I have a three dimensional array of integers, each dimension represents a colour R, G or B, and the integer represents the number of times a colour value of R, G, B has occurred in an image.
Now I need to find the X colours that have occurred the most in this image.
So far, I have created a list of structs, with the struct holding both colour and count. I iterate through the three dimensions of the original array, and for each value that is greater than the first item in the list, I replace the item in the list, after which I sort the list.
This of course takes a lot of time, nearly 2 seconds per run.
How can I find the X colours that have occurred the most in this array while doing this in a time-efficient way? At the moment it's costing me 2 seconds per run on my computer, yet this piece of code will need to be running on a Raspberry Pi, on which it takes a lot longer. Ideally, I'm looking for a 100ms time on the Pi, but I have no idea how to get there or what algorithm to use.
Edit: added the code on Jerry's request
// Struct holding colours & colour count
struct ColourCount
{
public byte red;
public byte green;
public byte blue;
public Int32 count;
}
List<Int32> matrix = new List<Int32>();
ColourCount[] count = new ColourCount[100];
int idx = 0;
for (int r = 0; r < 256; r++)
{
for (int g = 0; g < 256; g++)
{
for (int b = 0; b < 256; b++)
{
if (idx < count.Length)
{
count[idx].red = (byte)r;
count[idx].green = (byte)g;
count[idx].blue = (byte)b;
count[idx].count = colours[r, g, b];
idx++;
}
else
{
Array.Sort<ColourCount>(count, (x, y) => (x.count.CompareTo(y.count)));
if (colours[r, g, b] > count[0].count)
{
count[0].red = (byte)r;
count[0].green = (byte)g;
count[0].blue = (byte)b;
count[0].count = colours[r, g, b];
}
}
}
}
}
for (int i = 0; i < count.Length; i++)
{
matrix.Add((count[i].red * 1000 * 1000) + (count[i].green * 1000) + count[i].blue);
}
Move your sort. you only need to sort your array if you've changed the minimum value.
Avg time: 48ms, i7-6700k
ColourCount[] count = new ColourCount[100];
int idx = 0;
for (int r = 0; r < 256; r++)
{
for (int g = 0; g < 256; g++)
{
for (int b = 0; b < 256; b++)
{
if (idx < count.Length)
{
count[idx].red = (byte)r;
count[idx].green = (byte)g;
count[idx].blue = (byte)b;
count[idx].count = colours[r, g, b];
idx++;
}
else
{
if (colours[r, g, b] > count[0].count)
{
count[0].red = (byte)r;
count[0].green = (byte)g;
count[0].blue = (byte)b;
count[0].count = colours[r, g, b];
Array.Sort(count, (x, y) => (x.count.CompareTo(y.count)));
}
}
}
}
}
I have to calculate the spectrum values of an audio.
I used aForge's FFT in Sources/Math/FourierTransform.cs and I used an example of sampling with 16 samples as used in this video to check the results with excel (I tested the results in a spreadsheet like in the video).
FFT:
public enum Direction
{
Forward = 1,
Backward = -1
};
private const int minLength = 2;
private const int maxLength = 16384;
private const int minBits = 1;
private const int maxBits = 14;
private static int[][] reversedBits = new int[maxBits][];
private static Complex[,][] complexRotation = new Complex[maxBits, 2][];
static void Main(string[] args)
{
var Data = new Complex[16];
Data[0] = new Complex(0, 0);
Data[1] = new Complex((float)0.998027, 0);
Data[2] = new Complex((float)0.125333, 0);
Data[3] = new Complex((float)-0.98229, 0);
Data[4] = new Complex((float)-0.24869, 0);
Data[5] = new Complex((float)0.951057, 0);
Data[6] = new Complex((float)0.368125, 0);
Data[7] = new Complex((float)-0.90483, 0);
Data[8] = new Complex((float)-0.48175, 0);
Data[9] = new Complex((float)0.844328, 0);
Data[10] = new Complex((float)0.587785, 0);
Data[11] = new Complex((float)-0.77051, 0);
Data[12] = new Complex((float)-0.68455, 0);
Data[13] = new Complex((float)0.684547, 0);
Data[14] = new Complex((float)0.770513, 0);
Data[15] = new Complex((float)-0.58779, 0);
FFT(Data, Direction.Forward);
for (int a = 0; a <= Data.Length - 1; a++)
{
Console.WriteLine(Data[a].Re.ToString());
}
Console.ReadLine();
}
public static void FFT(Complex[] data, Direction direction)
{
int n = data.Length;
int m = Tools.Log2(n);
// reorder data first
ReorderData(data);
// compute FFT
int tn = 1, tm;
for (int k = 1; k <= m; k++)
{
Complex[] rotation = GetComplexRotation(k, direction);
tm = tn;
tn <<= 1;
for (int i = 0; i < tm; i++)
