Related
I have a 2D array I have the
number of rows = H
number of columns = W
the 2d array itself = arr
they are all integers
I am tasked to return the lowest sum of the vertical line starts with each point at first row
input
1 2 3
4 5 6
7 8 9
output 12
I thought about a way to solve this by using recursion but i don't get right results.
the function takes the array, the position i want to calculate minimum sum at (should be on the first-row coz it's a line) number of columns and rows and res is to return the sum res is initialized in the main function by arr at row x and column y
I am sure about the idea but the way I am summing is probably wrong
static int Summ(int [,]arr,int x,int y,int W,int H,int res)
{
if (x == H - 1)
res += arr[x, y];
else
if (y == 0)
res += Math.Min(Summ(arr, x + 1, y, W, H, res), Summ(arr, x + 1, y + 1, W, H, res));
else if (y== W-1)
res += Math.Min(Summ(arr, x + 1, y, W, H, res), Summ(arr, x + 1, y - 1, W, H, res));
else
res+= res += Math.Min(Math.Min(Summ(arr, x + 1, y, W, H, res), Summ(arr, x + 1, y + 1, W, H, res)),Summ(arr,x+1,y-1,W,H,res));
return res;
}
An iterative solution may be simpler:
static int MinSum(int [,]arr)
{
int min = Int32.MaxValue;
for (int j = 0; j < arr.GetLength(1); ++j)
{
int sum = 0;
for (int i = 0; i < arr.GetLength(0); ++i)
{
sum += arr[i, j];
}
min = Math.min(min, sum);
}
return min;
}
EDIT:
Given the clarification in the comment below, this solution is not correct.
Instead, you could iterate over the rows and sum the minimal value in each row:
static int MinSum(int [,]arr)
{
int sum = 0;
for (int i = 0; j < arr.GetLength(0); ++i)
{
int minCol = arr[i, 0];
for (int j = 1; j < arr.GetLength(1); +_j)
{
minCoal = Math.min(minCol, arr[i, j]);
}
min += minCol;
}
return sum;
}
I'm currently trying to make a code, that will help to count efficient reactors set for game StarMade.
I'm using recursive method to explore the 3d tree of elements and find all related groups. For ex. group - it's a cluster of elements, that stay close to each other.
On picture something like this:
XOX
OOX
XXO
where O is nothing, and X is reactor (element).
On this picture there are 3 groups of elements. [0,0], [2,0]-[2,1], [0,2]-[1,2]
Another variant:
XXX
OOX
XXX
Here is only one group, because all elements stay close to each other.
Here is my code:
void CheckGroup(int x, int y, int z, Group group)
{
if(x >= maxz || x < 0 || y >= maxy || y < 0 || z >= maxz || z < 0)
{
return;
}
if (reactorsChecked[x, y, z])
{
return;
}
reactorsChecked[x, y, z] = true;
if (reactors[x, y, z])
{
if (group == null)
{
group = new Group();
group.MaxX = x;
group.MaxY = y;
group.MaxZ = z;
group.MinX = x;
group.MinY = y;
group.MinZ = z;
group.Blocks = 1;
}
else
{
group.MaxX = Math.Max(group.MaxX, x);
group.MaxY = Math.Max(group.MaxY, y);
group.MaxZ = Math.Max(group.MaxZ, z);
group.MinX = Math.Min(group.MinX, x);
group.MinY = Math.Min(group.MinY, y);
group.MinZ = Math.Min(group.MinZ, z);
group.Blocks += 1;
}
CheckGroup(x + 1, y, z, group);
CheckGroup(x - 1, y, z, group);
CheckGroup(x, y + 1, z, group);
CheckGroup(x, y - 1, z, group);
CheckGroup(x, y, z + 1, group);
CheckGroup(x, y, z - 1, group);
if (!groups.Contains(group))
{
groups.Add(group);
}
}
}
group - is simple class for cluster, that store data about elements count in this cluster and bounding box of this cluster.
reactorsChecked - is simple bool[,,] array, that store information about elements, that we have checked, to avoid doubles
reactor - simple bool[,,] array of random elements.
At first I insert random values to reactors array, and then call CheckGroup(x,y,z,null). If reactors array size less then 25x25x25, then all ok. In single thread size of array could be 100x100x100 and all would be ok. But if I try to use Parallel.For, then I got StackOverflow after near 9000 recursions...
