Project Euler Problem 62 Solution
The cube, (), can be permuted to produce two other cubes: () and (). In fact, is the smallest cube which has exactly three permutations of its digits which are also cube.
Find the smallest cube for which exactly five permutations of its digits are cube.
import Data.List (sort) import qualified Data.Map as Map cubes :: Map.Map String [Integer] cubes = Map.fromListWith (++) [(sort (show cube), [cube]) | x <- [1..10000], let cube = x^3] main :: IO () main = print $ minimum [minimum ns | (_, ns) <- Map.toList cubes, length ns == 5]
$ ghc -O2 -o cubic-permutations cubic-permutations.hs $ time ./cubic-permutations real 0m0.046s user 0m0.044s sys 0m0.000s
#!/usr/bin/env python from collections import defaultdict def cube(x): return x**3 def main(): cubes = defaultdict(list) for i in range(10000): c = cube(i) digits = ''.join(sorted([d for d in str(c)])) cubes[digits].append(c) print(min([min(v) for k, v in list(cubes.items()) if len(v) == 5])) if __name__ == "__main__": main()
$ time python3 cube-permutations.py real 0m0.070s user 0m0.064s sys 0m0.000s
Questions? Comments? Send me an email: firstname.lastname@example.org