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])))
Related
I'm trying to implement FFT, using butterfly notation with decimation in spatial domain. I have two separated functions, one to calculate the FFT on 1D array, and the other that applies the first function over columns of the image, puts these columns in a temporary array, and then computes the FFT of rows in the temporary array.
Here is the code
public static Complex[] FFT(Complex[] input)
{
// Get the length of the input array
int n = input.Length;
Complex I = new Complex(0, 1);
// Check if the input has a length of 1
if (n == 1)
{
return input;
}
// Split the input into even and odd elements
Complex[] even = new Complex[n / 2];
Complex[] odd = new Complex[n / 2];
for (int i = 0; i < n; i++)
{
if (i % 2 == 0)
{
even[i / 2] = input[i];
}
else
{
odd[(i - 1) / 2] = input[i];
}
}
// Compute the FFT of the even and odd elements
Complex[] evenFFT = FFT(even);
Complex[] oddFFT = FFT(odd);
// Combine the FFT of the even and odd elements using the butterfly notation
Complex[] output = new Complex[n];
for (int i = 0; i < n / 2; i++)
{
Complex w = Complex.Exp(-2 * I * Math.PI * i / n);
output[i] = evenFFT[i] + w * oddFFT[i];
output[i + n / 2] = evenFFT[i] - w * oddFFT[i];
}
return output;
}
public static Complex[,] FFT2D(Complex[,] input)
{
int N = input.GetLength(0);
int M = input.GetLength(1);
Complex[,] output = new Complex[128, 128];
Complex[,] columnsFFT = new Complex[128, 128];
//Perform FFT over the columns of the input
for(int i = 0; i < M; i++)
{
//Put all the values from i'th column in the tempColumn variable
var tempColumn = new Complex[128];
for(int j = 0; j < N; j++)
{
tempColumn[j] = input[j, i];
}
//Calculate the FFT of i'th column
tempColumn = FFT(tempColumn);
//Assign the column to columnsFFT, after calculating its FFT
for(int z = 0; z < 128; z++)
{
columnsFFT[z, i] = tempColumn[z];
}
}
//Perform FFT over the rows of the columnsFFT
for(int i = 0; i < N; i++)
{
//Put the values from i'th row in tempRow, so we can perform the FFT on its entirety
var tempRow = new Complex[128];
for(int j = 0; j < M; j++)
{
tempRow[j] = columnsFFT[i, j];
}
//Calculate the FFT on tempRow
tempRow = FFT(tempRow);
//Assign the tempRow to the output
for(int z = 0; z < M; z++)
{
if (i == 0)
Console.WriteLine(tempRow[z].Real);
output[i, z] = tempRow[z];
}
}
return output;
}
The problem that i have is that it simply doesn't work, or i think it doesn't, based on the image output i get.
Input image
Output
What is it that i'm doing wrong here?
The problem is to make a program the flips a 2D (2 by 2) array either horizontally(H) or vertically(V), depending on the input the user puts. For example, if the user puts "HHHVVHVHV", it means to flip 5 times horizontally, and 4 times vertically. The code I attached below works, but I am not sure if there is a built-in method in C# to flip arrays or an easier way to do it.
using System;
using System.Linq;
namespace _2019JuniorQ4
{
class Program
{
static void Main(string[] args)
{
//Varables
int[,] grid = new int[2, 2] { { 1, 2 }, { 3, 4 } };
int[,] flippedGrid = new int[2, 2];
string input = "";
int h = 0;
int v = 0;
int flipHorizontal = 0;
int flipVertical = 0;
//Input
input = Console.ReadLine();
//Count number of times
for (int i=0; i<input.Length; i++)
{
if (input[i] == Convert.ToChar("H"))
{
h++;
}
else if(input[i] == Convert.ToChar("V"))
{
v++;
}
}
//Process
flipHorizontal = h % 2;
flipVertical = v % 2;
//Vertical flip
if (flipVertical == 1)
{
for (int y = 0; y < 2; y++)
{
for (int x = 1; x >= 0; x--)
{
flippedGrid[y, Math.Abs(x - 1)] = grid[y,x];
}
}
grid = flippedGrid.Clone() as int[,];
}
//Horizontal Flip
if (flipHorizontal == 1)
{
for (int x = 0; x < 2; x++)
{
for (int y = 1; y >= 0; y--)
{
flippedGrid[Math.Abs(y - 1), x] = grid[y,x];
}
}
grid = flippedGrid.Clone() as int[,];
}
//Output
for (int x=0; x<2; x++)
{
for (int y=0; y<2; y++)
{
Console.Write(grid[x,y]+" ");
}
Console.WriteLine();
}
}
}
}
I have to calculate the spectrum values of an audio.
