I do not know how to generate the Blanking and Sync signal for my PAL signal. I am working directly with pixel values from an image, and putting them into the formulas from the standard.
I took the pixel R, G and B values and computed the luminance Y.
I tried to make the sync and blanking signal manual but with no results.
//load values
float[,] rValues = new float[picture1.Width, picture1.Height];
float[,] gValues = new float[picture1.Width, picture1.Height];
float[,] bValues = new float[picture1.Width, picture1.Height];
using (Bitmap bmp = new Bitmap(picture1))
{
for (int i=0;i<bmp.Width;i++)
{
for(int j=0;j<bmp.Height;j++)
{
Color clr = bmp.GetPixel(i, j);
rValues[i, j] = clr.R;
gValues[i, j] = clr.G;
bValues[i, j] = clr.B;
}
}
}
//changing the matrices into 1d arrays
----------
double[] r = new double[picture1.Height * picture1.Width];
double[] g = new double[picture1.Height * picture1.Width];
double[] b = new double[picture1.Height * picture1.Width];
int k = 0;
for (int i = 0; i < picture1.Height; i++)
{
for (int j = 0; j < picture1.Width; j++)
{
r[k] = rValues[j, i];
g[k] = gValues[j, i];
b[k] = bValues[j, i];
k++;
}
}
// calculating the luminance Y
double[] Y = new double[picture1.Height * picture1.Width];
for (int i=0; i < Y.Length; i++)
{
Y[i] = 0.3 * r[i] + 0.59 * g[i] + 0.11 * b[i];
}
// trying to make a manual signal
double[] sync = new double[135];
for (int i = 0; i < 135; i++)
{
if (i < 17)
{
sync[i] = 0;
}
if (i > 16 && i < 70)
{
sync[i] = -0.3;
}
if (i > 69)
{
sync[i] = 0;
}
}
I want to "stretch" a one-dimensional float array into a bigger array.
//expected behaviour
float[] initialArray = {2.0, 6.5, 2.0}
float[] biggerArray = resample(initialArray, 7 /*new size*/)
//output: {2.0, 3.5, 5.0, 6.5, 5.0, 3.5, 2.0}
The new values should propobaly be calculated from linear interpolation of the previous array values but i can't figure out how to achieve that.
Any hint ?
Lets the length of a source array is N, and the length of a destination array is M where N < M and N > 1.
You can calculate the new index of the source i-th element by the formula:
j = i * (M - 1)/(N - 1);
When i == 0 then j == 0; and when i == N - 1 then j == M - 1. The external loop can looks like this:
float[] source = ...;
float[] destination = ...;
destination[0] = source[0];
for (int i = 1; i < source.Length; i++)
{
int j = i * (destination.Length - 1)/(source.Length - 1);
destination[j] = source[i];
// interpolation
}
To interpolation you should calculate intermediate values for each pair (source[i - 1], source[i]). You'll need to store previous value of j:
destination[0] = source[0];
int jPrevious = 0;
for (int i = 1; i < source.Length; i++)
{
int j = i * (destination.Length - 1)/(source.Length - 1);
Interpolate(destination, jPrevious, j, source[i - 1], source[i]);
jPrevious = j;
}
private static void Interpolate(float[] destination, int destFrom, int destTo, float valueFrom, float valueTo)
{
int destLength = destTo - destFrom;
float valueLength = valueTo - valueFrom;
for (int i = 0; i <= destLength; i++)
destination[destFrom + i] = valueFrom + (valueLength * i)/destLength;
}
This one works for both if source size is larger than destination and vice versa.
private double[] Resample(double[] source, int n)
{
//n destination length
int m = source.Length; //source length
double[] destination = new double[n];
destination[0] = source[0];
destination[n-1] = source[m-1];
for (int i = 1; i < n-1; i++)
{
double jd = ((double)i * (double)(m - 1) / (double)(n - 1));
int j = (int)jd;
destination[i] = source[j] + (source[j + 1] - source[j]) * (jd - (double)j);
}
return destination;
}
You can use List<float>:
float[] initialArray = { 2.0f, 6.5f, 2.0f };
List<float> initialArrayTemp = ToListFloat(initialArray);
private List<float> ToListFloat(float[] array)
{
List<float> list = new List<float>();
for (int i = 0; i < array.Length; i++)
{
list.Add(array[i]);
}
return list;
}
Now your array is a dynamic array and you can add your new nodes anywhere of your Array by using Insert() method.
