def pmf_dict(colors, nsample):
"""
Returns a dictionary of the PMF of the multivariate hypergeometric distr.
The keys are the points in the support, and the values are the
corresponding probabilities.
Example
-------
>>> pmf = multivariate_hypergeometric.pmf_dict([4, 5, 6], 6)
>>> len(pmf)
24
>>> float(pmf[2, 2, 2])
0.1798201798201798
"""
_validate_params(colors, nsample)
total = sum(colors)
denom = mpmath.binomial(total, nsample)
pmf = {}
for coords in support(colors, nsample):
numer = 1
for color, k in zip(colors, coords):
numer *= mpmath.binomial(color, k)
prob = numer/denom
pmf[tuple(coords)] = prob
return pmf
def getNthDelannoyNumber( n ):
result = 0
for k in arange( 0, fadd( real( n ), 1 ) ):
result = fadd( result, fmul( binomial( n, k ), binomial( fadd( n, k ), k ) ) )
return result
def mhyper(self,q,m,n,k):
u = min(self.tag_length ,n)-1
A = binomial(m,q)*binomial(n,k-q)
B = hyper([1,q-k,q-m],[1+q,1-k+n+q], 1)
C = -binomial(m,1+u)*binomial(n,k-1-u)
D = hyper([1,1-k+u,1-m+u],[2+u,2-k+n+u], 1)
return mathlog(float((A*B + C*D) / binomial(m+n, k)))
def __compute_surprise_weighted_by_quadrature(self, mi, pi, m, p):
f = lambda i: mpmath.binomial(
self.pi, i) * mpmath.binomial(self.p - self.pi, self.m - i) / \
mpmath.binomial(self.p, self.m)
# +- 0.5 is because of the continuity correction
return float(
-mpmath.log10(mpmath.quad(f, [self.mi - 0.5, self.m + 0.5])))
def getNthAperyNumber( n ):
result = 0
for k in arange( 0, real( n ) + 1 ):
result = fadd( result, fmul( power( binomial( n, k ), 2 ),
power( binomial( fadd( n, k ), k ), 2 ) ) )
return result
def getNthMotzkinNumber( n ):
result = 0
for j in arange( 0, floor( fdiv( real( n ), 3 ) ) + 1 ):
result = fadd( result, fprod( [ power( -1, j ), binomial( fadd( n, 1 ), j ),
binomial( fsub( fmul( 2, n ), fmul( 3, j ) ), n ) ] ) )
return fdiv( result, fadd( n, 1 ) )
def __compute_surprise_weighted(self, mi, pi, m, p):
b1 = mpmath.binomial(pi, mi)
b2 = mpmath.binomial(p - pi, m - mi)
b3 = mpmath.binomial(p, m)
h3f2 = mpmath.hyp3f2(1, mi - m, mi - pi, mi + 1, -m + mi + p - pi + 1,
1)
#h3f2 = hyper1([mpf(1),mi-m, mi-pi], [mpf(1),mpf(1)+mi, mi+p-pi-m+mpf(1)], 10, 10)
log10cdf = mpmath.log10(b1) + mpmath.log10(b2) + mpmath.log10(
h3f2) - mpmath.log10(b3)
return -float(log10cdf)
def binomial(A, B):
"""
A ! B
"""
try:
if isinstance(A, int) and isinstance(B, int):
return int(mpmath.binomial(B, A))
return float(mpmath.binomial(B, A))
except ValueError:
assertError("DOMAIN ERROR")
def sf(k, ntotal, ngood, nsample):
"""
Survival function of the hypergeometric distribution.