I used aForge's FFT in Sources/Math/FourierTransform.cs and I used an example of sampling with 16 samples as used in this video to check the results with excel (I tested the results in a spreadsheet like in the video).
FFT:
public enum Direction
{
Forward = 1,
Backward = -1
};
private const int minLength = 2;
private const int maxLength = 16384;
private const int minBits = 1;
private const int maxBits = 14;
private static int[][] reversedBits = new int[maxBits][];
private static Complex[,][] complexRotation = new Complex[maxBits, 2][];
static void Main(string[] args)
{
var Data = new Complex[16];
Data[0] = new Complex(0, 0);
Data[1] = new Complex((float)0.998027, 0);
Data[2] = new Complex((float)0.125333, 0);
Data[3] = new Complex((float)-0.98229, 0);
Data[4] = new Complex((float)-0.24869, 0);
Data[5] = new Complex((float)0.951057, 0);
Data[6] = new Complex((float)0.368125, 0);
Data[7] = new Complex((float)-0.90483, 0);
Data[8] = new Complex((float)-0.48175, 0);
Data[9] = new Complex((float)0.844328, 0);
Data[10] = new Complex((float)0.587785, 0);
Data[11] = new Complex((float)-0.77051, 0);
Data[12] = new Complex((float)-0.68455, 0);
Data[13] = new Complex((float)0.684547, 0);
Data[14] = new Complex((float)0.770513, 0);
Data[15] = new Complex((float)-0.58779, 0);
FFT(Data, Direction.Forward);
for (int a = 0; a <= Data.Length - 1; a++)
{
Console.WriteLine(Data[a].Re.ToString());
}
Console.ReadLine();
}
public static void FFT(Complex[] data, Direction direction)
{
int n = data.Length;
int m = Tools.Log2(n);
// reorder data first
ReorderData(data);
// compute FFT
int tn = 1, tm;
for (int k = 1; k <= m; k++)
{
Complex[] rotation = GetComplexRotation(k, direction);
tm = tn;
tn <<= 1;
for (int i = 0; i < tm; i++)
{
Complex t = rotation[i];
for (int even = i; even < n; even += tn)
{
int odd = even + tm;
Complex ce = data[even];
Complex co = data[odd];
double tr = co.Re * t.Re - co.Im * t.Im;
double ti = co.Re * t.Im + co.Im * t.Re;
data[even].Re += tr;
data[even].Im += ti;
data[odd].Re = ce.Re - tr;
data[odd].Im = ce.Im - ti;
}
}
}
if (direction == Direction.Forward)
{
for (int i = 0; i < n; i++)
{
data[i].Re /= (double)n;
data[i].Im /= (double)n;
}
}
}
private static int[] GetReversedBits(int numberOfBits)
{
if ((numberOfBits < minBits) || (numberOfBits > maxBits))
throw new ArgumentOutOfRangeException();
// check if the array is already calculated
if (reversedBits[numberOfBits - 1] == null)
{
int n = Tools.Pow2(numberOfBits);
int[] rBits = new int[n];
// calculate the array
for (int i = 0; i < n; i++)
{
int oldBits = i;
int newBits = 0;
for (int j = 0; j < numberOfBits; j++)
{
newBits = (newBits << 1) | (oldBits & 1);
oldBits = (oldBits >> 1);
}
rBits[i] = newBits;
}
reversedBits[numberOfBits - 1] = rBits;
}
return reversedBits[numberOfBits - 1];
}
private static Complex[] GetComplexRotation(int numberOfBits, Direction direction)
{
int directionIndex = (direction == Direction.Forward) ? 0 : 1;
// check if the array is already calculated
if (complexRotation[numberOfBits - 1, directionIndex] == null)
