<< problem 485 - Maximum number of divisors Under The Rainbow - problem 493 >>

# Problem 491: Double pandigital number divisible by 11

We call a positive integer double pandigital if it uses all the digits 0 to 9 exactly twice (with no leading zero).
For example, 40561817703823564929 is one such number.

How many double pandigital numbers are divisible by 11?

# My Algorithm

bruteForce counts all permutations which are divisible by 11.
It finds the correct result in a reasonable amount of time when the highest digit is 6.
And of course that simple approach is way to slow when the highest digit is 9.

A number is divisible by 11 if the difference of the sum of the digits at odd and the even positions is divisible by 11.
40561817703823564929 is not divisible by 11 because 4+5+1+1+7+0+2+5+4+2 = 31 and 0+6+8+7+0+8+3+6+9+9 = 56 and 56 - 31= 25 is not divisible by 11.
There is actually an astonishing variety of rules for divisibility by 11: en.wikipedia.org/wiki/Divisibility_rule

All double pandigital numbers are permutations of the string "00112233445566778899" (excluding leading zeros).
Ten of these digits will be at odd positions and ten at even positions.
If I create a bitmask which digits will be at odd positions then I need a 20-bit integer where exactly 10 bits are set.

All bitmasks with exactly 10 bits set are between 00000000001111111111b (minBitmask) and 11111111110000000000b (maxBitmask).
The nice function nextNumberWithSameBits(x) creates the next number with the same number of bits set as x and was first published in HAKMEM 175 (often called "snoob").

However, there are a few bordercases:
I can choose a digit once, twice or not at all to appear at an odd position. A bitmask might be:
99887766554433221100
01001101010111000101
→ exactly 10 bits set
For each digit I must only accept the bit patterns:
01 (=> use that digit once, e.g. for 9,6,...),
11 (twice, e.g. for 7 and 3) and
00 (none, e.g. for 8 and 2)
I must not accept bit pattern 10 because it's the same as 01 (see my loop with the variable reduce).

The sum of all digits is 2 * T(9) where T is the triangular number:
digitSum = 2 * dfrac{9 * (9 + 1)}{2} = 90

The sum of all digits at even positions can be computed when I know the sum of all digits at odd positions:
sumEven = digitSum - sumOdd

And the difference of sumEven and sumOdd has to be divisible by 11:
(sumEven - sumOdd) == 0 mod 11
((digitSum - sumOdd) - sumOdd) == 0 mod 11
(digitSum - 2 * sumOdd) == 0 mod 11

Whenever I know that the bitmasks generate matching numbers I have to figure out how many different numbers are possible.
If x_i represents how often x appears at odd positions then x_i can be either 0, 1 or 2.
The total number of permutations will be:
p(x) = dfrac{10!}{x_0!x_1!x_2!...x_9!}

If y_i represents how often y appears at even positions then y_i can be either 0, 1 or 2. The formula remains the same:
p(y) = dfrac{10!}{y_0!y_1!y_2!...y_9!}

And the sum of all p(x)p(y) will be the result.
Interesting is that the number of digits that appear twice is the same for p(x) and p(y). Even though a few x_i != y_i, it's still p(x) = p(y).
There my code simplifies to result += permutationsRepeated[repeated] * permutationsRepeated[repeated];

Excluding leading zeros is simple: just multiply with frac{9}{10}.

# 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 3 | ./491

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)

# My code

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

       #include <iostream>
#include <string>
#include <algorithm>
#include <cstdlib>

// count all numbers, where highest digit is maxDigit
unsigned long long bruteForce(unsigned int maxDigit)
{
unsigned long long result = 0;

// keep only a few digits
std::string digits = "10012233445566778899"; // smallest number without a leading zero
digits.resize(maxDigit * 2 + 2);

// count numbers divisible by 11
do
{
if (std::stoll(digits) % 11 == 0)
result++;
} while (std::next_permutation(digits.begin(), digits.end()));

return result;
}

// return next higher number with same number of bits set
unsigned int nextNumberWithSameBits(unsigned int x)
{
// from HAKMEM 175
// see http://www.hackersdelight.org/hdcodetxt/snoob.c.txt
auto smallest = x & -x;
auto ripple = x + smallest;
auto ones = ripple ^ x;
return ((ones >> 2) / smallest) | ripple;
}

// count numbers by using permutations
unsigned long long fast(unsigned int maxDigit)
{
unsigned long long result = 0;

int  digitSum  = 2 * (maxDigit + 1) * maxDigit / 2; // = (maxDigit + 1) * maxDigit = 90
auto numDigits = 2 * (maxDigit + 1);                // = 20

// precompute number of permutations with repeated elements
unsigned long long factorial = 1;
for (unsigned int i = 1; i <= maxDigit + 1; i++)
factorial *= i; // 10! = 3628800
// 10!, 10!/2!, 10!/2!2!, 10!/2!2!2!, ...
unsigned long long permutationsRepeated[10];
for (unsigned int i = 0; i <= maxDigit; i++)
permutationsRepeated[i] = factorial >> i;

// smallest bitmask where maxDigits bits are set
auto minBitmask = (1 << (maxDigit + 1)) - 1;
// largest  bitmask where maxDigits bits are set
// process all bitmasks where maxDigits bits are set
{
// when picking a number for odd positions:
// choose a digit once, twice or not at all
// e.g. bits represent             99887766554433221100
// and the current choice might be 01001101010111000101
// (exactly 10 bits set)
// for each digit only accept the bit patterns 01 (once, e.g. for 9,6,...),
//                                             11 (twice, e.g. for 7 and 3) and
//                                             00 (none, e.g. for 8 and 2)
// DO NOT accept 10 because it's the same as 01
// ok will be false if bit pattern 10 is found
bool ok = true;
while (reduce > 0)
{
// lowest two bits are 10b (=2 decimal) ?
if ((reduce & 3) == 2)
{
ok = false;
break;
}
// next two bits
reduce >>= 2;
}
if (!ok)
continue;

// add all digits at odd positions and count how many digits are repeatedly present at odd positions
int sumOdd = 0;
auto repeated = 0; // it's actually the same value for odd and even positions
for (unsigned int pos = 0; pos < numDigits; pos++)
{
// bit set ? use that digit
if (bitmask & (1 << pos))
{
sumOdd += pos / 2;

// use that digit twice ? (bit pattern 11b)
if (pos & 1) // same as repeated += pos & 1;
repeated++;
}
}

// divisible by 11 ?
if ((digitSum - 2*sumOdd) % 11 == 0)
result += permutationsRepeated[repeated] * permutationsRepeated[repeated];
}

return result * maxDigit / (maxDigit + 1);
}

int main()
{
unsigned int numDigits = 9;
std::cin >> numDigits;

//std::cout << bruteForce(numDigits) << std::endl;
std::cout << fast(numDigits) << std::endl;
return 0;
}


This solution contains 15 empty lines, 30 comments and 4 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 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

August 16, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 20% (out of 100%).

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