<< problem 31 - Coin sums | Digit cancelling fractions - problem 33 >> |

# 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:

This is equivalent to`echo 5 | ./32`

Output:

*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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<< problem 31 - Coin sums | Digit cancelling fractions - problem 33 >> |