I'm trying to create a function which will return string in Triangle Sine-Wave format:
but currently, my format is only in Wave format:
Code below:
public static void printWave(string str)
{
int height = 3;
// Get length of the string
int len = str.Length;
// Create a 2d character array
char[,] matrixArray = new char[height, len];
char[] charArray = str.ToCharArray();
// for counting the
// rows of the ZigZag
int row = 0;
bool down = true;
for (int i = 0; i < len; i++)
{
// put characters
// in the matrix
matrixArray[row, i] = charArray[i];
// You have reached the bottom
if (row == height - 1)
down = false;
else if (row == 0)
down = true;
if (down)
row++;
else
row--;
}
// Print the Zig-Zag String
for (int i = 0; i < height; i++)
{
for (int j = 0; j < len; j++)
{
Console.Write(matrixArray[i, j] + " ");
}
Console.Write("\n");
}
}
Can you please help me modify my code to it will return triangle sin wave format?
We can use 3 separate StringBuilders to append to depending on our boolean top and a simple even value comparison. A quick TL;DR is that anything at an even index goes in the middle row, and then we flip between appending to the top or bottom row:
public static void printWave(string str)
{
//for use to determine top or bottom StringBuilder
bool top = true;
//will be used to generate each row of the output
StringBuilder topString = new StringBuilder();
StringBuilder middleString = new StringBuilder();
StringBuilder bottomString = new StringBuilder();
//iterate through paramter string
for (int i = 0; i < str.Length; i++)
{
//if char is at an even index, it goes in the middle StringBuilder, blank spaces in top and bottom builders
if (i%2 == 0)
{
topString.Append(" ");
middleString.Append(str[i]);
bottomString.Append(" ");
}
//if not even index, determine top or bottom row
else
{
//simply check our boolean and then flip it after use
if (top)
{
topString.Append(str[i]);
middleString.Append(" ");
bottomString.Append(" ");
top = false;
}
else
{
topString.Append(" ");
middleString.Append(" ");
bottomString.Append(str[i]);
top = true;
}
}
}
//write each row of strings on new lines
Console.WriteLine(topString.ToString());
Console.WriteLine(middleString.ToString());
Console.WriteLine(bottomString.ToString());
}
For a variable height:
public static void printWave(string str)
{
//height we want the wave to reach
int height = 5;
//determine "middle" row
int startingRow = height / 2;
int currentRow = startingRow; //this one is for modifying inside loop
bool up = true;
//2D array to hold the rows
char[,] arr = new char[height, str.Length];
for (int i = 0; i < str.Length; i++)
{
for (int j = 0; j < height; j++)
{
if (j == currentRow)
{
arr[j, i] = str[i];
}
else
arr[j, i] = ' ';
}
//could probably break this into more digestible pieces if time to think about it
if (up)
{
if (currentRow == 0)
{
up = false;
currentRow++;
}
else
{
currentRow--;
}
}
else
{
if (currentRow == height - 1)
{
up = true;
currentRow--;
}
else
{
currentRow++;
}
}
}
for (int k = 0; k < height; k++)
{
for (int l = 0; l < str.Length; l++)
{
Console.Write(arr[k, l]);
}
Console.WriteLine();
}
}
Examples of height = 5 and height = 6 output:
And finally, height = 7
This version works, but it's hard-coded to just the 3 rows like the question shows. If larger waves are needed, or especially if the size of the wave depends on the input string, then this may be hard to adjust to the requirements.
public static void PrintWave(string str)
{
printWithRowLogic(str, (i) => (i - 1) % 4 == 0);
Console.WriteLine();
printWithRowLogic(str, (i) => i % 2 == 0);
Console.WriteLine();
printWithRowLogic(str, (i) => (i - 3) % 4 == 0);
}
private static void printWithRowLogic(string str, Func<int, bool> checkLogic)
{
for (int i = 0; i < str.Length; i++)
Console.Write(checkLogic(i) ? str[i] : ' ');
}
I am working on an A* Pathfinding method that uses a custom class instead of nodes, but am having issues with my loops. The first for loop using int i is able to go up to 3 (Player1.instance.movement = 3), but I need to use an if statement inside of that loop to check if the target position has already been found. I am wondering if it is possible to break my for loop when my If statement is false.
public void GetNeighbors(Tile originTile)
{
Tile originalTile = originTile;
nextTile.Clear();
int minX = 0;
int minY = 0;
var originCostFunc = Mathf.Infinity;
for (int i = 0; i < Player1.instance.movement; i++)
{
for (int x = -1; x <= 1; x++)
{
for (int y = -1; y <= 1; y++)
{
if (x != y && y != x)
{
var costX = Mathf.Abs((originTile.transform.position.x + x) - originalTile.transform.position.x);
var costY = Mathf.Abs((originTile.transform.position.y + y) - originalTile.transform.position.y);
var distanceX = Mathf.Abs(targetPos.transform.position.x - (originTile.transform.position.x + x));
var distanceY = Mathf.Abs(targetPos.transform.position.y - (originTile.transform.position.y + y));
var costFunc = costX + costY + distanceX + distanceY;
if (costFunc <= originCostFunc)
{
originCostFunc = costFunc;
minX = x;
minY = y;
Debug.Log($"x: {x}, y: {y}");
}
}
}
}
nextTile.Add(GridManagerHandPlaced.instance.GetTileAtPosition(new Vector2(originTile.transform.position.x + minX, originTile.transform.position.y + minY)));
if (nextTile[i] != targetPos)
{
originTile = nextTile[i];
}
else
{
break;
}
}
DisplayPath();
}
You can break loop several times by condition.