{
int n = 1 << (numberOfBits - 1);
double uR = 1.0;
double uI = 0.0;
double angle = System.Math.PI / n * (int)direction;
double wR = System.Math.Cos(angle);
double wI = System.Math.Sin(angle);
double t;
Complex[] rotation = new Complex[n];
for (int i = 0; i < n; i++)
{
rotation[i] = new Complex(uR, uI);
t = uR * wI + uI * wR;
uR = uR * wR - uI * wI;
uI = t;
}
complexRotation[numberOfBits - 1, directionIndex] = rotation;
}
return complexRotation[numberOfBits - 1, directionIndex];
}
// Reorder data for FFT using
private static void ReorderData(Complex[] data)
{
int len = data.Length;
// check data length
if ((len < minLength) || (len > maxLength) || (!Tools.IsPowerOf2(len)))
throw new ArgumentException("Incorrect data length.");
int[] rBits = GetReversedBits(Tools.Log2(len));
for (int i = 0; i < len; i++)
{
int s = rBits[i];
if (s > i)
{
Complex t = data[i];
data[i] = data[s];
data[s] = t;
}
}
}
These are the results after the transformation:
Output FFT results: Excel FFT results:
0,0418315622955561 0,669305
0,0533257974328085 0,716163407
0,137615673627316 0,908647001
0,114642731070279 1,673453043
0,234673940537634 7,474988602
0,0811255020953362 0,880988382
0,138088891589122 0,406276784
0,0623766891658306 0,248854492
0,0272978749126196 0,204227
0,0124250144575261 0,248854492
0,053787064184711 0,406276784
0,00783331226557493 0,880988382
0,0884368745610118 7,474988602
0,0155431246384978 1,673453043
0,0301093757152557 0,908647001
0 0,716163407
The results are not at all similar. Where is it wrong?
Is the implementation of complex (Data) wrong or is the FFT method wrong or other?
Thanks in advance!
First, the resulting FFT is a complex function in general. You're only displaying the real parts in your code but the thing you're comparing to is displaying the magnitudes, so of course they're going to be different: you're comparing apples to oranges.
When you use magnitudes and compare apples to apples, you should get this:
for (int a = 0; a <= Data.Length - 1; a++)
{
Console.WriteLine(Data[a].Magnitude.ToString());
}
...
0.0418315622955561
0.0447602132472683
0.0567904388057513
0.104590813761862
0.46718679147454
0.0550617784710375
0.025392294285886
0.0155534081359397
0.0127641875296831
0.0155534081359397
0.025392294285886
0.0550617784710375
0.46718679147454
0.104590813761862
0.0567904388057513
0.0447602132472683
That looks a little better -- it has the same symmetry property as the Excel output and there appear to be peaks in the same locations.
It almost looks like the scale is off. If I divide each element by the corresponding element from the Excel output, I get:
16
16
16
16
16
16
16
16
16
16
16
16
16
16
16
16
So your results are pretty much correct, just off by a scaling factor.
You're dividing everything by n in the last step of your FFT:
if (direction == Direction.Forward)
{
for (int i = 0; i < n; i++)
{
data[i].Re /= (double)n;
data[i].Im /= (double)n;
}
}
This is conventionally done for the inverse transform, not the forward transform.
