Scinawa

#!/usr/bin/python3

import matplotlib.pyplot as plt
import sys
import argparse


parser = argparse.ArgumentParser()
parser.add_argument("--nqubit", required=False, help="number of qubits", type=int)
parser.add_argument("--precision", required=False, default=32, type=int)

Scinawa

# http://stackoverflow.com/questions/10244180/python-generating-integer-partitions (checked)
def accelAsc(n):
    a = [0 for i in range(n + 1)]
    k = 1
    a[0] = 0
    y = n - 1
    while k != 0:
        x = a[k - 1] + 1
        k -= 1
        while 2*x <= y:

Scinawa

def stirling_second(x,n):
    if n <= 1 or x == n: return 1
    if n > x or x <= 0: return 0

    return stirling_second(x-1, n-1) + n * stirling_second(x-1, n)

Scinawa

#!/usr/bin/env python

def comp_enum ( x, n ):

#******************************************************************************/
#
## COMP_ENUM returns the number of compositions of the integer x into n parts.
#
#  Discussion:
#

Scinawa

#!/usr/bin/env python

def i4_choose ( x, n ):

#*****************************************************************************
#
## I4_CHOOSE computes the binomial coefficient C(x,n) as an I4.
#
#  Discussion:
#

Scinawa

from math import factorial
x=10
n=3
res = math.factorial(x+n-1) / ( math.factorial(n) * math.factorial(x+n-1-n))
>>> len([i for i in itertools.combinations_with_replacement(range(0,x), n)]) == res
True

Scinawa

@memo  # can be memoized, otherwise comment this line
def stirling1(n,k):
    """Returns the stirling number of the first kind using recursion.."""
    if n == 0 and k == 0:
       return 1
    if k == 0 and n >= 1:
       return 0
    if k > n:
       return 0
    return stirling1(n-1, k-1) - (n - 1) * stirling1(n-1, k)

Scinawa

import itertools
s="ABCD"
rep=3
a = [i for i in itertools.combinations(s, rep)]                                                
len(a)  == math.factorial(len(s)) / (math.factorial(rep) * math.factorial(len(s)-rep))                    

Scinawa

>>> s="ABCD"
>>> repetition=3
>>> a = [i for i in itertools.permutations(s, repetition)]                                               
>>> len(a)  == math.factorial(len(s)) / math.factorial(len(s)-repetition)                          
True


def sample_function(n=0, s=set() ):
  f = {}
  for i in range(0, n):

Scinawa

>>> x = len("ABCDEF") # x= 6,
>>> n=3

>>> for i in product("ABCDEF", repeat=n):
...     print(i)
('A', 'A', 'A')
 ...
('F', 'F', 'F')

>>> len( [i for i in product("ABCDEF", repeat=n)] ) == x**n