diff --git a/DIRECTORY.md b/DIRECTORY.md index f25b0c6ff4e3..e1ce44eedce1 100644 --- a/DIRECTORY.md +++ b/DIRECTORY.md @@ -334,6 +334,7 @@ ## Electronics * [Builtin Voltage](electronics/builtin_voltage.py) * [Carrier Concentration](electronics/carrier_concentration.py) + * [Circular Convolution](electronics/circular_convolution.py) * [Coulombs Law](electronics/coulombs_law.py) * [Electric Conductivity](electronics/electric_conductivity.py) * [Electric Power](electronics/electric_power.py) @@ -936,6 +937,8 @@ * [Sol1](project_euler/problem_097/sol1.py) * Problem 099 * [Sol1](project_euler/problem_099/sol1.py) + * Problem 100 + * [Sol1](project_euler/problem_100/sol1.py) * Problem 101 * [Sol1](project_euler/problem_101/sol1.py) * Problem 102 diff --git a/project_euler/problem_100/__init__.py b/project_euler/problem_100/__init__.py new file mode 100644 index 000000000000..e69de29bb2d1 diff --git a/project_euler/problem_100/sol1.py b/project_euler/problem_100/sol1.py new file mode 100644 index 000000000000..367378e7ab17 --- /dev/null +++ b/project_euler/problem_100/sol1.py @@ -0,0 +1,48 @@ +""" +Project Euler Problem 100: https://projecteuler.net/problem=100 + +If a box contains twenty-one coloured discs, composed of fifteen blue discs and +six red discs, and two discs were taken at random, it can be seen that +the probability of taking two blue discs, P(BB) = (15/21) x (14/20) = 1/2. + +The next such arrangement, for which there is exactly 50% chance of taking two blue +discs at random, is a box containing eighty-five blue discs and thirty-five red discs. + +By finding the first arrangement to contain over 10^12 = 1,000,000,000,000 discs +in total, determine the number of blue discs that the box would contain. +""" + + +def solution(min_total: int = 10**12) -> int: + """ + Returns the number of blue discs for the first arrangement to contain + over min_total discs in total + + >>> solution(2) + 3 + + >>> solution(4) + 15 + + >>> solution(21) + 85 + """ + + prev_numerator = 1 + prev_denominator = 0 + + numerator = 1 + denominator = 1 + + while numerator <= 2 * min_total - 1: + prev_numerator += 2 * numerator + numerator += 2 * prev_numerator + + prev_denominator += 2 * denominator + denominator += 2 * prev_denominator + + return (denominator + 1) // 2 + + +if __name__ == "__main__": + print(f"{solution() = }")