def subrange_exercise(mult, lb, ub):
"""Compare filter-based and more optimized subrange implementations
Helper for tests, called with both small and larger multisets.
"""
m = MultisetPartitionTraverser()
assert m.count_partitions(mult) == \
m.count_partitions_slow(mult)
# Note - multiple traversals from the same
# MultisetPartitionTraverser object cannot execute at the same
# time, hence make several instances here.
ma = MultisetPartitionTraverser()
mc = MultisetPartitionTraverser()
md = MultisetPartitionTraverser()
# Several paths to compute just the size two partitions
a_it = ma.enum_range(mult, lb, ub)
b_it = part_range_filter(multiset_partitions_taocp(mult), lb, ub)
c_it = part_range_filter(mc.enum_small(mult, ub), lb, sum(mult))
d_it = part_range_filter(md.enum_large(mult, lb), 0, ub)
for sa, sb, sc, sd in zip_longest(a_it, b_it, c_it, d_it):
assert compare_multiset_states(sa, sb)
assert compare_multiset_states(sa, sc)
assert compare_multiset_states(sa, sd)
def test_multiset_partitions_versions():
"""Compares Knuth-based versions of multiset_partitions"""
multiplicities = [5, 2, 2, 1]
m = MultisetPartitionTraverser()
for s1, s2 in zip_longest(m.enum_all(multiplicities),
multiset_partitions_taocp(multiplicities)):
assert compare_multiset_states(s1, s2)
def nT(n, k=None):
"""Return the number of ``k``-sized partitions of ``n`` items.
Possible values for ``n``::
integer - ``n`` identical items
sequence - converted to a multiset internally
multiset - {element: multiplicity}
Note: the convention for ``nT`` is different than that of ``nC`` and
``nP`` in that
here an integer indicates ``n`` *identical* items instead of a set of
length ``n``; this is in keeping with the ``partitions`` function which
treats its integer-``n`` input like a list of ``n`` 1s. One can use
``range(n)`` for ``n`` to indicate ``n`` distinct items.
If ``k`` is None then the total number of ways to partition the elements
represented in ``n`` will be returned.
Examples
========
>>> from sympy.functions.combinatorial.numbers import nT
Partitions of the given multiset:
>>> [nT('aabbc', i) for i in range(1, 7)]
[1, 8, 11, 5, 1, 0]
>>> nT('aabbc') == sum(_)
True
>>> [nT("mississippi", i) for i in range(1, 12)]
[1, 74, 609, 1521, 1768, 1224, 579, 197, 50, 9, 1]
Partitions when all items are identical:
>>> [nT(5, i) for i in range(1, 6)]
[1, 2, 2, 1, 1]
>>> nT('1'*5) == sum(_)
True
When all items are different:
>>> [nT(range(5), i) for i in range(1, 6)]
[1, 15, 25, 10, 1]
>>> nT(range(5)) == sum(_)
True
References
==========
.. [1] http://undergraduate.csse.uwa.edu.au/units/CITS7209/partition.pdf
See Also
========
sympy.utilities.iterables.partitions
sympy.utilities.iterables.multiset_partitions
"""
from sympy.utilities.enumerative import MultisetPartitionTraverser
if isinstance(n, SYMPY_INTS):
# assert n >= 0
# all the same
if k is None:
return sum(_nT(n, k) for k in range(1, n + 1))
return _nT(n, k)
if not isinstance(n, _MultisetHistogram):
try:
# if n contains hashable items there is some
# quick handling that can be done
u = len(set(n))
if u == 1:
return nT(len(n), k)
elif u == len(n):
n = range(u)
raise TypeError
except TypeError:
n = _multiset_histogram(n)
N = n[_N]
if k is None and N == 1:
return 1
if k in (1, N):
return 1
if k == 2 or N == 2 and k is None:
m, r = divmod(N, 2)
rv = sum(nC(n, i) for i in range(1, m + 1))
if not r:
rv -= nC(n, m) // 2
if k is None:
rv += 1 # for k == 1
return rv
if N == n[_ITEMS]:
# all distinct
if k is None:
return bell(N)
return stirling(N, k)
m = MultisetPartitionTraverser()
if k is None:
return m.count_partitions(n[_M])
# MultisetPartitionTraverser does not have a range-limited count
# method, so need to enumerate and count
tot = 0
for discard in m.enum_range(n[_M], k - 1, k):
tot += 1
return tot