In summary, changing the output from Data[a].Re to Data[a].Magnitude and changing the condition at the end of FFT from if (direction == Direction.Forward) to if (direction == Direction.Backward), I get this output:
0.669304996728897
0.716163411956293
0.908647020892022
1.67345302018979
7.47498866359264
0.880988455536601
0.406276708574176
0.248854530175035
0.20422700047493
0.248854530175035
0.406276708574176
0.880988455536601
7.47498866359264
1.67345302018979
0.908647020892022
0.716163411956293
which matches the Excel output.
I am creating a library for Neural Networks; I have successfully implemented the Back Propagation algorithm, but I'm having trouble with Resilient Propagation.
I have been using the XOR scenario to test my implementation; the error seems to decrease after a few epochs, then it increases and eventually stops changing.
The network I am testing on has 3 layers - 2 input, 2 hidden and 1 output neuron/s. The synapse/dendrite weights are on the following layer's neurons, for example, the input-hidden weights would be stored on the hidden neurons and the hidden-output weights would be stored on the output neurons.
This is my gradient calculation (EDIT 4):
private void CalculateGradient()
{
for (int t = 0; t < TrainingInputs.Count; ++t) //loop through training data
{
Network.Run(TrainingInputs[t]);
for (int l = Network.Layers.Count - 1; l > 0; l--)
{
for (int i = 0; i < Network.Layers[l].Neurons.Count; i++)
{
Network.Layers[l].Neurons[i].Delta = l < Network.Layers.Count - 1
? CalculateNonLastGradient(l + 1, i, Network.Layers[l].Neurons[i].Value)
: CalculateLastGradient(TrainingOutputs[t][i], Network.Layers[l].Neurons[i].Value);
}
}
for (int l = Network.Layers.Count - 1; l > 0; l--)
{
for (int j = 0; j < Network.Layers[l].Neurons.Count; ++j)
{
double grad = Network.Layers[l].Neurons[j].Delta;
biasGrads[l][j] += grad;
for (int i = 0; i < Network.Layers[l - 1].Neurons.Count; ++i)
{
grad = Network.Layers[l].Neurons[j].Delta * Network.Layers[l - 1].Neurons[i].Value;
grads[l][j][i] += grad;
}
}
}
int o = 0;
Layer layer = Network.Layers[Network.Layers.Count - 1];
errors.Add(layer.Neurons.Sum(n => Math.Abs(TrainingOutputs[t][o++] - n.Value)));
}
Error = errors.Average();
}
private double CalculateLastGradient(double ideal, double nValue)
{
return Network.Activation.Derivitive(nValue) * (ideal - nValue);
}
private double CalculateNonLastGradient(int nextLayer, int j, double nValue)
{
double sum = 0.0;
for (int i = 0; i < Network.Layers[nextLayer].Neurons.Count; i++)
{
sum += Network.Layers[nextLayer].Neurons[i].Delta * Network.Layers[nextLayer].Neurons[i].Dendrites[j].Weight;
}
return Network.Activation.Derivitive(nValue) * sum;
}
My RProp implementation (EDIT 4):
public bool Algorithm()
{
//initialise matrices
if (!initializedMatrices)
{
InitializeMatrices();
}
ZeroOut();
//calculate gradients
CalculateGradient();
for (int l = 1; l < Network.Layers.Count; l++) //layers
{
for (int i = 0; i < Network.Layers[l - 1].Neurons.Count; ++i) //prev layer neurons
{
for (int j = 0; j < Network.Layers[l].Neurons.Count; ++j) //current layer neurons
{
double delta = prevDeltas[l][j][i];
int change = Math.Sign(prevGrads[l][j][i] * grads[l][j][i]);
if (change > 0)
{
delta = Math.Min(delta * etaPlus, deltaMax);
double deltaWeight = -Math.Sign(grads[l][j][i]) * delta;
