I want to generate a matrix of cubes with x volume, but I want only the cubes on the surface (Cubes I can see).
The next code does what it does, a whole matrix. How I get what I need?
public class World : MonoBehaviour
{
public GameObject cube;
public int volume;
private void Awake()
{
for (int i = 0, x = 0; x < volume; x++)
{
for (int y = 0; y < volume; y++)
{
for (int z = 0; z < volume; z++)
{
i++;
var go = Instantiate(cube);
go.name = "Cube " + i;
go.transform.position = new Vector3
{
x = x,
y = y,
z = z
};
}
}
}
}
}
How about only using the outer limits and ignore anything between
// ignores the cubes that are not placed on the limits
if (x != 0 && x != volume - 1 && y != 0 && y != volume - 1 && z != 0 && z != volume - 1) continue;
i++;
var go = Instantiate(cube);
go.name = "Cube " + i;
go.transform.position = new Vector3(x, y, z);
or if it's easier to understand
// only spawns cubes that are placed on the limits
if (x == 0 || x == volume - 1 || y == 0 || y == volume - 1 || z == 0 || z == volume - 1)
{
i++;
var go = Instantiate(cube);
go.name = "Cube " + i;
go.transform.position = new Vector3(x, y, z);
}
as Eliasar mentioned I also would recommend to use a better variable name than i e.g. as yourselve said index. In the end it's just a name but it's cleaner. However I would also recommend to move it outside of the for definition like
int index = 0;
for(int x = 0; ...)
instead of
for(int index = 0 , x = 0; ...)
which is very hard to read
I don't know if key is the right word for this but I implemented a version of the Diamond-Square algorithm and I was wondering if there is any way to "save" the current output and recreate it using a simple (6 to 10 characters) key. The only way I tought so far is to save it as a single number/string containing the output value of each cell but for a 17X17 grid I would have at least a 289 characters long sequence (assuming the output is a single digit per cell).
Since the algorithm is random I can't think of a better way to do it. If it is impossible can you tell me about better algorithm to use. Thank you :)
PS: I'm using unity 3D and C#
the following class is from the unity tutorial for procedural cave generation, but ive included it because it includes the core elements you need. it uses a seed to generate a map, the same seed will regenerate the same map.... you could use a seed in your generation process, even if it is random, then you could always reuse that seed to return to that map.
using System.Collections;
using System.Collections.Generic;
using UnityEngine;
using System;
public class MapGenerator : MonoBehaviour {
public int width;
public int height;
public string seed;
public bool useRandomSeed;
[Range(0,100)]
public int randomFillPercent;
int[,] map;
private void Start()
{
GenerateMap();
}
void GenerateMap()
{
map = new int[width, height];
RandomFillMap();
for (int i = 0; i < 5; i++)
{
SmoothMap();
}
}
void RandomFillMap()
{
if (useRandomSeed)
{
seed = Time.time.ToString();
}
System.Random prng = new System.Random(seed.GetHashCode());
for (int x = 0; x < width; x++)
{
for (int y = 0; y < height; y++)
{
if (x == 0 || x == width - 1 || y == 0 || y == height - 1)
{
map[x, y] = 1;
}
else
{
map[x, y] = (prng.Next(0, 100) < randomFillPercent) ? 1 : 0;
}
}
}
}
private void Update()
{
if (Input.GetMouseButtonDown(0))
{
GenerateMap();
}
}
void SmoothMap()
{
for (int x = 0; x < width; x++)
{
for (int y = 0; y < height; y++)
{
int neighborWallTiles = GetSurroundingWallCount(x, y);
if (neighborWallTiles > 4)
{
map[x, y] = 1;
}
else if (neighborWallTiles<4)
{
map[x, y] = 0;
}
}
}
}
int GetSurroundingWallCount(int gridx, int gridy)
{
int wallcount = 0;
for(int neighborx=gridx-1; neighborx<=gridx + 1; neighborx++)
{
for (int neighbory = gridy - 1; neighbory <= gridy + 1; neighbory++)
{
if (neighborx >= 0 && neighborx < width && neighbory >= 0 && neighbory < height)
{
if (neighborx != gridx || neighbory != gridy)
{
wallcount += map[neighborx, neighbory];
}
}
else
{
wallcount++;
}
}
}
return wallcount;
}
void OnDrawGizmos()
{
if (map != null)
{
for (int x = 0; x < width; x++)
{
for (int y = 0; y < height; y++)
{
Gizmos.color = (map[x, y] == 1) ? Color.black : Color.white;
Vector3 pos = new Vector3(-width / 2 + x + .5f, 0, -height / 2 + y + .5f);
Gizmos.DrawCube(pos, Vector3.one);
}
}
}
}
}
I did like the comments said I used Random.InitState provided with unity and everything is working fine. Thanks Again!
UnityEngine.Random.InitState(seed);
I can control the seed as I wish and it gives me the same map for the same seed.