"""
_validate(ntotal, ngood, nsample)
h = mpmath.hyp3f2(1, k + 1 - ngood, k + 1 - nsample, k + 2,
ntotal + k + 2 - ngood - nsample, 1)
num = (mpmath.binomial(nsample, k + 1) *
mpmath.binomial(ntotal - nsample, ngood - k - 1))
den = mpmath.binomial(ntotal, ngood)
sf = (num / den) * h
return sf
def getNthAperyNumber( n ):
'''
http://oeis.org/A005259
a(n) = sum(k=0..n, C(n,k)^2 * C(n+k,k)^2 )
'''
result = 0
for k in arange( 0, real( n ) + 1 ):
result = fadd( result, fmul( power( binomial( n, k ), 2 ),
power( binomial( fadd( n, k ), k ), 2 ) ) )
return result
def getNthMotzkinNumber( n ):
'''
http://oeis.org/A001006
a(n) = sum((-1)^j*binomial(n+1, j)*binomial(2n-3j, n), j=0..floor(n/3))/(n+1)
'''
result = 0
for j in arange( 0, floor( fdiv( real( n ), 3 ) ) + 1 ):
result = fadd( result, fprod( [ power( -1, j ), binomial( fadd( n, 1 ), j ),
binomial( fsub( fmul( 2, n ), fmul( 3, j ) ), n ) ] ) )
return fdiv( result, fadd( n, 1 ) )
def pmf(k, ntotal, ngood, untilnbad):
"""
Probability mass function of the negative hypergeometric distribution.
"""
_validate(ntotal, ngood, untilnbad)
if k < 0 or k > ngood:
return mpmath.mp.zero
with mpmath.extradps(5):
b1 = mpmath.binomial(k + untilnbad - 1, k)
b2 = mpmath.binomial(ntotal - untilnbad - k, ngood - k)
b3 = mpmath.binomial(ntotal, ngood)
return b1 * (b2 / b3)
def test_Wigner_coefficient():
import mpmath
mpmath.mp.dps = 4 * sf.ell_max
i = 0
for twoell in range(2*sf.ell_max + 1):
for twomp in range(-twoell, twoell + 1, 2):
for twom in range(-twoell, twoell + 1, 2):
tworho_min = max(0, twomp - twom)
a = sf._Wigner_coefficient(twoell, twomp, twom)
b = float(mpmath.sqrt(mpmath.fac((twoell + twom)//2) * mpmath.fac((twoell - twom)//2)
/ (mpmath.fac((twoell + twomp)//2) * mpmath.fac((twoell - twomp)//2)))
* mpmath.binomial((twoell + twomp)//2, tworho_min//2)
* mpmath.binomial((twoell - twomp)//2, (twoell - twom - tworho_min)//2))
assert np.allclose(a, b), (twoell, twomp, twom, i, sf._Wigner_index(twoell, twomp, twom))
i += 1
def getNthDelannoyNumber(n):
if n == 1:
return 3
precision = int(fmul(n, 0.8))
if mp.dps < precision:
mp.dps = precision
result = 0
for k in arange(0, fadd(n, 1)):
result = fadd(result, fmul(binomial(n, k), binomial(fadd(n, k), k)))
return result
def binomial(n, k):
if abs(k) > 1e8*(abs(n) + 1):
# The binomial is rapidly oscillating in this region,
# and the function is numerically ill-defined. Don't
# compare values here.
return np.nan
return mpmath.binomial(n, k)
def round5_fp_fast(n, h, q, p, t, f, mu, B, add_coeff=False, verbose=False):
## use this for full precision
### dif = mp_zeros(q)
### dif[0] = mp.mpf(1.0)
## use this for double precision
dif = np.zeros(q)
dif[0] = 1.0
# determine number of times to convolve the term differences in term s*e.
#If prime cyclotomic, it is h, if ring swithing or scaler, just h/2.
# Only h(2h) convolutions as we start with probability gen function
# for difference between two random variables with q-p rounding noise
# This can be done as we consider balanced secrets
w = h
if add_coeff:
w = 2 * w
# computation of convolution of w variables (we have the sum of w variables)
for i in range(w):
dif = spec_conv_comb_qp(dif, q, p)
# convert to mpf if we were double
dif = np.array([mp.mpf(dif[i]) for i in range(len(dif))])
# normalize, just in case
dif /= dif.sum()
# second convolution with the p to t noise. Single time to account for coefficient compression from p to t.
dif = spec_conv_comb_pt(dif, q, p, t)
# Decryption failure ranges for case when parameter B>1
up_fail = (q + (1 << B)) // (1 << (B + 1))
down_fail = (q * ((1 << (B + 1)) - 1) + (1 << B)) // (1 << (B + 1))
# Bit failure probability (bfp). Distribution goes from 0 to q-1, highly provable values are centered around 0, thus, we sum the central part.
