def combos_entropy(string):
# where string is an integer string corresp to a binary matrix
# calc #combos = #int_perms * #matrix_perms
int_perms = round(fct(sum(string)), ROUND)
for s in string:
int_perms = int_perms // round(fct(s), ROUND)
unq_s = {}
for s in string:
if s in unq_s.keys():
unq_s[s] += 1
else:
unq_s[s] = 1
matrix_perms = round(fct(len(string)), ROUND) #(sum(unq_s))
H = 0
for s in unq_s.keys():
matrix_perms = matrix_perms // round(fct(unq_s[s]), ROUND)
for s in string:
pr = s / sum(string)
assert (pr >= 0 and pr <= 1)
if pr != 0: H -= pr * log(pr, 2)
total2 = pow(2, sum(string) * log(len(string), 2))
#print(string,int_perms,matrix_perms,total2,H)
prob_perm = int_perms * matrix_perms / total2
assert (prob_perm >= 0 and prob_perm <= 1)
return prob_perm * H
def compute_possibilities(self):
len_row = self.length
len_vec, sum_vec = len(self.vector), sum(self.vector)
# translate current problem to "how many ways to distribute spaces(blank boxes)
# to the slots between the blocks of filled squares?"
# [compulsory space of at least 1 between the blocks]
# =>
# classic combinatorics problem: how many ways to put b X balls into c X containers?
balls = len_row - sum_vec - len_vec + 1
containers = len_vec + 1
elements = balls + containers - 1
partitions = containers - 1
return int(fct(elements) / (fct(balls)*fct(partitions)))
def pdf(self, k):
"""Probability density function calculator for the binomial distribution.
Args:
k (float): point for calculating the probability density function
Returns:
float: probability density function output
"""
n = self.n
p = self.p
return (1.0 * ((fct(n) / (fct(k) * fct(n - k))) * (p**k) * ((1 - p)**(n - k))))
def get_percent(ppl: int) -> Decimal:
"""Will calculate the percent chance of matching birthdays.
Args:
ppl (integer): The amount of people.
Returns:
float: The percent chance of a match.
"""
if ppl > 365:
return Decimal(1 - 0)
frac_top: Decimal = Decimal(fct(365))
frac_bot: Decimal = Decimal((365**ppl) * fct(365 - ppl))
return 1 - Decimal(frac_top / frac_bot)
def zernike_poly(Y, X, n, l):
"""
Computes the Zernike polynomial
for order n
"""
y, x = Y[0], X[0]
poly = np.zeros(Y.size, dtype=complex)
index = 0
for x, y in zip(X, Y):
Vnl = 0.
for m in range(int((n - l) // 2) + 1):
Vnl += (-1.)**m * fct(n-m) / \
(fct(m) * fct((n - 2*m + l) // 2) * fct((n - 2*m - l) // 2) ) * \
(np.sqrt(x**2 + y**2)**(n - 2*m) * getpolar(l*atan2(y,x)))
poly[index] = Vnl
index = index + 1
return poly
def main():
words = "Why sometimes I have believed as many as six impossible things before breakfast".split(
)
pp(words)
pp([len(word) for word in words])
pp(sorted([fct(i) for i in range(20)]))
pp(sorted({fct(i) for i in range(20)}))
cnt_to_cap = {
'Odisha': "Bhubaneswar",
"MP": "Bhopal",
"Gujrat": "Gandhinagar",
"India": "New Delhi",
"Pakistan": "Islamabad"
}
cap_to_cnt = {cap: con for con, cap in cnt_to_cap.items()}
pp(cnt_to_cap)
pp(cap_to_cnt)
pass
from math import sqrt
def is_prime(x):
if x < 2:
return False
for i in range(2, int(sqrt(x) + 1)):
if x % i == 0:
return False
return True
print([x for x in range(101) if is_prime(x)])
prime_square_divisors = {
x * x: (1, x, x * x)
for x in range(101) if is_prime(x)
}
pp(prime_square_divisors)
from math import factorial as fct
n = 4
ar = []
for i in range(n + 1):
num = int(fct(n) / (fct(i) * fct(n - i)))
ar.append(num)
print(ar)
from math import factorial as fct, log, e
def higher(l):
h=l[0]
for j in range(len(l)):
if l[j]>h:
h=l[j]
