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I have a matrix
and I should write code using Gauss-Seidel and coupled gradient methods taking the structure of the matrix.
Ax = e where A is matrix and e is vector with values of 1
I don't know how to write code using coupled gradient methods and my Gauss-Seidel algorithm don't have main part where I add this all thinks
//part of gauss-seidel method
const int N = 128; //size of array
const int no_of_iter = 128; //iterations
int main() {
double result[N]; //array for result
double result_pom[N]; //temporary result array
double sum = 0.0;
int x, y;
for (int i = 0; i < no_of_iter; i++) {
for (y = 0; y < N; y++) {
result_pom[y] = result[y]; //set values of result to result_pom
}
for (x = 0; x < N; x++) {
sum = 0.0;
//for functions where I add (x,y) el
result[x] = 0.25 * (1 - sum); //because 4 is dominant el of matrix and
//1 is value of vector e
}
}
}
I am trying to implement Laplace sharpening using C++ , here's my code so far:
img = imread("cow.png", 0);
Mat convoSharp() {
//creating new image
Mat res = img.clone();
for (int y = 0; y < res.rows; y++) {
for (int x = 0; x < res.cols; x++) {
res.at<uchar>(y, x) = 0.0;
}
}
//variable declaration
int filter[3][3] = { {0,1,0},{1,-4,1},{0,1,0} };
//int filter[3][3] = { {-1,-2,-1},{0,0,0},{1,2,1} };
int height = img.rows;
int width = img.cols;
int filterHeight = 3;
int filterWidth = 3;
int newImageHeight = height - filterHeight + 1;
int newImageWidth = width - filterWidth + 1;
int i, j, h, w;
//convolution
for (i = 0; i < newImageHeight; i++) {
for (j = 0; j < newImageWidth; j++) {
for (h = i; h < i + filterHeight; h++) {
for (w = j; w < j + filterWidth; w++) {
res.at<uchar>(i,j) += filter[h - i][w - j] * img.at<uchar>(h,w);
}
}
}
}
//img - laplace
for (int y = 0; y < res.rows; y++) {
for (int x = 0; x < res.cols; x++) {
res.at<uchar>(y, x) = img.at<uchar>(y, x) - res.at<uchar>(y, x);
}
}
return res;
}
I don't really know what went wrong, I also tried different filter (1,1,1),(1,-8,1),(1,1,1) and the result is also same (more or less). I don't think that I need to normalize the result because the result is in range of 0 - 255. Can anyone explain what really went wrong in my code?
Problem: uchar is too small to hold partial results of filerting operation.
You should create a temporary variable and add all the filtered positions to this variable then check if value of temp is in range <0,255> if not, you need to clamp the end result to fit <0,255>.
By executing below line
res.at<uchar>(i,j) += filter[h - i][w - j] * img.at<uchar>(h,w);
partial result may be greater than 255 (max value in uchar) or negative (in filter you have -4 or -8). temp has to be singed integer type to handle the case when partial result is negative value.
Fix:
for (i = 0; i < newImageHeight; i++) {
for (j = 0; j < newImageWidth; j++) {
int temp = res.at<uchar>(i,j); // added
for (h = i; h < i + filterHeight; h++) {
for (w = j; w < j + filterWidth; w++) {
temp += filter[h - i][w - j] * img.at<uchar>(h,w); // add to temp
}
}
// clamp temp to <0,255>
res.at<uchar>(i,j) = temp;
}
}
You should also clamp values to <0,255> range when you do the subtraction of images.
The problem is partially that you’re overflowing your uchar, as rafix07 suggested, but that is not the full problem.
The Laplace of an image contains negative values. It has to. And you can’t clamp those to 0, you need to preserve the negative values. Also, it can values up to 4*255 given your version of the filter. What this means is that you need to use a signed 16 bit type to store this output.
But there is a simpler and more efficient approach!
You are computing img - laplace(img). In terms of convolutions (*), this is 1 * img - laplace_kernel * img = (1 - laplace_kernel) * img. That is to say, you can combine both operations into a single convolution. The 1 kernel that doesn’t change the image is [(0,0,0),(0,1,0),(0,0,0)]. Subtract your Laplace kernel from that and you obtain [(0,-1,0),(-1,5,-1),(0,-1,0)].
