<< problem 286 - Scoring probabilities An enormous factorial - problem 288 >>

# Problem 287: Quadtree encoding (a simple compression algorithm)

The quadtree encoding allows us to describe a 2^N * 2^N black and white image as a sequence of bits (0 and 1).
Those sequences are to be read from left to right like this:

• the first bit deals with the complete 2^N * 2^N region;
• "0" denotes a split: the current 2^n * 2^n region is divided into 4 sub-regions of dimension 2^{n-1} * 2^{n-1},
the next bits contains the description of the top left, top right, bottom left and bottom right sub-regions - in that order;
• "10" indicates that the current region contains only black pixels;
• "11" indicates that the current region contains only white pixels.
Consider the following 4x4 image (colored marks denote places where a split can occur):

This image can be described by several sequences, for example : "001010101001011111011010101010", of length 30, or
"0100101111101110", of length 16, which is the minimal sequence for this image.

For a positive integer N, define D_N as the 2^N * 2^N image with the following coloring scheme:

the pixel with coordinates x = 0, y = 0 corresponds to the bottom left pixel,
if (x - 2^{N-1})^2 + (y - 2^{N-1})^2 <= 2^{2N-2} then the pixel is black,
otherwise the pixel is white.

What is the length of the minimal sequence describing D_24 ?

# My Algorithm

The given formula can be easily translated to a function isBlack(x,y) that returns true if the pixel at (x,y) is black.
It looks like the equation of a circle - and when printing D_4 on screen I get (see #ifdef DRAW_IMAGE):
.....BBBBBBB....
...BBBBBBBBBBB..
..BBBBBBBBBBBBB.
..BBBBBBBBBBBBB.
.BBBBBBBBBBBBBBB
.BBBBBBBBBBBBBBB
.BBBBBBBBBBBBBBB
BBBBBBBBBBBBBBBB
.BBBBBBBBBBBBBBB
.BBBBBBBBBBBBBBB
.BBBBBBBBBBBBBBB
..BBBBBBBBBBBBB.
..BBBBBBBBBBBBB.
...BBBBBBBBBBB..
.....BBBBBBB....
........B.......

The combination of a circle and rectangles has some nice properties:

• if all four corners of the rectangle are inside the circle (= black) then the whole rectangle is black
• if the rectangle is much smaller than the circle and all fours corners of the rectangle are outside the circle then the whole rectangle is outside the circle
My function encode returns the size of an optimal encoding and performs the following tasks in a recursive manner:
• if the current square covers only 1 pixel then it needs 2 bits
• check all four corners of the current square, if they are all inside or outside then its encoding requires 2 bits
• subdivide the current square into 4 equally-sized squares and determine their encoding size plus 1 bit for the split

## Note

When the current square is 2x2 and some of its corner are black and some are white then I know that these four pixels need 1 + 2+2+2+2 bits, saving one recursion step (about 10% faster).

The encoded image has a compression ratio of about 900000 to 1. That's excellent - photos usually compress to about 10:1.

# 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 4 | ./287

Output:

Note: the original problem's input 24 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.

       #include <iostream>

// D24 => 2^24
unsigned int size = 1 << 24;

// return true if pixel at (x,y) is black
bool isBlack(unsigned int x, unsigned int y)
{
// 2^(N-1)
long long middle = size >> 1;
// right side of the equation: 2^(2N - 2) = 2^(N-1) * 2^(N-1) = middle * middle
auto threshold = middle * middle;

// be a bit careful with negative differences
auto dx = (long long)x - middle;
auto dy = (long long)y - middle;
return dx*dx + dy*dy <= threshold;
}

// return size of optimal encoding
// note: I expect only valid input, such that (toX - fromX) = (toY - fromY) = a power of two
unsigned int encode(unsigned int fromX, unsigned int fromY, unsigned int toX, unsigned int toY, bool isFirst = true)
{
// a single pixel ?
if (fromX == toX) // implies fromY == toY
return 2; // doesn't matter whether black or white, both encodings need two bits

// isBlack() will produce a black circle
// checking all four corners is sufficient to know when to split
bool a = isBlack(fromX, fromY);
bool b = isBlack(toX,   fromY);
bool c = isBlack(toX,   toY);
bool d = isBlack(fromX, toY);

// same color on all four corner => the whole area is covered by the same color
// however, this assumption doesn't hold on the first level
if (a == b && b == c && c == d && !isFirst)
return 2; // again: the color doesn't matter, both need two bits to fill the entire area

// speed optimization: if a 2x2 area needs to be split, then it always requires 9 bits
if (fromX + 1 == toX)
return 1 + 4*2; // a split marker and four single pixels (2 bits each)

// split evenly
auto half = (toX - fromX + 1) / 2;
return encode(fromX,        fromY + half, toX - half, toY,        false) + // upper-left
encode(fromX + half, fromY + half, toX,        toY,        false) + // upper-right
encode(fromX,        fromY,        toX - half, toY - half, false) + // lower-left
encode(fromX + half, fromY,        toX,        toY - half, false) + // lower-right
1; // don't forget: there's one bit for the split marker
}

int main()
{
// D24 => 24
unsigned int shift = 4;
std::cin >> shift;

// length of an edge of the image in pixels
size = 1 << shift;

// draw on screen (only useful for very small values of "shift" 1 ... 5)
#define DRAW_IMAGE
#ifdef DRAW_IMAGE
if (size <= 5)
{
for (unsigned int y = 0; y < size; y++)
{
auto flipY = (size - 1) - y; // problem states that lower-left corner is (0,0), must flip image upside-down
for (unsigned int x = 0; x < size; x++)
std::cout << (isBlack(x, flipY) ? "B" : ".");
std::cout << std::endl;
}
}
#endif

// let's compress to infinity and beyond !
std::cout << encode(0, 0, size - 1, size - 1) << std::endl;
return 0;
}


This solution contains 12 empty lines, 18 comments and 4 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.8 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

August 29, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 40% (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
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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 286 - Scoring probabilities An enormous factorial - problem 288 >>
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