bfp = dif[up_fail:down_fail].sum()
# Failure probability after error correction
ffp = mp.mpf(0)
for j in range(f + 1, mu + 1):
ffp += (mp.binomial(mu, j) * mp.power(bfp, j) *
mp.power(mp.mpf(1) - bfp, (mu - j)))
return float(mp.log(ffp, 2))
def daubechies(N):
# make polynomial
q_y = [mpmath.binomial(N - 1 + k, k) for k in reversed(range(N))]
# get polynomial roots y[k]
y = mpmath.mp.polyroots(q_y, maxsteps=200, extraprec=64)
z = []
for yk in y:
# subustitute y = -1/4z + 1/2 - 1/4/z to factor f(y) = y - y[k]
f = [mpmath.mpf('-1/4'), mpmath.mpf('1/2') - yk, mpmath.mpf('-1/4')]
# get polynomial roots z[k] within unit circle
z += [zk for zk in mpmath.mp.polyroots(f) if mpmath.fabs(zk) < 1]
# make polynomial using the roots
h0z = mpmath.sqrt('2')
for zk in z:
h0z *= sympy.sympify('(z-zk)/(1-zk)').subs('zk', zk)
# adapt vanishing moments
hz = (sympy.sympify('(1+z)/2')**N * h0z).expand()
# get scaling coefficients
return [sympy.re(hz.coeff('z', k)) for k in reversed(range(N * 2))]
def sign_test_precision(preds_system_A, preds_system_B, ground_truth, q=0.5):
ties = 0
plus = 0
minus = 0
for i, gt in enumerate(ground_truth):
if preds_system_A[i] == preds_system_B[i]:
ties += 1
continue
else:
if preds_system_A[i] == gt:
plus += 1
continue
else:
minus += 1
continue
all_samples = len(ground_truth)
k = np.ceil(ties / 2) + min(plus, minus)
N = 2 * np.ceil(ties / 2) + plus + minus
print("The k is: ", k)
print("The N is: ", N)
res = 0
for idx in range(0, int(k)):
res += (mpmath.binomial(N, idx) * mpmath.power(q, idx) * mpmath.power(
(1 - q), (N - idx)))
#res += (nCr(N, idx)*pow(q, idx)*pow((1-q), (N-idx)))
return (2 * res)
def getNthSchroederNumber( n ):
if real( n ) == 1:
return 1
if n < 0:
raise ValueError( '\'nth_schroeder\' expects a non-negative argument' )
n = fsub( n, 1 )
result = 0
for k in arange( 0, fadd( n, 1 ) ):
result = fadd( result, fdiv( fprod( [ power( 2, k ), binomial( n, k ),
binomial( n, fsub( k, 1 ) ) ] ), n ) )
return result
def daubechies(N):