So, simply compute the convolution with that kernel, and do it using int as intermediate type, which you then clamp to the uchar output range as shown by rafix07.
I want to generate an array of integers where the total sum of each row and column in the array is known , for example if I create a 4 by 4 array in c++ and then populate it pseudo randomly with numbers between 1 and 100:
int array[4][4] = {} ;
for(int x = 0 ; x<4 ; x++){
for(int y = 0 ; y<4 ; y++){
array[x][y] = rand() % 100 + 1 ;
}
}
the array would be :
8, 50, 74, 59
31, 73, 45, 79
24, 10, 41, 66
93, 43, 88, 4
then if I sum each row and each column by :
int rowSum[4] = {} ;
int columnSum[4] = {} ;
for(int x = 0 ; x < 4; x++){
for(int y = 0 ; y < 4; y++){
rowSum[x] += array[x][y] ;
columnSum[y] += array[x][y] ;
}
}
the rowSum would be {191,228,141,228} and the columnSum = {156,176,248,208}
what I'm trying to do at this point is to generate any random 4x4 1~100 array that will satisfy rowSum and columnSum I understand there is thousands of different arrays that will sum up to the same row and column sum ,and I've been trying to write the part of the code that will generate it , I would really appreciate it if anyone can give me a clue .
It is very easy to find some solution.
Start with generating row that sum to given values. It could be as simple as making all values in each row approximately equal to rowSum[i]/n, give or take one. Of course sums of columns will not match at this point.
Now fix the columns from the leftmost to the rightmost. To fix i th column, distribute the difference between the desired sum and the actual sum equally between column entries, and then fix each row by distributing the added value equally between items i+1...n of the row.
It is easier done than said:
void reconstruct (int array[4][4], int rows[4], int cols[4])
{
// build an array with each row adding up to the correct row sum
for (int x = 0; x < 4; x++){
int s = rows[x];
for(int y = 0; y < 4 ; y++){
array[x][y] = s / (4 - y);
s -= array[x][y];
}
}
// adjust columns
for(int y = 0; y < 4 ; y++){
// calculate the adjustment
int s = 0;
for (int x = 0; x < 4; x++){
s += array[x][y];
}
int diff = s - cols[y];
// adjust the column by diff
for (int x = 0; x < 4; x++){
int k = diff / (4 - x);
array[x][y] -= k;
diff -= k;
// adjust the row by k
for (int yy = y + 1; yy < 4; ++yy)
{
int corr = k / (4 - yy);
array[x][yy] += corr;
k -= corr;
}
}
}
}
This array won't be random of course. One can randomise it by selecting x1, x2, y1, y2 and d at random and executing:
array[x1][y1] += d
array[x1][y2] -= d
array[x2][y1] -= d
array[x2][y2] += d
taking care that the resulting values won't spill out of the desired range.
Here's the quick and dirty brute force search mentioned in comments. It ought to give you a starting point. This is C, not C++.
You never said it, but I'm assuming you want the matrix elements to be non-negative. Consequently, this searches the space where each element a[i][j] can have any value in [0..min(rowsum[i], colsum[j])] with the search cut off when assigning the next array element value would admit no possible future solution.
#include <stdio.h>
int a[4][4] = {
{-1, -1, -1, -1},
{-1, -1, -1, -1},
{-1, -1, -1, -1},
{-1, -1, -1, -1}};
int rs[] = {191, 228, 141, 228};
int cs[] = {156, 176, 248, 208};
long long n_solutions = 0;
void research(int i, int j, int ii, int jj, int val);
void print_a(void);
void search(int i, int j) {
if (j < 3) {
if (i < 3) {
int m = rs[i] < cs[j] ? rs[i] : cs[j];
for (int val = 0; val <= m; ++val) research(i, j, i, j + 1, val);
} else {
if (rs[3] >= cs[j]) research(i, j, i, j + 1, cs[j]);
}
} else {
if (i < 3) {
if (cs[j] >= rs[i]) research(i, 3, i + 1, 0, rs[i]);
} else {
if (rs[3] == cs[3]) {
a[3][3] = rs[i];
if (++n_solutions % 100000000 == 0) {
printf("\n%lld\n", n_solutions);
print_a();
}
a[3][3] = -1;
}
}
}
}
void research(int i, int j, int ii, int jj, int val) {
a[i][j] = val; rs[i] -= val; cs[j] -= val;
search(ii, jj);
rs[i] += val; cs[j] += val; a[i][j] = -1;
}
void print_a(void) {
for (int i = 0; i < 4; ++i) {
for (int j = 0; j < 4; ++j)
printf("%4d", a[i][j]);
printf("\n");
}
}
int main(void) {
search(0, 0);
printf("Total solutions: %lld\n", n_solutions);
return 0;
}
For example, if you replace the simple for loop with this, you won't get so many zeros in the upper left hand corner:
int b = m / 2; // m/2 can be replaced with any int in [0..m], e.g. a random value.
