Add solution for Project Euler problem 86 (#4025)

* Added solution for Project Euler problem 86

* updating DIRECTORY.md

Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>

DIRECTORY.md

@@ -729,6 +729,8 @@
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_081/sol1.py)
   * Problem 085
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_085/sol1.py)
+  * Problem 086
+    * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_086/sol1.py)
   * Problem 087
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_087/sol1.py)
   * Problem 089

project_euler/problem_086/sol1.py

@@ -0,0 +1,105 @@
+"""
+Project Euler Problem 86: https://projecteuler.net/problem=86
+
+A spider, S, sits in one corner of a cuboid room, measuring 6 by 5 by 3, and a fly, F,
+sits in the opposite corner. By travelling on the surfaces of the room the shortest
+"straight line" distance from S to F is 10 and the path is shown on the diagram.
+
+However, there are up to three "shortest" path candidates for any given cuboid and the
+shortest route doesn't always have integer length.
+
+It can be shown that there are exactly 2060 distinct cuboids, ignoring rotations, with
+integer dimensions, up to a maximum size of M by M by M, for which the shortest route
+has integer length when M = 100. This is the least value of M for which the number of
+solutions first exceeds two thousand; the number of solutions when M = 99 is 1975.
+
+Find the least value of M such that the number of solutions first exceeds one million.
+
+Solution:
+    Label the 3 side-lengths of the cuboid a,b,c such that 1 <= a <= b <= c <= M.
+    By conceptually "opening up" the cuboid and laying out its faces on a plane,
+    it can be seen that the shortest distance between 2 opposite corners is
+    sqrt((a+b)^2 + c^2). This distance is an integer if and only if (a+b),c make up
+    the first 2 sides of a pythagorean triplet.
+
+    The second useful insight is rather than calculate the number of cuboids
+    with integral shortest distance for each maximum cuboid side-length M,
+    we can calculate this number iteratively each time we increase M, as follows.
+    The set of cuboids satisfying this property with maximum side-length M-1 is a
+    subset of the cuboids satisfying the property with maximum side-length M
+    (since any cuboids with side lengths <= M-1 are also <= M). To calculate the
+    number of cuboids in the larger set (corresponding to M) we need only consider
+    the cuboids which have at least one side of length M. Since we have ordered the
+    side lengths a <= b <= c, we can assume that c = M. Then we just need to count
+    the number of pairs a,b satisfying the conditions:
+        sqrt((a+b)^2 + M^2) is integer
+        1 <= a <= b <= M
+
+    To count the number of pairs (a,b) satisfying these conditions, write d = a+b.
+    Now we have:
+        1 <= a <= b <= M  =>  2 <= d <= 2*M
+                                   we can actually make the second equality strict,
+                                   since d = 2*M => d^2 + M^2 = 5M^2
+                                              => shortest distance = M * sqrt(5)
+                                              => not integral.
+        a + b = d => b = d - a
+                 and a <= b
+                  => a <= d/2
+                also a <= M
+                  => a <= min(M, d//2)
+
+        a + b = d => a = d - b
+                 and b <= M
+                  => a >= d - M
+                also a >= 1
+                  => a >= max(1, d - M)
+
+        So a is in range(max(1, d - M), min(M, d // 2) + 1)
+
+    For a given d, the number of cuboids satisfying the required property with c = M
+    and a + b = d is the length of this range, which is
+        min(M, d // 2) + 1 - max(1, d - M).
+
+    In the code below, d is sum_shortest_sides
+                   and M is max_cuboid_size.
+
+
+"""
+
+
+from math import sqrt
+
+
+def solution(limit: int = 1000000) -> int:
+    """
+    Return the least value of M such that there are more than one million cuboids
+    of side lengths 1 <= a,b,c <= M such that the shortest distance between two
+    opposite vertices of the cuboid is integral.
+    >>> solution(100)
+    24
+    >>> solution(1000)
+    72
+    >>> solution(2000)
+    100
+    >>> solution(20000)
+    288
+    """
+    num_cuboids: int = 0
+    max_cuboid_size: int = 0
+    sum_shortest_sides: int
+
+    while num_cuboids <= limit:
+        max_cuboid_size += 1
+        for sum_shortest_sides in range(2, 2 * max_cuboid_size + 1):
+            if sqrt(sum_shortest_sides ** 2 + max_cuboid_size ** 2).is_integer():
+                num_cuboids += (
+                    min(max_cuboid_size, sum_shortest_sides // 2)
+                    - max(1, sum_shortest_sides - max_cuboid_size)
+                    + 1
+                )
+
+    return max_cuboid_size
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")