Programming Praxis

12 Responses to “Partitions”

1. uberlazy said

   May 11, 2012 at 10:18 AM

   A recursive solution in Clojure. The result is in the reverse order than what is shown in the problem description, because it’s slightly more convenient to check the subtrahend against 0 than against the original minuend, which would have to be remembered somehow. It shouldn’t matter as the problem calls for a set.

   (defn- parts-helper [num subtr acc]
     (cond
      (= num 0) [acc]
      (or (= subtr 0) (< num 0)) []
      :else (concat
   	  (intpart (- num subtr) subtr (conj acc subtr))
   	  (intpart num (dec subtr) acc))))
   
   (defn parts [num]
     (parts-helper num num []))
   
2. uberlazy said

   May 11, 2012 at 10:20 AM

   A copy-paste blunder. The code should read:

   [sourceoode=”css”]
   (defn- parts-helper [num subtr acc]
   (cond
   (= num 0) [acc]
   (or (= subtr 0) (< num 0)) []
   :else (concat
   (parts-helper (- num subtr) subtr (conj acc subtr))
   (parts-helper num (dec subtr) acc))))

   (defn parts [num]
   (parts-helper num num []))

   [/sourcecode]
3. Programming Praxis – Partitions « Bonsai Code said

   May 11, 2012 at 4:50 PM

   […] today’s Programming Praxis exercise, our goal is to write a function that gives all unique sums that […]
4. Remco Niemeijer said

   May 11, 2012 at 4:51 PM

   My Haskell solution (see http://bonsaicode.wordpress.com/2012/05/11/programming-praxis-partitions/ for a version with comments):

   part :: Int -> [[Int]]
   part 0 = [[]]
   part n = [x:y | x <- [1..n], y <- part (n-x), [x] >= take 1 y]
   
5. Mike said

   May 11, 2012 at 6:27 PM

   Python 2.7

   def partitionsOf(n):
       if n == 0:
           return {()}
       else:
           return {tuple(sorted(t + (x,))) for x in range(1, n + 1) for t in partitionsOf(n - x)}
   
6. Graham said

   May 11, 2012 at 10:56 PM

   It strikes me that many recursive solutions will probably be inefficient
   (except for Haskell ones, perhaps), since solutions to subproblems are
   recomputed every time they’re needed, similar to SICP’s discussion of Fibonacci
   number generation.
   First up, a memoizing version (based on a Python version that Programming
   Praxis pointed me to on StackOverflow) that stores previous answers.
   Next are zs1 and zs2, Python versions of two
   algorithms found in Fast Algorithms for Generating Integer
Paritions by Zoghbi and Stojmenovic (1998). They’re iterative and more
   efficient than the first solution, if uglier. I’ve made them generators instead
   of lists, but that’s not really relevant to the algorithms.

   from functools import update_wrapper
   
   def memoize(func):
       func.memo = {}
       def wrapper(arg):
           try:
               return func.memo[arg]
           except KeyError:
               func.memo[arg] = result = func(arg)
               return result
       return update_wrapper(wrapper, func)
   
   @memoize
   def partitions(n):
       parts = set([tuple([n])])
       for k in xrange(1, n):
           for p in partitions(n - k):
               parts.add(tuple(sorted([k] + p)))
       return map(list, parts)
   
   def zs1(n):
       x = [1] * n
       x[0], m, h = n, 0, 0
       yield [x[0]]
       while x[0] != 1:
           if x[h] == 2:
               m, x[h] = m + 1, 1
               h -= 1
           else:
               r = x[h] - 1
               t, x[h] = m - h + 1, r
               while t >= r:
                   h += 1
                   x[h], t = r, t - r
               if t == 0:
                   m = h
               else:
                   m = h + 1
                   if t > 1:
                       h += 1
                       x[h] = t
           yield x[:m + 1]
   
   def zs2(n):
       x = [1] * (n + 1)
       yield x[1:]
       x[:2] = -1, 2
       h, m = 1, n - 1
       yield x[1: m + 1]
       while x[1] != n:
           if m - h > 1:
               h += 1
               x[h], m = 2, m - 1
           else:
               j = m - 2
               while x[j] == x[m - 1]:
                   x[j] = 1
                   j -= 1
               h = j + 1
               x[h], r, x[m] = x[m - 1] + 1, x[m] + x[m - 1] * (m - h - 1), 1
               if m - h > 1:
                   x[m - 1] = 1
               m = h + r - 1
           yield x[1: m + 1]
   
   if __name__ == "__main__":
       print sorted(partitions(6))
       print list(zs1(6))
       print list(zs2(6))
   
