integer;
function PartiDiffDiff(n: Integer): Integer;
begin
if (n and 1) = 1 then result := ((n + 1) div 2)
else Result := n + 1;
end;
function partDiff(n: Integer): Integer;
begin
if n < 2 then result := 1
else begin
pd[n] := pd[n - 1] + PartiDiffDiff(n - 1);
Result := pd[n];
end;
end;
procedure partitionP(n: Integer);
begin
if n < 2 then exit;
var psum := 0bi;
for var i := 1 to n do
begin
var pdi := partDiff(i);
if pdi > n then Break;
var sign := -1bi;
if (i - 1) mod 4 < 2 then sign := 1;
psum += p[n - pdi] * sign;
end;
p[n] := psum;
end;
begin
SetLength(p, n + 1);
SetLength(pd, n + 1);
p[0] := 1bi;
pd[0] := 1;
p[1] := 1bi;
pd[1] := 1;
for var i := 2 to n do partitionP(i);
println(p[n]);
println('Took', milliseconds, 'milliseconds');
end.
- Output:
193655306161707661080005073394486091998480950338405932486880600467114423441282418165863 Took 62 milliseconds
Perl
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
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
/* 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
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
/* 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
Python: Mathloger
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
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
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)
Quackery
0 partitions returns 1 as per oeis.org/A000041 (Partitions of n).
This is a naive recursive solution, so computing the partitions of 6666 would take a hideously long time.
[ 1 swap
dup 0 = iff drop done
[ 2dup = iff [ 2drop 1 ] done
2dup > iff [ 2drop 0 ] done
2dup dip 1+ recurse
unrot over - recurse + ] ] is partitions ( n --> n )
say "Partitions of 0 to 29" cr
30 times [ i^ partitions echo sp ]- Output:
Partitions of 0 to 29 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 2436 3010 3718 4565
Racket
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
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
These three REXX versions are recursive.
version 1
/*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
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
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
// [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
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
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
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
Categories:
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