else:
continue
return h
list=[]
for i in range(10):
if i%2==0:
list.append(3**i + 7*fct(i))
else:
list.append(2**i+4*log(i, e))
print(list)
print(sum(list)/len(list), higher(list))
math.sqrt(144) math.degrees(math.pi/2) from math import factorial factorial(12) from math import sqrt, degrees, pi sqrt(144) degrees(pi/2) import math as mt from math import factorial as fct fct(12) # Numpy import numpy as np my_list = [1, 2, 3] array = np.array(my_list) type(array) np.arange(0, 11, 2) np.zeros(5) np.zeros((3, 4)) np.ones((3, 4))
def ncr(n, r):
if r > n:
return None
return fct(n) // (fct(r) * fct(n - r))
# FINISHED in 2 lines from math import factorial as fct print sum([int (i) for i in list(str(fct(100)))])
def permute(listin, bigarray): for i in range(fct(len(listin))): listin.insert(2,listin[0]) listin.pop(0) listin.insert(0,listin[-1]) listin.pop(-1)
def p_n(C, a1, a2):
p_0 = p_zero(C, a1, a2)
return [[(a1**n1 / fct(n1)) * (a2**n2 / fct(n2)) * fct(n1 + n2) * p_0
for n2 in range(C + 1) if (n1 + n2 <= C)] for n1 in range(C + 1)]
import math
print(math.factorial(32))
n = 7
k = 3
t = math.factorial(n) / (math.factorial(k) * math.factorial(n - k))
print(t)
from math import factorial as fct
t = fct(n) // (fct(k) * fct(n - k))
print(t)
# print(len(str(fct(90000))))
print(10)
print(0b10)
print(0o10)
print(0x12)
print(int(23.8))
print(int("23423"))
print(int("23423", 5))
print(float("3e52"))
print(float("1.6522e-12"))
print(float("inf"))
print(float("-inf"))
print(float("nan"))
def fact():
for el in count(1):
yield fct(el)
def p_n(C, r1, r2):
p_00 = p_zero(C, r1, r2)
return [[
p_00 * ((r1**n1) / fct(n1)) * ((r2**n2) / fct(n2))
for n2 in range(C + 1) if (n1 + n2 <= C)
] for n1 in range(g + 1)]
def p_zero(C, r1, r2):
return sum([((r1**n1) / fct(n1)) * ((r2**n2) / fct(n2))
for n1 in range(g + 1) for n2 in range(C + 1)
if (n1 + n2 <= C)])**(-1)
def test_fact2(self): # Testing error scenario
n = 'K'
factorial = fact(n)
self.assertEqual(factorial, fct(n), "Should be {}".format(fct(n)))
def test_fact1(self): # Testing usual scenario
n = 8
factorial = fact(n)
self.assertEqual(factorial, fct(n), "Should be {}".format(fct(n)))
def combin(i):
return fct(i[1]) // (fct(i[0]) * fct(i[1] - i[0]))
#Method1:
from math import fct as fct
fcts = [
fct(0),
fct(1),
fct(2),
fct(3),
fct(4),
fct(5),
fct(6),
fct(7),
fct(8),
fct(9)
]
def fct_digits(n):
s = 0
while n:
s += fcts[n % 10]
n //= 10
return s
res = 0
for i in range(10, 1854721):
if fct_digits(i) == i: res += i
print(res)
#Method2:
def p_zero(C, a1, a2):
return sum([
fct(n1 + n2) * (a1**n1 / fct(n1)) * (a2**n2 / fct(n2))
for n2 in range(C + 1) for n1 in range(C + 1) if (n1 + n2 <= C)
])**-1
def flush_prob():
from math import factorial as fct
result = (fct(13) * fct(4)) / (fct(8) * fct(5) * fct(3) * fct(1))
return result
def c(n, r):
return fct(n) / (fct(r) * fct(n-r))
def flush_prob():
from math import factorial as fct
result = (fct(13)*fct(4))/(fct(8)*fct(5)*fct(3)*fct(1))
return result
def catlan(num):
if num < 2:
return 1
return fct(2 * num) // (fct(num + 1) * fct(num))
print('\n','Effective Frequency (\u03C9_eff)','\n','\u03C9_eff =', round(weff/c,4),'cm-\N{SUPERSCRIPT ONE}')
lv0=np.sum(Si*hbar*W)
lv1=np.sum(Si1*hbar*W1)
lv=lv0+lv1
print('\n',' Vibronic Internal Reorganization Energy (\u03BBv) ','\n','\u03BB_v =',round(lv,4),'eV')
print('Calling CATNIP to compute transfer integral (J_eff) between '+orb_ty_1+' and '+orb_ty_2) #orbitals defined in begining of this file.
