Partitions of n into k parts with restrictions - algorithm

Question

I need an algorithm that produces a partition of the number n into k parts with the added restrictions that each element of the partition must be between a and b. Ideally, all possible partitions satisfying the restrictions should be equally likely. Partitions are considered the same if they have the same elements in different order.
For example, with n=10, k=3, a=2, b=4 one has only {4,4,2} and {4,3,3} as possible outcomes.
Is there a standard algorithm for such a problem? One can assume that at least one partition satisfying the restrictions always exists.

Answer 1

You can implement this as a recursive algorithm. Basically, the recurrence is like this:
if k == 1 and a <= n <= b, then the only partition is [n], otherwise none
otherwise, combine all the elements x from a to b with all the partitions for n-x, k-1
to prevent duplicates, also substitute the lower bound a with x
Here's some Python (aka executable pseudo-code):
def partitions(n, k, a, b):
if k == 1 and a <= n <= b:
yield [n]
elif n > 0 and k > 0:
for x in range(a, b+1):
for p in partitions(n-x, k-1, x, b):
yield [x] + p
print(list(partitions(10, 3, 2, 4)))
# [[2, 4, 4], [3, 3, 4]]
This could be further improved by checking (k-1)*a and (k-1)*b for the lower and upper bounds for the remaining elements, respectively, and restricting the range for x accordingly:
min_x = max(a, n - (k-1) * b)
max_x = min(b, n - (k-1) * a)
for x in range(min_x, max_x+1):
For partitions(110, 12, 3, 12) with 3,157 solutions, this reduces the number of recursive calls from 638,679 down to 24,135.

Answer 2

Here's a sampling algorithm that uses conditional probability.
import collections
import random
countmemo = {}
def count(n, k, a, b):
assert n >= 0
assert k >= 0
assert a >= 0
assert b >= 0
if k == 0:
return 1 if n == 0 else 0
key = (n, k, a, b)
if key not in countmemo:
countmemo[key] = sum(
count(n - c, k - 1, a, c) for c in range(a, min(n, b) + 1))
return countmemo[key]
def sample(n, k, a, b):
partition = []
x = random.randrange(count(n, k, a, b))
while k > 0:
for c in range(a, min(n, b) + 1):
y = count(n - c, k - 1, a, c)
if x < y:
partition.append(c)
n -= c
k -= 1
b = c
break
x -= y
else:
assert False
return partition
def test():
print(collections.Counter(
tuple(sample(20, 6, 2, 5)) for i in range(10000)))
if __name__ == '__main__':
test()

Answer 3

If k and b - a are not too big you can try a randomized depth-first search:
import random
def restricted_partition_rec(n, k, min, max):
if k <= 0 or n < min:
return []
ps = list(range(min, max + 1))
random.shuffle(ps)
for p in ps:
if p > n:
continue
elif p < n:
subp = restricted_partition(n - p, k - 1, min, max)
if subp:
return [p] + subp
elif k == 1:
return [p]
return []
def restricted_partition(n, k, min, max):
return sorted(restricted_partition_rec(n, k, min, max), reverse=True)
print(restricted_partition(10, 3, 2, 4))
>>>
[4, 4, 2]
Although I'm not sure if all the partitions have exactly the same probability in this case.