Created problem_45 in project_euler and Speed Boost for problem_34/sol1.py (#2349)

* Create __init__.py

* Add files via upload

* 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>

project_euler/problem_34/sol1.py

@@ -4,37 +4,7 @@ Find the sum of all numbers which are equal to the sum of the factorial of their
 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:
@@ -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:
@@ -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()} = ")

project_euler/problem_45/__init__.py

@@ -0,0 +1 @@
+#

project_euler/problem_45/sol1.py

@@ -0,0 +1,57 @@
+"""
+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)} = ")