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Created problem_45 in project_euler and Speed Boost for problem_34/so…
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…l1.py (TheAlgorithms#2349)

* Create __init__.py

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* Update sol1.py

* Update sol1.py

* Update project_euler/problem_45/sol1.py

Co-authored-by: Christian Clauss <cclauss@me.com>

* Update sol1.py

* Update sol1.py

* Update project_euler/problem_34/sol1.py

Co-authored-by: Christian Clauss <cclauss@me.com>

* Update project_euler/problem_34/sol1.py

Co-authored-by: Christian Clauss <cclauss@me.com>

* Update sol1.py

* Update project_euler/problem_34/sol1.py

Co-authored-by: Christian Clauss <cclauss@me.com>

* Update sol1.py

* Update project_euler/problem_34/sol1.py

Co-authored-by: Christian Clauss <cclauss@me.com>

Co-authored-by: Christian Clauss <cclauss@me.com>
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Kush1101 and cclauss authored Aug 25, 2020
1 parent 5cfc017 commit 402ba7f
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43 changes: 5 additions & 38 deletions project_euler/problem_34/sol1.py
Original file line number Diff line number Diff line change
Expand Up @@ -4,37 +4,7 @@
Note: As 1! = 1 and 2! = 2 are not sums they are not included.
"""


def factorial(n: int) -> int:
"""Return the factorial of n.
>>> factorial(5)
120
>>> factorial(1)
1
>>> factorial(0)
1
>>> factorial(-1)
Traceback (most recent call last):
...
ValueError: n must be >= 0
>>> factorial(1.1)
Traceback (most recent call last):
...
ValueError: n must be exact integer
"""

if not n >= 0:
raise ValueError("n must be >= 0")
if int(n) != n:
raise ValueError("n must be exact integer")
if n + 1 == n: # catch a value like 1e300
raise OverflowError("n too large")
result = 1
factor = 2
while factor <= n:
result *= factor
factor += 1
return result
from math import factorial


def sum_of_digit_factorial(n: int) -> int:
Expand All @@ -45,7 +15,7 @@ def sum_of_digit_factorial(n: int) -> int:
>>> sum_of_digit_factorial(0)
1
"""
return sum(factorial(int(digit)) for digit in str(n))
return sum(factorial(int(char)) for char in str(n))


def compute() -> int:
Expand All @@ -56,12 +26,9 @@ def compute() -> int:
>>> compute()
40730
"""
return sum(
num
for num in range(3, 7 * factorial(9) + 1)
if sum_of_digit_factorial(num) == num
)
limit = 7 * factorial(9) + 1
return sum(i for i in range(3, limit) if sum_of_digit_factorial(i) == i)


if __name__ == "__main__":
print(compute())
print(f"{compute()} = ")
1 change: 1 addition & 0 deletions project_euler/problem_45/__init__.py
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#
57 changes: 57 additions & 0 deletions project_euler/problem_45/sol1.py
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"""
Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
Triangle T(n) = (n * (n + 1)) / 2 1, 3, 6, 10, 15, ...
Pentagonal P(n) = (n * (3 * n − 1)) / 2 1, 5, 12, 22, 35, ...
Hexagonal H(n) = n * (2 * n − 1) 1, 6, 15, 28, 45, ...
It can be verified that T(285) = P(165) = H(143) = 40755.
Find the next triangle number that is also pentagonal and hexagonal.
All trinagle numbers are hexagonal numbers.
T(2n-1) = n * (2 * n - 1) = H(n)
So we shall check only for hexagonal numbers which are also pentagonal.
"""


def hexagonal_num(n: int) -> int:
"""
Returns nth hexagonal number
>>> hexagonal_num(143)
40755
>>> hexagonal_num(21)
861
>>> hexagonal_num(10)
190
"""
return n * (2 * n - 1)


def is_pentagonal(n: int) -> bool:
"""
Returns True if n is pentagonal, False otherwise.
>>> is_pentagonal(330)
True
>>> is_pentagonal(7683)
False
>>> is_pentagonal(2380)
True
"""
root = (1 + 24 * n) ** 0.5
return ((1 + root) / 6) % 1 == 0


def compute_num(start: int = 144) -> int:
"""
Returns the next number which is traingular, pentagonal and hexagonal.
>>> compute_num(144)
1533776805
"""
n = start
num = hexagonal_num(n)
while not is_pentagonal(num):
n += 1
num = hexagonal_num(n)
return num


if __name__ == "__main__":
print(f"{compute_num(144)} = ")

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