<< problem 118 - Pandigital prime sets Square remainders - problem 120 >>

# Problem 119: Digit power sum

The number 512 is interesting because it is equal to the sum of its digits raised to some power: 5 + 1 + 2 = 8, and 8^3 = 512.
Another example of a number with this property is 614656 = 28^4.

We shall define a_n to be the nth term of this sequence and insist that a number must contain at least two digits to have a sum.

You are given that a_2 = 512 and a_10 = 614656.

Find a_30.

# My Algorithm

A simple but extremely slow brute-force program revealed that a_30 has more than 10 digits:
I was iterating through numbers from 10 to 10,000,000,000 and didn't find enough terms a_i:
81, 512, 2401, 4913, 5832, 17576, 19683, 234256, 390625, 614656, 1679616, 17210368, 34012224, 52521875, 60466176, 205962976, 612220032, 8303765625, ...

Then I turned the whole problem upside-down: there are 90 two-digit numbers but only 18 different digit sums.
Even better: there are 9000 four-digit numbers but only 36 different digit sums !
Actually each digit is always between 0 and 9 therefore the maximum digit sum is sum_i{digit_i} = 9i.

My program looks at all potential digit sums from 1 to 9*20 (I thought that maybe a_30 < 10^20, and that's actually true).
For each potential digit sum, I call them base in my code, all exponents are computed until base^{exponent} overflows.
If digitSum(base^{exponent}) = base then a solution has been found.

Solutions will be found in a "random" order, so I add them to a sorted std::set and read its 30th element.

## Modifications by HackerRank

The search space is much bigger: all numbers up to 10^100 (that's a Googol) have to be printed.
Even worse, numbers can be encoded in any base from 2 to 1000.

My BigNum class was already able to store its digits in arbirary bases - but I always had powers of 10 in my mind.
Addition and multiplication is reduced to in-place operations to avoid slow memory allocations.
Pretty simple but effective is BigNum::convert which converts a BigNum to a different base.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

This live test is based on the Hackerrank problem.

Input data (separated by spaces or newlines):
Note: Enter a base, e.g. 2 stands for the binary number system

This is equivalent to
echo 2 | ./119

Output:

Note: the original problem's input 10 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.

The code contains #ifdefs to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL to produce the result for the original problem (default setting for most problems).

       #include <iostream>
#include <set>
#include <climits>

//#define ORIGINAL
#ifdef ORIGINAL

// add all digits (base 10)
unsigned int digitSum(unsigned long long x)
{
unsigned int result = 0;
while (x > 0)
{
result += x % 10;
x      /= 10;
}
return result;
}

int main()
{
std::set<unsigned long long> solutions;

// based on my experiments with BigNum I knew that the result fits in 64 bit
// which is about 10^20, each digit <= 9, therefore their sum <= 20*9
unsigned int MaxBase = 20 * 9;
// base is the sum of all digits
for (unsigned int base = 2; base <= MaxBase; base++)
{
unsigned int exponent = 1;
// step-by-step: base^exponent, then base^(exponent+1), ...
unsigned long long current = base;

while (current < ULLONG_MAX / base) // 1^64 - 1
{
auto sum = digitSum(current);
// digit sum equals base ? (note: single-digit numbers excluded)
if (sum == base && current >= 10)
solutions.insert(current);

// next iteration
current *= base;
exponent++;
}
}

// sorted results
auto i = solutions.begin();
// skip 30 - 1 = 29 elements
// print the 30th
std::cout << *i << std::endl;
return 0;
}

#else

#include <vector>

// store single digits with lowest digits first
// e.g. 1024 is stored as { 4,2,0,1 } (when base = 10)
struct BigNum : public std::vector<unsigned int>
{
unsigned int maxDigit;

// store a non-negative number
BigNum(unsigned long long x = 0, unsigned int base = 10)
: maxDigit(base)
{
do
{
push_back(x % maxDigit);
x /= maxDigit;
} while (x > 0);
}

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 (unsigned int 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);
}

// multiply a big number by an integer
void operator*=(unsigned int factor)
{
unsigned long long carry = 0;
for (size_t i = 0; i < size(); i++)
{
carry += operator[](i) * (unsigned long long)factor;
operator[](i) = carry % maxDigit;
carry /= maxDigit;
}
// store remaining carry in new digits
while (carry > 0)
{
push_back(carry % maxDigit);
carry /= maxDigit;
}
}

// compare two big numbers
bool operator<(const BigNum& other) const
{
if (size() < other.size())
return true;
if (size() > other.size())
return false;
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 a different radix
{
for (auto i = rbegin(); i != rend(); i++)
{
result *= maxDigit;
result += *i;
}
return result;
}
};

unsigned int digitSum(const BigNum& x)
{
unsigned int result = 0;
for (auto digit : x)
result += digit;
return result;
}

int main()
{
std::set<BigNum> solutions;

// do not exceed 10^100
BigNum googol(1, 10);
for (unsigned int digits = 1; digits <= 100; digits++)
googol *= 10;

// analyze all base^exponent below 10^100
// base is the sum of digits
for (unsigned int base = 2; base < (radix-1)*max.size(); base++)
{
unsigned int exponent = 1;
// step-by-step: base^exponent, then base^(exponent+1), ...
// still below a googol
while (current < max)
{
auto sum = digitSum(current);
// digit sum equals base ? single-digit numbers excluded
if (sum == base && current.size() >= 2)
{
// store result in decimal notation
BigNum decimal = current.convert(10);
solutions.insert(decimal);
}

// next iteration
current *= base;
exponent++;
}
}

// sorted results
for (auto i : solutions)
{
// print digits (note: they are stored in reverse)
for (auto j = i.rbegin(); j != i.rend(); j++)
std::cout << *j;
std::cout << " ";
}

return 0;
}
#endif


This solution contains 27 empty lines, 34 comments and 7 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

May 13, 2017 submitted solution

# Hackerrank

My code solves 20 out of 20 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 30% (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
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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 118 - Pandigital prime sets Square remainders - problem 120 >>
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