<< problem 214 - Totient Chains Investigating the primality of numbers of the ... - problem 216 >>

# Problem 215: Crack-free Walls

Consider the problem of building a wall out of 2x1 and 3x1 bricks (horizontal x vertical dimensions) such that,
for extra strength, the gaps between horizontally-adjacent bricks never line up in consecutive layers, i.e. never form a "running crack".

For example, the following 9x3 wall is not acceptable due to the running crack shown in red:

There are eight ways of forming a crack-free 9x3 wall, written W(9,3) = 8.

Calculate W(32,10).

# My Algorithm

My solution can be divided in 3 steps:
1. generateRows: find every possible sequence of "long" and "short" bricks, store them in allRows
2. checkCompatibility: compare each possible row to each other to figure out which rows can be placed next to each other without running cracks, store in compatible
3. count: determine how many rows can be below the current row, based on information in compatible

My rows contain the positions of the edges of two bricks (the "cracks"). The first and last are omitted because they are always 0 and 32 and not considered to be a crack.
The first row of the 9x3 wall would be { 3, 5, 7 }, the second { 2, 4, 7 } and the bottom { 3, 6 }.

Step 1 was implemented as a recursive function that adds a "long" and a "short" brick to a row until its width exceeds 32.
Step 2 compares all rows and contains a stripped-down version of std::intersection to find values contained in two containers.
Step 3 requires a bit of memoization but is the most simple step: it calls itself with all compatible rows.

# 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 width and the height of the wall.

This is equivalent to
echo "9 3" | ./215

Output:

Note: the original problem's input 32 10 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 <vector>

// a single row of brick, store the distance of each crack from the left border
typedef std::vector<unsigned char> Row;
// all unique rows
std::vector<Row> allRows;

// for each row of allRows store the indices of potential neighboring rows
std::vector<std::vector<unsigned int>> compatible;

// generate all distinct row patterns and store them in allRows
void generateRows(Row row, unsigned char maxWidth)
{
// last element contains the width of current row
auto width = row.empty() ? 0 : row.back();

// just one brick left ?
if (width + 2 == maxWidth || width + 3 == maxWidth)
{
allRows.push_back(row);
return;
}
// gap is too small for a brick
if (width + 1 == maxWidth)
return;

row.push_back(width + 2);
generateRows(row, maxWidth);

row.back()++; // replace last 2-brick by a 3-brick
generateRows(row, maxWidth);
}

// find all rows that can be neighbors
void checkCompatibility()
{
compatible.resize(allRows.size());

for (size_t i = 0; i < allRows.size(); i++)
for (size_t j = i + 1; j < allRows.size(); j++)
{
// verify that both rows share no crack
bool valid = true;

// similar to implementation of std::intersection
// note: all rows are already sorted
auto current1 = allRows[i].begin();
auto current2 = allRows[j].begin();
while (current1 != allRows[i].end() && current2 != allRows[j].end())
{
if (*current1 < *current2)
current1++;
else
if (*current2 < *current1)
current2++;
else
{
valid = false;
break;
}
}

// yes, both rows are compatible to each other
if (valid)
{
compatible[i].push_back(j);
compatible[j].push_back(i);
}
}
}

// count crack-free walls
unsigned long long count(unsigned int rowId, unsigned int rowsLeft)
{
// last row ?
if (rowsLeft == 1)
return 1;

static std::vector<std::vector<unsigned long long>> cache(allRows.size());
const unsigned long long Invalid = 0;
// try to look up memoized result
if (cache[rowId].size() <= rowsLeft)
cache[rowId].resize(rowsLeft + 1, Invalid); // not known yet, allocate memory
else
if (cache[rowId][rowsLeft] != Invalid)
return cache[rowId][rowsLeft];

// process all compatible walls
unsigned long long result = 0;
for (auto x : compatible[rowId])
result += count(x, rowsLeft - 1);

cache[rowId][rowsLeft] = result;
return result;
}

int main()
{
unsigned int width  = 32;// 9;
unsigned int height = 10;// 3;
std::cin >> width >> height;

// create all 3329 distinct rows
Row empty;
generateRows(empty, width);

// set up compatibility relationships
checkCompatibility();

unsigned long long result = 0;
for (unsigned int i = 0; i < allRows.size(); i++)
result += count(i, height);

std::cout << result << std::endl;
return 0;
}


This solution contains 19 empty lines, 21 comments and 2 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.06 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 3 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

July 22, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 50% (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.

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