<< problem 67 - Maximum path sum II Totient maximum - problem 69 >>

# Problem 68: Magic 5-gon ring

Consider the following "magic" 3-gon ring, filled with the numbers 1 to 6, and each line adding to nine.

Working clockwise, and starting from the group of three with the numerically lowest external node (4,3,2 in this example),
each solution can be described uniquely. For example, the above solution can be described by the set: 4,3,2; 6,2,1; 5,1,3.

It is possible to complete the ring with four different totals: 9, 10, 11, and 12. There are eight solutions in total.
TotalSolution Set
94,2,3; 5,3,1; 6,1,2
94,3,2; 6,2,1; 5,1,3
102,3,5; 4,5,1; 6,1,3
102,5,3; 6,3,1; 4,1,5
111,4,6; 3,6,2; 5,2,4
111,6,4; 5,4,2; 3,2,6
121,5,6; 2,6,4; 3,4,5
121,6,5; 3,5,4; 2,4,6

By concatenating each group it is possible to form 9-digit strings; the maximum string for a 3-gon ring is 432621513.

Using the numbers 1 to 10, and depending on arrangements, it is possible to form 16- and 17-digit strings. What is the maximum 16-digit string for a "magic" 5-gon ring?

# My Algorithm

I split the graph into two parts:

• an "inner ring"
• and an "outer ring"
In the initial example, the inner ring consists of 1,2,3 while the outer ring contains 4,5,6.

Each cell of the inner ring is part of two lines. It is indexed from 0 to size - 1 (size is 3 for a 3-gon, 5 for a 5-gon, etc.).
Each cell of the outer ring is part of only one line. It is indexed from size to 2*size - 1.

The idea is to fill the inner ring using backtracking.
Whenever two out of three cells of a line are known, the third can be directly computed.
If that third number is already used, then we have to backtrack.

My function fillLine fills the inner ring: it tries all combination of available numbers for the cell at index pos.
Whenever it succeeds, it calls itself (recursion) to fill pos + 1.

All available numbers are stored in a bitmask called used. Number x is available if (used & (1 << x)) == 0.
At the end, result contains all valid n-gons.

Project Euler asks for the maximum string while Hackerrank asks for all strings in ascending order.
My solution prints the Hackerrank result and I manually have to choose the larger one (there are just two solutions).

The correct result will be printed when size = 5 and tripletSum = 14, that means the program input is "5 14".
There are solutions for "5 16", "5 17" and "5 19" but they are either lexicographically smaller or have 17 instead of 16 digits.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

This live test is based on the Hackerrank problem.

Input data (separated by spaces or newlines):

This is equivalent to
echo "3 9" | ./68

Output:

Note: the original problem's input 5 14 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).

       #include <iostream>
#include <vector>
#include <string>
#include <set>

// sorted container of valid n-gon strings
std::set<std::string> result;
// size of the n-gon (3,4,5)
unsigned int size;
// sum of all three cells alongs a diagonal/edge
unsigned int tripletSum;

void fillLine(unsigned int pos, std::vector<unsigned int> inner, std::vector<unsigned int> outer, unsigned int used)
{
// inner ring completely filled, just one cell of the outer ring left
if (pos == size - 1)
{
// check last line
outer[size - 1] = tripletSum - (inner[0] + inner[size - 1]);
unsigned int mask = 1 << outer[size - 1];
if ((used & mask) != 0)
return;

// first element of outer ring must be the smallest
for (auto x : outer)
if (x < outer[0])
return;

// build string
std::string id;
for (unsigned int i = 0; i < size; i++)
id += std::to_string(outer[i]) + std::to_string(inner[i]) + std::to_string(inner[(i + 1) % size]);

// will be alphabetically ordered
result.insert(id);
return;
}

// move a number between 1 and 2*size into one of the inner cells of the n-gon
for (unsigned int i = 1; i <= 2*size; i++)
{
// fill a cell of the inner ring
unsigned int innerMask = 1 << i;
// is that number still available ?
if ((innerMask & used) != 0)
continue;

// occupy cell
inner[pos + 1] = i;
unsigned int nextUsed = used | innerMask;

// compute the according cell in the outer ring
outer[pos] = tripletSum - (inner[pos] + i);
unsigned int outerMask = 1 << outer[pos];
// is that number still available ?
if ((nextUsed & outerMask) != 0)
continue;

// next line
fillLine(pos + 1, inner, outer, nextUsed);
}
}

int main()
{
std::cin >> size >> tripletSum;

// generate the inner and outer ring
std::vector<unsigned int> inner(size);
std::vector<unsigned int> outer(size);
// a triplet consists of inner[a], inner[(a+1) % (2*a)], outer[a]

// generate a bitmask of allowed numbers (0 = still available, 1 = already used / disallowed)
unsigned int allowed = 0;
for (unsigned int i = 1; i <= 2 * size; i++)
allowed |= 1 << i;
allowed = ~allowed;

// fill first cell of inner ring
for (unsigned int i = 1; i <= 2*size; i++)
{
inner[0] = i;
// fill remaining cells
fillLine(0, inner, outer, allowed | (1 << i));
}

//#define ORIGINAL
#ifdef ORIGINAL
std::cout << *result.rbegin() << std::endl;
#else
for (auto r : result)
std::cout << r << std::endl;
#endif
}


This solution contains 14 empty lines, 21 comments and 7 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in less than 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 13, 2017 submitted solution

# Hackerrank

My code solves 22 out of 22 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 25% (out of 100%).

Hackerrank describes this problem as hard.

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 the flashing problem is the one I solved most recently

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The 289 solved problems (level 11) had an average difficulty of 32.1% at Project Euler and
I scored 13,486 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 67 - Maximum path sum II Totient maximum - problem 69 >>
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