{
Complex t = rotation[i];
for (int even = i; even < n; even += tn)
{
int odd = even + tm;
Complex ce = data[even];
Complex co = data[odd];
double tr = co.Re * t.Re - co.Im * t.Im;
double ti = co.Re * t.Im + co.Im * t.Re;
data[even].Re += tr;
data[even].Im += ti;
data[odd].Re = ce.Re - tr;
data[odd].Im = ce.Im - ti;
}
}
}
if (direction == Direction.Forward)
{
for (int i = 0; i < n; i++)
{
data[i].Re /= (double)n;
data[i].Im /= (double)n;
}
}
}
private static int[] GetReversedBits(int numberOfBits)
{
if ((numberOfBits < minBits) || (numberOfBits > maxBits))
throw new ArgumentOutOfRangeException();
// check if the array is already calculated
if (reversedBits[numberOfBits - 1] == null)
{
int n = Tools.Pow2(numberOfBits);
int[] rBits = new int[n];
// calculate the array
for (int i = 0; i < n; i++)
{
int oldBits = i;
int newBits = 0;
for (int j = 0; j < numberOfBits; j++)
{
newBits = (newBits << 1) | (oldBits & 1);
oldBits = (oldBits >> 1);
}
rBits[i] = newBits;
}
reversedBits[numberOfBits - 1] = rBits;
}
return reversedBits[numberOfBits - 1];
}
private static Complex[] GetComplexRotation(int numberOfBits, Direction direction)
{
int directionIndex = (direction == Direction.Forward) ? 0 : 1;
// check if the array is already calculated
if (complexRotation[numberOfBits - 1, directionIndex] == null)
{
int n = 1 << (numberOfBits - 1);
double uR = 1.0;
double uI = 0.0;
double angle = System.Math.PI / n * (int)direction;
double wR = System.Math.Cos(angle);
double wI = System.Math.Sin(angle);
double t;
Complex[] rotation = new Complex[n];
for (int i = 0; i < n; i++)
{
rotation[i] = new Complex(uR, uI);
t = uR * wI + uI * wR;
uR = uR * wR - uI * wI;
uI = t;
}
complexRotation[numberOfBits - 1, directionIndex] = rotation;
}
return complexRotation[numberOfBits - 1, directionIndex];
}
// Reorder data for FFT using
private static void ReorderData(Complex[] data)
{
int len = data.Length;
// check data length
if ((len < minLength) || (len > maxLength) || (!Tools.IsPowerOf2(len)))
throw new ArgumentException("Incorrect data length.");
int[] rBits = GetReversedBits(Tools.Log2(len));
for (int i = 0; i < len; i++)
{
int s = rBits[i];
if (s > i)
{
Complex t = data[i];
data[i] = data[s];
data[s] = t;
}
}
}
These are the results after the transformation:
Output FFT results: Excel FFT results:
0,0418315622955561 0,669305
0,0533257974328085 0,716163407
0,137615673627316 0,908647001
0,114642731070279 1,673453043
0,234673940537634 7,474988602
0,0811255020953362 0,880988382
0,138088891589122 0,406276784
0,0623766891658306 0,248854492
0,0272978749126196 0,204227
0,0124250144575261 0,248854492
0,053787064184711 0,406276784
0,00783331226557493 0,880988382
0,0884368745610118 7,474988602
0,0155431246384978 1,673453043
0,0301093757152557 0,908647001
0 0,716163407
The results are not at all similar. Where is it wrong?
Is the implementation of complex (Data) wrong or is the FFT method wrong or other?
Thanks in advance!
First, the resulting FFT is a complex function in general. You're only displaying the real parts in your code but the thing you're comparing to is displaying the magnitudes, so of course they're going to be different: you're comparing apples to oranges.
When you use magnitudes and compare apples to apples, you should get this:
for (int a = 0; a <= Data.Length - 1; a++)
{
Console.WriteLine(Data[a].Magnitude.ToString());
}
...
0.0418315622955561
0.0447602132472683
0.0567904388057513
0.104590813761862
0.46718679147454
0.0550617784710375
0.025392294285886
0.0155534081359397
0.0127641875296831
0.0155534081359397
0.025392294285886
0.0550617784710375
0.46718679147454
0.104590813761862
0.0567904388057513
0.0447602132472683
That looks a little better -- it has the same symmetry property as the Excel output and there appear to be peaks in the same locations.
It almost looks like the scale is off. If I divide each element by the corresponding element from the Excel output, I get:
16
16
16
16
16
16
16
16
16
16
16
16
16
16
16
16
So your results are pretty much correct, just off by a scaling factor.
You're dividing everything by n in the last step of your FFT:
if (direction == Direction.Forward)
{
for (int i = 0; i < n; i++)
{
data[i].Re /= (double)n;
data[i].Im /= (double)n;
}
}
This is conventionally done for the inverse transform, not the forward transform.