As soon as you need a new static array, use below:
float[] newInitialArray = initialArrayTemp.ToArray();
Below is the code I wrote for clustering using genetic algorithm. Points are from a picturebox, generated randomly (X,Y) before calling this class. However, the result of this algorithm is much worse than k-means or lbg I'm comparing it to. Can someone take a look for any errors in the algorithm, maybe I omitted something. Thanks.
I did this using arrays, the 2 other I did using lists, but I don't think that should have any impact on result.
public class geneticAlgorithm
{
static int pom = 0;
static PictureBox pb1;
public geneticAlgorithm(PictureBox pb)
{
pb1 = pb;
}
public static void doGA(PointCollection points, int clusterCounter)
//points is a list of points,
//those point have (X,Y) coordinates generated randomly from pictureBox
//coordinates. clusterCounter is how many clusters I want to divide the points into
{
//this part converts list of points into array testtab,
//where each array field hold X,Y of a point
Point[] test = new Point[points.Count];
test = points.ToArray();
double[][] testtab = new double[test.Length][];
for (int i = 0; i < testtab.GetLength(0); i++)
{
testtab[i] = new double[2];
testtab[i][0] = test[i].X;
testtab[i][1] = test[i].Y;
}
//end of converting
int n = testtab.GetLength(0);
int k = clusterCounter;
int chromosomeCount = 500;
int dimensions = 2;
double[][] testClusters = new double[k][];
for (int i = 0; i < k; i++)
{
testClusters[i] = new double[dimensions];
}
double[][] testChromosomes = new double[chromosomeCount][];
for (int i = 0; i < chromosomeCount; i++)
{
testChromosomes[i] = new double[2 * k];
}
int[][] testChromosomesInt = new int[chromosomeCount][];
for (int i = 0; i < chromosomeCount; i++)
{
testChromosomesInt[i] = new int[2 * k];
}
double[] partner = new double[chromosomeCount];
double[][] roulette = new double[chromosomeCount][];
for (int i = 0; i < chromosomeCount; i++)
{
roulette[i] = new double[1];
}
double[][] errors = new double[chromosomeCount][];
for (int i = 0; i < chromosomeCount; i++)
{
errors[i] = new double[1];
}
double crossingPossibility = 0.01;
double mutationPossibility = 0.0001;
int maxIterations = 10000;
//here I create chromosomes and initial clusters
for (int j = 0; j < chromosomeCount; j++)
{
for (int i = 0; i < k; i++)
{
Random rnd = new Random();
int r = rnd.Next(n);
for (int q = 0; q < dimensions; q++)
{
testClusters[i][q] = testtab[r][q];
}
}
int help = 0;
for (int i = 0; i < k; i++)
for (int l = 0; l < dimensions; l++) // here is creation of chromosome
{
testChromosomes[j][help] = testClusters[i][l];
help++;
}
//end
//here I call accomodation function to see which of them are good
errors[j][0] = accomodationFunction(testClusters, testtab, n, k);
//end
//cleaning of the cluster table
testClusters = new double[k][];
for (int i = 0; i < k; i++)
{
testClusters[i] = new double[dimensions];
}
}
//end
for (int counter = 0; counter < maxIterations; counter++)
{
//start of the roulette
double s = 0.0;
for (int i = 0; i < chromosomeCount; i++)
s += errors[i][0];
for (int i = 0; i < chromosomeCount; i++)
errors[i][0] = chromosomeCount * errors[i][0] / s;
int idx = 0;
for (int i = 0; i < chromosomeCount; i++)
