<< problem 280 - Ant and seeds Scoring probabilities - problem 286 >>

The 3-digit number 376 in the decimal numbering system is an example of numbers with the special property that its square ends with the same digits: 376^2 = 141376.
Let's call a number with this property a steady square.

Steady squares can also be observed in other numbering systems. In the base 14 numbering system, the 3-digit number c37 is also a steady square: c37^2 = aa0c37,
and the sum of its digits is c+3+7=18 in the same numbering system.
The letters a, b, c and d are used for the 10, 11, 12 and 13 digits respectively, in a manner similar to the hexadecimal numbering system.

For 1 <= n <= 9, the sum of the digits of all the n-digit steady squares in the base 14 numbering system is 2d8 (582 decimal).

Find the sum of the digits of all the n-digit steady squares in the base 14 numbering system for
1 <= n <= 10000 (decimal) and give your answer in the base 14 system using lower case letters where necessary.

# My Algorithm

This problem bugged me for a while ... the amount of code written for my solution was astonishing (and only a fraction of it survived ...).

Nevertheless, here's how I did it:
Within a few seconds I found exactly three single steady digits:
1^2 mod 14 = 1
7^2 mod 14 = 49 mod 14 == 7
8^2 mod 14 = 64 mod 14 == 8
(zero was disallowed by the problem statement)

Continuing with two digits caused a problem:
I couldn't find a digit that can be prepended to 1 - but there are such digits for 7 and 8.
As it turns out, I can "extend" 7 and 8 basically to infinitely long steady squares.
And there is always exactly one digit I can prepend to a steady square such that it becomes a larger steady square.
The BigNum class from my toolbox already supported different bases.
Adding a toString() method for base 14 was straightforward.
Then I wrote isSteady(): it multiplies the number with itself and in each step checks whether the last digits remain unchanged.

The obvious brute force algorithm is as follows:

• prepend every possible digit d = 0,1,2,3,...9,a,... to a steady square s of length m so that s' = d * 14^m + s
• compute its square t = s' * s'
• if the last m+1 digits of t are identical to s' then I found the correct digit
It works pretty well for up to 100 digits.

The Wikipedia page en.wikipedia.org/wiki/Automorphic_number focusses on an algorithm based on the Chinese Remainder Theorem.
And I don't really like it ... so it looked at the output and saw that there is a certain relationship between the first steady squares of 7:
s_1 = 7_14 = 7_10
s_2 = 37_14 = 49_10 = 7^2
s_3 = c37_14 = 2401_10 = 49^2 = 7^4
s_4 = 0c37_14 = 2401_10 = 49^2 = 7^4
s_5 = a0c37_14 = 386561_10 = 7^5 * 23
Unfortunately it doesn't work for s_4 and s_5 anymore, so I stopped working and solved other Project Euler problems.
Problem 455 isn't related to the current problem but it reminded my that modulo often works in strange ways.
And indeed:
s_4 = s^2_3 mod 14^4 = 2401^2 mod 14^4 = 5764801 mod 38416 == 2401
s_5 = s^2_4 mod 14^5 = 2401^2 mod 14^5 = 2401 mod 537824 == 386561
So I have a simple way to generate all digits step-by-step.

My multiplication code was extremely slow: the square of a number with m digits has 2m digits.
But I only need m + 1 of those - so I wrote multiplyLow() which stops after enough digits were computed and discards the rest.

The Wikipedia mentions a smart way to generate the second automorphic number: "the sum of two automorphic numbers is 10^k + 1".
If I have one automorphic number s then the other one is S = 10^k + 1 - s. My findOther() function does exactly that.
The solution for 1000 digits takes half a second and the solution for 10000 digits is found after about 8 minutes.

But another formula on the Wikipedia page caught my attention (I didn't realize its importance when I read it the first time):
n' = (3n^2 - 2n^3) mod 10^{2k}
Starting with an automorphic number n I get a new automorphic number n' that has twice as many digits.
Even though this formula is intended for base 10 numbers it still works for base 14 (I have no idea about other bases).

My BigNum class supports only positive numbers and 3n^2 - 2n^3 is always negative for n > 1.
So I had to rewrite the formula using the relationship -a mod b = (b-a) mod b:
n' = (3n^2 - 2n^3) mod 10^{2k}
n' = -(2n^3 - 3n^2) mod 10^{2k}
n' = (10^{2k} - (2n^3 - 3n^2)) mod 10^{2k}

See my fastDoubling() function → it finishes after 1.1 seconds.