Network.Layers[l].Neurons[j].Dendrites[i].Weight += deltaWeight;
}
else if (change < 0)
{
delta = Math.Max(delta * etaMinus, deltaMin);
Network.Layers[l].Neurons[j].Dendrites[i].Weight -= prevDeltas[l][j][i];
prevGrads[l][j][i] = 0;
}
else if (change == 0)
{
double deltaWeight = -Math.Sign(grads[l][j][i]) * delta;
Network.Layers[l].Neurons[j].Dendrites[i].Weight += deltaWeight;
}
prevGrads[l][j][i] = grads[l][j][i];
prevDeltas[l][j][i] = delta;
} //j
} //i
for (int i = 1; i < Network.Layers[l].Neurons.Count; ++i)
{
double delta = prevBiasDeltas[l][i];
int change = Math.Sign(prevBiasGrads[l][i] * biasGrads[l][i]);
if (change > 0)
{
delta = Math.Min(prevBiasDeltas[l][i] * etaPlus, deltaMax);
double biasDeltaWeight = -Math.Sign(biasGrads[l][i]) * delta;
Network.Layers[l].Neurons[i].Bias += biasDeltaWeight;
}
else if (change < 0)
{
delta = Math.Max(prevBiasDeltas[l][i] * etaMinus, deltaMin);
Network.Layers[l].Neurons[i].Bias -= prevBiasDeltas[l][i];
prevBiasGrads[l][i] = 0;
}
else if (change == 0)
{
double biasDeltaWeight = -Math.Sign(biasGrads[l][i]) * delta;
Network.Layers[l].Neurons[i].Bias += biasDeltaWeight;
}
prevBiasGrads[l][i] = biasGrads[l][i];
prevBiasDeltas[l][i] = delta;
}
}
return true;
}
Could anyone point me as to what could be going wrong?
EDIT:
The issue seems to be that the delta is not changing.
EDIT 2:
I have fixed the delta not changing by initializing the first previous deltas to 0.01 and Zero'ing the gradients before each call. The error increases very fast, then decreases very slowly with TANH; with Sigmoid it does the opposite.
EDIT 3:
The loop for the bias' was starting at 0, when it should have been 1. Fixing this fixed the original issue, but has brought up at new one - the error stops decreasing after a certain point.
EDIT 4: I realised that I had not accumulated the gradients for each neuron over the training set. The error now decreases, but very slowly.
I need to sort a list of points by distance.
So for e.g.
input : [[1,2],[5,10],[2,4]...]
output : [[1,2],[2,4],[5,10]...]
(assuming that geometrically [1,2] and [2,4] are nearest & [2,4] & [5,10] are nearest.
I need them to sort it so they are ordered by distance i.e. on the geometrical graph, point a is nearest to point b , point b is nearest to c and so on.
Any idea?
Edit: Code example
public class Point
{
public double X {get;set;}
public double Y {get;set;}
}
List<Point> points = new List<Point>();
let say my points list is getting populated in random order (not by geometrical distance).
So points would look something like...
point ~ [[1,2],[5,10],[2,4]...]
Now my charting control just take the first & second point and draw a line between them. Which means it does not care which geometrical order is it in.
If I simply supply the "points" list as above its going to draw lines between each points and from charting point of view it would not look correct as they would be "zig-zag".
To make sure the charting control draws a straight line (& and not zig-zag) I have to pass the points in the right order which would look something like ...
destination points ~ [[1,2],[2,4],[5,10]...]
So my question is how to achieve this.
Note: Changing chart control is not an option here.
Thanks
The code first takes the nearest point to (0, 0) at the '0' index then start sorting the points by distance from the last spotted point..