for (int j = 0; i < errors[i][0] && idx < chromosomeCount; j++)
{
roulette[idx++][0] = i;
}
double[][] newTab = new double[chromosomeCount][];
for (int i = 0; i < chromosomeCount; i++)
{
newTab[i] = new double[2 * k];
}
Random rnd = new Random();
for (int i = 0; i < chromosomeCount; i++)
{
int q = rnd.Next(chromosomeCount);
newTab[i] = testChromosomes[(int)roulette[q][0]];
}
testChromosomes = newTab;
//end of roulette
//start of crossing chromosomes
for (int i = 0; i < chromosomeCount; i++)
partner[i] = (rnd.NextDouble() < crossingPossibility + 1) ? rnd.Next(chromosomeCount) : -1;
for (int i = 0; i < chromosomeCount; i++)
if (partner[i] != -1)
testChromosomes[i] = crossing(testChromosomes[i], testChromosomes[(int)partner[i]], rnd.Next(2 * k), k);
//end of crossing
//converting double to int
for (int i = 0; i < chromosomeCount; i++)
for (int j = 0; j < 2 * k; j++)
testChromosomes[i][j] = (int)Math.Round(testChromosomes[i][j]);
//end of conversion
//start of mutation
for (int i = 0; i < chromosomeCount; i++)
if (rnd.NextDouble() < mutationPossibility + 1)
testChromosomesInt[i] = mutation(testChromosomesInt[i], rnd.Next(k * 2), rnd.Next(10));
//end of mutation
}
//painting of the found centre on the picture box
int centrum = max(errors, chromosomeCount);
Graphics g = pb1.CreateGraphics();
SolidBrush brush = new SolidBrush(Color.Red);
for (int i = 0; i < 2 * k - 1; i += 2)
{
g.FillRectangle(brush, testChromosomesInt[centrum][i], testChromosomesInt[centrum][i + 1], 20, 20);
}
return;
}
//end of painting
public static int max(double[][] tab, int chromosomeCount)
{
double max = 0;
int k = 0;
for (int i = 0; i < chromosomeCount; i++)
{
if (max < tab[i][0])
{
max = tab[i][0];
k = i;
}
}
return k;
}
public static int[] mutation(int[] tab, int elem, int bit)
{
int mask = 1;
mask <<= bit;
tab[elem] = tab[elem] ^ mask;
return tab;
}
public static double[] crossing(double[] tab, double[] tab2, int p, int k)
{
double[] hold = new double[2 * k];
for (int i = 0; i < p; i++)
hold[i] = tab[i];
for (int i = p; i < 2 * k; i++)
hold[i] = tab2[i];
return hold;
}
//accomodation function, checks to which centre which point belongs based on distance
private static double accomodationFunction(double[][] klastry, double[][] testtab, int n, int k)
{
double Error = 0;
for (int i = 0; i < n; i++)
{
double min = 0;
int ktory = 0;
for (int j = 0; j < k; j++)
{
double dist = distance(klastry[j], testtab[i]);
if (j == 0)
{
min = dist;
}
if (min > dist)
{
min = dist;
ktory = j;
}
}
Error += min;
}
pom++;
return 1 / Error;
}
public static double distance(double[] tab, double[] tab2)
{
double dist = 0.0;
for (int i = 0; i < tab.GetLength(0); i++)
dist += Math.Pow(tab[i] - tab2[i], 2);
return dist;
}
}
The algorithm should work like so: (excuse me if not the best english)
1. Get random points (let's say 100)
2. Check into how many clusters we want to split them (the same thing I would do using k-means for example
3. Get starting population of chromosomes
4. Throu cutting on the roulette, crossing and mutation pick the best x centres, where x is the amount of clusters we want
5. Paint them.
Here are some results, and why I think it's wrong: (it's using 100 points, 5 clusters)
k-means:
lbg:
genetic(without colors now):
I hope this clarifies a bit.