The final step is to compute the digit sum of my steady squares:
• the last digit is part of all steady squares, that means it appears 10000 times
• the second-to-last digit is part of all steady squares, except the steady square with only one digit, that means it appears 9999 times
• ... and so on ...
• but any steady square with a leading zero must be discarded → I subtract all digits in a simple loop (looks slow, but it surprisingly fast)

## Alternative Approaches

Dedicated large integer library such as GMP are much faster than my simple BigNum class.
With a clever use of residues you can get away with simple integer arithmetic as well and solve the problem in less than 0.1 seconds.

## Note

This solution currently shares the "first place" of most code I needed to solve a problem, see my ranking.
Despite all the code (and time) needed to crack the problem, I felt that this was a really nice one.

# 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 9 | ./284

Output:

Note: the original problem's input 10000 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. Or just jump to my GitHub repository.

       #include <iostream>
#include <vector>
#include <string>

// ---------- code taken from my toolbox, with a few modifications / extensions, see explanations above ----------

// store single digits with lowest digits first
// e.g. 1024 (in base 14) is stored as { 4,2,0,1 }
// only non-negative numbers supported
struct BigNum : public std::vector<unsigned char>
{
static const unsigned int MaxDigit = 14;

// store a non-negative number
BigNum(unsigned int x = 0)
{
do
{
push_back(x % MaxDigit);
x /= MaxDigit;
} while (x > 0);
}

// convert from base 14 to string
std::string toString() const
{
const char digits[] = "0123456789abcd";

std::string result;
for (auto x : *this)
{
// process a bucket
for (unsigned int shift = 1; shift < MaxDigit; shift *= MaxDigit)
{
auto digit = (x / shift) % MaxDigit;
result.insert(0, 1, digits[digit]);
}
}

while (result.size() > 1 && result.front() == '0')
result.erase(0, 1);

return result;
}

BigNum operator+(const BigNum& other) const
{
auto result = *this;
// add in-place, make sure it's big enough
if (result.size() < other.size())
result.resize(other.size(), 0);

unsigned int carry = 0;
for (size_t i = 0; i < result.size(); i++)
{
carry += result[i];
if (i < other.size())
carry += other[i];
else
if (carry == 0)
return result;

if (carry < MaxDigit)
{
// no overflow
result[i] = carry;
carry     = 0;
}
else
{
// yes, we have an overflow
result[i] = carry - MaxDigit;
carry = 1;
}
}

if (carry > 0)
result.push_back(carry);

return result;
}

// multiply a big number by an integer
BigNum operator*(unsigned int factor) const
{
// faster multiplication possible ?
if (factor == 0)
return 0;
if (factor == 1)
return *this;
if (factor == MaxDigit)
{
if (size() == 1 && operator[](0) == 0)
return 0;

auto result = *this;
result.insert(result.begin(), 0);
return result;
}

unsigned int carry = 0;
auto result = *this;
for (auto& i : result)
{
carry += i * factor;
i      = carry % MaxDigit;
carry /= MaxDigit;
}
// store remaining carry in new digits
while (carry > 0)
{
result.push_back(carry % MaxDigit);
carry /= MaxDigit;
}

return std::move(result);
}

// subtract a smaller-or-equal number
BigNum operator-(const BigNum& other) const
{
BigNum result = *this;
int borrow = 0;
for (size_t i = 0; i < result.size(); i++)
{
int diff = (int)result[i] - borrow;
if (i < other.size())
diff -= other[i];
else
if (borrow == 0)
break;

if (diff < 0)
{
borrow = 1;
diff += MaxDigit;
}
else
borrow = 0;

result[i] = diff;
}

// no high zeros
while (result.size() > 1 && result.back() == 0)
result.pop_back();

return result;
}

// multiply two big numbers
BigNum operator*(const BigNum& other) const
{
if (size() < other.size())
return other * *this;

// multiply single digits of "other" with the current object
BigNum result = 0;
for (int i = (int)other.size() - 1; i >= 0; i--)
result = result * MaxDigit + (*this * other[i]);

return std::move(result);
}

// ---------- problem-specific code (still part of BigNum class) ----------

// multiply by a big number and keep only the lowest digits
void multiplyLow(const BigNum& other, size_t numDigits)
{
BigNum result;
result.resize(numDigits, 0);

// multiply single digits of "other" with the current object
unsigned int carry = 0;
for (size_t i = 0; i < other.size() && i < numDigits; i++)
{
carry = 0;
for (size_t j = 0; i + j < numDigits; j++)
{
// multiply two digits (if possible) and add carry
if (j >= size())
carry += result[i + j];
else
carry += result[i + j] + other[i] * operator[](j);

// store lowest digit of product and keep "carrying on" :-)
result[i + j] = carry % MaxDigit;
carry /= MaxDigit;
}
}

