<< problem 158 - Exploring strings for which only one character ... Factorial trailing digits - problem 160 >>

# Problem 159: Digital root sums of factorisations

A composite number can be factored many different ways. For instance, not including multiplication by one, 24 can be factored in 7 distinct ways:
24 = 2 * 2 * 2 * 3
24 = 2 * 3 * 4
24 = 2 * 2 * 6
24 = 4 * 6
24 = 3 * 8
24 = 2 * 12
24 = 24

Recall that the digital root of a number, in base 10, is found by adding together the digits of that number,
and repeating that process until a number is arrived at that is less than 10. Thus the digital root of 467 is 8.

We shall call a Digital Root Sum (DRS) the sum of the digital roots of the individual factors of our number.
The chart below demonstrates all of the DRS values for 24.

FactorisationDigital Root Sum
2x2x2x39
2x3x49
2x2x610
4x610
3x811
2x125
246

The maximum Digital Root Sum of 24 is 11.
The function mdrs(n) gives the maximum Digital Root Sum of n. So mdrs(24)=11.
Find sum{mdrs(n)} for 1 < n < 1000000.

# My Algorithm

The problem statement explains that mdrs(a * b) = mdrs(a) + mdrs(b).

I needed a bit to realize that if x = a * b * c then mdrs(a * b * c) = max(mdrs(a) + mdrs(b * c), mdrs(a * b) + mdrs(c))

That means that mdrs(2 * 2 * 6) = max(mdrs(2) + mdrs(12), mdrs(4) + mdrs(6)) = max(5, 10) = 10.

Right at the start, my program fills the mdrs container with the "non-factorized" digitRoot(x) (same as the last line in the problem statement's table).

Then for each a I find all its multiples a * b and update mdrs(a * b) if mdrs(a * b) < mdrs(a) + mdrs(b).

The sum of all relevant mdrs is a simple loop.

## Modifications by HackerRank

The program must be able to handle tons of test cases. That's why I precompute all results in step 1 and then simply look them up in step 2.
And there's a tiny difference in the way the upper limit is handled: the original problem excludes it while Hackerrank includes it.

## Note

After I finished my program I read the Wikipedia page en.wikipedia.org/wiki/Digital_root and saw that a much simpler, faster,
non-iterative and/or recursive formula for digitRoot(x) is possible: result = ((x - 1) % 9) + 1

# Interactive test

This feature is not available for the current problem.

# 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.

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 <vector>
#include <algorithm>

//#define ORIGINAL

// find digit root of a number
unsigned int digitRoot(unsigned int x)
{
// taken from https://en.wikipedia.org/wiki/Digital_root
//return ((x - 1) % 9) + 1;

// and my original code:
unsigned int result = 0;
while (x > 0)
{
result += x % 10;
x      /= 10;
}

// repeat again ?
if (result >= 10)
result = digitRoot(result);

return result;
}

int main()
{
unsigned int limit = 1000000;

#ifndef ORIGINAL
limit = 1;
unsigned int tests = 1;
std::cin >> tests;
std::vector<unsigned int> input(tests);
for (auto& x : input)
{
std::cin >> x;
if (limit < x)
limit = x;
}
#endif

// step 1: precompute sum(mdrs) for every x

std::vector<unsigned char> mdrs(limit + 1, 0);

// digit root without any factorization
for (unsigned int i = 2; i <= limit; i++)
mdrs[i] = digitRoot(i);

// for each number a ...
for (unsigned int a = 2; a <= limit; a++)
// ... adjust all its multiples a*b
for (unsigned int b = 2; a * b <= limit && b <= a; b++)
// improved ?
if (mdrs[a * b] < mdrs[a] + mdrs[b])
mdrs[a * b] = mdrs[a] + mdrs[b];

// step 2: display result(s)

#ifdef ORIGINAL
// sum of mdrs(2..limit-1)
unsigned int sum = 0;
for (unsigned int i = 2; i < limit; i++)
sum += mdrs[i];

std::cout << sum << std::endl;
#else
// sum of mdrs(2..i) for every 2 <= i <= limit
// note: unlike the original problem, it is including limit
std::vector<unsigned int> sums(mdrs.size(), 0);
for (unsigned int i = 2; i < sums.size(); i++)
sums[i] = sums[i-1] + mdrs[i];

// display results
for (auto x : input)
std::cout << sums[x] << std::endl;
#endif

return 0;
}


This solution contains 16 empty lines, 17 comments and 8 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.05 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

June 22, 2017 submitted solution

# Hackerrank

My code solves 11 out of 11 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 60% (out of 100%).

Hackerrank describes this problem as medium.

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 158 - Exploring strings for which only one character ... Factorial trailing digits - problem 160 >>
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