def siemens_B(alpha, beta, x1, y1, z1, R0):
''' Calculate displacement field from Siemens coefficients
'''
nmax = alpha.shape[0] - 1
x1 = x1 + 0.0001 # hack to avoid singularities at R=0
# convert to spherical coordinates
r = np.sqrt(x1 * x1 + y1 * y1 + z1 * z1)
theta = np.arccos(z1 / r)
phi = np.arctan2(y1 / r, x1 / r)
b = np.zeros(x1.shape)
for n in xrange(0, nmax + 1):
f = np.power(r / R0, n)
for m in xrange(0, n + 1):
f2 = alpha[n, m] * np.cos(m * phi) + beta[n, m] * np.sin(m * phi)
_ptemp = utils.legendre(n, m, np.cos(theta))
#_ptemp = scipy.special.lpmv(m, n, np.cos(theta))
normfact = 1
# this is Siemens normalization
if m > 0:
normfact = math.pow(-1, m) * \
math.sqrt(float((2 * n + 1) * factorial(n - m)) \
/ float(2 * factorial(n + m)))
_p = normfact * _ptemp
b = b + f * _p * f2
return b
def siemens_B(alpha, beta, x1, y1, z1, R0):
''' Calculate displacement field from Siemens coefficients
'''
nmax = alpha.shape[0] - 1
x1 = x1 + 0.0001 # hack to avoid singularities at R=0
# convert to spherical coordinates
r = np.sqrt(x1 * x1 + y1 * y1 + z1 * z1)
theta = np.arccos(z1 / r)
phi = np.arctan2(y1 / r, x1 / r)
b = np.zeros(x1.shape)
for n in xrange(0, nmax + 1):
f = np.power(r / R0, n)
for m in xrange(0, n + 1):
f2 = alpha[n, m] * np.cos(m * phi) + beta[n, m] * np.sin(m * phi)
_ptemp = utils.legendre(n, m, np.cos(theta))
#_ptemp = scipy.special.lpmv(m, n, np.cos(theta))
normfact = 1
# this is Siemens normalization
if m > 0:
normfact = math.pow(-1, m) * \
math.sqrt(float((2 * n + 1) * factorial(n - m)) \
/ float(2 * factorial(n + m)))
_p = normfact * _ptemp
b = b + f * _p * f2
return b
def p34():
target = factorial(9) * 7
digit_fact = [factorial(i)for i in range(0, 10)]
result = 0
for num in range(4, target):
sum_of_fact = sum([digit_fact[int(digit)]
for digit in str(num)])
if sum_of_fact != num:
continue
result += sum_of_fact
return result
def run():
num = 20
max = utils.factorial(num)
for x in xrange(num, max + 1):
if not any([y for y in range(1, num + 1) if x % y != 0]):
print x
break
def _calc_fact_term(m, n, dn, s, eta):
denom1 = 2 * s + 1
denom2 = denom1 + dn
denom3 = m - 2 * s
if denom3 > denom1:
tmp = denom1
denom1 = denom3
denom3 = tmp
if m > denom1:
num = permutation(n, denom1) * permutation(m, denom1)
denom = factorial(denom2)**2 * factorial(denom3)**2
else:
num = permutation(n, denom1)
denom = (permutation(denom1, m) * factorial(denom2)**2 *
factorial(denom3)**2)
res = sqrt(num / denom)
return res
def test_profiler():
profiler = TracingProfiler(top_frames=[sys._getframe()])
assert isinstance(profiler.stats, RecordingStatistics)
stats, cpu_time, wall_time = profiler.result()
assert len(stats) == 0
with profiler:
factorial(1000)
factorial(10000)
stats1 = find_stats(profiler.stats, 'factorial')
stats2 = find_stats(profiler.stats, '__enter__')
stats3 = find_stats(profiler.stats, '__exit__')
assert stats1.deep_time != 0
assert stats1.deep_time == stats1.own_time
assert stats1.own_time > stats2.own_time
assert stats1.own_time > stats3.own_time
assert stats1.own_hits == 2
assert stats2.own_hits == 0 # entering to __enter__() wasn't profiled.
assert stats3.own_hits == 1
def test_profiler():
profiler = TracingProfiler(top_frames=[sys._getframe()])
assert isinstance(profiler.stats, RecordingStatistics)
stats, cpu_time, wall_time = profiler.result()
assert len(stats) == 0
with profiler:
factorial(1000)
factorial(10000)
stats1 = find_stats(profiler.stats, 'factorial')
stats2 = find_stats(profiler.stats, '__enter__')
stats3 = find_stats(profiler.stats, '__exit__')
assert stats1.deep_time != 0
assert stats1.deep_time == stats1.own_time
assert stats1.own_time > stats2.own_time
assert stats1.own_time > stats3.own_time
assert stats1.own_hits == 2
assert stats2.own_hits == 0 # entering to __enter__() wasn't profiled.
