<< problem 102 - Triangle containment Pandigital Fibonacci ends - problem 104 >>

# Problem 103: Special subset sums: optimum

Let S(A) represent the sum of elements in set A of size n. We shall call it a special sum set if for any two non-empty disjoint subsets,
B and C, the following properties are true:

i. S(B) != S(C); that is, sums of subsets cannot be equal.
ii. If B contains more elements than C then S(B) > S(C).

If S(A) is minimised for a given n, we shall call it an optimum special sum set.
The first five optimum special sum sets are given below.

n = 1: { 1 }
n = 2: { 1, 2 }
n = 3: { 2, 3, 4 }
n = 4: { 3, 5, 6, 7 }
n = 5: { 6, 9, 11, 12, 13 }

It seems that for a given optimum set, A = {a_1, a_2, ... , a_n}, the next optimum set is of the form B = {b, a_1+b, a_2+b, ... ,a_n+b}, where b is the "middle" element on the previous row.

By applying this "rule" we would expect the optimum set for n = 6 to be A = { 11, 17, 20, 22, 23, 24 }, with S(A) = 117.
However, this is not the optimum set, as we have merely applied an algorithm to provide a near optimum set.
The optimum set for n = 6 is A = { 11, 18, 19, 20, 22, 25 }, with S(A) = 115 and corresponding set string: 111819202225.

Given that A is an optimum special sum set for n = 7, find its set string.

NOTE: This problem is related to Problem 105 and Problem 106.

# My Algorithm

search recursively produces all ascending sequences where all elements are between minElement and maxElement.

check returns true if the set is "special" by analyzing all subsets:
each number is either part or not part of a subset. That can be encoding by a single bit which is 0 (no) or 1 (yes).
Then there are 2^size combinations → I just run a counter named mask from 0000000 to 1111111
(actually, 0000000 represents the empty set and can be skipped).

For each sum I encounter along the way, I set a bit in sums. It must never be set twice (first rule).
Moreover, I track the lower and highest sum for each number of elements.
If the highest sum of all n-element subsets is higher than then lowest sum of (n+1)-subsets then rule 2 was violated.

All successfully verified sequences are stored in an ordered set solution and the first one is printed.

## Modifications by HackerRank

My simple brute-force approach produces only time-outs.
There is a certain generating function I haven't fully figured out. All I see is that x[2n+1] = x[2n] and x[2m] = x[2m-1] - ?.

# 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):
Note: Enter the size of the set

This is equivalent to
echo 6 | ./103

Output:

Note: the original problem's input 7 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++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

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).

       #define ORIGINAL
#ifdef  ORIGINAL

#include <iostream>
#include <vector>
#include <map>

typedef std::vector<unsigned int> Sequence;
std::map<unsigned int, Sequence> solutions;

// number of elements in full set
unsigned int finalSize;
// lower limit for the smallest element
unsigned int minElement;
// upper limit for the biggest element
unsigned int maxElement;

// return true if sequence is special
bool check(const Sequence& sequence)
{
// sum of all elements
unsigned int fullSum = 0;
for (auto x : sequence)
fullSum += x;

// mark each generated sum as true, no collisions allowed
std::vector<bool> sums(fullSum + 1, false);

// track the lowest and highest sum for each subset size
std::vector<unsigned int> maxSum(sequence.size() + 1, 0);
std::vector<unsigned int> minSum(sequence.size() + 1, fullSum + 1);
minSum[0] = maxSum[0] = 0; // empty set

unsigned int fullMask = (1 << sequence.size()) - 1;

// 2^elements iterations (actually, I ignore the empty set)
{
unsigned int sum  = 0;
unsigned int size = 0;
for (unsigned int element = 0; element < sequence.size(); element++)
{
// use that element ?
unsigned int bit = 1 << element;
if ((mask & bit) == 0)
continue;

sum += sequence[element];
// count subset size
size++;
}

// two subsets share the same sum ?
if (sums[sum])
return false;
sums[sum] = true;

// adjust lowest and highest sum of current subset
if (minSum[size] > sum)
minSum[size] = sum;
if (maxSum[size] < sum)
maxSum[size] = sum;
}

// make sure that no set will fewer elements has a higher sum
for (size_t i = 1; i < sequence.size(); i++)
if (maxSum[i] > minSum[i + 1])
return false;

// yes, have another solution
solutions[fullSum] = sequence;
return true;
}

// enumerate all sequences where each number is between minElement and maxElement and a_i > a_j if i > j
void search(Sequence& sequence) // manipulate in-place
{
// enough elements ? check whether it is "special"
if (sequence.size() == finalSize)
{
check(sequence);
return;
}

// all numbers which are bigger than the predecessor
auto last = sequence.empty() ? minElement - 1 : sequence.back(); // minus 1 because last always adds 1
for (unsigned int i = last + 1; i <= maxElement; i++)
{
sequence.push_back(i);

// go deeper ...
search(sequence);

sequence.pop_back();
}
}

int main()
{
std::cin >> finalSize;

// crude heuristics for the lower and upper limits
minElement = 1;
maxElement = 10;
if (finalSize >= 5)
{
maxElement = finalSize * finalSize;
minElement = maxElement / 4;
}

Sequence empty;
search(empty);

// n = 1: {  1 }
// n = 2: {  1,  2 }
// n = 3: {  2,  3,  4 }
// n = 4: {  3,  5,  6,  7 }
// n = 5: {  6,  9, 11, 12, 13 }
// n = 6: { 11, 18, 19, 20, 22, 25 }
for (auto s : solutions)
{
for (auto x : s.second)
std::cout << x;
break;
}

return 0;
}

#else

// heavily modified Hackerrank problem

#include <iostream>
#include <vector>

typedef std::vector<unsigned int> Sequence;

// apply near-optimal algorithm to increase sequence by one element
// my algorithm builds the sequence in reverse
void next(Sequence& sequence) // modify in-place
{
auto middle = sequence[(sequence.size() - 1) / 2];
for (auto& x : sequence)
x = (x + middle) % 1000000007;
sequence.push_back(middle);
}

int main()
{
unsigned int length;
std::cin >> length;

Sequence sequence = { 1 };
sequence.reserve(length);
for (unsigned int i = 2; i <= length; i++)
next(sequence);

for (auto i = sequence.rbegin(); i != sequence.rend(); i++)
std::cout << *i << " ";
std::cout << std::endl;
return 0;
}

#endif


This solution contains 31 empty lines, 29 comments and 9 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 4.3 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

May 16, 2017 submitted solution

# Hackerrank

My code solves 1 out of 31 test cases (score: 0%)

I failed 0 test cases due to wrong answers and 30 because of timeouts

# Difficulty

Project Euler ranks this problem at 45% (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 102 - Triangle containment Pandigital Fibonacci ends - problem 104 >>
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