Partition function P

Partition function P is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

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. For 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

• 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.

Factor

Works with: Factor version 0.99 2020-08-14

<lang 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 ;</lang>

Output:

IN: scratchpad [ 6666 partitions ] time .

Running time: 0.084955341 seconds

193655306161707661080005073394486091998480950338405932486880600467114423441282418165863

Go

Translation of: Julia

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

}</lang>

Output:

p[6666)] = 193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
Took 54.82683ms

Julia

Recursive

<lang Julia>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))</lang>

Output:

p(6666) = 193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
  0.260310 seconds (3.58 M allocations: 77.974 MiB, 8.54% gc time)

Phix

Not exactly super-fast, but this sort of stuff is not really what Phix does best. <lang Phix>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)})</lang>

Output:

p(6666) = 193655306161707661080005073394486091998480950338405932486880600467114423441282418165863 (0.8s)

Python

This follows the algorithm from the Mathloger video closely

<lang python>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)])</lang>

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 [2]: %time part(6666)
Wall time: 243 ms
Out[2]: 193655306161707661080005073394486091998480950338405932486880600467114423441282418165863

REXX

This REXX version is recursive. <lang rexx>/*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. */ w= length(hi) /*W: is used for aligning the index. */

            do j=lo  to hi                      /*compute a range of  partitionsP.     */
            say right(j, w)  partitionsP(j)     /*display index and it's  partitionsP. */
            end     /*j*/

exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ partitionsP: procedure expose @.; parse arg n /*obtain number (index) for computation*/

            if @.n\==0  then return @.n         /*Is it already computed?   Return it. */
            #= 0
                   do k=1  for n;  _= n - (k*3 - 1) * k % 2;    if _<0  then leave
                   if @._==0  then x= partitionsP(_)                    /*use recursion*/
                              else x= @._                               /*use found #. */
                   _= _ - k
                   if _<0     then y= 0                                 /*Negative?    */
                              else if @._==0  then y= partitionsP(_)    /*use recursion*/
                                              else y= @._               /*use found #. */
                   if k//2    then #= # + x + y          /*Odd?  Then   sum    X and Y */
                              else #= # - x - y          /*Even?   "  subtract "  "  " */
                   end   /*k*/
            @.n= #                              /*define the computed partitionsP of N.*/
            return #                            /*return  "      "          "      " " */</lang>

output   when using the input of:     6666

6666 193655306161707661080005073394486091998480950338405932486880600467114423441282418165863

output   when using the input of:     66666

66666 931283431869095717238416063534148471363928685651267074563360050977549251436058076515262033
789845457356081278451246751375974038318319810923102769109446980055567090089060954748065131666952
830623286286824837240058805177319865230913417587625830803675380262765598632742903192085254824621

Wren

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. <lang ecmascript>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")</lang>

Output:

p[6666] = 193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
Took 1.428762 seconds