bool breakLoop = false;
for (int i = 0; i < length; i++)
{
for (int j = 0; j < length; j++)
{
for (int k = 0; k < length; k++)
{
breakLoop = nextTile == target;
if (breakLoop)
break;
}
if (breakLoop)
break;
}
if (breakLoop)
break;
}
Or move search logic to separated method and return a value from any number of nested loops
string path = FindPath();
Display(path);
string FindPath()
{
for (int i = 0; i < length; i++)
{
for (int j = 0; j < length; j++)
{
for (int k = 0; k < length; k++)
{
if (nextTile == target)
return nextTile;
}
}
}
return null;
}
Never use goto operator.
This is one of the few valid cases where I'd use goto. In-fact, this is the example given in the docs for when it should be used.
void CheckMatrices(Dictionary<string, int[][]> matrixLookup, int target)
{
foreach (var (key, matrix) in matrixLookup)
{
for (int row = 0; row < matrix.Length; row++)
{
for (int col = 0; col < matrix[row].Length; col++)
{
if (matrix[row][col] == target)
{
goto Found;
}
}
}
Console.WriteLine($"Not found {target} in matrix {key}.");
continue;
Found:
Console.WriteLine($"Found {target} in matrix {key}.");
}
}
Note the syntax for the label is simply myLabel: and you can place it anywhere in procedurally executable code.
For sake of covering other ways of handling this situation, here is the boolean solution.
bool breakLoops = false;
for (int i = 0; i < length1; i++)
{
for (int ii = 0; ii < length2; ii++)
{
for (int iii = 0; iii < length3; iii++)
{
if (breakingCondition)
{
breakLoops = true;
break;
}
}
if (breakLoops) break;
}
if (breakLoops) break;
}
Simple and straightforward, but requires a break condition check at the end of each loop that you want to break out of.
Hello i have a realy strange behavior in this piece of code:
public class IGraphics
{
public int[,] screen;
private int[,] world;
private int[,] entitys;
private int[,] buffer;
private int screenW;
private int screenH;
public IGraphics(int screenW, int screenH) {
this.screenH = screenH;
this.screenW = screenW;
screen = new int[screenW + 1, screenH];
buffer = new int[screenW + 1, screenH];
}
public void loadWorld(int[,] world) {
this.world = world;
}
public void clear() {
screen = new int[screenW + 1, screenH];
world = new int[screenW, screenH];
for (int y = 0; y < world.GetLength(1); y++) {
for (int x = 0; x < world.GetLength(0); x++) {
world[x, y] = 0;
}
}
}
private void loadScreen() {
}
private void updateEntitys()
{
entitys = new int[screenW, screenH];
List<GameObject> EntRow = Common.world.getEntitys();
for (int i = 0; i < EntRow.Count(); i++)
{
entitys[EntRow[i].x, EntRow[i].y] = EntRow[i].Icon;
}
}
public void draw() {
updateEntitys();
for (int y = 0; y < screen.GetLength(1); y++)
{
for (int x = 0; x < screen.GetLength(0) - 1; x++)
{
if (entitys[x, y] == 0)
{
screen[x, y] = world[x, y];
}
else
{
screen[x, y] = entitys[x, y];
}
}
screen[screen.GetLength(0) - 1, y] = 123;
}
if (buffer.Cast<int>().SequenceEqual(screen.Cast<int>()))
{
return;
}
Console.Clear();
buffer = screen;
for (int y = 0; y < screen.GetLength(1); y++) {
for (int x = 0; x < screen.GetLength(0); x++) {
if (screen[x, y] == 123)
{
Console.WriteLine();
}
else {
Console.Write(objectStore.getIcon(screen[x, y]));
}
}
}
}
}
the problem comes in the Draw() function where i set the value of the screen[,] array for some reason it also change the value of the buffer[,] array before the control also tried moving the buffer[,] in a seperate class but i had the same problem.
Someone as an explanation?
When you assign a reference variable to another variable, you copy the pointer to that variable, instead of copying the content, so what you end up with is two variables that point to the same array.
Try to instead copy the array using Clone or copy or something. I think its screen.CopyTo
screen.CopyTo(buffer, 0);
If you look at the body of the method called draw, you will notice this assignment:
buffer = screen;
This is might the cause of the change you noticed.
See the other answers for possible reasons for your problem. However, the following code shows that changing screen[] does not change buffer[]. This kind of effort on your part would allow you to focus your investigation elsewhere. The principle is to first simply.
int counter = 1;
public void draw()
{
for (int y = 0; y < screen.GetLength(1); y++)
{
for (int x = 0; x < screen.GetLength(0) - 1; x++)
{
screen[x, y] = counter++;
}
screen[screen.GetLength(0) - 1, y] = 123;
}
if (buffer.Cast<int>().SequenceEqual(screen.Cast<int>()))
{
MessageBox.Show("Help!");
return;
}
// check again
for (int y = 0; y < screen.GetLength(1); y++)
{
for (int x = 0; x < screen.GetLength(0) - 1; x++)
{
if (screen[x, y] == buffer[x, y])
{
MessageBox.Show("Help two!");
return;
}
}
}
Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 8 years ago.
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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])))