Partition Numbers
April 15, 2011
The partition-number function P(n) gives the number of ways of writing n as the sum of positive integers, irrespective of the order of the addends. For instance P(4) = 5, since 4 = 4 = 3 + 1 = 2 + 2 = 2 + 1 + 1 = 1 + 1 + 1 + 1. Sloane’s A000041 gives the first ten partition numbers as 1, 2, 3, 5, 7, 11, 15, 22, 30, and 42; the numerous references to that sequence point to many fascinating facts about partition numbers, including their close association with pentagonal numbers. By convention, P(0) = 1 and P(n) = 0 for negative n. Partition numbers are normally calculated by the formula, which was discovered by Leonhard Euler:
![P(n) = \sum_{k=1}^n (-1)^{k+1} \left[ P\left( n - \frac{k\left(3k-1\right)}{2} \right) + P\left( n - \frac{k\left(3k+1\right)}{2} \right) \right]](https://s0.wp.com/latex.php?latex=P%28n%29+%3D+%5Csum_%7Bk%3D1%7D%5En+%28-1%29%5E%7Bk%2B1%7D+%5Cleft%5B+P%5Cleft%28+n+-+%5Cfrac%7Bk%5Cleft%283k-1%5Cright%29%7D%7B2%7D+%5Cright%29+%2B+P%5Cleft%28+n+-+%5Cfrac%7Bk%5Cleft%283k%2B1%5Cright%29%7D%7B2%7D+%5Cright%29+%5Cright%5D&bg=ffffff&fg=555555&s=0&c=20201002)
Your task is to write a function that computes P(n), and to calculate P(1000). When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Programming Praxis 
My Haskell solution (see http://bonsaicode.wordpress.com/2011/04/15/programming-praxis-partition-numbers/ for a version with comments):
import Data.IntMap ((!), fromList, insert, findWithDefault) partition :: Int -> Integer partition x = if x < 0 then 0 else foldl p (fromList [(0,1)]) [1..x] ! x where p s n = insert n (sum [(-1)^(k+1) * (r pred + r succ) | k <- [1..n], let r f = findWithDefault 0 (n - div (k * f (3 * k)) 2) s]) s main :: IO () main = print $ partition 1000 == 24061467864032622473692149727991[…] today’s Programming Praxis exercise, our goal is to write a function to calculate partition numbers and to […]
A Python solution:
class Memoize: """decorator to memoise a function""" def __init__(self, f): self.f = f self.cache = {} def __call__(self, *args): if not args in self.cache: self.cache[args] = self.f(*args) return self.cache[args] @Memoize def bigp(n): if n < 0: return 0 if n == 0: return 1 # run k from n to 1 to avoid excessive recursion depth return sum((-1) ** (k + 1) * (bigp(n - k * (3 * k - 1) / 2) + bigp(n - k * (3 * k + 1) / 2)) for k in range(n,0,-1)) print bigp(1000)My non-recursive solution in Python:
#!/usr/bin/python aMem = {0 : 1} aStackNeedto = [] def Update(nParent, nSum): if ((aStackNeedto[nParent]['Num'] * 2 - aStackNeedto[nParent]['Count'] - 1) // 2) % 2 != 0: nSum *= -1 aStackNeedto[nParent]['Count'] += 1 aStackNeedto[nParent]['Sum'] += nSum aStackNeedto.pop() def Partition(n): aStackNeedto.append({'Num' : n, 'Count' : 0, 'Sum' : 0, 'Parent' : 0}) while aStackNeedto: nNewParent = len(aStackNeedto) - 1 aCur = aStackNeedto[nNewParent] if aCur['Num'] < 0: Update(aCur['Parent'], 0) elif aCur['Num'] in aMem: Update(aCur['Parent'], aMem[aCur['Num']]) elif aCur['Count'] == aCur['Num'] * 2: aMem[aCur['Num']] = aCur['Sum'] Update(aCur['Parent'], aMem[aCur['Num']]) else: for k in range(1, aCur['Num'] + 1): aStackNeedto.append( {'Num' : (aCur['Num'] - (k * (3 * k - 1) / 2)), 'Count' : 0, 'Sum' : 0, 'Parent' : nNewParent} ) aStackNeedto.append( {'Num' : (aCur['Num'] - (k * (3 * k + 1) / 2)), 'Count' : 0, 'Sum' : 0, 'Parent' : nNewParent} ) return aMem[n] print Partition(1000)My Python solution, similar to Dave Webb’s:
