Problem 571: Super Pandigital Numbers

(see projecteuler.net/problem=571)

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:

(please click 'Go !')

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
September 22, 2017 added comments

Difficulty

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

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The 289 solved problems (level 11) had an average difficulty of 32.1% at Project Euler and
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