*this = std::move(result);
}

// try to find the square and check whether its lower digits remain unchanged
{
// multiply the number with itself and abort as soon as possible
BigNum square = 0;

for (size_t pos = 0; pos < size(); pos++)
{
// multiply with next digit
auto digit = operator[](pos);
if (digit > 0)
{
auto product = *this * digit;

for (size_t appendZeros = 0; appendZeros < pos; appendZeros++)
product.insert(product.begin(), 0);
square = square + product;
}

// mismatch ?
if (digit != square[pos])
return false;
}

return true;
}
};

// ---------- problem-specific code (not part of BigNum class) ----------

// prepend every possible digit until a steady square is fouund
BigNum bruteForce(const BigNum& number)
{
auto next = number;
next.push_back(0);

// prepend a digit (remember: digits are stored in reverse order)
for (unsigned int digit = 0; digit < BigNum::MaxDigit; digit++)
{
next.back() = digit;
break;
}
return next;
}

// compute steady square where the right-most digits are "number", result has exactly "numDigits" digits
BigNum fastDoubling(const BigNum& number, unsigned int numDigits)
{
// https://en.wikipedia.org/wiki/Automorphic_number
// n' = (3n^2 - 2n^3) mod 14^2k          => a negative number, which my simple BigNum class doesn't support yet
// n' = 14^2k - (2n^3 - 3n^2) mod 14^2k

// after each iteration, the number of steady digits will double
auto current = number;
while (current.size() < numDigits)
{
// new number will have twice as many digits
auto twiceDigits = 2 * current.size();

auto square = current * current;     // n^2
auto cube   = square  * current;     // n^3
auto diff   = cube * 2 - square * 3; // 3n^2 - 2n^3

diff.resize(twiceDigits);            // (3n^2 - 2n^3) mod 14^twiceDigits

BigNum largeOne = 0;
largeOne.resize(twiceDigits, 0);     // 14*twiceDigits zeros
largeOne.push_back(1);               // and prepend a "1" => 14^twiceDigits

current = largeOne - diff;           // 14^twiceDigits - (3n^2 - 2n^3) mod 14^twiceDigits
}

// remove superfluous leading digits (current has always 2^iterations digits, e.g. 16384 which will be reduced to 10000)
current.resize(numDigits);
return current;
}

// given an automorphic number, find its sibling
BigNum findOther(const BigNum& number)
{
// eight = (14^maxDigits + 1) - seven
BigNum one0one = 1;
one0one.resize(number.size(), 0);
one0one.push_back(1);
return one0one - number;
}

int main()
{
unsigned int maxDigits = 10000;
std::cin >> maxDigits;

// manually found those two number to be the only steady digits mod 14 (except for 1, see below)
BigNum seven = 7;
BigNum eight = 8;

//#define BRUTEFORCE
#ifdef  BRUTEFORCE
// brute-force
for (unsigned int i = 2; i < maxDigits; i++)
{
seven = bruteForce(seven);
eight = bruteForce(eight);
}
#endif

//#define POWER
#ifdef  POWER
for (unsigned int numDigits = 2; numDigits <= maxDigits; numDigits++)
{
// seven = seven * seven, reduced to numDigits
seven.multiplyLow(seven, numDigits);
//std::cout << numDigits << " " << std::flush;
}
eight = findOther(seven);
#endif

#define DOUBLING
#ifdef  DOUBLING
// use equation (3n^2 - 2n^3) mod 14^k until I have 10000 digits
seven = fastDoubling(7, maxDigits);
eight = findOther(seven); // faster than eight = fastDoubling(8, maxDigits);
#endif

// now I have the 10000 steady digits ending with 7 and 8, let's find their digit sum !

// 1 is a steady digit, too, but a strange special case: 1^2 % 14 = 1
unsigned int sum = 1;
// prepending any digit to 1 doesn't produce a steady square

// okay, let's proceed with ...7 and ...8
for (unsigned int i = 0; i < maxDigits; i++)
{
// last digit appears in 10000 steady squares, next to last in 9999 steady squares, ...
auto howOften = maxDigits - i;
sum += howOften * seven[i];
sum += howOften * eight[i];

// but don't count those numbers having a leading zero
if (seven[i] == 0)
for (unsigned int j = 0; j < i; j++) // could be more efficient ... but it's still too fast to be measurable
sum -= seven[j];

if (eight[i] == 0)
for (unsigned int j = 0; j < i; j++)
sum -= eight[j];
}

// print digit sum in base 14
BigNum sumBase14 = sum;
std::cout << sumBase14.toString() << std::endl;
return 0;
}


This solution contains 56 empty lines, 56 comments and 10 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 1.1 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

October 13, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 55% (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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