#!/usr/bin/env python from functools import update_wrapper def memoize(func): func.memo = {} def memoizer(arg): if arg in func.memo: return func.memo[arg] else: func.memo[arg] = result = func(arg) return result return update_wrapper(memoizer, func) @memoize def partitions(n): if n in xrange(11): return (1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42)[n] else: return sum(pow(-1, k + 1) * (partitions(n - k * (3 * k - 1) // 2) + partitions(n - k * (3 * k + 1) // 2)) for k in xrange(n, 0, -1)) if __name__ == "__main__": print partitions(1000)A version that runs faster on my laptop, based on a formula given on the Mathworld page:
#!/usr/bin/env python from functools import update_wrapper def memoize(func): func.memo = {} def memoizer(arg): if arg in func.memo: return func.memo[arg] else: func.memo[arg] = result = func(arg) return result return update_wrapper(memoizer, func) @memoize def sigma_1(n): return sum(filter(lambda k: n % k == 0, xrange(1, n + 1))) @memoize def partitions(n): if n in xrange(11): return (1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42)[n] else: return sum(sigma_1(n - k) * partitions(k) for k in xrange(n)) // n if __name__ == "__main__": print partitions(1000)Here’s a ruby version. The PN array is used for caching. I actually tried using inject, but couldn’t get the caching working with it. I’d love to see someone else’s version with that in there.
PN = [] def p(n) return 1 if n == 0 return 0 if n < 0 return PN[n] if PN[n] sum = 0 (1..n).each do |k| sum += ((-1)**(k+1)) * (p(n-(k*(3*k-1))/2) + p(n-(k*(3*k+1))/2)) PN[n] = sum end sum end puts "Partition of 1000: #{p(1000)}"Also @Dave’s python version memoizing the function is pretty cool. I haven’t tried anything like that in Ruby. Anyone know if it’s doable?
Here’s my Ruby implementation:
$partitions = [1,1,2,3,5,7,11,15,22,30,42]
def partition(n)
return 0 if n < 0
return $partitions[n] if $partitions[n] != nil
partition = 0
1.upto(n) do |k|
partition += (-1) ** (k + 1) * (partition(n – (k * (3 * k – 1)) / 2) + partition(n – (k * (3 * k + 1)) / 2))
end
$partitions[n] = partition
return $partitions[n]
end
p partition(1000)
Here’s the output:
24061467864032622473692149727991
I was inspired by @Remco’s slick Haskell solutions to try my own. After little
luck with IntMaps (perhaps due to overflow?) I came up with the following. Note:
I’m indebted to @Remco for nearly all of this, either in this exercise or
previous ones.
import Data.List (foldl', nub, sort, subsequences) import Data.Map ((!), findWithDefault, fromList, insert) import Data.Numbers.Primes (primeFactors) sigma1 :: (Integral a) => a -> a sigma1 = sum . nub . map product . subsequences . primeFactors partitions :: (Integral a) => a -> a partitions n = foldl' p' (fromList [(0, 1)]) [1 .. n] ! n where p' m i = insert i (sum [sigma1 (i - j) * p'' j | j <- [0 .. i - 1], let p'' k = findWithDefault 1 k m] `div` i) m main :: IO () main = print $ partitions 1000inspired by you graham
open System open System.Collections.Generic open Microsoft.FSharp.Collections open Microsoft.FSharp.Math let memoize (f: 'a -> 'b) = let dict = new Dictionary<'a,'b>() let memoizedFunc (input : 'a) = match dict.TryGetValue(input) with | true, x -> x | false, _ -> let ans = f input dict.Add(input, ans) ans memoizedFunc let rec P = let aux (n: int) = match n with | x when x < 0 -> 0.0 | 0 -> 1.0 | _ -> [1..n] |> List.fold (fun acc k -> let a = n - (k*(3*k-1))/2 let b = n - (k*(3*k+1))/2 acc + float(pown -1 (k+1))*((P a) + (P b))) 0.0 memoize aux P 1000Straight Scheme with delayed and forced promises.
