Project euler problem 66 complete

Author: ParthKGarg

project_euler/problem_066/__init__.py

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+"""
+Project Euler Problem 61: https://projecteuler.net/problem=66
+
+Problem statement-
+Consider quadratic Diophantine equations of the form:
+x^2 - D * y^2 = 1
+
+For example, when D = 13 , the minimal solution in x is 649^2 - 13* 180^2 = 1.
+
+It can be assumed that there are no solutions in positive integers when D is square.
+
+By finding minimal solutions in x for D = {2, 3, 5, 6, 7} , we obtain the following:
+3^2 - 2 * 2^2 = 1
+2^2 - 3 * 1^2 = 1
+9^2 - 5 * 4^2 = 1 (highest value of x)
+5^2 - 6 * 2^2 = 1
+8^2 - 7 * 3^2 = 1
+
+Hence, by considering minimal solutions in x for D<= 7 ,
+the largest x is obtained when D = 5.
+
+Find the value of D<=1000  in minimal solutions of x for
+which the largest value of  is obtained.
+
+
+Solution explanation -
+The above given equation is known as Pell's equation. The solution uses the standard
+method for solving pell's equation. Which is available on the wikipedia page.
+The solution uses the continued fraction method for finding close approximations
+of sqrt(D) which are of the form h/k. Because for these solutions the h and k are
+solutions for x and y respectively. But we dont want an exact approximation because
+there is a 1 on the RHS. We consider upto 2 periodic cycles, which guarantee a
+solution. We check in the continued fraction considering newer fractions and checking
+if numerator(x) and denominator(y) satisfy the diophantine equation.
+
+
+References-
+Wikipedia - https://en.wikipedia.org/wiki/Pell%27s_equation
+
+
+"""
+
+import math
+
+
+def is_perfect_square(num: int) -> bool:
+    """
+    Checks if the number inputted is a perfect square or not.
+    Returns True if perfect square else False.
+
+    >>> is_perfect_square(25)
+    True
+    >>> is_perfect_square(3)
+    False
+
+    Time complexity - O(log2(num))
+
+    """
+
+    if num < 0:
+        return False
+    root = math.isqrt(num)
+    return root * root == num
+
+
+def continued_fraction(num: int) -> list[int]:
+    """
+    Computes and returns the continued fraction of the square root of the given integer.
+    Note for irrational numbers, the continued fraction stops when there is repetition.
+    The returned list is of the form -> [a0;a1,a2,a3...an]
+    where a1,a2,a3...an represents the minimumperiodic part of the continued fraction.
+
+    Time complexity - O(sqrt(num))
+
+    >>> continued_fraction(2)
+    [1, 2]
+    >>> continued_fraction(13)
+    [3, 1, 1, 1, 1, 6]
+    """
+
+    ans = []
+
+    # Initialize variables for continued fraction
+    m = 0
+    d = 1
+    a0 = math.isqrt(num)
+
+    # this is the end point from where the sequence starts to repeat
+    terminate_point = 2 * a0
+
+    ans.append(a0)
+    an = a0
+
+    # loop till termination point is reached, storing each an value
+    while an != terminate_point:
+        m = d * an - m
+        d = (num - m**2) / d
+        an = math.floor((a0 + m) / d)
+        ans.append(an)
+
+    return ans
+
+
+def diophantine_equation(var_x: int, var_y: int, var_d: int) -> int:
+    """
+    Returns the LHS of the diophantine equation using x, y and D.
+
+    >>> diophantine_equation(3, 2, 2)
+    1
+    >>> diophantine_equation(8, 3, 7)
+    1
+    >>> diophantine_equation(3, 4, 5)
+    -71
+    """
+
+    return var_x**2 - var_d * var_y**2
+
+
+def solution(max_d: int = 1000) -> int:
+    """
+    This function provides a solution to the problem 66 from project Euler.
+    The original problem is for D<= 1000. But there is a optional parameter
+    max_d, by changing it you can find solutions for any D theoretically.
+
+    Time complexity - O(max_d * (log2(max_d) + sqrt(max_d)))
+    But since sqrt(max_d) dominates log2(max_d)
+    Final time complexity is - O(max_d^3/2)
+
+    >>> solution()
+    661
+    >>> solution(10000)
+    9949
+    """
+
+    # Initialise answer which stored max value of D.
+    answer = -1
+    # Initialise max_x which stores max value of x.
+    max_x = -1
+
+    # loop over all values of D
+    for d in range(2, max_d + 1):
+        # skip if d is prefect square
+        if is_perfect_square(d):
+            continue
+
+        # Get continued fraction and then
+        # Add the periodic part of the fraction to itself because we are
+        # guaranteed to find the solution in 2 cycles of the periodic part.
+        # example -
+        # fraction = [3,1,1,2]
+        # First element is the integer part, rest is periodic fractional part.
+        # Add the fractional part on itself.
+        # fraction becomes [3,1,1,2,1,1,2].
+        # This contains our solution for sure.
+        fraction = continued_fraction(d)
+        fraction += fraction[1:]
+
+        num_0 = 0
+        num_1 = 1
+        den_0 = 1
+        den_1 = 0
+
+        # This is a recursive relation to find the next numerator and denominator
+        num_2 = fraction[0] * num_1 + num_0
+        den_2 = fraction[0] * den_1 + den_0
+
+        n = 1
+        # Use the recursive relation to get new, more accurate numerator and
+        # denominator, then check if they satisfy the given diophantine equation.
+        while diophantine_equation(num_2, den_2, d) != 1:
+            num_0 = num_1
+            num_1 = num_2
+            num_2 = fraction[n] * num_1 + num_0
+
+            den_0 = den_1
+            den_1 = den_2
+            den_2 = fraction[n] * den_1 + den_0
+
+            n += 1
+
+        # If we see a new larger value of x, we store the corresponding D
+        # value as our answer.
+        if num_2 > max_x:
+            answer = d
+            max_x = num_2
+
+    return answer
+
+
+if __name__ == "__main__":
+    print(f"{solution(10000) = }")