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Problem 284: Steady Squares
(see projecteuler.net/problem=284)
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).
Steady squares with leading 0's are not allowed.
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
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:
This is equivalent toecho 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.
#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]);
}
}
// remove leading zeros
while (result.size() > 1 && result.front() == '0')
result.erase(0, 1);
return result;
}
// add two big numbers
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
bool isSteady() const
{
// 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;
}
// yes, a steady square
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;
// steady ?
if (next.isSteady())
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
October 13, 2017 added comments
Difficulty
Project Euler ranks this problem at 55% (out of 100%).
Links
projecteuler.net/thread=284 - the best forum on the subject (note: you have to submit the correct solution first)
Code in various languages:
Python github.com/evilmucedin/project-euler/blob/master/euler284/euler284.py (written by Den Raskovalov)
Python github.com/Meng-Gen/ProjectEuler/blob/master/284.py (written by Meng-Gen Tsai)
Python github.com/roosephu/project-euler/blob/master/284.py (written by Yuping Luo)
C++ github.com/evilmucedin/project-euler/blob/master/euler284/euler284.cpp (written by Den Raskovalov)
C++ github.com/roosephu/project-euler/blob/master/284.cpp (written by Yuping Luo)
Those links are just an unordered selection of source code I found with a semi-automatic search script on Google/Bing/GitHub/whatever.
You will probably stumble upon better solutions when searching on your own. Maybe not all linked resources produce the correct result and/or exceed time/memory limits.
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 | |
the flashing problem is the one I solved most recently |
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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.
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 !!!
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