In summary, changing the output from Data[a].Re to Data[a].Magnitude and changing the condition at the end of FFT from if (direction == Direction.Forward) to if (direction == Direction.Backward), I get this output:
0.669304996728897
0.716163411956293
0.908647020892022
1.67345302018979
7.47498866359264
0.880988455536601
0.406276708574176
0.248854530175035
0.20422700047493
0.248854530175035
0.406276708574176
0.880988455536601
7.47498866359264
1.67345302018979
0.908647020892022
0.716163411956293
which matches the Excel output.
So, I am trying to wrap my head around understanding how you can use a variable to denote how many times a loop is nested.
Here is an example I write up to simulate the output of dimensions = 4:
static void Main(string[] args)
{
int dimensions = 4; // e.g. for (1, 2, 3, 4), dimensions = 4
Console.WriteLine($"{addNumbers(dimensions)}");
Console.ReadKey();
}
static long addNumbers(int dimensions)
{
long number = 0;
// hard coded to be dimensions = 4
for (int h = 0; h <= dimensions; h++)
for (int i = 0; i <= dimensions; i++)
for (int j = 0; j <= dimensions; j++)
for (int k = 0; k <= dimensions; k++)
number += h + i + j + k; // just some random math
return number;
}
This will present the expected output of:
5000
So to readdress the problem, how can I code to allow this for n dimensions? Thanks for your help!
For arbitrary n dimensions you can loop with a help of array int[] address which represents n dimensions:
static long addNumbers(int dimensions) {
int[] address = new int[dimensions];
// size of each dimension; not necessary equals to dimensions
// + 1 : in your code, int the loops you have i <= dimensions, j <= dimensions etc.
int size = dimensions + 1;
long number = 0;
do {
//TODO: some math here
// i == address[0]; j = address[1]; ... etc.
number += address.Sum();
// next address: adding 1 to array
for (int i = 0; i < address.Length; ++i) {
if (address[i] >= size - 1)
address[i] = 0;
else {
address[i] += 1;
break;
}
}
}
while (!address.All(index => index == 0)); // all 0 address - items're exhausted
return number;
}
Finally, let's add some Linq to look at the results:
int upTo = 5;
string report = string.Join(Environment.NewLine, Enumerable
.Range(1, upTo)
.Select(i => $"{i} -> {addNumbers(i),6}"));
Console.Write(report);
Outcome:
1 -> 1
2 -> 18
3 -> 288
4 -> 5000 // <- We've got it: 5000 for 4 dimensions
5 -> 97200
I don't understand why the messagebox keeps displaying 0. For each sequence there is a direction. The purpose of the random function is to find the best point to start a new sequence. There seems to be a problem with my howfree function, I can't understand what the problem is, please help me.
public int howfree(int x, int y)
{
int freenum = 0;
int counter = 0;
foreach (GameForm.direction dirs in (GameForm.direction[]) Enum.GetValues(typeof(GameForm.direction)))
{
for (int j = 0; j < 5; j++)
{
y += Directions[(int)dirs, 0];
x += Directions[(int)dirs, 1];
if ( InBoard(y, x) && cells[y,x].cellType == Type.EMPTY)
{
counter++;
}
else
break;
}
if (counter == 5)
{
freenum++;
}
counter = 0;
}
return freenum;
}
///////////////////////////////////////////////////////////////////////////////
public Cell Randomize()
{
int row=0;
int col=0;
Random random = new Random();
int rand = 0;
//bool Found = false;
int max = 0;
int fff=0;
List<Cell> Options = new List<Cell>();
foreach (Cell CCC in cells)
{
fff=howfree(CCC.row,CCC.col);
if (fff > max)
max = fff;
}
foreach (Cell c in cells)
{
if (howfree(c.row, c.col) == max)
{
Options.Add(new Cell(c.row, c.col));
}
}
// while (!Found)
// {
rand = (int)random.NextDouble() * (Options.Count - 1);
//row = random.Next() * (Settings.Rows);
//col = random.Next() * (Settings.Cols);
MessageBox.Show(rand.ToString());
row = Options[rand].row;
col = Options[rand].col;
//}
return new Cell(row, col);
}
Why not use the overload that's designed for integers?
rand = random.Next(Options.Count);
From the MSDN documentation:
Returns a nonnegative random integer that is less than the specified maximum.
You're doing it wrong.
rand = (int)random.NextDouble() * (Options.Count - 1);
Random.NextDouble() will produce number between 0 and less than 1.0.
That means it can get up to 0.999 (You get my point) but will never be 1.
When you use explicit int cast on a fraction less bigger than 0 and less than 1 you will always get 0.
You should have done it like this:
rand = (int)(random.NextDouble() * (Options.Count - 1));
So now it will int cast after the count of options.
Please see my own answer, I think I did it!
Hi,
An example question for a programming contest was to write a program that finds out how much polyominos are possible with a given number of stones.
So for two stones (n = 2) there is only one polyominos:
XX
You might think this is a second solution:
X
X
But it isn't. The polyominos are not unique if you can rotate them.
