<< problem 60 - Prime pair sets Cubic permutations - problem 62 >>

# Problem 61: Cyclical figurate numbers

Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are all figurate (polygonal) numbers and are generated by the following formulae:

TriangleP_{3,n}=n(n+1)/21, 3, 6, 10, 15, ...
SquareP_{4,n}=n^21, 4, 9, 16, 25, ...
PentagonalP_{5,n}=n(3n-1)/21, 5, 12, 22, 35, ...
HexagonalP_{6,n}=n(2n-1)1, 6, 15, 28, 45, ...
HeptagonalP_{7,n}=n(5n-3)/21, 7, 18, 34, 55, ...
OctagonalP_{8,n}=n(3n-2)1, 8, 21, 40, 65, ...

The ordered set of three 4-digit numbers: 8128, 2882, 8281, has three interesting properties.

The set is cyclic, in that the last two digits of each number is the first two digits of the next number (including the last number with the first).
Each polygonal type: triangle (P_{3,127}=8128), square (P_{4,91}=8281), and pentagonal (P_{5,44}=2882), is represented by a different number in the set.
This is the only set of 4-digit numbers with this property.

Find the sum of the only ordered set of six cyclic 4-digit numbers for which each polygonal type: triangle, square, pentagonal, hexagonal, heptagonal, and octagonal,
is represented by a different number in the set.

# My Algorithm

The container all contains a simple bitmask for each number x:

• if the 3rd bit is set, then x is a triangle number
• if the 4th bit is set, then x is a square number
• if the 5th bit is set, then x is a pentagonal number
• if the 6th bit is set, then x is a hexagonal number
• if the 7th bit is set, then x is a heptagonal number
• if the 8th bit is set, then x is a octagonal number
Function deeper recursively builds a sequence of number, where each number's highest two digits are equal to its predecessor's lowest two digits.
Even more, all numbers represent differetn categories (triangle, square, pentagonal, ...).

Some numbers belong to multiple categories. In such a case, all combinations have to be tried.
A number must not occur twice in a sequence. And finally, the last number's lowest two digits have to match the first number's highest two digits, too.

# 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 "3 3 4 5" | ./61

Output:

(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. Or just jump to my GitHub repository.

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

// all sums of valid sequences
std::set<unsigned int> results;

// 4 digits => all numbers must be below 10^4 = 10000
const unsigned int Limit = 10000;
// all number sets per number as a bitmask
std::vector<unsigned int> all(Limit, 0);

// bit mask of all categories (e.g. 1<<3 for triangle numbers, 1<<4 for square numbers, ...)

// add a category to a number
void add(unsigned int x, unsigned int category)
{
// reject if not exactly 4 digits
if (x < 1000 || x >= 10000)
return;

// set one bit
auto bit = 1 << category;
}

void deeper(std::vector<unsigned int>& sequence, unsigned int mask = 0)
{
// all four digit-numbers
unsigned int from =  1000;
unsigned int to   = 10000;

if (!sequence.empty())
{
// we only have to look at those numbers where the highest two digits match
// the previous numbers lowest two digits
auto lowerTwoDigits = sequence.back() % 100;
from = lowerTwoDigits * 100;
to   = from + 100;
}

// try all relevant numbers
for (auto next = from; next < to; next++)
{
auto categories = all[next];
// not a member of any relevant set ?
if (categories == 0)
continue;

// must not use the same number twice
bool isUnique = true;
for (auto x : sequence)
if (x == next)
{
isUnique = false;
break;
}
if (!isUnique)
continue;

// extract all categories it belongs to
for (auto j = 3; j <= 8; j++)
{
auto thisCategory = 1 << j;
// not a member of that category ?
if ((categories & thisCategory) == 0)
continue;

// must be a new category we haven't seen yet in the current sequence
if ((mask & thisCategory) != 0)
continue;

// we have all categories ?
{
// must compare against first number, too
auto first = sequence.front();

auto lowerTwoDigits = next  % 100;
auto upperTwoDigits = first / 100;

if (lowerTwoDigits == upperTwoDigits)
{
// we found a result, store its sum
auto sum = next;
for (auto x : sequence)
sum += x;
results.insert(sum);
}
}
else
{
// go deeper
sequence.push_back(next);
sequence.pop_back();
}
}
}
}

int main()
{
// only some sets are relevant to current problem
unsigned int numSets;
std::cin >> numSets;
for (unsigned int i = 0; i < numSets; i++)
{
unsigned int x;
std::cin >> x;
}

// build sets
unsigned int n = 1;
while (true)
{
auto triangle = n * (n + 1) / 2;
// triangle numbers grow the slowest, once we have all of those then we are done
if (triangle >= 10000)
break;

auto square   = n * n;

auto pentagon = n * (3 * n - 1) / 2;

auto hexagon  = n * (2 * n - 1);

auto heptagon = n * (5 * n - 3) / 2;

auto octagon  = n * (3 * n - 2);

n++;
}

// start search with an empty sequence
std::vector<unsigned int> sequence;
deeper(sequence);

// print all results in ascending order
for (auto x : results)
std::cout << x << std::endl;

return 0;
}


This solution contains 26 empty lines, 32 comments and 3 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 6, 2017 submitted solution

# Hackerrank

My code solves 14 out of 14 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 20% (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 60 - Prime pair sets Cubic permutations - problem 62 >>
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