assert stats3.own_hits == 1
def test_profiler():
profiler = Profiler(top_frame=sys._getframe())
assert isinstance(profiler.stats, RecordingStatistics)
assert isinstance(profiler.result(), FrozenStatistics)
assert len(profiler.stats) == 0
with profiling(profiler):
factorial(1000)
factorial(10000)
stat1 = find_stat(profiler.stats, 'factorial')
stat2 = find_stat(profiler.stats, '__enter__')
stat3 = find_stat(profiler.stats, '__exit__')
assert stat1.total_time != 0
assert stat1.total_time == stat1.own_time
assert stat1.own_time > stat2.own_time
assert stat1.own_time > stat3.own_time
assert stat1.calls == 2
assert stat2.calls == 0 # entering to __enter__() wasn't profiled.
assert stat3.calls == 1
def test_profiler():
profiler = Profiler(top_frame=sys._getframe())
assert isinstance(profiler.stats, RecordingStatistics)
assert isinstance(profiler.result(), FrozenStatistics)
assert len(profiler.stats) == 0
with profiling(profiler):
factorial(1000)
factorial(10000)
stat1 = find_stat(profiler.stats, 'factorial')
stat2 = find_stat(profiler.stats, '__enter__')
stat3 = find_stat(profiler.stats, '__exit__')
assert stat1.total_time != 0
assert stat1.total_time == stat1.own_time
assert stat1.own_time > stat2.own_time
assert stat1.own_time > stat3.own_time
assert stat1.calls == 2
assert stat2.calls == 0 # entering to __enter__() wasn't profiled.
assert stat3.calls == 1
def curiousNumber(num):
l = list(map(int, str(num)))
sum = 0
for i in l:
sum += factorial(i)
if (num == sum):
return True
else:
return False
def main():
factorials = [utils.factorial(i) for i in range(10)]
upper_limit = sum(factorials) + 1
summation = 0
for i in range(3, upper_limit):
if factorial_digits(i, factorials) == i:
summation += i
return summation
def num_permutations(l):
"""returns the number of permutations possible from iterable l. Copied from 74, with fixes"""
counts = defaultdict(int)
for i in l:
counts[i] += 1
# Can't have leading digit 0
invalid = 0
if 0 in counts:
n = list(l)
n.remove(0) # just get rid of one
invalid = num_permutations(['A' if i == 0 else i for i in n])
return factorial(len(l)) / reduce(operator.mul, map(factorial, counts.values()), 1) - invalid
def main(NUMBERS, TARGET):
n = len(NUMBERS)
progress = 0
solutionsGiven = 0
SOL_SPACE_SIZE = n * (factorial(n - 1)**2) * (2**(n - 1)
) # See attached PDF
countdown.spinner(NUMBERS, TARGET)
INDENT = "\t\t\t"
with open("solutions.txt", "w") as file:
solution_generator = solver(NUMBERS, TARGET)
for solution in solution_generator:
progress += 1
if solution != "":
if solutionsGiven <= 100:
file.write(solution + "\n")
solutionsGiven += 1
progressFractional = progress / SOL_SPACE_SIZE
countdown.spinner(NUMBERS, TARGET)
progressBar(INDENT, progressFractional, solution)
# Uncomment line below to see number of operations that ran.
# This number is exactly equal to the solution
# space calculated!
# file.write(str(progress))
with open("solutions.txt", "r") as file:
solutions = file.readlines()
if solutions == []:
print("This problem is impossible!")
return
[print(line) for line in solutions[:100]]
def run():
"""
Solution: Brute force search with a heuristic upper bound.
What's the largest number beyond which this property cannot hold?
I don't know enough number theory to procure a tight bound, but we can get
one that's computationally good enough.
Heuristic: Since every digit can at most contribute 9! = 362880 to the sum,
beyond 6 digits the fac sum becomes increasingly unlikely to match the
magnitude of the number since every additional digit only adds a relatively
small 6 digit number to the sum.