So, for 4 stones (n = 4), there are 7 solutions:
X
X XX X X X X
X X XX X XX XX XX
X X X XX X X XX
The application has to be able to find the solution for 1 <= n <=10
PS: Using the list of polyominos on Wikipedia isn't allowed ;)
EDIT: Of course the question is: How to do this in Java, C/C++, C#
I started this project in Java. But then I had to admit I didn't know how to build polyominos using an efficient algorithm.
This is what I had so far:
import java.util.ArrayList;
import java.util.List;
public class Main
{
private int countPolyminos(int n)
{
hashes.clear();
count = 0;
boolean[][] matrix = new boolean[n][n];
createPolyominos(matrix, n);
return count;
}
private List<Integer> hashes = new ArrayList<Integer>();
private int count;
private void createPolyominos(boolean[][] matrix, int n)
{
if (n == 0)
{
boolean[][] cropped = cropMatrix(matrix);
int hash = hashMatrixOrientationIndependent(matrix);
if (!hashes.contains(hash))
{
count++;
hashes.add(hash);
}
return;
}
// Here is the real trouble!!
// Then here something like; createPolyominos(matrix, n-1);
// But, we need to keep in mind that the polyominos can have ramifications
}
public boolean[][] copy(boolean[][] matrix)
{
boolean[][] b = new boolean[matrix.length][matrix[0].length];
for (int i = 0; i < matrix.length; ++i)
{
System.arraycopy(matrix[i], 0, b, 0, matrix[i].length);
}
return b;
}
public boolean[][] cropMatrix(boolean[][] matrix)
{
int l = 0, t = 0, r = 0, b = 0;
// Left
left: for (int x = 0; x < matrix.length; ++x)
{
for (int y = 0; y < matrix[x].length; ++y)
{
if (matrix[x][y])
{
break left;
}
}
l++;
}
// Right
right: for (int x = matrix.length - 1; x >= 0; --x)
{
for (int y = 0; y < matrix[x].length; ++y)
{
if (matrix[x][y])
{
break right;
}
}
r++;
}
// Top
top: for (int y = 0; y < matrix[0].length; ++y)
{
for (int x = 0; x < matrix.length; ++x)
{
if (matrix[x][y])
{
break top;
}
}
t++;
}
// Bottom
bottom: for (int y = matrix[0].length; y >= 0; --y)
{
for (int x = 0; x < matrix.length; ++x)
{
if (matrix[x][y])
{
break bottom;
}
}
b++;
}
// Perform the real crop
boolean[][] cropped = new boolean[matrix.length - l - r][matrix[0].length - t - b];
for (int x = l; x < matrix.length - r; ++x)
{
System.arraycopy(matrix[x - l], t, cropped, 0, matrix[x].length - t - b);
}
return cropped;
}
public int hashMatrix(boolean[][] matrix)
{
int hash = 0;
for (int x = 0; x < matrix.length; ++x)
{
for (int y = 0; y < matrix[x].length; ++y)
{
hash += matrix[x][y] ? (((x + 7) << 4) * ((y + 3) << 6) * 31) : ((((x+5) << 9) * (((y + x) + 18) << 7) * 53));
}
}
return hash;
}
public int hashMatrixOrientationIndependent(boolean[][] matrix)
{
int hash = 0;
hash += hashMatrix(matrix);
for (int i = 0; i < 3; ++i)
{
matrix = rotateMatrixLeft(matrix);
hash += hashMatrix(matrix);
}
return hash;
}
public boolean[][] rotateMatrixRight(boolean[][] matrix)
{
/* W and H are already swapped */
int w = matrix.length;
int h = matrix[0].length;
boolean[][] ret = new boolean[h][w];
for (int i = 0; i < h; ++i)
{
for (int j = 0; j < w; ++j)
{
ret[i][j] = matrix[w - j - 1][i];
}
}
return ret;
}
public boolean[][] rotateMatrixLeft(boolean[][] matrix)
{
/* W and H are already swapped */
int w = matrix.length;
int h = matrix[0].length;
boolean[][] ret = new boolean[h][w];
for (int i = 0; i < h; ++i)
{
for (int j = 0; j < w; ++j)
{
ret[i][j] = matrix[j][h - i - 1];
}
}
return ret;
}
}
There are only 4,461 polynominoes of size 10, so we can just enumerate them all.
Start with a single stone. To expand it by one stone, try add the new stone in at all empty cells that neighbour an existing stone. Do this recursively until reaching the desired size.
To avoid duplicates, keep a hash table of all polynominoes of each size we've already enumerated. When we put together a new polynomino, we check that its not already in the hash table. We also need to check its 3 rotations (and possibly its mirror image). While duplicate checking at the final size is the only strictly necessary check, checking at each step prunes recursive branches that will yield a new polynomino.