research(i, j, i, j + 1, b);
for (int d = 1; b + d <= m || b - d >= 0; ++d) {
if (b + d <= m) research(i, j, i, j + 1, b + d);
if (b - d >= 0) research(i, j, i, j + 1, b - d);
}
Here's the 2-billionth solution:
78 56 28 29
39 20 84 85
28 34 61 18
11 66 75 76
The problem becomes interesting if we place condition that the matrix elements must be non-negative integers. Here's an O(mn) JAVA solution based on greedy algorithm.
int m=rowSum.length;
int n=colSum.length;
int mat[][] = new int[m][n];
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
int tmp=Math.min(rowSum[i],colSum[j]);
mat[i][j]=tmp;
rowSum[i]-=tmp;
colSum[j]-=tmp;
}
}
return mat;
i want to transport the follow codes into c++:
gaussFilter = fspecial('gaussian', 2*neighSize+1, 0.5*neighSize);
pointFeature = imfilter(pointFeature, gaussFilter, 'symmetric');
where the pointFeature is a [height, width, 24] array.
i try to use filter2D, but it only support the 2D array.
so i want to know if there are functions in opencv that can filtering the multi-dimensional array?
You can use separable kernel filters for make anydimentional filter.
If you are using OpenCV, you could try this for a 3 Dimensional MatND:
void Smooth3DHist(cv::MatND &hist, const int& kernDimension)
{
assert(hist.dims == 3);
int x_size = hist.size[0];
int y_size = hist.size[1];
int z_size = hist.size[2];
int xy_size = x_size*y_size;
cv::Mat kernal = cv::getGaussianKernel(kernDimension, -1, CV_32F);
// Filter XY dimensions for every Z
for (int z = 0; z < z_size; z++)
{
float *ind = (float*)hist.data + z * xy_size; // sub-matrix pointer
cv::Mat subMatrix(2, hist.size, CV_32F, ind);
cv::sepFilter2D(subMatrix, subMatrix, CV_32F, kernal.t(), kernal, Point(-1,-1), 0.0, cv::BORDER_REPLICATE);
}
// Filter Z dimension
float* kernGauss = (float *)kernal.data;
unsigned kernSize = kernal.total();
int kernMargin = (kernSize - 1)/2;
float* lineBuffer = new float[z_size + 2*kernMargin];
for (int y = 0; y < y_size; y++)
{
for (int x = 0; x < x_size; x++)
{
// Copy along Z dimension into a line buffer
float* z_ptr = (float*)hist.data + y * x_size + x;//same as hist.ptr<float>(0, y, x)
for (int z = 0; z < z_size; z++, z_ptr += xy_size)
{
lineBuffer[z + kernMargin] = *z_ptr;
}
// Replicate borders
for (int m = 0; m < kernMargin; m++)
{
lineBuffer[m] = lineBuffer[kernMargin];// replicate left side
lineBuffer[z_size + 2*kernMargin - 1 - m] = lineBuffer[kernMargin + z_size - 1];//replicate right side
}
// Filter line buffer 1D - convolution
z_ptr = (float*)hist.data + y * x_size + x;
for (int z = 0; z < z_size; z++, z_ptr += xy_size)
{
*z_ptr = 0.0f;
for (unsigned k = 0; k < kernSize; k++)
{
*z_ptr += lineBuffer[z+k]*kernGauss[k];
}
}
}
}
delete [] lineBuffer;
}
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])))