Here is full code:
Parallel.For(0, Environment.ProcessorCount, (i) =>
{
Calculator calc = new Calculator(x, y, z, max, cycles);
calcs.Add(calc);
});
public class Calculator
{
Random rnd = new Random();
//List<Group> groups = new List<Group>();
HashSet<Group> groups = new HashSet<Group>();
bool[, ,] reactors;
public bool[, ,] reactorsMax;
bool[, ,] reactorsChecked;
public double maxEnergy = 0;
public string result = "";
public string resultPic = "";
int maxx, maxy, maxz;
public Calculator(int x, int y, int z, int max, int cycles)
{
maxx = x;
maxy = y;
maxz = z;
maxEnergy = max;
for (int i = 0; i < cycles; i++)//check few variants per thread
{
Calculate(x,y,z);
}
}
private void Calculate(int X, int Y, int Z)
{
//groups = new List<Group>();
groups = new HashSet<Group>();
reactors = new bool[X, Y, Z];
for (int x = 0; x < X; x++)
{
for (int y = 0; y < Y; y++)
{
for (int z = 0; z < Z; z++)
{
reactors[x, y, z] = rnd.Next(2)==1;//fill array with random values
}
}
}
reactorsChecked = new bool[X, Y, Z];
for (int x = 0; x < X; x++)
{
for (int y = 0; y < Y; y++)
{
for (int z = 0; z < Z; z++)
{
CheckGroup(x, y, z, null);//start calculations
}
}
}
double sum = 0;
int blocks = 0;
foreach(Group g in groups)
{
float dims = g.MaxX - g.MinX + g.MaxY - g.MinY + g.MaxZ - g.MinZ + 3;
sum += (2000000.0f / (1.0f + Math.Pow(1.000696f, (-0.333f * Math.Pow((dims / 3.0f), 1.7)))) - 1000000.0f + 25.0f * g.Blocks);
blocks += g.Blocks;
}
if (sum > maxEnergy)
{
maxEnergy = sum;
reactorsMax = reactors;
}
}
void CheckGroup(int x, int y, int z, Group group)
{
if(x >= maxz || x < 0 || y >= maxy || y < 0 || z >= maxz || z < 0)
{
return;
}
if (reactorsChecked[x, y, z])
{
return;
}
reactorsChecked[x, y, z] = true;
if (reactors[x, y, z])
{
if (group == null)
{
group = new Group();
group.MaxX = x;
group.MaxY = y;
group.MaxZ = z;
group.MinX = x;
group.MinY = y;
group.MinZ = z;
group.Blocks = 1;
}
else
{
group.MaxX = Math.Max(group.MaxX, x);
group.MaxY = Math.Max(group.MaxY, y);
group.MaxZ = Math.Max(group.MaxZ, z);
group.MinX = Math.Min(group.MinX, x);
group.MinY = Math.Min(group.MinY, y);
group.MinZ = Math.Min(group.MinZ, z);
group.Blocks += 1;
}
CheckGroup(x + 1, y, z, group);
CheckGroup(x - 1, y, z, group);
CheckGroup(x, y + 1, z, group);
CheckGroup(x, y - 1, z, group);
CheckGroup(x, y, z + 1, group);
CheckGroup(x, y, z - 1, group);
if (!groups.Contains(group))
{
groups.Add(group);
}
}
}
}
So the main question - is it possible to avoid stackOverflow in Parallel.For, or to rewrite it to iteration loop?
Parallel.For using default stackSize value even if you will use
Thread(()=>
{
Parallel.For(...);
},stackSize).Start()
it will use default values...
I don't like variant like this:
for(int i = 0; i < cpuCount; i++)
{
Thread t = new Thread(()=>{calculate();},stackSize).Start()
}
because I have to manage all threads, wait while all finishes, so it makes code very complicated... May be there are easier things?
There are two options:
to use recursion and try to increase the stack size (by using the Thread(ThreadStart, maxStackSize) constructor). The stack in applications is usually set to 1MB (see this link for details). Especially in DEBUG mode without optimizations (no inlining optimization done) this is a very limited value. Having a thread with separate stack for every Paralllel.For() statement
might help.
Use a iteration look instead of recursion to handle the stack depth by yourself.
I personally would go with option 1. (with or without separate stack) only in case I known the maximum depth of my recursion.
My preferred solution in most cases like yours will be the iteration approach.
Edit by #LordXaosa:
I tried this, and all works fine
int stackSize = 1024*1024*1024;//1GB limit
ManualResetEvent[] mre = new ManualResetEvent[Environment.ProcessorCount];
Parallel.For(0, Environment.ProcessorCount, (i) =>
{
mre[i] = new ManualResetEvent(false);
Thread t = new Thread((object reset) =>
{
Calculator calc = new Calculator(x, y, z, max, cycles);
calcs.Add(calc);
ManualResetEvent m = (ManualResetEvent)reset;
m.Set();
}, stackSize / (Environment.ProcessorCount * 4));
t.Start(mre[i]);
});
WaitHandle.WaitAll(mre);
But there also a limit... 50x50x50 array works fine, but more - stack overflow... In original game it can process 1000x1000x1000 sets, so may be there is another algorithm.