Here's some pseudo-code:
polynomino = array of n hashtables
function find_polynominoes(n, base):
if base.size == n:
return
for stone in base:
for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
new_stone.x = stone.x + dx
new_stone.y = stone.y + dy
if new_stone not in base:
new_polynomino = base + new_stone
is_new = true
for rotation in [0, 90, 180, 270]:
if new_polynomino.rotate(rotation) in polynomino[new_polynomino.size]:
is_new = false
break
if is_new:
polynomino[new_polynomino.size].add(new_polynomino)
Just solved this as well in java. Since all here appear to have performance issues. I give you mine as well.
Board reprsentation:
2 arrays of integers. 1 for the rows and 1 for the columns.
Rotation: column[i]=row[size-(i+1)], row[i] = reverse(column[i]) where reverse is the bits reversed according to the size (for size = 4 and first 2 bits are taken: rev(1100) = 0011)
Shifting block: row[i-1] = row[i], col[i]<<=1
Check if bit is set: (row[r] & (1<<c)) > 0
Board uniqueness: The board is unique when the array row is unique.
Board hash: Hashcode of the array row
..
So this makes all operations fast. Many of them would have been O(size²) in the 2D array representation instead of now O(size).
Algorithm:
Start with the block of size 1
For each size start from the blocks with 1 stone less.
If it's possible to add the stone. Check if it was already added to the set.
If it's not yet added. Add it to the solution of this size.
add the block to the set and all its rotations. (3 rotations, 4 in total)
Important, after each rotation shift the block as left/top as possible.
+Special cases: do the same logic for the next 2 cases
shift block one to the right and add stone in first column
shift block one to the bottom and add stone in first row
Performance:
N=5 , time: 3ms
N=10, time: 58ms
N=11, time: 166ms
N=12, time: 538ms
N=13, time: 2893ms
N=14, time:17266ms
N=15, NA (out of heapspace)
Code:
https://github.com/Samjayyy/logicpuzzles/tree/master/polyominos
The most naive solution is to start with a single X, and for each iteration, build the list of unique possible next-states. From that list, build the list of unique states by adding another X. Continue this until the iteration you desire.
I'm not sure if this runs in reasonable time for N=10, however. It might, depending on your requirements.
I think I did it!
EDIT: I'm using the SHA-256 algorithm to hash them, now it works correct.
Here are the results:
numberOfStones -> numberOfPolyominos
1 -> 1
2 -> 1
3 -> 2
4 -> 7
5 -> 18
6 -> 60
7 -> 196
8 -> 704
9 -> 2500
10 -> terminated
Here is the code (Java):
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
/* VPW Template */
public class Main
{
/**
* #param args
*/
public static void main(String[] args) throws IOException
{
new Main().start();
}
public void start() throws IOException
{
/* Read the stuff */
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] input = new String[Integer.parseInt(br.readLine())];
for (int i = 0; i < input.length; ++i)
{
input[i] = br.readLine();
}
/* Process each line */
for (int i = 0; i < input.length; ++i)
{
processLine(input[i]);
}
}
public void processLine(String line)
{
int n = Integer.parseInt(line);
System.out.println(countPolyminos(n));
}
private int countPolyminos(int n)
{
hashes.clear();
count = 0;
boolean[][] matrix = new boolean[n][n];
matrix[n / 2][n / 2] = true;
createPolyominos(matrix, n - 1);
return count;
}
private List<BigInteger> hashes = new ArrayList<BigInteger>();
private int count;
private void createPolyominos(boolean[][] matrix, int n)
{
if (n == 0)
{
boolean[][] cropped = cropMatrix(matrix);
BigInteger hash = hashMatrixOrientationIndependent(cropped);
if (!hashes.contains(hash))
{
// System.out.println(count + " Found!");
// printMatrix(cropped);
// System.out.println();
count++;
hashes.add(hash);
}
return;
}
for (int x = 0; x < matrix.length; ++x)
{
for (int y = 0; y < matrix[x].length; ++y)
{
if (matrix[x][y])
{
if (x > 0 && !matrix[x - 1][y])
{
boolean[][] clone = copy(matrix);
clone[x - 1][y] = true;
createPolyominos(clone, n - 1);
}
if (x < matrix.length - 1 && !matrix[x + 1][y])
{
boolean[][] clone = copy(matrix);
clone[x + 1][y] = true;
createPolyominos(clone, n - 1);
}
if (y > 0 && !matrix[x][y - 1])
{
boolean[][] clone = copy(matrix);
clone[x][y - 1] = true;
createPolyominos(clone, n - 1);
}
if (y < matrix[x].length - 1 && !matrix[x][y + 1])
{
boolean[][] clone = copy(matrix);
clone[x][y + 1] = true;
createPolyominos(clone, n - 1);
}
}
}
}
}
public boolean[][] copy(boolean[][] matrix)
{
boolean[][] b = new boolean[matrix.length][matrix[0].length];
