<< problem 235 - An Arithmetic Geometric sequence Twenty-two Foolish Primes - problem 239 >>

# Problem 237: Tours on a 4 x n playing board

Let T(n) be the number of tours over a 4 * n playing board such that:

• The tour starts in the top left corner.
• The tour consists of moves that are up, down, left, or right one square.
• The tour visits each square exactly once.
• The tour ends in the bottom left corner.
The diagram shows one tour over a 4 * 10 board:

T(10) is 2329. What is T(10^12) mod 10^8?

# My Algorithm

A nice problem that you can easily understand within a few seconds. But it took a few days to come up with a solution ...

Of course I immediately wrote a bruteForce algorithm and it solves the T(10) case. Anything beyond that is impossible.

The main realization was to split the whole board in its columns.
I identified 15 different columns (A - O) which can have a unique "flow" on their left and right border.
The term "flow" means the chronological way how a piece moves across the board.

The arrows symbolize the "flow" in and out of a column while hash signs stand for "no border crossing":

ABCDE
→ →→ →→ →→ →→ ##
← ←← #### ←← ←← ##
→ →→ #### →→ ##→ →
← ←← ←← ←← ##← ←

FGHIJ
→ →## →→ #### →→ →
← ←## ←## →→ ##← ##
## →→ →## ←← #### ##
## ←← ←← #### ←## ←

KLMNO
→ ##→ →## →→ ##→ ##
## #### ←## ##← #### ##
## →## ##→ ##→ #### ##
← ←← ##← ←← ##← ##

Columns N and O can only be found at the right edge of the board.

Unfortunately it's not sufficient to represent the flow by arrows because they are still ambigious:
there are three different chronological orders how the "flow" can pass through column A.

If I look at each column's left and right border then there are just 6 patterns for these borders.
With a proper labelling of the flow's chronological order (indicated by 1,2,3,4 and a hash means "no crossing") I get 8 different borders:

111##13#
#221#42#
##32131#
2#4#224#

The function fill() stores the borders of each column, e.g. column C is
neighbors.insert( { "1##2", "1234" } );
Trust me, getting all this stuff right was a lot of work: I made tons of mistakes !

I wrote two algorithms: a simple one that verifies T(10) and a much faster one to solve T(10^12).
slow() linearly goes through all borders that are allowed on the right side of the current border and stops if it reaches the right side of the board.
Assuming that there are about 3 borders that are compatible in such a way, the routine analyzes 3^width combinations.
There's no way it can solve T(10^12) - but I really needed this algorithm to get my borders right.
When the output finally matched the results of bruteForce I went on to write a faster (and more complex) algorithm.

fast() is a divide-and-conquer approach:

• I treat a group of columns as a blackbox where I only knows its left and right border
• if I cut through this blackbox at an arbitrary point then any of the 8 borders could be found
• well, that's not quite right, since the 8th border is reserved for the right-most border of the board → only 7 borders "inside" the blackbox
• then the number of combinations of a blackbox is the product of its left and right half
• if I keep doing this until the blackbox contains only a single column then I check whether this type of column is valid
This isn't much faster than what slow() does ... but when the blackbox becomes smaller, I process the same kinds of blackboxes over and over again.
Thus memoization drastically reduced the number of different blackboxes. At the end, cache contains 3417 values.

## Alternative Approaches

I was blown away by the simple solutions found by others: they discovered a relationship between T(n) and T(n-1), ..., T(n-4).
Incredible stuff - or maybe just looked up in OEIS A181688.

## Note

Even though the result is found within about 0.02 seconds, I felt that dividing each blackbox in the middle isn't optimal:
if I try to divide the blackbox in such a way that at least one half's size is a power of two (that means 2^i) then cache contains only 2031 values.
Moreover, the program runs about 50% faster.

Replacing the std::string by plain integers would be still faster but I think it would be much harder to understand the code.

# 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 10 | ./237

Output:

Note: the original problem's input 1000000000000 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. Or just jump to my GitHub repository.