J_eff=CATNIP(pun_file_1,orb_ty_1,pun_file_2,orb_ty_2,pun_file)
### MLJ calculation
Had=float(J_eff[1])**2
SOMA = 0
C = spc.pi/(hbar*np.sqrt(spc.pi*kb*T*ls))
S = lv/(hbar*weff)
S1 = np.exp(-S)
for ni in range(len(W)):
S2 = (S**ni)/fct(ni)
S3n = (-G0 + ls + ni*hbar*weff)**2
S3d = 4*ls*kb*T
S3 = np.exp(-S3n/S3d)
SOMA += S2*S3
K_mlj=C*Had*S1*SOMA
### Semi-Classical Marcus (SCM) calculation
lamb=lv+ls
Cm = 2*spc.pi/(hbar*np.sqrt(4*spc.pi*lamb*kb*T))
Smn=(lamb-G0)**2
Smd=(4*lamb*kb*T)
Sm=np.exp(-Smn/Smd)
K_scm=Cm*Had*Sm
### Array with transfer rates
ket=np.array([K_scm,K_mlj])
print('\n','SCM and MJL rates respectivelly','\n',ket,'s-\N{SUPERSCRIPT ONE}')
def noofcomb(n, r):
res = fct(n) // (fct(r) * fct(n - r))
return res
past=fct(i)/(fct((i-j))*fct(j),end=" ")
from math import factorial as fct
# input n
n = int(input())
arr=[]
for i in range(n):
for j in range(n-i+1):
# for left spacing
#print(end=" ")
for j in range(i+1):
# nCr = n!/((n-r)!*r!)
#arr.append((fct(i)//(fct(j)*fct(i-j))))
print(fct(i)//(fct(j)*fct(i-j)), end=" ")
# for new line
print() '''
from math import factorial as fct
n = int(input())
for i in range(1, n + 1):
for j in range(1, n - i + 1):
print(fct(i) / (fct(j) * fct(i - j)), end=" ")
def factorial(n):
return fct(n)
'''
A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we call it lexicographic order.
The lexicographic permutations of 0, 1 and 2 are:
012 021 102 120 201 210
What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
'''
from math import factorial as fct
numbers = ['0','1','2','3','4','5','6','7','8','9']
solution = ''
n = 0
for d in range(9, 0, -1):
f = fct(d)
i = 1
while f*(i+1) + n < 1000000:
i += 1
solution += numbers[i]
numbers.pop(i)
n = f*i + n
solution += numbers[0]
print('Answer:', solution)
from math import factorial as fct
while True:
try:
L = input().split()
M = int(L[0])
Mfct = fct(M)
N = int(L[1])
Nfct = fct(N)
print(Nfct + Mfct)
except (EOFError):
break
# ex02.module.py # 방법 1 import threading import time import os import threading as th # OR import threading, time, os, sys, math # 모듈 안에 있는 모든 메소드를 메모리에 올려 놓는다. # 방법2 from math import factorial from math import factorial as fct # 모듈의 특정 메소드만 가져온다 # 방법3 from math import * # 1번과 동일 # 방법4 from math import (factorial as ff, acos as ac) n = fct(5) / fct(3) print(n) n2 = ff(5) + ac(1) print(n2)
# + Создать текстовый файл (не программно),
# сохранить в нём несколько строк, выполнить подсчёт строк и слов в каждой строке.
from math import factorial as fct
with open('for_task_5.2.txt', 'a', encoding='utf8') as f_obj:
for i in range(1, 7):
for j in range(0, i):
f_obj.write(str(fct(i)) + ' ')
f_obj.write('\n')
with open('for_task_5.2.txt', 'r', encoding='utf8') as f_obj:
out_dict = dict({})
for i, line in enumerate(f_obj, start=1):
words = len(line.split())
out_dict.update({f'в строке {i} слов: ': [words]})
for key in out_dict:
print(key, out_dict[key])
def analyse_dataset(dataset_file, grid_size):
# open the file containing the data and create array with results
unique_grids = dict()
counter = 0
with open(dataset_file, 'r') as grid_dataset:
# Use a dictionary to accumulate unique grids and their frequencies to be used for data analysis. Dict is a
# good option as searching for unique grids is fast. Each grid, which is a in the dataset (txt file), is used
# as a key for the dictionary. If there is no value associated with it yet, then the value is set to 1. If
# there is, then increment the value by 1. Thus, every unique grid from the dataset is a key in the dictionary,
# and the associated value is the frequency of that grid being generated.
for line in grid_dataset:
# if no key in dict for that grid, add it and set value to 1 for the frequency
if unique_grids.get(line) is None:
unique_grids[line] = 1
else: # if key is already present, increase value by 1
unique_grids[line] += 1
counter += 1
grid_dataset.close() # think with closes the file but kept this anyway...
