<< problem 79 - Passcode derivation Path sum: two ways - problem 81 >>

# Problem 80: Square root digital expansion

It is well known that if the square root of a natural number is not an integer, then it is irrational.
The decimal expansion of such square roots is infinite without any repeating pattern at all.

The square root of two is 1.41421356237309504880..., and the digital sum of the first one hundred decimal digits is 475.

For the first one hundred natural numbers, find the total of the digital sums of the first one hundred decimal digits for all the irrational square roots.

# My Algorithm

Dealing with floating-point numbers is hard, especially with the limited precision of C++.
Therefore my program performs all computations on integers only.
The main idea is: sqrt{2} = sqrt{2 * frac{1000000}{1000000}} = frac{sqrt{2 * 1000000}}{1000} = 1414
→ to find the first three fractional digits of sqrt{2} we need to "shift" 2 by 10^{2 * 3} = 10^6 = 1000000.

Computing the first 100 digits of sqrt{2} means that 2 must be converted to an integer with about 200 digits,
which is obviously too much for C++'s native data types.
It's showtime for my BigNum class - again ! About 2/3 of this solution is just BigNum code.

My first experiments were based on the Newton square root algorithm (see en.wikipedia.org/wiki/Newton's_method).
The code worked but was too slow (mainly because of several BigNum divisions).
Browsing through the internet people recommended a surprisingly simple and efficient digit-by-digit square root algorithm
nicely explained in a paper by Frazer Jarvis (see www.afjarvis.staff.shef.ac.uk/maths/jarvisspec02.pdf).
That paper is quite accessible - read it if you know to know more about it.

My code can be found in the function jarvis. Its parameter precision is a huge number 10^{digits+extra}.
(due to the way this algorithm works, precision has 100 digits, not 2*100=200 digits).
I deliberately compute more digits than required because I use a trick for faster calculations:
only the square roots of prime numbers are actually computed.

For any composite number c = a * b where a and b are arbitrary factors of c, the square is
sqrt{c} = sqrt{a * b} = sqrt{a} * sqrt{b}
A single multiplication of two BigNums is much faster than a full square root computation.

## Modifications by HackerRank

I worked really hard to get 100% on this problem but I am stuck at 77.78%. Apparently only one person solved it in C++ so far.

My class BigNum isn't the fastest code. I used operator+= and operator*= instead of operator+ and operator* which helped a bit
but I still need to be about twice as fast for the two remaining test cases.

# 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 "2 100" | ./80

Output:

Note: the original problem's input 100 100 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>

// store single digits with lowest digits first
// e.g. 1024 is stored as { 4,2,0,1 }
// only non-negative numbers supported
struct BigNum : public std::vector<unsigned int>
{
// string conversion works only properly when MaxDigit is a power of 10
static const unsigned int MaxDigit = 1000000000;

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

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

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

while (carry > 0)
{
push_back(carry % MaxDigit);
carry /= MaxDigit;
}
}

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

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

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

if (carry > 0)
push_back(carry);
}

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

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

operator[](i) = diff;
}

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

// multiply a big number by an integer
void operator*=(unsigned int factor)
{
// nulled
if (factor == 0)
{
clear();
push_back(0);
return;
}
// unchanged
if (factor == 1)
return;

// append an empty block
if (factor == MaxDigit)
{
if (size() > 1 || operator[](0) > 0)
insert(begin(), 0);
return;
}

// multiply all blocks with the factor
unsigned long long carry = 0;
for (auto& i : *this)
{
carry += i * (unsigned long long)factor;
i      = carry % MaxDigit;
carry /= MaxDigit;
}
// store remaining carry in new digits
while (carry > 0)
{
push_back(carry % MaxDigit);
carry /= MaxDigit;
}
}

// 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;
result.reserve(size() + other.size());
for (int i = (int)other.size() - 1; i >= 0; i--)
{
BigNum temp = *this;
temp   *= other[i];

result *= MaxDigit;
result += temp;
}

return result;
}

// compare two big numbers
bool operator<(const BigNum& other) const
{
// different number of digits/buckets ?
if (size() < other.size())
return true;
if (size() > other.size())
return false;
// compare all digits/buckets, beginning with the most significant
for (int i = (int)size() - 1; i >= 0; i--)
{
if (operator[](i) < other[i])
return true;
if (operator[](i) > other[i])
return false;
}
return false;
}

