/
multinomial.py
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/
multinomial.py
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def binomial_coefficients(n):
"""Return a dictionary containing pairs `{(k1,k2) : C_{kn}}` where
`C_{kn}` are binomial coefficients and `n=k1+k2`.
Examples
========
>>> binomial_coefficients(9)
{(0, 9): 1, (1, 8): 9, (2, 7): 36,
(3, 6): 84, (4, 5): 126, (5, 4): 126, (6, 3): 84,
(7, 2): 36, (8, 1): 9, (9, 0): 1}
See Also
========
binomial_coefficients_list, multinomial_coefficients
"""
d = {(0, n): 1, (n, 0): 1}
a = 1
for k in range(1, n//2 + 1):
a = (a * (n - k + 1))//k
d[k, n - k] = d[n - k, k] = a
return d
def binomial_coefficients_list(n):
"""Return a list of binomial coefficients as rows of the Pascal's triangle.
Examples
========
>>> binomial_coefficients_list(9)
[1, 9, 36, 84, 126, 126, 84, 36, 9, 1]
See Also
========
binomial_coefficients, multinomial_coefficients
"""
d = [1] * (n + 1)
a = 1
for k in range(1, n//2 + 1):
a = (a * (n - k + 1))//k
d[k] = d[n - k] = a
return d
def multinomial_coefficients(m, n):
r"""Return a dictionary containing pairs ``{(k1,k2,..,km) : C_kn}``
where ``C_kn`` are multinomial coefficients such that ``n=k1+k2+..+km``.
Examples
========
>>> multinomial_coefficients(2, 5)
{(0, 5): 1, (1, 4): 5,
(2, 3): 10, (3, 2): 10, (4, 1): 5, (5, 0): 1}
Notes
=====
The algorithm is based on the following result:
.. math::
\binom{n}{k_1, \ldots, k_m} = \frac{k_1 + 1}{n - k_1}
\sum_{i=2}^m \binom{n}{k_1 + 1, \ldots, k_i - 1, \ldots}
See Also
========
binomial_coefficients_list, binomial_coefficients
"""
if not m:
if n:
return {}
return {(): 1}
if m == 2:
return binomial_coefficients(n)
if m >= 2*n and n > 1:
return dict(multinomial_coefficients_iterator(m, n))
t = [n] + [0] * (m - 1)
r = {tuple(t): 1}
if n:
j = 0 # j will be the leftmost nonzero position
else:
j = m
# enumerate tuples in co-lex order
while j < m - 1:
# compute next tuple
tj = t[j]
if j:
t[j] = 0
t[0] = tj
if tj > 1:
t[j + 1] += 1
j = 0
start = 1
v = 0
else:
j += 1
start = j + 1
v = r[tuple(t)]
t[j] += 1
# compute the value
# NB: the initialization of v was done above
for k in range(start, m):
if t[k]:
t[k] -= 1
v += r[tuple(t)]
t[k] += 1
t[0] -= 1
r[tuple(t)] = (v * tj) // (n - t[0])
return r
def multinomial_coefficients_iterator(m, n, _tuple=tuple):
"""Multinomial coefficient iterator.
This routine has been optimized for `m` large with respect to `n` by taking
advantage of the fact that when the monomial tuples `t` are stripped of
zeros, their coefficient is the same as that of the monomial tuples from
``multinomial_coefficients(n, n)``. Therefore, the latter coefficients are
precomputed to save memory and time.
>>> m53, m33 = multinomial_coefficients(5, 3), multinomial_coefficients(3, 3)
>>> (m53[(0, 0, 0, 1, 2)] == m53[(0, 0, 1, 0, 2)] ==
... m53[(1, 0, 2, 0, 0)] == m33[(0, 1, 2)])
True
Examples
========
>>> it = multinomial_coefficients_iterator(20, 3)
>>> next(it)
((3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), 1)
"""
if m < 2*n or n == 1:
mc = multinomial_coefficients(m, n)
for k, v in mc.items():
yield(k, v)
else:
mc = multinomial_coefficients(n, n)
mc1 = {}
for k, v in mc.items():
mc1[_tuple(filter(None, k))] = v
mc = mc1
t = [n] + [0] * (m - 1)
t1 = _tuple(t)
b = _tuple(filter(None, t1))
yield (t1, mc[b])
if n:
j = 0 # j will be the leftmost nonzero position
else:
j = m
# enumerate tuples in co-lex order
while j < m - 1:
# compute next tuple
tj = t[j]
if j:
t[j] = 0
t[0] = tj
if tj > 1:
t[j + 1] += 1
j = 0
else:
j += 1
t[j] += 1
t[0] -= 1
t1 = _tuple(t)
b = _tuple(filter(None, t1))
yield (t1, mc[b])