"""
N = 100000
total = 0
for n in xrange(3, N):
if sum(map(lambda x: factorial(int(x)), str(n))) == n:
total += n
# sidenote: it turns out that only one other number besides the example
#fits the given criteria.
return total
from utils import factorial print(sum(map(int, list(str(factorial(100))))))
def factorialDigitSum(n):
fact = factorial(n)
return sumDigit(fact)
# How many chains, with a starting number below one million, contain exactly sixty non-repeating terms?
# brute force? ok!
import sys
sys.path.append('../utils')
sys.path.append('utils')
from utils import factorial
fact_dict = {0:1}
for k in range(1,10):
fact_dict[k] = factorial(k)
def digit_wise_factorial(n):
return(sum([fact_dict[int(x)] for x in str(n)]))
stopping_criteria = {}
stopping_criteria[169] = [363601, 1454]
stopping_criteria[363601] = [1454, 169]
stopping_criteria[1454] = [363601, 169]
stopping_criteria[871] = [45361]
stopping_criteria[45361] = [871]
stopping_criteria[872] = [45362]
stopping_criteria[45362] = [872]
chain_of_60 = 0
for x in range(10,1000000):
if x % 10000 == 0:
print('workig on: ' + str(x))
y = x
chain = set()
while y not in chain:
def solve(self):
sum = 0
strValue = str(factorial(self.num))
for item in strValue:
sum += int(item)
return sum
def n_C_r(n, r, fact):
return factorial(fact, n) / (factorial(fact, r) * factorial(fact, (n-r)))
#
# Solution to Project Euler Problem 34
#
import utils
for i in range(3, 9999999):
if sum([utils.factorial(int(x)) for x in str(i)]) == i:
print(i)
def ncr(n, r):
return factorial(n) / (factorial(r) * factorial(n-r))
def heavy():
factorial(10000)
from utils import factorial print(sum([int(d) for d in str(next(factorial(100)))]))
def light():
factorial(10)
sleep(0.1)
factorial(10)
def factorialDigitSum(n):
fact = factorial(n)
return sumDigit(fact)
def run_factorial():
assert_true(utils.factorial(5), 120, 'Factorial 5')
def count_lattice_path(n, m):
return factorial(n + m) // (factorial(n) * factorial(m))
def main():
factorial = utils.factorial(100)
return sum(int(s) for s in str(factorial))
# n! means n x (n - 1) x ... x 3 x 2 x 1
#
# For example, 10! = 10 x 9 x ... x 3 x 2 x 1 = 3628800,
# and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
#
# Find the sum of the digits in the number 100!
import time
from utils import factorial
if __name__ == '__main__':
t0 = time.time()
value = str(factorial(100))
result = 0
for c in value:
result += int(c)
runtime = time.time() - t0
print 'Result = {}'.format(result)
print 'Runtime = {}'.format(runtime)
def problem_020(n: int = 100) -> int:
"""
Straightforward solution.
"""
return sum_of_digits_in_number(factorial(n))
def run():
"""
Pretty straight forward. Python saves the day again.
"""
return sum(map(int, str(factorial(100))))
# n! means n x (n - 1) x ... x 3 x 2 x 1
#
# For example, 10! = 10 x 9 x ... x 3 x 2 x 1 = 3628800,
# and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
#
# Find the sum of the digits in the number 100!
import time
from utils import factorial
if __name__ == '__main__':
t0 = time.time()
value = str(factorial(100))
result = 0
for c in value:
result += int(c)
runtime = time.time() - t0
print 'Result = {}'.format(result)
print 'Runtime = {}'.format(runtime)
def comb(n,k):
return factorial(n)/(factorial(k)*factorial(n-k))
def e20():
return sum(int(c) for c in str(factorial(100)))
def _operation(self, x, y):
return factorial(x)
def light():
factorial(10)
sleep(0.1)
factorial(10)
""" Project Euler Problem 15 ======================== Starting in the top left corner of a 2 * 2 grid, there are 6 routes (without backtracking) to the bottom right corner. How many routes are there through a 20 * 20 grid? """ # The number of "North-East" paths through a lattice from (0,0) to (x,y) is # [https://en.wikipedia.org/wiki/Lattice_path] # the number of combinations of 'a' objects out of a set of 'a + b' objects, # i.e. # / x + y \ (x + y)! # | | = ------- # \ x / (x!)^2 # # In the case of a square grid: # # / 2x \ (2x)! # | | = ------- # \ x / 2 * x! from utils import factorial GRIDSIZE = 20 routes = factorial(2 * GRIDSIZE) / ((factorial(GRIDSIZE)) ** 2) print routes
def run_factorial():
assert_true(utils.factorial(5), 120, 'Factorial 5')
from utils import factorial, sum_digits print(sum_digits(factorial(100)))
def heavy():
factorial(10000)
def comb(n,r):
return factorial(n)/(factorial(n-r)*factorial(r))
from utils import factorial
strFac = str( factorial(100) )
digitTotal = 0
for c in strFac:
digitTotal += int(c)
if __name__ == '__main__':
print digitTotal
import pyperclip
from utils import factorial
import pyinputplus as pyip
num = pyip.inputInt(prompt="Enter a number: ")
factorial_num = factorial(num)
print(factorial_num)
pyperclip.copy(factorial_num)
print("Factorial Copied to Clipboard")