# grid_list = [grid for grid in unique_grids]
# grid_indexes = [index for index in range(len(unique_grids))]
# frequency_list = [unique_grids[key] for key in unique_grids]
# make a dataframe (used to display data in plots) with two columns :
# "grid", gives a unique grid as a string of the numbers that make up the grid
# "frequency", gives the frequency of the unique grid in the dictionary (how many were generated)
df = pd.DataFrame({
"grid": [grid for grid in unique_grids],
"frequency": [unique_grids[key] for key in unique_grids]
})
# TODO | NOTE: DataFrame has method from_dict to convert, but essentially performs the above operations to do so
# df = pd.DataFrame.from_dict(data=unique_grids, orient='index') # keeps names of grid, slow to load scatter
# df.columns = ['frequency']
# print out some stats regarding the dataset: number of possible unique (solvable) grids, number of unique grids
# generated, mean/median/std dev/min/max for frequency of unique grids
num_solvable_combinations = int(fct(grid_size) /
2) # factorials always even...use int
num_unique_grids = len(unique_grids)
percent_grids_created = num_unique_grids / num_solvable_combinations * 100
percent_unique = num_unique_grids / counter
print(
"\nThere are {:d} possible (solvable) combinations \nof {:d} consecutive"
" integers in a grid.".format(num_solvable_combinations, grid_size))
print(
"{:.3f}% ({:d}) of possible unique grids were \nfound from a total of {:d}"
" generated.\n".format(percent_grids_created, num_unique_grids,
counter))
print(df.describe())
# Sorted unique grid frequency values in ascending order and displayed the data in a scatter plot and histogram.
# These clearly display how many of each unique grid was generated by the randomiser, which can show whether or not
# the grids are generated in an appropriately random manner. If every possible unique grid is created, with
# relatively small spread of frequencies (not favouring any particular grid), then it is sufficiently random.
# freq_sorted_series = df['frequency'].sort_values()
# plt.scatter(df.index, df['frequency'], 0.5)
# # plt.scatter(df.index, freq_sorted_series, 1)
# plt.title("Scatter plot for frequency of unique lists, in a dataset of\n "
# + str(counter) + " randomly generated lists of values 1 through " + str(grid_size))
# plt.xlabel("Index of list in dataset")
# plt.ylabel("Frequency of unique lists")
# plt.show()
# plt.hist(freq_sorted_series, bins=30)
# plt.title("Histogram to display the spread of different frequencies of\n unique lists from a dataset"
# " of " + str(counter) + " generated lists")
# plt.xlabel("Unique frequency values")
# plt.ylabel("Frequency of frequency unique lists")
# plt.grid = True
# plt.show()
# Determined the unique grids with the maximum, minimum, and median frequency in the 10mil dataset (largest)
# Created multiple datasets of same size (1mil x 10), then determined the frequency of each of the aforementioned
# grids in each dataset, to calculate the average frequency and check it is still approx. 49.6 (1mil/20.1k)
# Get the min, max, and median frequency values, and determine the grids these correspond to
# min_freq_grid = df.loc[df['frequency'] == 402]
# max_freq_grid = df.loc[df['frequency'] == 585]
# med_freq_grid = df.loc[df['frequency'] == 496]
# print("\nlow:\n" + str(min_freq_grid) + "\nhigh:\n" + str(max_freq_grid) +
# "\nmedian:\n" + str(med_freq_grid))
# For subsequent datasets, slice dataframe to get the data for each of the unique grids chosen (min,max,med) to
# track the frequency across different datasets. Can then assess if there is any preference for specific grids
# even if the distribution looks random
min_freq_df = df.loc[df['grid'] == "0,6,4,3,5,1,2,7,8,\n"]
max_freq_df = df.loc[df['grid'] == "7,4,5,6,2,3,1,0,8,\n"]
med_freq_df = df.loc[df['grid'] == "4,5,3,0,2,6,7,1,8,\n"]
# print("\nmin:\n" + str(min_freq_df) + "\nmax:\n" + str(max_freq_df)
# + "\nmedian:\n" + str(med_freq_df))
min_freq_val = min_freq_df.iloc[0]['frequency']
max_freq_val = max_freq_df.iloc[0]['frequency']
med_freq_val = med_freq_df.iloc[0]['frequency']
# Add frequency values to a lists to create a table for the data
table_min_freq.append(min_freq_val)
table_max_freq.append(max_freq_val)
table_med_freq.append(med_freq_val)
# TODO: can test how a change in function might affect randomness for instance, can swap a random cell with random
# neighbour or any other cell then check if it has changed inversions. I think swapping wih 1 beside changes
# inversions by set amount? think of an unsolvable puzzle in closest to solved
# position, there will be 13, 15, 14, _ in last row, meaning 1 is out of place and only requires one swap
# Testing Chi Squared goodness of fit statistic for frequency of unique grids:
# X^2 = sum((observed-expected)^2 / expected)
chi_squared = 0
expected_frequency = counter / num_solvable_combinations # num grids generated/num possible unique grids
print(expected_frequency)
for grid in unique_grids:
chi_squared += (
(unique_grids[grid] - expected_frequency)**2) / expected_frequency
print("\nchi squared: " + str(chi_squared))