(define (generate-vector n g) (let ((v (make-vector n))) (do ((k 0 (+ k 1))) ((= k n)) (vector-set! v k (g v k))) v)) (define world (generate-vector 1001 (lambda (world n) ;delayed numbers (delay (let sum ((k 1) (partial 0)) (if (<= k n) (sum (+ k 1) ((if (odd? k) + -) ; partial (+ (P (- n (* k (- (* 3 k) 1) 1/2))) (P (- n (* k (+ (* 3 k) 1) 1/2)))))) partial)))))) (define (P n) ;forces the delayed numbers (cond ((negative? n) 0) ((zero? n) 1) ((positive? n) (force (vector-ref world n))))) (write (P 1000)) (newline)(The promise at 0 remains an empty sum. The rest are good.)
Hey there! New to the site, but already loving it. Here’s my Python solution. Much like the above, with one VERY BIG time saving feature. (I’m not sure if the “regular crowd” here is in to optimizations… but this one seemed worthy of note.)
@memoize
def partition_number_opt1(n):
if n < 0: return 0
if n == 0: return 1
sum = 0
for k in range(1,n+1):
# slight optimization:
# once the inner expressions get below zero, then all remaining
# partition numbers (in the rest of the sum) are zero,
# thus they contribute nothing to the sum.
# so: break out of the loop
if n-(k*(3*k-1)//2) < 0: break
sum += ((-1) ** (k+1)) * (partition_number_opt1(n-(k*(3*k-1)//2))+partition_number_opt1(n-(k*(3*k+1)//2)))
return sum
Just trying my post once again, after I “read the manual”. Mea culpa!
Using Dan’s observation as an excuse for a little refactoring that I wished
I did in the first place: making sum a procedure of its own. And yes, it
does feel faster with the early completion of the sums.
(define (generate-vector n g) (do ((k 0 (+ k 1)) (v (make-vector n) v)) ((= k n) v) (vector-set! v k (g v k)))) (define (sum b e g) ;sum (g k) from b until (e k) (do ((k b (+ k 1)) (s 0 (+ s (g k)))) ((e k) s))) (define world (generate-vector 1001 (lambda (world n) ;delayed numbers (delay (sum 1 (lambda (k) (< n (* k (- (* 3 k) 1) 1/2))) (lambda (k) (* (if (odd? k) +1 -1) (+ (P (- n (* k (- (* 3 k) 1) 1/2))) (P (- n (* k (+ (* 3 k) 1) 1/2))))))))))) (define (P n) ;forces the delayed numbers (cond ((negative? n) 0) ((zero? n) 1) ((positive? n) (force (vector-ref world n))))) (write (P 1000)) (newline)The recursive solution is very slow because redundant calculation of partition numbers.
So, The trick is to remember the previously calculated value of partition numbers.
Checkout my implementation in C using red-black trees.
http://codepad.org/vKEivBGq
def p(n, L = [[1]]): for i in range(1, n + 1): L.append([L[x][0] for x in range(i - 1, -1, -1)]) for j in range((i + 2)/2, i): L[j][0] = L[j].pop(0) + L[j][0] return sum(L[n]) print p(1000)Here is my non-recursive python solution, simple and efficient!
Since Euler’s formula is complex for calculation, I used another algorithm,
instead, which can easily be discovered from the process of partition.
def p(n, L = [[1]]): for i in range(1, n + 1): L.append([L[x][0] for x in range(i - 1, -1, -1)]) for j in range((i + 2)/2, i): L[j][0] = L[j].pop(0) + L[j][0] return sum(L[n]) print p(1000)I found this quite interesting, as there is an enormous range of performance between different solutions. The Perl version below is combinatorial and, at least for me, is quite a bit faster and less memory than the previously shown ones (barring the C double solution which isn’t correct for large values). However it is about 300 times slower than the GMP combinatorial solution below. In turn, that is about 1,000 times slower than the Rademacher formula used in Pari or SymPy. Pari in turn is quite a bit slower than Jonathan Bober’s MPFR Rademacher solution. The current state of the art implementation seems to be Fredrik Johansson’s FLINT/MPFR/Arb implementation which looks to be over 1000x faster than Pari, computing p(10^12) in under 12s. The growth rates differ (especially between the combinatorial and Rademacher solutions), so the “x faster” is simplistic.