# p vanishing moments.
p = int(N / 2)
# make polynomial; see Mallat, 7.96
Py = [sm.binomial(p - 1 + k, k) for k in reversed(range(p))]
# get polynomial roots y[k]
Py_roots = sm.mp.polyroots(Py, maxsteps=200, extraprec=64)
z = []
for yk in Py_roots:
# substitute y = -1/4z + 1/2 - 1/4/z to factor f(y) = y - y[k]
# We've found the roots of P(y). We need the roots of Q(z) = P((1-z-1/z)/4)
f = [sm.mpf('-1/4'), sm.mpf('1/2') - yk, sm.mpf('-1/4')]
# get polynomial roots z[k]
z += sm.mp.polyroots(f)
# make polynomial using the roots within unit circle
h0z = sm.sqrt('2')
for zk in z:
if sm.fabs(zk) < 1:
# This calculation is superior to Mallat, (equation between 7.96 and 7.97)
h0z *= sympy.sympify('(z-zk)/(1-zk)').subs('zk', zk)
# adapt vanishing moments
hz = (sympy.sympify('(1+z)/2')**p * h0z).expand()
# get scaling coefficients
return [sympy.re(hz.coeff('z', k)) for k in reversed(range(p * 2))]
def get_prob_correct_sync(k, pc):
"""prob = \sum_{l=0}^{ceil(k/2)-1} (k_choose_l) * (pc)^l * (1 - pc)^(k-l)
"""
retval = 0
lmax = mpmath.ceil(float(k)/2) - 1
for l in range(0, lmax + 1):
retval += mpmath.binomial(k,l) * mpmath.power(pc, l) * \
mpmath.power(1 - pc, k - l)
return retval
def getNthDelannoyNumber( n ):
if real_int( n ) == 1:
return 3
if n < 0:
raise ValueError( '\'nth_delannoy\' expects a non-negative argument' )
precision = int( fmul( n, 0.8 ) )
if ( mp.dps < precision ):
mp.dps = precision
result = 0
for k in arange( 0, fadd( real( n ), 1 ) ):
result = fadd( result, fmul( binomial( n, k ), binomial( fadd( n, k ), k ) ) )
return result
def get_prob_correct_async(k, pc, rprime):
"""prob = \sum_{l=0}^{k-rprime} (k_choose_l) * (pc)^l * (1 - pc)^(k-l)
"""
retval = 0
lmax = k - rprime
for l in range(0, lmax + 1):
retval += mpmath.binomial(k,l) * mpmath.power(pc, l) * \
mpmath.power(1 - pc, k - l)
return retval
def round5_fp_fast(d, n, h, q, p, t, f, mu, B, add_coeff=False, verbose=False):
# dif = mp_zeros(q) # full precision
# dif[0] = mp.mpf(1.0)
dif = np.zeros(q) # double precision approximation
dif[0] = 1.0
scale = 0
for w in h: # w = weight of +-scale
scale += 1
if add_coeff:
w = 2 * w # closer to 2 * w - 1 if h ~= w/2
w *= 2 # double it here; can use floats
if verbose:
print "\nw" + str(scale), "=", w
for i in range(w):
dif = conv_comb_qp(dif, q, p, scale)
if verbose:
sys.stdout.write('.')
sys.stdout.flush()
if verbose:
print "done"
# convert to mpf if we were double
dif = np.array([mp.mpf(dif[i]) for i in range(len(dif))])
dif /= dif.sum() # normalize, just in case
if verbose:
print "eqp*R + eqp'*S".ljust(16) + str_stats(dif)
dif = conv_comb_pt(dif, q, p, t)
if verbose:
print "rounded p->t".ljust(16) + str_stats(dif)
# Decryption failure ranges for case when parameter B>1
up_fail = (q + (1 << B)) // (1 << (B + 1))
down_fail = (q * ((1 << (B + 1)) - 1) + (1 << B)) // (1 << (B + 1))
bfp = dif[up_fail:down_fail].sum()
if verbose:
print "per bit failure probability:", str_log2(bfp)
print bfp
ffp = mp.mpf(0)
for j in range(f + 1, mu + 1):
ffp += (mp.binomial(mu, j) * mp.power(bfp, j) *
mp.power(mp.mpf(1) - bfp, (mu - j)))
if verbose:
print "more than", f, "errors in", mu, "bits:", str_log2(ffp)
print ffp
return float(mp.log(ffp, 2))
def pmf(k, n, p):
"""
Probability mass function of the binomial distribution.
