<< problem 87 - Prime power triples Roman numerals - problem 89 >>

# Problem 88: Product-sum numbers

A natural number, N, that can be written as the sum and product of a given set of at least two natural numbers, {a1, a2, ... , ak}
is called a product-sum number: N = a_1 + a_2 + ... + a_k = a_1 * a_2 * ... * a_k.

For example, 6 = 1 + 2 + 3 = 1 * 2 * 3.

For a given set of size, k, we shall call the smallest N with this property a minimal product-sum number.
The minimal product-sum numbers for sets of size, k = 2, 3, 4, 5, and 6 are as follows.

k=2: 4 = 2 * 2 = 2 + 2
k=3: 6 = 1 * 2 * 3 = 1 + 2 + 3
k=4: 8 = 1 * 1 * 2 * 4 = 1 + 1 + 2 + 4
k=5: 8 = 1 * 1 * 2 * 2 * 2 = 1 + 1 + 2 + 2 + 2
k=6: 12 = 1 * 1 * 1 * 1 * 2 * 6 = 1 + 1 + 1 + 1 + 2 + 6

Hence for 2 <= k <= 6, the sum of all the minimal product-sum numbers is 4+6+8+12 = 30; note that 8 is only counted once in the sum.

In fact, as the complete set of minimal product-sum numbers for 2 <= k <= 12 is {4, 6, 8, 12, 15, 16}, the sum is 61.

What is the sum of all the minimal product-sum numbers for 2 <= k <= 12000?

# My Algorithm

My program is based on a Dynamic Programming approach, the core routine is getMinK:

• try all divisors i of a number n, remove them from the product and sum and call getMinK recursively: getMinK(n, product / i, sum - i)
• initially product = n and sum = n
• if the product is reduced to 1 and the sum is still bigger than 1 then remaining sum must be a sequence of 1s because 1 * product = product
• if product == sum then the last factor/summand was found, but we can't do that in the first iteration (because the product / sum need at least two terms)
• if I removed i, then I already tried all potential divisors smaller than i → keep track of i with the parameter minFactor

# 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 50 | ./88

Output:

Note: the original problem's input 50000000 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++ 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>

// initial values of the generated sequence:
// oeis.org/A104173

// max. number of factors / summands
const unsigned int Limit = 200000; // Project Euler: 12000
// [k] => [size of set]
std::vector<unsigned int> minK(Limit, 9999999); // initialize with an extremely high value

// found a better solution with k terms for number n ?
bool valid(unsigned int n, unsigned int k)
{
// too many terms ? (more than 12000)
if (k >= minK.size())
return false;

// found a smaller number n with the same number of terms k ?
if (minK[k] > n)
{
// note: the default value of minK[] is 9999999
minK[k] = n;
return true;
}

return false;
}

// return number of minimum k found (a number may be responsible for multiple minimum k, e.g. 8 => k=4 and k=5)
// n: the initial number
// product:   n divided by removed numbers
// sum:       n minus      removed numbers
// depth:     count        removed numbers
// minFactor: skip checking factors smalled than this
unsigned int getMinK(unsigned int n, unsigned int product, unsigned int sum,
unsigned int depth = 1, unsigned int minFactor = 2)
{
// already removed all factors > 1 ?
// => add a bunch of 1s to the sum
if (product == 1)
return valid(n, depth + sum - 1) ? 1 : 0;

unsigned int result = 0;
if (depth > 1)
{
// perfect match ?
if (product == sum)
return valid(n, depth) ? 1 : 0;

// try to finish sequence
if (valid(n, depth + sum - product))
result++;
}

// and now all remaining factors
for (unsigned int i = minFactor; i*i <= product; i++)
if (product % i == 0) // found a factor ? let's dig deeper ...
result += getMinK(n, product / i, sum - i, depth + 1, i);

return result;
}

int main()
{
unsigned int limit;
std::cin >> limit;
minK.resize(limit + 1);

// k(2) = 4
unsigned int n = 4;

// result
unsigned int sum = 0;

// 0 and 1 are not used, still 11999 to go ...
unsigned int todo = limit - 1;
while (todo > 0)
{
// analyze n
unsigned int found = getMinK(n, n, n);
// at least one new k(n) found ?
if (found > 0)
{
todo -= found;
sum  += n;
}

// next number
n++;
}

// print result
std::cout << sum << std::endl;
return 0;
}


This solution contains 16 empty lines, 26 comments and 2 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 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 19, 2017 submitted solution

# Hackerrank

My code solves 41 out of 41 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 40% (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 87 - Prime power triples Roman numerals - problem 89 >>
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