add solution for project_euler/problem_009

Author: AksharGoyal

project_euler/problem_009/sol4.py

@@ -0,0 +1,74 @@
+"""
+Project Euler Problem 9: https://projecteuler.net/problem=9
+
+Special Pythagorean triplet
+
+A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
+
+    a^2 + b^2 = c^2
+
+For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
+
+There exists exactly one Pythagorean triplet for which a + b + c = 1000.
+Find the product a*b*c.
+
+Solution:
+Let's consider the constraint a + b + c = n (n = 1000) and think of the values
+we can get for each variable while satisfying the constraint. We can have:
+    a = 333, b = 333, c = 334
+    a = 500, b = 500, c = 0
+    a = 5, b = 990, c = 5
+and various other combinations. Note that at least one value has to be
+at least 333 (or n//3 as n = 1000). Raising at least one variable to
+nearly n//2 value decreases another variable's value.
+
+When we introduce the constraint a < b < c, we will have combinations of
+three distinct values. The triplet cannot form an isoceles triangle.
+Thus, we observe:
+    a = 167, b = 333, c = 500
+    a = 331, b = 333, c = 336
+    a = 1, b = 499, c = 500
+If n is even, only two variables will be odd or all will have even values.
+Furthermore, the constraint a**2 + b**2 = c**2 suggest the if 'a' or 'b' is odd
+then 'c' is odd too and vice versa.
+
+Therefore, our solution will:
+- use "a + b + c = 1000" to elimiate 'c' (and hence remove a third for loop)
+  by having "c = 1000 - a - b" and hence comparing a**2 + b**2 = (1000-a-b)**2
+- have an odd number and an even number to ensure 'a' and 'b' are distinct
+- iterate for values of 'b' from (n//2 - 1) to (n//3 + 1) to ensure
+  that minimum value of 'c' can be (n//2) and 'a' can be (n//3) at maximum,
+  thus satisfying the constraint a < b < c and a + b + c = n
+
+References:
+    - https://en.wikipedia.org/wiki/Pythagorean_triple
+"""
+
+
+def solution(n: int = 1000) -> int:
+    """
+    Returns the product of a,b,c which are Pythagorean Triplet that satisfies
+    the following:
+      1. a < b < c
+      2. a**2 + b**2 = c**2
+      3. a + b + c = 1000
+
+    >>> solution()
+    31875000
+    >>> solution(910)
+    11602500
+    >>> solution(2002)
+    123543420
+    """
+
+    for b in range(n // 2 - 1, n // 3, -2):
+        for a in range(b - 1, 0, -2):
+            c = n - b - a
+            if b > c: # constraint a < b < c should be satisfied
+                continue
+            if a**2 + b**2 == c**2: # is it a pythagorean triplet
+                return a * b * c
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")