"""
_validate_np(n, p)
with mpmath.extradps(5):
p = mpmath.mpf(p)
return (mpmath.binomial(n, k) *
mpmath.power(p, k) *
mpmath.power(1 - p, n - k))
def u_binom(N):
out = np.zeros((len(x),len(t)))
out = out + 0.5*W(0)[...,None]*np.ones(len(t))
with mpmath.workdps(workdps):
for n in range(1,N):
coeff = np.zeros(len(x))
for m in range(n):
coeff = coeff + (-1)**((n-1)-m)*mpmath.binomial(n-1,m)*W(m+1)
out = out + coeff[...,None]*((t/(2*tau))**n)/mpmath.factorial(n)
return out
def getNthMenageNumber( n ):
if n < 0:
raise ValueError( '\'menage\' requires a non-negative argument' )
elif n in [ 1, 2 ]:
return 0
elif n in [ 0, 3 ]:
return 1
else:
return nsum( lambda k: fdiv( fprod( [ power( -1, k ), fmul( 2, n ), binomial( fsub( fmul( 2, n ), k ), k ),
fac( fsub( n, k ) ) ] ), fsub( fmul( 2, n ), k ) ), [ 0, n ] )
def getNthAperyNumber(n):
'''
http://oeis.org/A005259
a(n) = sum(k=0..n, C(n,k)^2 * C(n+k,k)^2 )
'''
precision = int(fmul(n, 1.6))
if mp.dps < precision:
mp.dps = precision
result = 0
for k in arange(0, n + 1):
result = fadd(
result,
fmul(power(binomial(n, k), 2), power(binomial(fadd(n, k), k), 2)))
return result
def estimate_S_from_counts(dfA, dfB):
"""
Give an unbiased estimate for the value of S given the counts.
dfA & dfB need to have a column "read_count" & "cdr3_nt"
"""
cAs = dfA.groupby("cdr3_nt").read_count.sum().to_frame()
cBs = dfB.groupby("cdr3_nt").read_count.sum().to_frame()
cs = cAs.merge(cBs, left_index=True, right_index=True,
how='outer', suffixes={'_A', '_B'})
cs = cs.fillna(0)
cs["read_count"] = cs["read_count_A"] + cs["read_count_B"]
M2 = len(cBs)
M1 = len(cAs)
# The formula below remove the bias associated with sampling
Sest = sum([1 + (mpmath.binomial(0, int(c))
- mpmath.binomial(M2, int(c))
- mpmath.binomial(M1, int(c)))
/mpmath.binomial(M1+M2, int(c))
for c in cs["read_count"].values if int(c) != 0])
return float(Sest)
def iterate(rest, n_rest):
if rest and n_rest > 0:
for k in range(n_rest + 1):
for coeff, next in iterate(rest[1:], n_rest - k):
if k == 0:
this_factor = []
else:
this_factor = [Expression('Power', rest[0], Integer(k))]
yield (int(mpmath.binomial(n_rest, k) * coeff), this_factor + next)
elif n_rest == 0:
yield (sympy.Integer(1), [])
def getNthAperyNumber( n ):
'''
http://oeis.org/A005259
a(n) = sum(k=0..n, C(n,k)^2 * C(n+k,k)^2 )
'''
if n < 0:
raise ValueError( '\'nth_apery\' expects a non-negative argument' )
precision = int( fmul( n, 1.6 ) )
if ( mp.dps < precision ):
mp.dps = precision
result = 0
for k in arange( 0, real( n ) + 1 ):
result = fadd( result, fmul( power( binomial( n, k ), 2 ),
power( binomial( fadd( n, k ), k ), 2 ) ) )
return result
def getNthSchroederNumber(n):
if n == 1:
return 1
n = fsub(n, 1)
result = 0
precision = int(fmul(n, 0.8))
if mp.dps < precision:
mp.dps = precision
for k in arange(0, fadd(n, 1)):
result = fadd(
result,
fdiv(fprod([power(2, k),
binomial(n, k),
binomial(n, fsub(k, 1))]), n))
return result
def apply_inexact(self, n, k, evaluation):
'Binomial[n_?InexactNumberQ, k_?NumberQ]'
with workprec(min_prec(n, k)):
n = gmpy2mpmath(n.value)
k = gmpy2mpmath(k.value)
result = mpmath.binomial(n, k)
try:
