<< problem 160 - Factorial trailing digits Hexadecimal numbers - problem 162 >>

Problem 161: Triominoes

A triomino is a shape consisting of three squares joined via the edges. There are two basic forms:

If all possible orientations are taken into account there are six:

Any n by m grid for which nxm is divisible by 3 can be tiled with triominoes.
If we consider tilings that can be obtained by reflection or rotation from another tiling as different
there are 41 ways a 2 by 9 grid can be tiled with triominoes:

In how many ways can a 9 by 12 grid be tiled in this way by triominoes?

My Algorithm

I wrote a typical Dynamic Programming solution:

• the grid is filled from top to bottom
• bit masks represent whether a cell of the grid is occupied (=1) or available/empty (=0)
• search attempts to place every shape in the top-most and left-most empty cell
There are a few observations:
1. All rows are either empty, partially filled or full
2. There are at most 3 partially filled rows (due to my strict "from-top-to-bottom" algorithm)

My memoization hash therefore consists only of the number of empty rows and the bitmasks of (up to) three partially filled rows.
For the 9x12 grid, it just fits into a 32 bit integer.

The number of solutions for a grid of size n x m is identical to a grid of size m x n.
The hash is "smaller" and creates more opportunities for memoization hits when the rows are small.

Modifications by HackerRank

The result has to be printed modulo 1000000007.

Note

I was surprised to find only 223255 values in cache (for the 9x12 grid).
On the other side, the rotated 12x9 grid would cause over 5 million values (and run about 70x slower).
However, my std::swap optimization automatically converts the 12x9 grid to a 9x12 grid.

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 width and height of the grid.

This is equivalent to
echo "2 9" | ./161

Output:

(please click 'Go !')

Note: the original problem's input 9 12 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.

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 <unordered_map>

#define ORIGINAL

// grid size
unsigned int height = 9;
unsigned int width  = 2;

// indicate an empty row
const unsigned int EmptyRow = 0;

// set a certain bit to one (= position is not available anymore), return true if bit was zero
// note: second parameter is an in/out parameter
bool use(unsigned int pos, unsigned int& row)
{
unsigned int mask = 1 << pos;
bool result = (row & mask) == 0;
row |= mask;
return result;
}

// recursive search
unsigned long long search(unsigned int rowsLeft,
unsigned int rowA, unsigned int rowB, unsigned int rowC)
{
// done ?
if (rowsLeft == 0)
return 1;

// filled another row ?
unsigned int fullRow = (1 << width) - 1;
if (rowA == fullRow)
return search(rowsLeft - 1, rowB, rowC, EmptyRow);

// find first gap in rowA
unsigned int pos = 0;
while ((rowA & (1 << pos)) != 0)
pos++;

// unique ID
unsigned long long hash; // unsigned int would suffice for grid with up to 9 columns, too
hash   = rowsLeft;
hash <<= width;
hash  |= rowA;
hash <<= width;
hash  |= rowB;
hash <<= width;
hash  |= rowC;

// about twice as fast as std::map
static std::unordered_map<unsigned long long, unsigned long long> cache;
auto i = cache.find(hash);
if (i != cache.end())
return i->second;

unsigned long long result = 0;

// shape: ##
//        #
unsigned int a = rowA;
unsigned int b = rowB;
unsigned int c = rowC;
if (rowsLeft >= 2 && pos < width - 1 &&
use(pos, a) && use(pos + 1, a) && use(pos, b))
result += search(rowsLeft, a, b, c);

// shape: ##
//         #
a = rowA; b = rowB; //c = rowC;
if (rowsLeft >= 2 && pos < width - 1 &&
use(pos, a) && use(pos + 1, a) && use(pos + 1, b))
result += search(rowsLeft, a, b, c);

// shape: #
//        ##
a = rowA; b = rowB; //c = rowC;
if (rowsLeft >= 2 && pos < width - 1 &&
use(pos, a) && use(pos, b) && use(pos + 1, b))
result += search(rowsLeft, a, b, c);

// shape:  #
//        ##
// note: this shape extends one "negative" unit to the left
a = rowA; b = rowB; //c = rowC;
if (rowsLeft >= 2 && pos > 0 && pos < width &&
use(pos, a) && use(pos - 1, b) && use(pos, b))
result += search(rowsLeft, a, b, c);

// shape:  #
//         #
//         #
a = rowA; b = rowB; //c = rowC;
if (rowsLeft >= 3 && pos < width &&
use(pos, a) && use(pos, b) && use(pos, c))
result += search(rowsLeft, a, b, c);

// shape: ###
a = rowA; b = rowB; c = rowC;
if (rowsLeft >= 1 && pos < width - 2 &&
use(pos, a) && use(pos + 1, a) && use(pos + 2, a))
result += search(rowsLeft, a, b, c);

#ifndef ORIGINAL
result %= 1000000007; // Hackerrank only
#endif

// memoize
cache[hash] = result;
return result;
}

int main()
{
// read grid size
std::cin >> width >> height;

// prefer fewer columns
if (width > height)
std::swap(width, height);

// let's go !
std::cout << search(height, EmptyRow, EmptyRow, EmptyRow) << std::endl;
return 0;
}


This solution contains 21 empty lines, 27 comments and 5 preprocessor commands.

Benchmark

The correct solution to the original Project Euler problem was found in 0.07 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 11 MByte.

(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

June 29, 2017 submitted solution
June 29, 2017 added comments

Hackerrank

My code solves 11 out of 11 test cases (score: 100%)

Difficulty

Project Euler ranks this problem at 70% (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 [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 160 - Factorial trailing digits Hexadecimal numbers - problem 162 >>
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