for (int i = 0; i < matrix.length; ++i)
{
System.arraycopy(matrix[i], 0, b[i], 0, matrix[i].length);
}
return b;
}
public void printMatrix(boolean[][] matrix)
{
for (int y = 0; y < matrix.length; ++y)
{
for (int x = 0; x < matrix[y].length; ++x)
{
System.out.print((matrix[y][x] ? 'X' : ' '));
}
System.out.println();
}
}
public boolean[][] cropMatrix(boolean[][] matrix)
{
int l = 0, t = 0, r = 0, b = 0;
// Left
left: for (int x = 0; x < matrix.length; ++x)
{
for (int y = 0; y < matrix[x].length; ++y)
{
if (matrix[x][y])
{
break left;
}
}
l++;
}
// Right
right: for (int x = matrix.length - 1; x >= 0; --x)
{
for (int y = 0; y < matrix[x].length; ++y)
{
if (matrix[x][y])
{
break right;
}
}
r++;
}
// Top
top: for (int y = 0; y < matrix[0].length; ++y)
{
for (int x = 0; x < matrix.length; ++x)
{
if (matrix[x][y])
{
break top;
}
}
t++;
}
// Bottom
bottom: for (int y = matrix[0].length - 1; y >= 0; --y)
{
for (int x = 0; x < matrix.length; ++x)
{
if (matrix[x][y])
{
break bottom;
}
}
b++;
}
// Perform the real crop
boolean[][] cropped = new boolean[matrix.length - l - r][matrix[0].length - t - b];
for (int x = l; x < matrix.length - r; ++x)
{
System.arraycopy(matrix[x], t, cropped[x - l], 0, matrix[x].length - t - b);
}
return cropped;
}
public BigInteger hashMatrix(boolean[][] matrix)
{
try
{
MessageDigest md = MessageDigest.getInstance("SHA-256");
md.update((byte) matrix.length);
md.update((byte) matrix[0].length);
for (int x = 0; x < matrix.length; ++x)
{
for (int y = 0; y < matrix[x].length; ++y)
{
if (matrix[x][y])
{
md.update((byte) x);
} else
{
md.update((byte) y);
}
}
}
return new BigInteger(1, md.digest());
} catch (NoSuchAlgorithmException e)
{
System.exit(1);
return null;
}
}
public BigInteger hashMatrixOrientationIndependent(boolean[][] matrix)
{
BigInteger hash = hashMatrix(matrix);
for (int i = 0; i < 3; ++i)
{
matrix = rotateMatrixLeft(matrix);
hash = hash.add(hashMatrix(matrix));
}
return hash;
}
public boolean[][] rotateMatrixRight(boolean[][] matrix)
{
/* W and H are already swapped */
int w = matrix.length;
int h = matrix[0].length;
boolean[][] ret = new boolean[h][w];
for (int i = 0; i < h; ++i)
{
for (int j = 0; j < w; ++j)
{
ret[i][j] = matrix[w - j - 1][i];
}
}
return ret;
}
public boolean[][] rotateMatrixLeft(boolean[][] matrix)
{
/* W and H are already swapped */
int w = matrix.length;
int h = matrix[0].length;
boolean[][] ret = new boolean[h][w];
for (int i = 0; i < h; ++i)
{
for (int j = 0; j < w; ++j)
{
ret[i][j] = matrix[j][h - i - 1];
}
}
return ret;
}
Here's my solution in Java to the same problem. I can confirm Martijn's numbers (see below). I've also added in the rough time it takes to compute the results (mid-2012 Macbook Retina Core i7). I suppose substantial performance improvements could be achieved via parallelization.
numberOfStones -> numberOfPolyominos
1 -> 1
2 -> 1
3 -> 2
4 -> 7
5 -> 18
6 -> 60
7 -> 196
8 -> 704 (3 seconds)
9 -> 2500 (46 seconds)
10 -> 9189 (~14 minutes)
.
/*
* This class is a solution to the Tetris unique shapes problem.
* That is, the game of Tetris has 7 unique shapes. These 7 shapes
* are all the possible unique combinations of any 4 adjoining blocks
* (i.e. ignoring rotations).
*
* How many unique shapes are possible with, say, 7 or n blocks?
*
* The solution uses recursive back-tracking to construct all the possible
* shapes. It uses a HashMap to store unique shapes and to ignore rotations.
* It also uses a temporary HashMap so that the program does not needlessly
* waste time checking the same path multiple times.
*
* Even so, this is an exponential run-time solution, with n=10 taking a few
* minutes to complete.
*/
package com.glugabytes.gbjutils;
import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
public class TetrisBlocks {
private HashMap uShapes;
private HashMap tempShapes;
/* Get a map of unique shapes for n squares. The keys are string-representations
* of each shape, and values are corresponding boolean[][] arrays.
* #param squares - number of blocks to use for shapes, e.g. n=4 has 7 unique shapes
*/
public Map getUniqueShapes(int squares) {
uShapes = new HashMap();
tempShapes = new HashMap();
boolean[][] data = new boolean[squares*2+1][squares*2+1];
data[squares][squares] = true;
make(squares, data, 1); //start the process with a single square in the center of a boolean[][] matrix
return uShapes;
}
/* Recursivelly keep adding blocks to the data array until number of blocks(squares) = required size (e.g. n=4)
* Make sure to eliminate rotations. Also make sure not to enter infinite backtracking loops, and also not
* needlessly recompute the same path multiple times.