I come looking for general tips about the program I'm writing now.
The goal is:
Use neural network program to recognize 3 letters [D,O,M] (or display "nothing is recognized" if i input anything other than those 3).
Here's what I have so far:
A class for my single neuron
public class neuron
{
double[] weights;
public neuron()
{
weights = null;
}
public neuron(int size)
{
weights = new double[size + 1];
Random r = new Random();
for (int i = 0; i <= size; i++)
{
weights[i] = r.NextDouble() / 5 - 0.1;
}
}
public double output(double[] wej)
{
double s = 0.0;
for (int i = 0; i < weights.Length; i++) s += weights[i] * wej[i];
s = 1 / (1 + Math.Exp(s));
return s;
}
}
A class for a layer:
public class layer
{
neuron[] tab;
public layer()
{
tab = null;
}
public layer(int numNeurons, int numInputs)
{
tab = new neuron[numNeurons];
for (int i = 0; i < numNeurons; i++)
{
tab[i] = new neuron(numInputs);
}
}
public double[] compute(double[] wejscia)
{
double[] output = new double[tab.Length + 1];
output[0] = 1;
for (int i = 1; i <= tab.Length; i++)
{
output[i] = tab[i - 1].output(wejscia);
}
return output;
}
}
And finally a class for a network
public class network
{
layer[] layers = null;
public network(int numLayers, int numInputs, int[] npl)
{
layers = new layer[numLayers];
for (int i = 0; i < numLayers; i++)
{
layers[i] = new layer(npl[i], (i == 0) ? numInputs : (npl[i - 1]));
}
}
double[] compute(double[] inputs)
{
double[] output = layers[0].compute(inputs);
for (int i = 1; i < layers.Length; i++)
{
output = layers[i].compute(output);
}
return output;
}
}
Now for the algorythm I chose:
I have a picture box, size 200x200, where you can draw a letter (or read one from jpg file).
I then convert it to my first array(get the whole picture) and 2nd one(cut the non relevant background around it) like so:
Bitmap bmp2 = new Bitmap(this.pictureBox1.Image);
int[,] binaryfrom = new int[bmp2.Width, bmp2.Height];
int minrow=0, maxrow=0, mincol=0, maxcol=0;
for (int i = 0; i < bmp2.Height; i++)
{
for (int j = 0; j < bmp2.Width; j++)
{
if (bmp2.GetPixel(j, i).R == 0)
{
binaryfrom[i, j] = 1;
if (minrow == 0) minrow = i;
if (maxrow < i) maxrow = i;
if (mincol == 0) mincol = j;
else if (mincol > j) mincol = j;
if (maxcol < j) maxcol = j;
}
else
{
binaryfrom[i, j] = 0;
}
}
}
int[,] boundaries = new int[binaryfrom.GetLength(0)-minrow-(binaryfrom.GetLength(0)-(maxrow+1)),binaryfrom.GetLength(1)-mincol-(binaryfrom.GetLength(1)-(maxcol+1))];
for(int i = 0; i < boundaries.GetLength(0); i++)
{
for(int j = 0; j < boundaries.GetLength(1); j++)
{
boundaries[i, j] = binaryfrom[i + minrow, j + mincol];
}
}
And convert it to my final array of 12x8 like so (i know I could shorten this a fair bit, but wanted to have every step in different loop so I can see what went wrong easier[if anything did]):
int[,] finalnet = new int[12, 8];
int k = 1;
int l = 1;
for (int i = 0; i < finalnet.GetLength(0); i++)
{
for (int j = 0; j < finalnet.GetLength(1); j++)
{
finalnet[i, j] = 0;
}
}
while (k <= finalnet.GetLength(0))
{
while (l <= finalnet.GetLength(1))
{
for (int i = (int)(boundaries.GetLength(0) / finalnet.GetLength(0)) * (k - 1); i < (int)(boundaries.GetLength(0) / finalnet.GetLength(0)) * k; i++)
{
for (int j = (int)(boundaries.GetLength(1) / finalnet.GetLength(1)) * (l - 1); j < (int)(boundaries.GetLength(1) / finalnet.GetLength(1)) * l; j++)
{