Thanks for your help!
I want to split a matrix into quadrants, upper and bottom left and right, according to the current position. Here is an example(to avoid writing T/U, for example, I placed the different characters with top and left priority, but they do overlap):
o T T U o o o
B T X U L L L
B B B L L L L
B B B L L L L
o o L L L L L
o o L L L L L
X is the central element. o is an element not included in any of the quadrants, T is top left, U is top right, B is lower left and L is lower right, I hope you get the idea. My solution includes checking in which quadrant of the matrix the starting position is, and then making the quadrants accordingly, which makes a total of 4 cases each with 4 quadrants = 16. I hope there's a better way, so can you please help me with this? I'm writing the program in C#, but pseudocode is fine.
You can access the different quadrant of cartesian system through their respective property.
public class Matrix<T>
{
public T[,] Elements { get; private set; }
public int Size { get; private set; }
public Point CentralPoint { get; set; }
public T this [int y, int x]
{
get { return Elements[y, x]; }
set { Elements[y, x] = value; }
}
public Matrix(int size)
{
this.Size = size;
this.Elements = new T[Size, Size];
}
public T[,] FirstQuadrant { get { return FromCentralTo(Size - 1, 0); } }
public T[,] SecondQuadrant { get { return FromCentralTo(0, 0); } }
public T[,] ThirdQuadrant { get { return FromCentralTo(0, Size - 1); } }
public T[,] FourthQuadrant { get { return FromCentralTo(Size - 1 , Size - 1); } }
private T[,] FromCentralTo(int x1, int y1, [CallerMemberName]string caller = "")
{
var translate = Math.Min(Math.Abs(CentralPoint.X - x1), Math.Abs(CentralPoint.Y - y1));
//fix the p1, so this results in a square array
if (Math.Abs(CentralPoint.X - x1) > translate)
x1 = CentralPoint.X + translate * Math.Sign(x1 - CentralPoint.X);
if (Math.Abs(CentralPoint.Y - y1) > translate)
y1 = CentralPoint.Y + translate * Math.Sign(y1 - CentralPoint.Y);
var size = translate + 1;
var start = new Point(Math.Min(CentralPoint.Y, y1), Math.Min(CentralPoint.X, x1));
var result = new T[size, size];
for (int x = 0; x < size; x++)
for (int y = 0; y < size; y++)
result[y, x] = this[start.Y + y, start.X + x];
return result;
}
}
public struct Point
{
public int X, Y;
public Point(int y, int x) { this.X = x; this.Y = y; }
}
void Main()
{
var matrix = new Matrix<int>(7);
matrix.CentralPoint = new Point(1, 2);
int i = 0;
for(int x = 0; x < 7; x++)
for (int y = 0; y < 7; y++)
matrix[x, y] = i++;
matrix.Elements.Dump("Matrix");
matrix[matrix.CentralPoint.Y, matrix.CentralPoint.X].Dump("Central");
matrix.FirstQuadrant.Dump("1");
matrix.SecondQuadrant.Dump("2");
matrix.ThirdQuadrant.Dump("3");
matrix.FourthQuadrant.Dump("4");
}
The above code produce the following in LINQPad :
I need help with my map editor, I'm stuck on saving it. When I save, after I put some grass on map, it gets grass everywhere in map file.
Here's variables:
mapMaximumX: maximum of the map in X (it is set as 500)
mapMaximumY: maximum of the map in Y (it is also set as 500)
mapTiles[index]: this is a list with class, each class has ID (0 = empty, 1 = grass, 2 = water), X and Y
if (Keyboard.GetState().IsKeyDown(Keys.F1))
{
for(int y = 0; y < mapMaximumY; y++)
{
for (int x = 0; x < mapMaximumX; x++)
{
if (MapTiles[i3].X == x && MapTiles[i3].Y == y)
{
}
else
{
MapTiles.Add(new Class1(0, x * 32, y * 32));
}
if (i3 < MapTiles.Count)
{
i3++;
}
}
}
TextWriter file = new StreamWriter("map1.MAP");
for (int y = 0; y < mapMaximumY; y++)
{
for (int x = 0; x < mapMaximumX; x++)
{
file.Write(MapTiles[i2].ID + ", ");
}
file.Write(file.NewLine);
}
i2 = 0;
System.Windows.Forms.MessageBox.Show("Saved!");
file.Close();
}
Full code is here, if u need it:
http://pastebin.com/qrWbuPtb
Thanx.
file.Write(MapTiles[i2].ID + ", ");
i2 never changes within your loop, so whatever i2 is will always be what's used to write your output.
You need to be using X and Y from your loops in determining which cell to write out.
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])))