       #include <iostream>
#include <set>
#include <map>
#include <vector>
#include <tuple>

// assuming that the route could exceed the left border I get these states for the left-most and right-most borders:
typedef std::string Border;
const Border LeftBorder  = "1##2";
const Border RightBorder = "####";

// all possible borders that can be found inside the grid
std::set<Border> borders;
// define which borders can be next to each other
std::set<std::pair<Border, Border>> neighbors;

// set up the containers "borders" and "neighbors"
void fill()
{
// the following lines are derived from my drawings above
//                   left    right      column
neighbors.insert( { "1234", "1234" } ); // A
neighbors.insert( { "1432", "1432" } ); // A
neighbors.insert( { "3214", "3214" } ); // A
neighbors.insert( { "1432", "1##2" } ); // B
neighbors.insert( { "3214", "1##2" } ); // B
neighbors.insert( { "1##2", "1234" } ); // C
neighbors.insert( { "1234", "12##" } ); // D
neighbors.insert( { "1234", "##12" } ); // E
neighbors.insert( { "12##", "1432" } ); // F
neighbors.insert( { "##12", "3214" } ); // G
neighbors.insert( { "1##2", "#12#" } ); // H
neighbors.insert( { "#12#", "1##2" } ); // I
neighbors.insert( { "12##", "1##2" } ); // J
neighbors.insert( { "1##2", "##12" } ); // K
neighbors.insert( { "1##2", "12##" } ); // L
neighbors.insert( { "##12", "1##2" } ); // M
neighbors.insert( { "1234", RightBorder } ); // N
neighbors.insert( { "1##2", RightBorder } ); // O

for (auto x : neighbors)
borders.insert(x.first);
}

// fast search in O(log n)
unsigned long long search(const Border& left, const Border& right, unsigned long long length, unsigned int modulo)
{
// reduced to a single column ?
if (length == 1)
// can these two borders be next to each other ?
return neighbors.count(std::make_pair(left, right));

// memoize
auto id = std::make_tuple(left, right, length);
// I don't add "modulo" to the key to keep it simple
static std::map<std::tuple<Border, Border, unsigned long long>, unsigned long long> cache;
auto lookup = cache.find(id);
if (lookup != cache.end())
return lookup->second;

// split region into two parts: every possible border can be at the splitting point
unsigned long long result = 0;
for (const auto& next : borders)
{
// prefer a "power of two"-splitting, causes less states than splitting 50:50
unsigned long long pow2 = 1;
while (pow2 < length / 2)
pow2 *= 2;
//pow2 = length / 2; // alternatively: less efficient 50:50 method

// process left  half
auto leftHalf  = search(left, next,  pow2,          modulo);
// process right half
auto rightHalf = search(next, right, length - pow2, modulo);

// each left half can be combined with each right half
auto combined = (leftHalf * rightHalf) % modulo;
result += combined;
}

result %= modulo;
cache[id] = result;
return result;
}

// slow search, no caching whatsoever
unsigned long long slow(const std::string& border, unsigned int length, unsigned int width, unsigned int modulo)
{
// walked across the whole board ?
if (length == width)
return (border == RightBorder) ? 1 : 0;

// proceed with each border that is compatible to the current one
unsigned long long result = 0;
for (auto x : neighbors)
if (x.first == border)
result += slow(x.second, length + 1, width, modulo);

return result % modulo;
}

// backtracking of possible paths, doesn't need the information about borders etc.
typedef std::vector<std::vector<unsigned int>> Grid;
unsigned int bruteForce(Grid& grid, unsigned int x, unsigned int y, unsigned int step)
{
// reached final position ?
if (x == 0 && y == 3)
return (step == grid.size() * grid[0].size()) ? 1 : 0;

// take a step
grid[x][y] = step;

// try to search deeper in each direction
unsigned int result = 0;
if (x > 0 && grid[x - 1][y] == 0)
result += bruteForce(grid, x - 1, y, step + 1);
if (x + 1 < grid.size() && grid[x + 1][y] == 0)
result += bruteForce(grid, x + 1, y, step + 1);
if (y > 0 && grid[x][y - 1] == 0)
result += bruteForce(grid, x, y - 1, step + 1);
if (y < 3 && grid[x][y + 1] == 0)
result += bruteForce(grid, x, y + 1, step + 1);

// undo step
grid[x][y] = 0;

return result;
}

int main()
{
// set up borders and their relationships
fill();

unsigned int       modulo = 100000000;
unsigned long long limit  = 1000000000000;
std::cin >> limit;

//#define BRUTEFORCE
#ifdef  BRUTEFORCE
// allocate memory
Grid grid(limit);
for (auto& column : grid)
column.resize(4, 0);
// start in upper left corner (0,0), that's the first step
std::cout << bruteForce(grid, 0, 0, 1) << std::endl;
#endif

//#define SLOW
#ifdef  SLOW
std::cout << slow(LeftBorder, 0, limit, modulo) << std::endl;
#endif

#define FAST
#ifdef  FAST
std::cout << search(LeftBorder, RightBorder, limit, modulo) << std::endl;
#endif

return 0;
}


This solution contains 24 empty lines, 30 comments and 12 preprocessor commands.

# Benchmark

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

October 17, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 70% (out of 100%).

# 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 235 - An Arithmetic Geometric sequence Twenty-two Foolish Primes - problem 239 >>
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