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

# Problem 88: Product-sum numbers

(see projecteuler.net/problem=88)

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`

I had to add a few optimizations:

- 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:

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

May 5, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler088

My code solved **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.

# Links

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