I have a function to check neighbors of an array and if that element is equal with 1. X is for each neighbor found and v[l] is the position for each 0. I have a problem with this code each time gives me "Index was outside the bounds of the array" and i don't know what to do else.
public int modificari(int i,int j,int n,int m)
{
int x = 0;
v = new int[n];
l=0;
if (mat[i, j] == 1)
{
if (j++ < m)
{
if (mat[i, j++] == 1)
x++;
else
{
v[l] = i * n + j + 2;
l++;
}
}
if (j++ < m && i++ < n)
{
if (mat[i++, j++] == 1)
x++;
else
{
v[l] = (i + 1) * n + j + 2;
l++;
}
}
if (i++ < n)
{
if (mat[i++, j] == 1)
x++;
else
{
v[l] = (i + 1) * n + j + 1;
l++;
}
}
if (j-- >= 0 && i++ < n)
{
if (mat[i++, j--] == 1)
x++;
else
{
v[l] = (i + 1) * n + j;
l++;
}
}
if (j-- >= 0)
{
if (mat[i, j--] == 1)
x++;
else
{
v[l] = i * n + j;
l++;
}
}
if (j-- >= 0 && i-- >= 0)
{
if (mat[i--, j--] == 1)
x++;
else
{
v[l] = (i - 1) * n + j;
l++;
}
}
if (i-- >= 0)
{
if (mat[i--, j] == 1)
x++;
else
{
v[l] = (i - 1) * n + j + 1;
l++;
}
}
if (j < n && i-- >= 0)
{
if (mat[i--, j++] == 1)
x++;
else
{
v[l] = (i - 1) * n + j + 2;
l++;
}
}
if (x < 2 && x > 3)
return 1;
else
return random();
}
return x;
}
That is a total mess. It is very hard to follow, even for an experienced coder. Use of one letter variable names and inline ++ operators is usually discouraged for the sake of readability.
I've quickly tried to rewrite your function from my best guess of what you're trying to achieve. I'm hoping you can see a different way to approach the problem that suits you better.
NOTE: I did not test this code at all, it probably has compile errors.
public struct Point
{
public int X;
public int Y;
public Point( int x, int y )
{
X = x;
Y = y;
}
}
public class Whatever
{
// ...
// Here is a list of the positions of all the neighbours whose values are
// zero.
List<Point> zeroPositions = new List<Point>();
// ...
public int Modificari(int pointX, int pointY)
{
// Determine dimensions of array.
int height = mat.GetLength(0);
int width = mat.GetLength(1);
// Find the minimum and maximum positions bounded by array size. (So we
// don't try to look at cell (-1, -1) when considering the neighbours of
// cell (0, 0) for instance.
int left = Math.Max( pointX - 1, 0 );
int right = Math.Min( pointX + 1, width );
int top = Math.Max( pointY - 1, 0 );
int bottom = Math.Min( pointY + 1, height );
// This is the number of neighbours whose value is 1.
int oneCount = 0;
zeroPositions.Clear();
for( int y = top; y <= bottom; y++ )
{
for( int x = left; x <= right; x++ )
{
if( mat[x, y] == 1 )
{
oneCount++;
}
else if( mat[x, y] == 0 )
{
zeroPositions.Add( new Point( x, y ) );
}
}
}
return oneCount;
}
//...
}
Also I'd really advise you to try not to do too many things in a function. Try making a different function for getting positions of ones and for returning the number of zeros.
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I'm trying to create a dungeon generator for a project I've been working on based off of this algorithm. I've gotten everything down, but my array (Fig. 1) doesn't seem to be holding giving the map data for some reason. I'm using three types of data to determine if a cell in the map is either empty (0), a space a character can be on (1), a hallway (2), or a wall (3).
I've gotten a bit stuck on this portion so any help is appreciated!
EDIT: The problem is the map object isn't storing the data in the loop shown in Fig. 1. Sorry for being so vague.
