Problem 32: Pandigital products

(see projecteuler.net/problem=32)

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.

Algorithm

We have to solve an equation a * b = c.
There are 9!=362880 permutations of { 1,2,3,4,5,6,7,8,9 } and for each of those permutation I try all possible combinations of a, b and c.

All digits (0..9) are stored in a plain std::vector and the STL's function std::next_permutation generates all permutations.
Then two nested loops split the sequence into three parts (a has length lenA, b has length lenB and c has length lenC).
If a * b = c then my program stores the product c in std::set named valid. Duplicates are automatically avoided by the std::set.

Finally all elements of that std::set are added.

Modifications by HackerRank

The digits are 1 to n instead of 1 to 9.

Note

There are many opportunities for optimizations but the code is already extremely fast.

My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
 
int main()
{
// read highest digit
unsigned int maxDigit;
std::cin >> maxDigit;
 
// all digits from 1..9
std::vector<unsigned int> digits = { 1,2,3,4,5,6,7,8,9 };
// remove higher numbers so there is only 1..n left
digits.resize(maxDigit);
 
// all pandigital products
std::set<unsigned int> valid;
 
// create all permutations
do
{
// let's say a * b = c
// each variable contains at least one digit
// the sum of their digits is limited by n (which should be 9)
// try all combinations of lengths with the current permutation of digits
for (unsigned int lenA = 1; lenA < maxDigit; lenA++)
for (unsigned int lenB = 1; lenB < maxDigit - lenA; lenB++)
{
unsigned int lenC = maxDigit - lenA - lenB;
 
// a*b=c => c>=a && c>=b => c has at least as many digits as a or b
if (lenC < lenA || lenC < lenB)
break;
 
// pos contains the currently used position in "digits"
unsigned int pos = 0;
 
// build "a" out of the first digits
unsigned int a = 0;
for (unsigned int i = 1; i <= lenA; i++)
{
a *= 10;
a += digits[pos++];
}
 
// next digits represent "b"
unsigned int b = 0;
for (unsigned int i = 1; i <= lenB; i++)
{
b *= 10;
b += digits[pos++];
}
 
// and the same for "c"
unsigned int c = 0;
for (unsigned int i = 1; i <= lenC; i++)
{
c *= 10;
c += digits[pos++];
}
 
// is a*b = c ?
if (a*b == c)
valid.insert(c);
}
} while (std::next_permutation(digits.begin(), digits.end()));
 
// find sum
unsigned int sum = 0;
for (auto x : valid)
sum += x;
std::cout << sum << std::endl;
 
return 0;
}

This solution contains 12 empty lines, 16 comments and 4 preprocessor commands.

Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Input data (separated by spaces or newlines):

This is equivalent to
echo 5 | ./32

Output:

(please click 'Go !')

Note: the original problem's input 9 cannot be entered
because just copying results is a soft skill reserved for idiots.

(this interactive test is still under development, computations will be aborted after one second)

Benchmark

The correct solution to the original Project Euler problem was found in 0.05 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=c++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

Changelog

February 23, 2017 submitted solution
April 6, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler032

My code solved 6 out of 6 test cases (score: 100%)

Difficulty

Project Euler ranks this problem at 5% (out of 100%).

Hackerrank describes this problem as easy.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is never an option.

Links

projecteuler.net/thread=32 - the best forum on the subject (note: you have to submit the correct solution first)

Code in various languages:

Python: www.mathblog.dk/project-euler-32-pandigital-products/ (written by Kristian Edlund)
Haskell: github.com/nayuki/Project-Euler-solutions/blob/master/haskell/p032.hs (written by Nayuki)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p032.java (written by Nayuki)
Mathematica: github.com/nayuki/Project-Euler-solutions/blob/master/mathematica/p032.mathematica (written by Nayuki)
C: github.com/eagletmt/project-euler-c/blob/master/30-39/problem32.c (written by eagletmt)
Go: github.com/frrad/project-euler/blob/master/golang/Problem032.go (written by Frederick Robinson)
Javascript: github.com/dsernst/ProjectEuler/blob/master/32 Pandigital products.js (written by David Ernst)
Scala: github.com/samskivert/euler-scala/blob/master/Euler032.scala (written by Michael Bayne)

Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.
yellow problems score less than 100% at Hackerrank (but still solve the original problem).
gray problems are already solved but I haven't published my solution yet.
blue problems are already solved and there wasn't a Hackerrank version of it (at the time I solved it) or I didn't care about it because it differed too much.

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The 133 solved problems had an average difficulty of 16.9% at Project Euler and I scored 11,174 points (out of 12300) at Hackerrank's Project Euler+.
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