*/
private void make(int squares, boolean[][] data, int size) {
if(size == squares) { //used the required number of squares
//get a trimmed version of the array
boolean[][] trimmed = trimArray(data);
if(!isRotation(trimmed)) { //if a unique piece, add it to unique map
uShapes.put(arrayToString(trimmed), trimmed);
}
} else {
//go through the grid 1 element at a time and add a block next to an existing block
//do this for all possible combinations
for(int iX = 0; iX < data.length; iX++) {
for(int iY = 0; iY < data.length; iY++) {
if(data[iX][iY] == true) { //only add a block next to an existing block
if(data[iX+1][iY] != true) { //if no existing block to the right, add one and recuse
data[iX+1][iY] = true;
if(!isTempRotation(data)) { //only recurse if we haven't already been on this path before
make(squares, data, size+1);
tempShapes.put(arrayToString(data), data); //store this path so we don't repeat it later
}
data[iX+1][iY] = false;
}
if(data[iX-1][iY] != true) { //repeat by adding a block on the left
data[iX-1][iY] = true;
if(!isTempRotation(data)) {
make(squares, data, size+1);
tempShapes.put(arrayToString(data), data);
}
data[iX-1][iY] = false;
}
if(data[iX][iY+1] != true) { //repeat by adding a block down
data[iX][iY+1] = true;
if(!isTempRotation(data)) {
make(squares, data, size+1);
tempShapes.put(arrayToString(data), data);
}
data[iX][iY+1] = false;
}
if(data[iX][iY-1] != true) { //repeat by adding a block up
data[iX][iY-1] = true;
if(!isTempRotation(data)) {
make(squares, data, size+1);
tempShapes.put(arrayToString(data), data);
}
data[iX][iY-1] = false;
}
}
}
}
}
}
/**
* This function basically removes all rows and columns that have no 'true' flags,
* leaving only the portion of the array that contains useful data.
*
* #param data
* #return
*/
private boolean[][] trimArray(boolean[][] data) {
int maxX = 0;
int maxY = 0;
int firstX = data.length;
int firstY = data.length;
for(int iX = 0; iX < data.length; iX++) {
for (int iY = 0; iY < data.length; iY++) {
if(data[iX][iY]) {
if(iY < firstY) firstY = iY;
if(iY > maxY) maxY = iY;
}
}
}
for(int iY = 0; iY < data.length; iY++) {
for (int iX = 0; iX < data.length; iX++) {
if(data[iX][iY]) {
if(iX < firstX) firstX = iX;
if(iX > maxX) maxX = iX;
}
}
}
boolean[][] trimmed = new boolean[maxX-firstX+1][maxY-firstY+1];
for(int iX = firstX; iX <= maxX; iX++) {
for(int iY = firstY; iY <= maxY; iY++) {
trimmed[iX-firstX][iY-firstY] = data[iX][iY];
}
}
return trimmed;
}
/**
* Return a string representation of the 2D array.
*
* #param data
* #return
*/
private String arrayToString(boolean[][] data) {
StringBuilder sb = new StringBuilder();
for(int iX = 0; iX < data.length; iX++) {
for(int iY = 0; iY < data[0].length; iY++) {
sb.append(data[iX][iY] ? '#' : ' ');
}
sb.append('\n');
}
return sb.toString();
}
/**
* Rotate an array clockwise by 90 degrees.
* #param data
* #return
*/
public boolean[][] rotate90(boolean[][] data) {
boolean[][] rotated = new boolean[data[0].length][data.length];
for(int iX = 0; iX < data.length; iX++) {
for(int iY = 0; iY < data[0].length; iY++) {
rotated[iY][iX] = data[data.length - iX - 1][iY];
}
}
return rotated;
}
/**
* Checks to see if two 2d boolean arrays are the same
* #param a
* #param b
* #return
*/
public boolean equal(boolean[][] a, boolean[][] b) {
if(a.length != b.length || a[0].length != b[0].length) {
return false;
} else {
for(int iX = 0; iX < a.length; iX++) {
for(int iY = 0; iY < a[0].length; iY++) {
if(a[iX][iY] != b[iX][iY]) {
return false;
}
}
}
}
return true;
}
public boolean isRotation(boolean[][] data) {
//check to see if it's a rotation of a shape that we already have
data = rotate90(data); //+90*
String str = arrayToString(data);
if(!uShapes.containsKey(str)) {
data = rotate90(data); //180*
str = arrayToString(data);
if(!uShapes.containsKey(str)) {
data = rotate90(data); //270*
str = arrayToString(data);
if(!uShapes.containsKey(str)) {
return false;
}
}
}
return true;
}
public boolean isTempRotation(boolean[][] data) {
//check to see if it's a rotation of a shape that we already have
data = rotate90(data); //+90*
String str = arrayToString(data);
if(!tempShapes.containsKey(str)) {
data = rotate90(data); //180*
str = arrayToString(data);
if(!tempShapes.containsKey(str)) {
data = rotate90(data); //270*
str = arrayToString(data);
if(!tempShapes.containsKey(str)) {
return false;
}
}
}
return true;
}
/**
* #param args the command line arguments
*/
public static void main(String[] args) {
TetrisBlocks tetris = new TetrisBlocks();
long start = System.currentTimeMillis();
Map shapes = tetris.getUniqueShapes(8);
long end = System.currentTimeMillis();
Iterator it = shapes.keySet().iterator();
while(it.hasNext()) {
String shape = (String)it.next();
System.out.println(shape);
}
System.out.println("Unique Shapes: " + shapes.size());
System.out.println("Time: " + (end-start));
}
}
Here's some python that computes the answer. Seems to agree with Wikipedia. It isn't terribly fast because it uses lots of array searches instead of hash tables, but it still takes only a minute or so to complete.
