<< problem 429 - Sum of squares of unitary divisors | Under The Rainbow - problem 493 >> |

# Problem 491: Double pandigital number divisible by 11

(see projecteuler.net/problem=491)

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:

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
auto maxBitmask = minBitmask << (maxDigit + 1);
// process all bitmasks where maxDigits bits are set
for (auto bitmask = minBitmask; bitmask <= maxBitmask; bitmask = nextNumberWithSameBits(bitmask))
{
// 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
auto reduce = bitmask;
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];
}
// exclude leading zeros
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 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

August 16, 2017 submitted solution

August 16, 2017 added comments

# Difficulty

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

# Links

projecteuler.net/thread=491 - **the** best forum on the subject (*note:* you have to submit the correct solution first)

# Heatmap

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

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I scored 13,386 points (out of 15600 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.

# Copyright

I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.

All of my solutions can be used for any purpose and I am in no way liable for any damages caused.

You can even remove my name and claim it's yours. But then you shall burn in hell.

The problems and most of the problems' images were created by Project Euler.

Thanks for all their endless effort !!!

<< problem 429 - Sum of squares of unitary divisors | Under The Rainbow - problem 493 >> |