<< problem 15 - Lattice paths Number letter counts - problem 17 >>

# Problem 16: Power digit sum

2^{15} = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.

What is the sum of the digits of the number 2^{1000}?

# Algorithm

These two equations describe the iterative computation of 2^x:
2^0 = 1 and 2^x = 2 * 2^{x-1}

These numbers grow pretty big and for x=64 exceed the range of unsigned long long.
That's why I store all decimal Digits in a std::vector, where the lowest index contains the least significant digits.

For x=0 - which represents 2^0 = 1 - this Digits container's elements are { 1 }
For x=15 - which represents 2^{15} = 1 - this Digits container's elements are { 8, 6, 7, 2, 3 }

Multiplying Digits by 2 follows the same rules as basic multiplication taught in school:
1. multiply each digit by 2, start at the lowest digit
2. if digit * 2 >= 10 then an overflow occurred: carry over (digit * 2) div 10 to the next digit and keep (digit * 2) mod 10.
Note: we must carry over at most 1 and instead of a computationally expensive modulo we can subtract 10 which is faster

## Alternative Approaches

Exponentation by squaring saves many steps:
The result of 2^1024 can be found in just 10 steps instead of 1024.

## Modifications by HackerRank

To avoid timeouts, I store all results (even intermediate steps) in a cache and re-use as much as possible from this cache.

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

       #include <vector>
#include <iostream>

// store single digits in an array, lowest digit come first
typedef std::vector<unsigned int> Digits;

int main()
{
// memoize powers of two
std::vector<Digits> cache;
cache.push_back({ 1 });

unsigned int tests;
std::cin >> tests;
while (tests--)
{
unsigned int exponent;
std::cin >> exponent;

// and compute the remaining exponents
for (unsigned int current = cache.size(); current <= exponent; current++)
{
auto power = cache.back();
unsigned int carry = 0;
for (auto& i : power)
{
// times two ...
i = 2 * i + carry;

// handle overflow
if (i >= 10)
{
i -= 10;
carry = 1;
}
else
{
carry = 0;
}
}

// still some carry left ?
if (carry != 0)
power.push_back(carry);

// memoize result
cache.push_back(power);
}

// sum of all digits
unsigned int sum = 0;
for (auto i : cache[exponent])
sum += i;
std::cout << sum << std::endl;
}

return 0;
}


This solution contains 9 empty lines, 9 comments and 2 preprocessor commands.

# Interactive test

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

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 15" | ./16

Output:

Note: the original problem's input 1000 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)

# 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

February 23, 2017 submitted solution

# Hackerrank

My code solves 10 out of 10 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 5% (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.

# Similar problems at Project Euler

Problem 20: Factorial digit sum

Note: I'm not even close to solving all problems at Project Euler. Chances are that similar problems do exist and I just haven't looked at them.

projecteuler.net/thread=16 - the best forum on the subject (note: you have to submit the correct solution first)

Code in various languages:

Python: www.mathblog.dk/project-euler-16/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p016.java (written by Nayuki)
Mathematica: github.com/nayuki/Project-Euler-solutions/blob/master/mathematica/p016.mathematica (written by Nayuki)
C: github.com/eagletmt/project-euler-c/blob/master/10-19/problem16.c (written by eagletmt)
Go: github.com/frrad/project-euler/blob/master/golang/Problem016.go (written by Frederick Robinson)
Javascript: github.com/dsernst/ProjectEuler/blob/master/16 Power digit sum.js (written by David Ernst)
Scala: github.com/samskivert/euler-scala/blob/master/Euler016.scala (written by Michael Bayne)

# Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.
yellow problems score less than 100% at Hackerrank (but still solve the original problem).
gray problems are already solved but I haven't published my solution yet.
blue problems are solved and there wasn't a Hackerrank version of it at the time I solved it or I didn't care about it because it differed too much.
red problems are solved but exceed the time limit of one minute.

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

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The 206 solved problems had an average difficulty of 27.5% at Project Euler and
I scored 12,626 points (out of 14300 possible points, top rank was 20 out ouf ≈60000 in July 2017) at Hackerrank's Project Euler+.
Look at my progress and performance pages to get more details.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

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