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Project Euler Problem 32

on project euler, erlang, python

Pandigital products

We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital.

The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.

Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.

HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.

Link to original description
Source code examples on Github

Erlang version 1

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#!/usr/bin/env escript
%% -*- erlang -*-
%%! -smp enable -sname p32
% vim:syn=erlang

-mode(compile).

main(_) ->
    Answer = lists:sum(lists:usort([X*Y||X<-lists:seq(1,99),Y<-lists:seq(100,9999), isPandigital(X,Y)])),
    io:format("Answer ~p ~n", [Answer]).

isPandigital(X,Y) ->
    P = X * Y,
    S = integer_to_list(X) ++ integer_to_list(Y) ++ integer_to_list(P),
    case {length(S), lists:usort(S)} of
        {9, "123456789"} -> true
        ;_               -> false
    end.

Python version

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#!/usr/bin/python

def isPandigital(x,y):
    p = x*y
    s = str(x)+str(y)+str(p)
    if len(s) != 9:
        return False
    if "".join(sorted(set(s))) == "123456789":
        return True
    else:
        return False

pd = [x*y for x in xrange(1,100) for y in xrange(100,10000) if isPandigital(x,y)]

print "Answer: %s" % sum(set(pd))

Another Python version

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#!/usr/bin/python

from combinatorics import permutations

def num(l):
    s = 0
    for n in l: s = s * 10 + n
    return s

product = {}
for perm in permutations(range(1,10)):
    for cross in range(1,4):            # Number can't be more than 4 digits
        for eq in range(cross+1, 6):    # Result can't be less than 4 digits
            a = num(perm[0:cross])
            b = num(perm[cross:eq])
            c = num(perm[eq:9])
            if a * b == c: product[c] = 1

print sum(p for p in product)