// convert to string, MaxDigit must be power of 10
std::string toString() const
{
std::string result;
for (auto x : *this)
{
// process a bucket
for (unsigned int shift = 1; shift < MaxDigit; shift *= 10)
{
auto digit = (x / shift) % 10;
result.insert(0, 1, (char)digit + '0');
}
}

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

return result;
}
};

// find square root of x
// see www.afjarvis.staff.shef.ac.uk/maths/jarvisspec02.pdf
BigNum jarvis(unsigned int x, const BigNum& precision)
{
static const BigNum Fortyfive = 45;

BigNum a = x * 5;
BigNum b = 5;

// avoid re-allocations when growing (plus a few bytes when exceeding target)
a.reserve(precision.size());
b.reserve(precision.size());

// until we have enough digits
while (b < precision)
{
if (!(a < b)) // same as a >= b but currently there is no operator >= in my BigNum class
{
a -=   b;
b +=  10;
}
else
{
a *= 100;
b *=  10;
b -= Fortyfive;
}
}

return b;
}

// for reference only:
// Newton's method
//BigNum newton(const BigNum& square, BigNum estimate)
//{
//  while (true)
//  {
//    // minimize error
//    auto next = (square + estimate * estimate) / (estimate * 2);
//    if (next == estimate)
//      return estimate;
//    estimate = next;
//  }
//}

// return the sum of the first digits
unsigned int countDigits(const BigNum& x, unsigned int numDigits)
{
unsigned int sum = 0;
for (auto i : x.toString().substr(0, numDigits))
sum += i - '0';
return sum;
}

int main()
{
// all square roots from 2 to maxNumber
unsigned int maxNumber = 100;
// the first digits include the integer part, too (first six digits of sqrt(2) are 1,4,1,4,2,1)
unsigned int digits    = 100;
std::cin >> maxNumber >> digits;

// precompute precision
// a single 1 and then 100 zeros
// => since BigNum stores everything in reverse it is technically 100 zeros and then a single 1
// actually we need a slightly higher precision because square roots of composite numbers will be computed as
// the product of the square roots of their prime factors
const unsigned int ExtraDigits = 15; // maybe I can get away with less extra digits, haven't tried it
BigNum precision = 10;
for (unsigned int i = 1; i < digits + ExtraDigits; i++)
precision *= 10;

// all square roots (initially 0)
std::vector<BigNum> roots(maxNumber + 1, 0);

// digits(sqrt(2)) + digits(sqrt(3)) + ...
unsigned int sum = 0;
for (unsigned int i = 1; i <= maxNumber; i++)
{
// handle perfect squares
unsigned int intSqrt = 1;
while (intSqrt * intSqrt < i)
intSqrt++;
// found a perfect square ?
if (intSqrt * intSqrt == i)
{
// compute its root, too (needed for roots of composite numbers)
roots[i] = precision * intSqrt;
continue;
}

// try to number split into a product (only possible for composite numbers)
auto factor = intSqrt - 1;
// find a factor close to the square root
while (i % factor != 0)
factor--;

// composite ?
if (factor > 1)
{
// sqrt(composite) = sqrt(factor) * sqrt(i / factor)
auto& current = roots[i] = roots[i / factor] * roots[factor];

// remove digits beyond required precision
if (current.size() > roots[i - 1].size())
current.erase(current.begin(), current.begin() + (current.size() - roots[i - 1].size()));
while (current < roots[i - 1])
current *= 10;
}
else
{
// number is prime ? must find the square root the hard way ...
roots[i] = jarvis(i, precision);
}

// compute digit sum of sqrt(i)
sum += countDigits(roots[i], digits);
}

// print result
std::cout << sum << std::endl;

return 0;
}


This solution contains 42 empty lines, 66 comments and 2 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

March 14, 2017 submitted solution

# Hackerrank

My code solves 7 out of 9 test cases (score: 77.78%)

I failed 0 test cases due to wrong answers and 2 because of timeouts

# Difficulty

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

Hackerrank describes this problem as easy.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.

# 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.

 << problem 79 - Passcode derivation Path sum: two ways - problem 81 >>
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