Project Euler 70 Solution

DIRECTORY.md

@@ -674,6 +674,8 @@
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_67/sol1.py)
   * Problem 69
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_69/sol1.py)
+  * Problem 70
+    * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_70/sol1.py)
   * Problem 71
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_71/sol1.py)
   * Problem 72

project_euler/problem_70/sol1.py

@@ -0,0 +1,112 @@
+"""
+Project Euler 70
+https://projecteuler.net/problem=70
+
+Euler's Totient function, φ(n) [sometimes called the phi function], is used to
+determine the number of positive numbers less than or equal to n which are
+relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than
+nine and relatively prime to nine, φ(9)=6.
+
+The number 1 is considered to be relatively prime to every positive number, so
+φ(1)=1.
+
+Interestingly, φ(87109)=79180, and it can be seen that 87109 is a permutation
+of 79180.
+
+Find the value of n, 1 < n < 10^7, for which φ(n) is a permutation of n and
+the ratio n/φ(n) produces a minimum.
+"""
+
+
+def solution(max: int = 10000000) -> int:
+    """
+    This is essentially brute force. Calculate all totients up to 10^7 and
+    find the minimum ratio of n/φ(n) that way. To minimize the ratio, we want
+    to minimize n and maximize φ(n) as much as possible, so we can store the
+    minimum fraction's numerator and denominator and calculate new fractions
+    with each totient to compare against. To avoid dividing by zero, I opt to
+    use cross multiplication.
+
+    >>> solution(100)
+    21
+
+    >>> solution(10000)
+    4435
+    """
+
+    min_num = 1  # i
+    min_den = 0  # φ(i)
+    totients = get_totients(max + 1)
+
+    for i in range(2, max + 1):
+        t = totients[i]
+
+        if i * min_den < min_num * t and has_same_digits(i, t):
+            min_num = i
+            min_den = t
+
+    return min_num
+
+
+def get_totients(max_one: int) -> list:
+    """
+    Calculates a list of totients from 0 to max_one exclusive, using the
+    definition of Euler's product formula:
+    https://en.wikipedia.org/wiki/Euler's_totient_function#Euler's_product_formula
+
+    >>> get_totients(5)
+    [0, 1, 1, 2, 2]
+
+    >>> get_totients(10)
+    [0, 1, 1, 2, 2, 4, 2, 6, 4, 6]
+    """
+    totients = [0] * max_one
+
+    for i in range(0, max_one):
+        totients[i] = i
+
+    for i in range(2, max_one):
+        if totients[i] == i:
+            for j in range(i, max_one, i):
+                totients[j] -= totients[j] // i
+
+    return totients
+
+
+def has_same_digits(num1: int, num2: int) -> bool:
+    """
+    Return True if num1 and num2 have the same frequency of every digit, False
+    otherwise.
+
+    digits[] is a frequency table where the index represents the digit from
+    0-9, and the element stores the number of appearances. Increment the
+    respective index every time you see the digit in num1, and decrement if in
+    num2. At the end, if the numbers have the same digits, every index must
+    contain 0.
+
+    >>> has_same_digits(123456789, 987654321)
+    True
+
+    >>> has_same_digits(123, 12)
+    False
+
+    >>> has_same_digits(1234566, 123456)
+    False
+    """
+    digits = [0] * 10
+
+    while num1 > 0 and num2 > 0:
+        digits[num1 % 10] += 1
+        digits[num2 % 10] -= 1
+        num1 //= 10
+        num2 //= 10
+
+    for digit in digits:
+        if digit != 0:
+            return False
+
+    return True
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")