if (boundaries[i, j] == 1) finalnet[k-1, l-1] = 1;
}
}
l++;
}
l = 1;
k++;
}
int a = boundaries.GetLength(0);
int b = finalnet.GetLength(1);
if((a%b) != 0){
k = 1;
while (k <= finalnet.GetLength(1))
{
for (int i = (int)(boundaries.GetLength(0) / finalnet.GetLength(0)) * finalnet.GetLength(0); i < boundaries.GetLength(0); i++)
{
for (int j = (int)(boundaries.GetLength(1) / finalnet.GetLength(1)) * (k - 1); j < (int)(boundaries.GetLength(1) / finalnet.GetLength(1)) * k; j++)
{
if (boundaries[i, j] == 1) finalnet[finalnet.GetLength(0) - 1, k - 1] = 1;
}
}
k++;
}
}
if (boundaries.GetLength(1) % finalnet.GetLength(1) != 0)
{
k = 1;
while (k <= finalnet.GetLength(0))
{
for (int i = (int)(boundaries.GetLength(0) / finalnet.GetLength(0)) * (k - 1); i < (int)(boundaries.GetLength(0) / finalnet.GetLength(0)) * k; i++)
{
for (int j = (int)(boundaries.GetLength(1) / finalnet.GetLength(1)) * finalnet.GetLength(1); j < boundaries.GetLength(1); j++)
{
if (boundaries[i, j] == 1) finalnet[k - 1, finalnet.GetLength(1) - 1] = 1;
}
}
k++;
}
for (int i = (int)(boundaries.GetLength(0) / finalnet.GetLength(0)) * finalnet.GetLength(0); i < boundaries.GetLength(0); i++)
{
for (int j = (int)(boundaries.GetLength(1) / finalnet.GetLength(1)) * finalnet.GetLength(1); j < boundaries.GetLength(1); j++)
{
if (boundaries[i, j] == 1) finalnet[finalnet.GetLength(0) - 1, finalnet.GetLength(1) - 1] = 1;
}
}
}
The result is a 12x8 (I can change it in the code to get it from some form controls) array of 0 and 1, where 1 form the rough shape of a letter you drawn.
Now my questions are:
Is this a correct algorythm?
Is my function
1/(1+Math.Exp(x))
good one to use here?
What should be the topology? 2 or 3 layers, and if 3, how many neurons in hidden layer? I have 96 inputs (every field of the finalnet array), so should I also take 96 neurons in the first layer? Should I have 3 neurons in the final layer or 4(to take into account the "not recognized" case), or is it not necessary?
Thank you for your help.
EDIT: Oh, and I forgot to add, I'm gonna train my network using Backpropagation algorythm.
You may need 4 layers at least to get accurate results using back propagation method. 1 input, 2 middle layers, and an output layer.
12 * 8 matrix is too small(and you may end up in data loss which will result in total failure) - try some thing 16 * 16. If you want to reduce the size then you have to peel out the outer layers of black pixels further.
Think about training the network with your reference characters.
Remember that you have to feed back the output back to the input layer again and iterate it multiple times.
A while back I created a neural net to recognize digits 0-9 (python, sorry), so based on my (short) experience, 3 layers are ok and 96/50/3 topology will probably do a good job. As for the output layer, it's your choice; you can either backpropagate all 0s when the input image is not a D, O or M or use the fourth output neuron to indicate that the letter was not recognized. I think that the first option would be the best one because it's simpler (shorter training time, less problems debugging the net...), you just need to apply a threshold under which you classify the image as 'not recognized'.
I also used the sigmoid as activation function, I didn't try others but it worked :)
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])))