(Fig. 1)
for (int i = 0; i < roomList.Count; i++)
{
for (int x = roomList[i].X; x < (roomList[i].X + roomList[i].W); x++)
{
for (int y = roomList[i].Y; y < (roomList[i].Y + roomList[i].H); y++)
{
map[x, y] = 1;
}
}
}
(All of my relevant code)
namespace Project
{
}
public class Room
{
int xValue, yValue, widthValue, heightValue;
public int X
{
get { return xValue; }
set { xValue = value; }
}
public int Y
{
get { return yValue; }
set { yValue = value; }
}
public int W
{
get { return widthValue; }
set { widthValue = value; }
}
public int H
{
get { return heightValue; }
set { heightValue = value; }
}
}
public class DungeonGenerate
{
public int baseWidth = 513;
public int baseHeight = 513;
public int width = 64;
public int height = 64;
Color[,] arrayColor;
Random rand = new Random();
Room room = new Room();
Rectangle[,] rectMap;
public void Generate()
{
rectMap = new Rectangle[baseWidth, baseHeight];
//Creates a 2-D Array/Grid for the Dungeon
int[,] map = new int[baseWidth, baseHeight];
//Determines all the cells to be empty until otherwise stated
for (int x = 0; x < width; x++)
{
for (int y = 0; y < height; y++)
{
map[x, y] = 0;
}
}
//Determines the amount of rooms in the dungeon
int minRooms = (width * height) / 300;
int maxRooms = (width * height) / 150;
int amountOfRooms = rand.Next(minRooms, maxRooms);
//Room dimensions
int widthRoot = Convert.ToInt32(Math.Round(Math.Sqrt(width * 2)));
int heightRoot = Convert.ToInt32(Math.Round(Math.Sqrt(height * 2)));
int minWidth = Convert.ToInt32(Math.Round((width * .5) / widthRoot));
int maxWidth = Convert.ToInt32((width * 2) / widthRoot);
int minHeight = Convert.ToInt32(Math.Round(height * .5) / heightRoot);
int maxHeight = Convert.ToInt32((height * 2) / heightRoot);
//Creates the rooms
List<Room> roomList = new List<Room>(amountOfRooms);
for (int i = 0; i < amountOfRooms; i++)
{
bool ok = false;
do
{
room.X = rand.Next(width);
room.Y = rand.Next(height);
room.W = (rand.Next(maxWidth)) + minWidth;
room.H = (rand.Next(maxHeight)) + minHeight;
if (room.X + room.W >= width && room.Y + room.H >= height)
{
continue;
}
for (int q = 0; q < roomList.Count; q++)
{
if (room.X > roomList[q].X && room.X < roomList[q].X + room.W && room.Y > roomList[q].Y && room.Y < roomList[q].Y + room.H)
{
ok = false;
break;
}
}
ok = true;
roomList.Add(room);
} while (!ok);
}
//This will create hallways that lead to and from the rooms
int connectionCount = roomList.Count;
List<Point> connectedCells = new List<Point>((width * height));
for (int i = 0; i < connectionCount; i++)
{
Room roomA = roomList[i];
int roomNum = i;
while (roomNum == i)
{
roomNum = rand.Next(roomList.Count);
}
Room roomB = roomList[roomNum];
//Increasing this will make the hallway more straight, decreasing it will make the hallway more skewed
int sidestepChance = 10;
Point pointA = new Point(x: (rand.Next(roomA.W)) + roomA.X, y: (rand.Next(roomA.H)) + roomA.Y);
Point pointB = new Point(x: (rand.Next(roomB.W)) + roomB.X, y: (rand.Next(roomB.H)) + roomB.Y);
while (pointA != pointB)
{
int num = rand.Next() * 100;
if (num < sidestepChance)
{
if (pointB.X != pointA.X)
{
if (pointB.X > pointA.X)
{
pointB.X--;
}
else
{
pointB.X++;
}
}
}
else if(pointB.Y != pointA.Y)
{
if (pointB.Y > pointA.Y)
{
pointB.Y--;
}
else
{
pointB.Y++;
}
}
}
if (pointB.X < width && pointB.Y < height)
{
connectedCells.Add(pointB);
}
}
//Fills the room with data
for (int i = 0; i < roomList.Count; i++)
{
for (int x = roomList[i].X; x < (roomList[i].X + roomList[i].W); x++)
{
for (int y = roomList[i].Y; y < (roomList[i].Y + roomList[i].H); y++)
{
map[x, y] = 1;
}
}
}
for (int y = 0; y < height; y++)
{
for (int x = 0; x < width; x++)
{
if (map[x, y] == 0)
{
bool wall = false;
for (int yy = y - 2; yy < y + 2; yy++)
{
for (int xx = x - 2; xx < x + 2; xx++)
{
if (xx > 0 && yy > 0 && xx < width && yy < height)
{
if (map[xx, yy] == 1 || map[xx, yy] == 2)
{
map[x, y] = 3;
wall = true;
}
}
}
if (wall)
{
break;
}
}
}
}
}
//Rendering the Map and giving it some Color (Sort of)!
int scaler = baseWidth / width;
for (int x = 0; x < baseWidth; x++)
{
for (int y = 0; y < baseHeight; y++)
{
rectMap[x, y] = new Rectangle(x, y, 1, 1);
arrayColor = new Color[baseWidth, baseHeight];
switch (map[x, y])
{
case 0:
arrayColor[x, y] = new Color(0,0,0);
break;
case 1:
arrayColor[x, y] = new Color(0,0,0);
break;
case 2:
arrayColor[x, y] = new Color(0,0,0);
break;
case 3:
arrayColor[x, y] = new Color (0,0,0);
break;
}
}
}
}
public Rectangle[,] GetMap()
{
return rectMap;
}
public Color[,] GetColors()
{
return arrayColor;
}
}
In the for-loop where you're populating roomList, you're not instantiating a new Room each time. You're simply manipulating the same Room object and re-adding it to the list, so roomList will just contain many references to the same Room object. Try removing the room field from your DungeonGenerate class and use a local variable instead:
for (int i = 0; i < amountOfRooms; i++)
{
bool ok = false;
do
{
var room = new Room();
...
roomList.Add(room);
} while (!ok);
}
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])))