In Perl, computes p(10000) in about 8s.
#!/usr/bin/env perl use warnings; use strict; use Math::BigInt try => "GMP"; my $n = shift || die "Usage: $0 <n> Returns the integer partition number of <n>\n"; print partitions($n), "\n"; sub partitions { my $n = shift; return 1 if defined $n && $n <= 0; my $d = int(sqrt($n+1)); my @part = (Math::BigInt->bone); my @pent = (1, map { (($_*(3*$_+1))>>1, (($_+1)*(3*$_+2))>>1) } 1 .. $d); foreach my $j (scalar @part .. $n) { my ($psum1, $psum2, $k) = (Math::BigInt->bzero, Math::BigInt->bzero, 1); foreach my $p (@pent) { last if $p > $j; if ((++$k) & 2) { $psum1->badd( $part[ $j - $p ] ); } else { $psum2->badd( $part[ $j - $p ] ); } } $part[$j] = $psum1 - $psum2; } return $part[$n]; }In C+GMP, computes p(200000) in about 5s. Compile with “gcc -O partitions.c -lgmp -lm”.
#include <stdio.h> #include <stdlib.h> #include <math.h> #include <gmp.h> static void partitions(unsigned long n, mpz_t result) { if (n <= 0) { mpz_set_ui(result, 1); } else if (n <= 3) { mpz_set_ui(result, n); } else { unsigned long i, j, k, d = (unsigned long) sqrt(n+1); unsigned long* pent = (unsigned long*) malloc((2*d+2) * sizeof(unsigned long)); mpz_t* part = (mpz_t*) malloc((n+1) * sizeof(mpz_t)); mpz_t psum; pent[0] = 0; pent[1] = 1; for (i = 1; i <= d; i++) { pent[2*i ] = ( i *(3*i+1)) / 2; pent[2*i+1] = ((i+1)*(3*i+2)) / 2; } mpz_init_set_ui(part[0], 1); mpz_init(psum); for (j = 1; j <= n; j++) { mpz_set_ui(psum, 0); for (k = 1; pent[k] <= j; k++) { if ((k+1) & 2) mpz_add(psum, psum, part[ j - pent[k] ]); else mpz_sub(psum, psum, part[ j - pent[k] ]); } mpz_init_set(part[j], psum); } mpz_clear(psum); free(pent); mpz_set(result, part[n]); for (i = 0; i <= n; i++) mpz_clear(part[i]); free(part); } } int main(int argc, char **argv) { unsigned long n; mpz_t res; if (argc != 2) { printf("Usage: %s <n> Returns the integer partition number of <n>\n", argv[0]); exit(-1); } n = strtoull(argv[1], 0, 10); mpz_init(res); partitions(n, res); gmp_printf("%Zd\n", res); mpz_clear(res); exit(0); }Here is a short Haskell solution that implicitly memoizes without using any non-Prelude data structures, but achieves quite good performances:
integerPartitions :: [Integer]
integerPartitions =
let
p = go (let kg k = (2*k):k:kg (k+1) in 0:kg 1) [] 0
go :: [Int] -> [[Integer]] -> Int -> [Integer]
go ks a !c = s:(if c>0 then go ks b (c-1) else (let (d:kt)=ks in go kt b d))
where
(s,b) = sf a
sf ((a1:t1):(a2:t2):r) = let (t,tr)=sf r in (a1+a2-t,t1:t2:tr)
sf ((a1:t1):_) = if c>0 then (a1,t1:[]) else (a1+1,t1:p:[])
sf _ = if c>0 then (0,[]) else (1,p:[])
in
1:p
Forgive me for screwing up the formatting in the previous post. This is better, I hope: http://codepad.org/wprw8CLb . FWIW, this code calculates integerPartitions!!100000 in a little over 10 seconds on my ancient workstation.
[…] I’ve written a Haskell function to count the number of partitions of an integer – it’s basically an implementation of the same algorithm as this one, using Euler’s Pentagonal Number Theorem: […]