C#:
List<Point> SortByDistance(List<Point> lst)
{
List<Point> output = new List<Point>();
output.Add(lst[NearestPoint(new Point(0, 0), lst)]);
lst.Remove(output[0]);
int x = 0;
for (int i = 0; i < lst.Count + x; i++)
{
output.Add(lst[NearestPoint(output[output.Count - 1], lst)]);
lst.Remove(output[output.Count - 1]);
x++;
}
return output;
}
int NearestPoint(Point srcPt, List<Point> lookIn)
{
KeyValuePair<double, int> smallestDistance = new KeyValuePair<double, int>();
for (int i = 0; i < lookIn.Count; i++)
{
double distance = Math.Sqrt(Math.Pow(srcPt.X - lookIn[i].X, 2) + Math.Pow(srcPt.Y - lookIn[i].Y, 2));
if (i == 0)
{
smallestDistance = new KeyValuePair<double, int>(distance, i);
}
else
{
if (distance < smallestDistance.Key)
{
smallestDistance = new KeyValuePair<double, int>(distance, i);
}
}
}
return smallestDistance.Value;
}
VB.Net:
Function SortByDistance(ByVal lst As List(Of Point)) As List(Of Point)
Dim out As New List(Of Point)
out.Add(lst(NearestPoint(New Point(0, 0), lst)))
lst.Remove(out(0))
Dim x As Integer = 0
For i As Integer = 0 To lst.Count - 1 + x
out.Add(lst(NearestPoint(out(out.Count - 1), lst)))
lst.Remove(out(out.Count - 1))
x += 1
Next
Return out
End Function
Function NearestPoint(ByVal srcPt As Point, ByVal lookIn As List(Of Point)) As Integer
Dim smallestDistance As KeyValuePair(Of Double, Integer)
For i As Integer = 0 To lookIn.Count - 1
Dim distance As Double = Math.Sqrt(Math.Pow(srcPt.X - lookIn(i).X, 2) + Math.Pow(srcPt.Y - lookIn(i).Y, 2))
If i = 0 Then
smallestDistance = New KeyValuePair(Of Double, Integer)(distance, i)
Else
If distance < smallestDistance.Key Then
smallestDistance = New KeyValuePair(Of Double, Integer)(distance, i)
End If
End If
Next
Return smallestDistance.Value
End Function
I think you want this, Dijkstra's algorithm. There is a c# project here.
What follows is unmodified code from that link
class Dijkstra
{
private int rank = 0;
private int[,] L;
private int[] C;
public int[] D;
private int trank = 0;
public Dijkstra(int paramRank,int [,]paramArray)
{
L = new int[paramRank, paramRank];
C = new int[paramRank];
D = new int[paramRank];
rank = paramRank;
for (int i = 0; i < rank; i++)
{
for (int j = 0; j < rank; j++) {
L[i, j] = paramArray[i, j];
}
}
for (int i = 0; i < rank; i++)
{
C[i] = i;
}
C[0] = -1;
for (int i = 1; i < rank; i++)
D[i] = L[0, i];
}
public void DijkstraSolving()
{
int minValue = Int32.MaxValue;
int minNode = 0;
for (int i = 0; i < rank; i++)
{
if (C[i] == -1)
continue;
if (D[i] > 0 && D[i] < minValue)
{
minValue = D[i];
minNode = i;
}
}
C[minNode] = -1;
for (int i = 0; i < rank; i++)
{
if (L[minNode, i] < 0)
continue;
if (D[i] < 0) {
D[i] = minValue + L[minNode, i];
continue;
}
if ((D[minNode] + L[minNode, i]) < D[i])
D[i] = minValue+ L[minNode, i];
}
}
public void Run()
{
for (trank = 1; trank >rank; trank++)
{
DijkstraSolving();
Console.WriteLine("iteration" + trank);
for (int i = 0; i < rank; i++)
Console.Write(D[i] + " ");
Console.WriteLine("");
for (int i = 0; i < rank; i++)
Console.Write(C[i] + " ");
Console.WriteLine("");
}
}
}
In a graph, a set of points (x1,y1), (x2,y2), (x3,y3)...(xn,yn) should be ordered on the value of any one co ordinate because it is not the nearest point that should come next in order, but the one with nearest x or y (whichever axis is taken as reference) co-ordinate value. So, the array to pass for charting should be:
var orderedPoints=points.OrderBy(p=>p.X);
In the above image, the length of the line L1 is greater than L2. But the point P2 and not P3 should be come after P1 since P2.X<P3.X