<< 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?

# 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

# My code

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

       #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.

# 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)

# Benchmark

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

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 never an option.

projecteuler.net/thread=88 - 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-88-minimal-product-sum-numbers/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p088.java (written by Nayuki)
Scala: github.com/samskivert/euler-scala/blob/master/Euler088.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.

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The 163 solved problems had an average difficulty of 22.2% at Project Euler and I scored 11,907 points (out of 13200) at Hackerrank's Project Euler+.
My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.
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