result = mpmath2gmpy(result)
except SpecialValueError, exc:
return Symbol(exc.name)
number = Number.from_mp(result)
return number
def apply_inexact(self, n, k, evaluation):
'Binomial[n_?InexactNumberQ, k_?NumberQ]'
with workprec(min_prec(n, k)):
n = gmpy2mpmath(n.value)
k = gmpy2mpmath(k.value)
result = mpmath.binomial(n, k)
try:
result = mpmath2gmpy(result)
except SpecialValueError, exc:
return Symbol(exc.name)
number = Number.from_mp(result)
return number
def test_Wigner_coefficient():
import mpmath
mpmath.mp.dps = 4 * sf.ell_max
i = 0
for twoell in range(2 * sf.ell_max + 1):
for twomp in range(-twoell, twoell + 1, 2):
for twom in range(-twoell, twoell + 1, 2):
tworho_min = max(0, twomp - twom)
a = sf._Wigner_coefficient(twoell, twomp, twom)
b = float(
mpmath.sqrt(
mpmath.fac((twoell + twom) // 2) * mpmath.fac(
(twoell - twom) // 2) /
(mpmath.fac((twoell + twomp) // 2) * mpmath.fac(
(twoell - twomp) // 2))) * mpmath.binomial(
(twoell + twomp) // 2, tworho_min // 2) *
mpmath.binomial((twoell - twomp) // 2,
(twoell - twom - tworho_min) // 2))
assert np.allclose(a,
b), (twoell, twomp, twom, i,
sf._Wigner_index(twoell, twomp, twom))
i += 1
def binomial(n, k, nonzero=False):
if abs(k) > 1e8*(abs(n) + 1):
# The binomial is rapidly oscillating in this region,
# and the function is numerically ill-defined. Don't
# compare values here.
return np.nan
if n < k and abs(float(n-k) - np.round(float(n-k))) < 1e-15:
# close to a zero of the function: mpmath and scipy
# will not round here the same, so the test needs to be
# run with an absolute tolerance
if nonzero:
bad_points.append((float(n), float(k)))
return np.nan
return mpmath.binomial(n, k)
def binomial(n, k, nonzero=False):
if abs(k) > 1e8 * (abs(n) + 1):
# The binomial is rapidly oscillating in this region,
# and the function is numerically ill-defined. Don't
# compare values here.
return np.nan
if n < k and abs(float(n - k) - np.round(float(n - k))) < 1e-15:
# close to a zero of the function: mpmath and scipy
# will not round here the same, so the test needs to be
# run with an absolute tolerance
if nonzero:
bad_points.append((float(n), float(k)))
return np.nan
return mpmath.binomial(n, k)
def getNthMotzkinNumber(n):
'''
http://oeis.org/A001006
a(n) = sum((-1)^j*binomial(n+1, j)*binomial(2n-3j, n), j=0..floor(n/3))/(n+1)
'''
precision = int(n)
if mp.dps < precision:
mp.dps = precision
result = 0
for j in arange(0, floor(fdiv(n, 3)) + 1):
result = fadd(
result,
fprod([
power(-1, j),
binomial(fadd(n, 1), j),
binomial(fsub(fmul(2, n), fmul(3, j)), n)
]))
return fdiv(result, fadd(n, 1))
def iterate(rest, n_rest):
if rest and n_rest > 0:
for k in range(n_rest + 1):
for coeff, next in iterate(
rest[1:], n_rest - k):
if k == 0:
this_factor = []
else:
this_factor = [
Expression('Power', rest[0],
Integer(k))
]
yield (int(
mpmath.binomial(n_rest, k) * coeff),
this_factor + next)
elif n_rest == 0:
yield (sympy.Integer(1), [])
def moment(a, i, j):
'''
Calculates the j-th moment of phi_i
Wavelet Analysis: The Scalable Structure of Information (2002)
Howard L. Resnikoff, Raymond O.Jr. Wells
page 262
'''
N = a.size
assert j > -1
if j == 0:
return 1 if i == 0 else 0
