<< problem 565 - Divisibility of sum of divisors Counting hexagons - problem 577 >>

# Problem 571: Super Pandigital Numbers

A positive number is pandigital in base b if it contains all digits from 0 to b - 1 at least once when written in base b.

A n-super-pandigital number is a number that is simultaneously pandigital in all bases from 2 to n inclusively.
For example 978 = 1111010010_2 = 1100020_3 = 33102_4 = 12403_5 is the smallest 5-super-pandigital number.
Similarly, 1093265784 is the smallest 10-super-pandigital number.
The sum of the 10 smallest 10-super-pandigital numbers is 20319792309.

What is the sum of the 10 smallest 12-super-pandigital numbers?

# My Algorithm

I made an assumption which turned out to be true:
there are at least ten 12-super-pandigital numbers with 12 digits.
Or in plain English: if I generate all permutations of the digits { 0,1,2,3,4,5,6,7,8,9,10,11 } in lexicographical order
and check whether they are 12-super-pandigital then I will find the result.

The smallest valid 12-pandigital number is { 1,0,2,3,4,5,6,7,8,9,10,11 } because leading zeros are forbidden.
Any 12-pandigital number fits nicely in a 64 bit integer: log_2 12^12 < 44 → I need at most 44 bits
The smallest such number is current = 1 * 12^11 + 0 * 12^10 + 2 * 12^9 + 3 * 12^8 + ... + 11 * 12^0 = 754777787027 (in decimal notation, base 10)

I rewrote the function isPandigital() from problem 170 such that it accepts an arbitrary base greater than 1 and tolerates if a digit appears more than once.
It basically chops off the right-most digit and sets a bit in a bitmask used.
If all 12 bits are set at the end, then the number is pandigital indeed.

Since all numbers are already pandigital in base 12 I only have to check bases 2,3,...,11.
It's much faster to check each number against a "high" base first because the "failure rate" is higher.
For example, all numbers except zero and 2^n-1 are pandigital in base 2, so pretty much all numbers are pandigital in base 2.
On the other side, a substantial amount of numbers fail the 11-pandigital test.
Consequently, my for-loop runs backwards from 11 to 2 and aborts if a isPandigital check fails.

Division and modulo operations are notoriously slow on modern CPUs - and my program basically does that non-stop.
Division and modulo operation can be quite fast if you divide by a power of two (and you have a smart compiler):
Division by 8 is just a right shift by 3 bits (because 2^3 = 8) and modulo 8 means taking the right-most three bits (number & (8-1)).
Checking isPandigital(current, 8) before anything else made my program more than twice as fast !

## Note

The inner loop of isPandigital skips the last iteration because then number is a single digit and I can save one division and one modulo. This trick is about 5% faster.

I was pretty sure that the for-loop which calls isPandigital(current, base) can be optimized:
if isPandigital(current, 8) was already tested outside the for-loop then there is no use in tested against it inside the for-loop.
To my surprise, to opposite is true and I have no idea why.

If a number passes the pandigital check for base=11,10,9,...,6 then it also passes the checks for base=5,4,3,2.

# 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):
Note: Enter the base and how many pandigital numbers should be found

This is equivalent to
echo "10 10" | ./571

Output:

Note: the original problem's input 12 10 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)

# 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 <algorithm>

const unsigned int MaxBase = 12;

// return true if decimal number is pandigital in a certain base
// note: a digit may found more than once unlike most other pandigital problems
//       where each digit must be found exactly once
bool isPandigital(unsigned long long number, unsigned int base)
{
// bitmask where the n-th bit is set if the digit n was observed in number
unsigned int used = 0;
// all bits set => all digits used
const unsigned int All = (1 << base) - 1;

// process right-most digit and remove it
while (number >= base) // skip last iteration
{
auto digit = number % base;
used |= 1 << digit;

number /= base;
}

// simplified last iteration
used |= 1 << number;

return used == All;
}

int main()
{
unsigned int base       = 12;
unsigned int numResults = 10;
std::cin >> base >> numResults;

// smallest 12-pandigital number
std::vector<unsigned char> twelve = { 1,0,2,3,4,5,6,7,8,9,10,11 };
// reduce for smaller bases
twelve.resize(base);

// look for the sum of the first 10 matches
unsigned int  numFound = 0;
unsigned long long sum = 0;
do
{
// convert from base 12 to an integer (technically that base-12-to-2 ?)
unsigned long long current = 0;
for (auto digit : twelve)
{
current *= base;
current += digit;
}

// an optimizing compiler can reduce mod/div by fast bit operations
if (base >= 8 && !isPandigital(current, 8))
continue;

// no need to check base 12 because all generated number are pandigital in base 12 by definition
// I'm pretty sure there are some relationships:
// - if pandigital in base 8 then always in base 4, too
// - if pandigital in base 9 then always in base 3, too
// my debugging output verified that no number is rejected by the tests in base 2,3,4,5
bool isGood = true;
for (auto i = base - 1; i >= 2; i--)
if (!isPandigital(current, i)) // I tried to prepend "i != 8" but it was slower !
{
isGood = false;
break;
}

// passed all tests
if (isGood)
{
sum += current;
numFound++;
// done ?
if (numFound == numResults)
break;
}
} while (std::next_permutation(twelve.begin(), twelve.end()));

// print result
std::cout << sum << std::endl;
return 0;
}


This solution contains 13 empty lines, 20 comments and 3 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 8.9 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++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

September 22, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 25% (out of 100%).

# Heatmap

Please click on a problem's number to open my solution to that problem:

 green solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily) gray problems are already solved but I haven't published my solution yet blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too black problems are solved but access to the solution is blocked for a few days until the next problem is published [new] the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.
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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Look at my progress and performance pages to get more details.

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