Problem 16: Power digit sum

(see projecteuler.net/problem=16)

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;
// add 2^0 = 1
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:

(please click 'Go !')

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
March 31, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler016

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.

Links

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)
Haskell: github.com/nayuki/Project-Euler-solutions/blob/master/haskell/p016.hs (written by Nayuki)
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.

more about me can be found on my homepage, especially in my coding blog.
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