Solution to problem 61 in Python

.pre-commit-config.yaml

@@ -1,6 +1,6 @@
 repos:
 -   repo: https://github.com/pre-commit/pre-commit-hooks
-    rev: v4.0.1
+    rev: v4.3.0
     hooks:
     -   id: trailing-whitespace
     -   id: end-of-file-fixer
@@ -11,11 +11,11 @@ repos:
     hooks:
     -   id: flake8
 -   repo: https://github.com/psf/black
-    rev: 21.12b0
+    rev: 22.6.0
     hooks:
       - id: black
 -   repo: https://github.com/pre-commit/mirrors-mypy
-    rev: v0.920
+    rev: v0.961
     hooks:
         - id: mypy
           additional_dependencies: [pydantic]  # add if use pydantic

src/Python/Problem061.py

@@ -0,0 +1,88 @@
+#!/usr/bin/env python3
+"""
+Created on 16 Jul 2022
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for problem 61 of Project Euler
+https://projecteuler.net/problem=61
+"""
+
+from utils import timeit
+
+
+@timeit("Problem 61")
+def compute():
+    """
+    Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers
+    are all figurate (polygonal) numbers and are generated by the following
+    formulae:
+        Triangle 	  	P3,n=n(n+1)/2 	  	1, 3, 6, 10, 15, ...
+        Square 	  	P4,n=n2 	  	1, 4, 9, 16, 25, ...
+        Pentagonal 	  	P5,n=n(3n-1)/2 	  	1, 5, 12, 22, 35, ...
+        Hexagonal 	  	P6,n=n(2n-1) 	  	1, 6, 15, 28, 45, ...
+        Heptagonal 	  	P7,n=n(5n-3)/2 	  	1, 7, 18, 34, 55, ...
+        Octagonal 	  	P8,n=n(3n-2) 	  	1, 8, 21, 40, 65, ...
+
+    The ordered set of three 4-digit numbers: 8128, 2882, 8281, has three
+    interesting properties.
+
+        The set is cyclic, in that the last two digits of each number is the
+        first two digits of the next number (including the last number with the
+        first).
+        Each polygonal type: triangle (P3,127=8128), square (P4,91=8281), and
+        pentagonal (P5,44=2882), is represented by a different number in the
+        set.
+        This is the only set of 4-digit numbers with this property.
+
+    Find the sum of the only ordered set of six cyclic 4-digit numbers for
+    which each polygonal type: triangle, square, pentagonal, hexagonal,
+    heptagonal, and octagonal, is represented by a different number in the set.
+    """
+
+    from itertools import permutations
+
+    P = {
+        "3": lambda n: n * (n + 1) // 2,
+        "4": lambda n: n * n,
+        "5": lambda n: n * (3 * n - 1) // 2,
+        "6": lambda n: n * (2 * n - 1),
+        "7": lambda n: n * (5 * n - 3) // 2,
+        "8": lambda n: n * (3 * n - 2),
+    }
+
+    # Set of values in each figurate set
+    dic = {
+        k: [str(v(i)) for i in range(1, 200) if 10**3 <= v(i) < 10**4]
+        for k, v in P.items()
+    }
+
+    perms = permutations(P.keys(), 6)
+    for perm in perms:  # 720 different permutations of the six polygonals
+        for a in dic.get(perm[0]):
+            lst = []
+            a1, a2 = a[:2], a[2:]
+            lst.append(a)
+            for b in filter(lambda n: n[:2] == a2, dic.get(perm[1])):
+                lst.append(b)
+                b2 = b[2:]
+                for c in filter(lambda n: n[:2] == b2, dic.get(perm[2])):
+                    lst.append(c)
+                    c2 = c[2:]
+                    for d in filter(lambda n: n[:2] == c2, dic.get(perm[3])):
+                        lst.append(d)
+                        d2 = d[2:]
+                        for e in filter(lambda n: n[:2] == d2, dic.get(perm[4])):
+                            lst.append(e)
+                            e2 = e[2:]
+                            for f in filter(
+                                lambda n: n[:2] == e2 and n[2:] == a1, dic.get(perm[5])
+                            ):
+                                lst.append(f)
+                                return sum(map(int, lst))
+
+
+if __name__ == "__main__":
+
+    print(f"Result for Problem 61: {compute()}")