if i == 0:
return moment_i0(a, j)
return mpmath.nsum(lambda k: binomial(j,k)*mpmath.power(i, (j-k))*moment_i0(a, k), [0, j])
def getNthMenageNumber(n):
'''https://oeis.org/A000179'''
if n == 1:
return -1
if n == 2:
return 0
if n in [0, 3]:
return 1
return nsum(
lambda k: fdiv(
fprod([
power(-1, k),
fmul(2, n),
binomial(fsub(fmul(2, n), k), k),
fac(fsub(n, k))
]), fsub(fmul(2, n), k)), [0, n])
def moment_i0(a, j):
'''
Calculates the j-th moment of phi_0
Wavelet Analysis: The Scalable Structure of Information (2002)
Howard L. Resnikoff, Raymond O.Jr. Wells
page 262
'''
N = a.size
if j == 0:
return 1
s = mpmath.nsum(lambda k: binomial(j,k)*moment_i0(a, k)*
mpmath.nsum(lambda i: a[int(i)]*mpmath.power(i, j-k),
[0, N-1]),
[0, j-1])
_2 = mp.mpmathify(2)
return s/(_2*(mpmath.power(_2, j)-1))
def AIntegrand(r,s,t,i):
f = mpmath.gamma
g = mpmath.factorial
x = mpmath.binomial(s,i)*(f(i+1/2)/g(r+i))*(f(1/2+r+i)/f(3/2+r+t+i))
return x
def getBinomial( n, k ):
return binomial( n, k )
def getBinomialOperator(n, k):
return binomial(n, k)
def summand(l):
return mp.binomial(s, l) * ( fallingFactorial(4, r+l) / mp.power(4, K+l) ) * lambdaMergerRate(b, K+l)
def getLahNumber( n, k ):
return fmul( power( -1, n ), fdiv( fmul( binomial( real( n ), real( k ) ), fac( fsub( n, 1 ) ) ), fac( fsub( k, 1 ) ) ) )
def getNarayanaNumber( n, k ):
return fdiv( fmul( binomial( n, k ), binomial( n, fsub( k, 1 ) ) ), n )
def multinomial_mpm(params):
"""High precision version of the multinomial function"""
if len(params) == 1:
return 1
return mpm.binomial(sum(params), params[-1]) * multinomial_mpm(params[:-1])
def getNthCatalanNumberOperator(n):
return fdiv(binomial(fmul(2, n), n), fadd(n, 1))
import mpmath as mp
s = 0
for r in range(1, 16):
s += mp.binomial(15, r) * r**2 / mp.binomial(15, r - 1)
print s
r = 15
print(-2 * r**3 + 45 * r**2 + 47 * r) / 6
from spherical_functions import ell_max
# The coefficients were originally produced with the following code,
# though obviously this doesn't need to be run each time.
import numpy as np
import mpmath
mpmath.mp.dps = 4 * ell_max
_binomial_coefficients = np.empty((((2 * ell_max + 1) * (2 * ell_max + 2)) // 2,), dtype=float)
i = 0
for n in range(2 * ell_max + 1):
for k in range(n + 1):
_binomial_coefficients[i] = float(mpmath.binomial(n, k))
i += 1
print(i, _binomial_coefficients.shape)
np.save('binomial_coefficients', _binomial_coefficients)
_ladder_operator_coefficients = np.empty((((2*ell_max + 3) * ell_max + 1),), dtype=float)
i = 0
for twoell in range(2*ell_max + 1):
for twom in range(-twoell, twoell + 1, 2):
_ladder_operator_coefficients[i] = mpmath.sqrt((twoell * (twoell + 2) - twom * (twom + 2))/4)
i += 1
print(i, _ladder_operator_coefficients.shape)
np.save('ladder_operator_coefficients', _ladder_operator_coefficients)
# _Wigner_coefficients = np.empty(((4 * ell_max ** 3 + 12 * ell_max ** 2 + 11 * ell_max + 3) // 3,), dtype=float)
# i = 0
# for ell in range(ell_max + 1):
def getNthPascalLine( n ):
for i in arange( 0, real( n ) ):
yield binomial( n - 1, i )
def getNthCatalanNumber( n ):
return fdiv( binomial( fmul( 2, real( n ) ), n ), fadd( n, 1 ) )