#!/usr/bin/python
# compute the canonical representation of polyomino p.
# (minimum x and y coordinate is zero, sorted)
def canonical(p):
mx = min(map(lambda v: v[0], p))
my = min(map(lambda v: v[1], p))
return sorted(map(lambda v: (v[0]-mx, v[1]-my), p))
# rotate p 90 degrees
def rotate(p):
return canonical(map(lambda v: (v[1], -v[0]), p))
# add one tile to p
def expand(p):
result = []
for (x,y) in p:
for (dx,dy) in ((-1,0),(1,0),(0,-1),(0,1)):
if p.count((x+dx,y+dy)) == 0:
result.append(canonical(p + [(x+dx,y+dy)]))
return result
polyominos = [[(0,0)]]
for i in xrange(1,10):
new_polyominos = []
for p in polyominos:
for q in expand(p):
dup = 0
for r in xrange(4):
if new_polyominos.count(q) != 0:
dup = 1
break
q = rotate(q)
if not dup: new_polyominos.append(q)
polyominos = new_polyominos
print i+1, len(polyominos)
Here is my full Python solution inspired by #marcog's answer. It prints the number of polyominos of sizes 2..10 in about 2s on my laptop.
The algorithm is straightforward:
Size 1: start with one square
Size n + 1: take all pieces of size n and try adding a single square to all possible adjacent positions. This way you find all possible new pieces of size n + 1. Skip duplicates.
The main speedup came from hashing pieces to quickly check if we've already seen a piece.
import itertools
from collections import defaultdict
n = 10
print("Number of Tetris pieces up to size", n)
# Times:
# n is number of blocks
# - Python O(exp(n)^2): 10 blocks 2.5m
# - Python O(exp(n)): 10 blocks 2.5s, 11 blocks 10.9s, 12 block 33s, 13 blocks 141s (800MB memory)
smallest_piece = [(0, 0)] # We represent a piece as a list of block positions
pieces_of_size = {
1: [smallest_piece],
}
# Returns a list of all possible pieces made by adding one block to given piece
def possible_expansions(piece):
# No flatMap in Python 2/3:
# https://stackoverflow.com/questions/21418764/flatmap-or-bind-in-python-3
positions = set(itertools.chain.from_iterable(
[(x - 1, y), (x + 1, y), (x, y - 1), (x, y + 1)] for (x, y) in piece
))
# Time complexity O(n^2) can be improved
# For each valid position, append to piece
expansions = []
for p in positions:
if not p in piece:
expansions.append(piece + [p])
return expansions
def rotate_90_cw(piece):
return [(y, -x) for (x, y) in piece]
def canonical(piece):
min_x = min(x for (x, y) in piece)
min_y = min(y for (x, y) in piece)
res = sorted((x - min_x, y - min_y) for (x, y) in piece)
return res
def hash_piece(piece):
return hash(tuple(piece))
def expand_pieces(pieces):
expanded = []
#[
# 332322396: [[(1,0), (0,-1)], [...]],
# 323200700000: [[(1,0), (0,-2)]]
#]
# Multimap because two different pieces can happen to have the same hash
expanded_hashes = defaultdict(list)
for piece in pieces:
for e in possible_expansions(piece):
exp = canonical(e)
is_new = True
if exp in expanded_hashes[hash_piece(exp)]:
is_new = False
for rotation in range(3):
exp = canonical(rotate_90_cw(exp))
if exp in expanded_hashes[hash_piece(exp)]:
is_new = False
if is_new:
expanded.append(exp)
expanded_hashes[hash_piece(exp)].append(exp)
return expanded
for i in range(2, n + 1):
pieces_of_size[i] = expand_pieces(pieces_of_size[i - 1])
print("Pieces with {} blocks: {}".format(i, len(pieces_of_size[i])))