7. Remco Niemeijer said

   May 11, 2012 at 11:26 PM

   @Graham: The Haskell version isn’t memoized. Memoizing the function is pretty simple and doesn’t add too much to the program length though:

   memopart = (memo !!) where
       memo = [[]] : [[x:y | x <- [1..n], y <- memo !! (n-x), [x] >= take 1 y] | n <- [1..]]
   
8. Graham said

   May 12, 2012 at 2:05 AM

   @Remco: thanks for that. I thought there might have been some sort of wizardry hidden in the lazy evaluation or something, but I appreciate the enlightenment.
9. Jussi Piitulainen said

   May 12, 2012 at 5:37 AM

   Find partitions where the least term is equal to or greater than another parameter. This way duplicates are not generated. All partitions are found by setting the other parameter to 1.

   (define integer-partitions
     (case-lambda
      ((n) (integer-partitions n 1))
      ((n m) (if (= n m) (list (list m))
                 (if (< n m) (list)
                     (append (map (lambda (p) (cons m p))
                                  (integer-partitions (- n m) m))
                             (integer-partitions n (+ m 1))))))))
   

   > (integer-partitions 4)
   ((1 1 1 1) (1 1 2) (1 3) (2 2) (4))
   
10. Jon Bodner said

    May 13, 2012 at 3:45 AM

    Solution in Go:

    
    

    	package main
    	import "fmt"
    	import "sort"
    	type EMPTY struct {}
    	var PRESENT EMPTY = EMPTY{}
    	func partition(val int) []string {
    	set := make(map[string]EMPTY)
    	for i := 1; i<=val;i++ {
    	for _,result := range worker([]int{i},val-i) {
    	set[fmt.Sprintf("%v",result)]=PRESENT
    	}
    	}
    	out := make([]string,len(set))
    	pos :=0
    	for key := range set {
    	out[pos] = key
    	pos++
    	}
    	sort.Strings(out)
    	return out
    	}
    	func worker(already []int, val int) [][]int {
    	if val == 0 {
    	sort.Ints(already)
    	return [][]int{already}
    	}
    	out := [][]int {}
    	for i := 1;i<=val;i++ {
    	out = append(out,worker(append(already,i),val-i)…)
    	}
    	return out
    	}
    	func main() {
    	for i := 0;i<10;i++ {
    	result := partition(i)
    	fmt.Printf("%d (%d): %v\n",i,len(result),result)
    	}
    	}

    view raw
    
    partition.go
    
    hosted with ❤ by GitHub

    This shows off some of the ugly things in Go: no built-in set type (use a map with an empty value), no easy way to get all the keys for a map (have to manually copy to a slice), no way to directly use a slice as the key for a map or define a custom hashcode or equals on a type (convert the slice to a string, and use that as the key in the map).
11. Benjamin Samson said

    May 13, 2012 at 11:21 AM

    My solution is not as neat and elegant as yours but all the same, I’m proud that I was able to do it.
    Here’s my solution in Python:

    def partitions(n):
    p = []
    # n is not 0
    if n:
    # from 1 to n
    for i in range(1, n+1):
    a = partitions(n-i)

    if len(a) == 0:
    p.append(i)
    else:
    for j in a:
    if isinstance(j, int):
    b = [j, i]
    b.sort()
    if b not in p:
    p.append(b)
    else:
    j.append(i)
    j.sort()
    if j not in p:
    p.append(j)
    return p

    # n is 0
    else:
    return []
12. Jan Van lent said

    August 16, 2012 at 11:31 PM

    Another Python version [1].
    It uses a sparse representation for the partitions, e.g., 4 = 4 x 1 = 2 x 1 + 2 = 1 x 1 + 3 = 2 x 2 = 1 x 4 instead of 4 = 1+1+1+1 = 1+1+2 = 1+3 = 2 + 2 = 4.

    [1] http://code.activestate.com/recipes/221132/