I have a list of points in 2D space that form an (imperfect) grid:
x x x x
x x x x
x
x x x
x x x x
What's the best way to fit these to a rigid grid (i.e. create a two-dimendional array and work out where each point fits in that array)?
There are no holes in the grid, but I don't know in advance what its dimensions are.
EDIT: The grid is not necessarily regular (not even spacing between rows/cols)
A little bit of an image processing approach:
If you think of what you have as a binary image where the X is 1 and the rest is 0, you can sum up rows and columns, and use a peak finding algorithm to identify peaks which would correspond to x and y lines of the grid:
Your points as a binary image:
Sums of row/columns
Now apply some smoothing technique to the signal (e.g. lowess):
I'm sure you get the idea :-)
Good luck
The best I could come up with is a brute-force solution that calculates the grid dimensions that minimize the error in the square of the Euclidean distance between the point and its nearest grid intersection.
This assumes that the number of points p is exactly equal to the number of columns times the number of rows, and that each grid intersection has exactly one point on it. It also assumes that the minimum x/y value for any point is zero. If the minimum is greater than zero, just subtract the minimum x value from each point's x coordinate and the minimum y value from each point's y coordinate.
The idea is to create all of the possible grid dimensions given the number of points. In the example above with 16 points, we would make grids with dimensions 1x16, 2x8, 4x4, 8x2 and 16x1. For each of these grids we calculate where the grid intersections would lie by dividing the maximum width of the points by the number of columns minus 1, and the maximum height of the points by the number of rows minus 1. Then we fit each point to its closest grid intersection and find the error (square of the distance) between the point and the intersection. (Note that this only works if each point is closer to its intended grid intersection than to any other intersection.)
After summing the errors for each grid configuration individually (e.g. getting one error value for the 1x16 configuration, another for the 2x8 configuration and so on), we select the configuration with the lowest error.
Initialization:
P is the set of points such that P[i][0] is the x-coordinate and
P[i][1] is the y-coordinate
Let p = |P| or the number of points in P
Let max_x = the maximum x-coordinate in P
Let max_y = the maximum y-coordinate in P
(minimum values are assumed to be zero)
Initialize min_error_dist = +infinity
Initialize min_error_cols = -1
Algorithm:
for (col_count = 1; col_count <= n; col_count++) {
// only compute for integer # of rows and cols
if ((p % col_count) == 0) {
row_count = n/col_count;
// Compute the width of the columns and height of the rows
// If the number of columns is 1, let the column width be max_x
// (and similarly for rows)
if (col_count > 1) col_width = max_x/(col_count-1);
else col_width=max_x;
if (row_count > 1) row_height = max_y/(row_count-1);
else row_height=max_y;
// reset the error for the new configuration
error_dist = 0.0;
for (i = 0; i < n; i++) {
// For the current point, normalize the x- and y-coordinates
// so that it's in the range 0..(col_count-1)
// and 0..(row_count-1)
normalized_x = P[i][0]/col_width;
normalized_y = P[i][1]/row_height;
// Error is the sum of the squares of the distances between
// the current point and the nearest grid point
// (in both the x and y direction)
error_dist += (normalized_x - round(normalized_x))^2 +
(normalized_y - round(normalized_y))^2;
}
if (error_dist < min_error_dist) {
min_error_dist = error_dist;
min_error_cols = col_count;
}
}
}
return min_error_cols;
Once you've got the number of columns (and thus the number of rows) you can recompute the normalized values for each point and round them to get the grid intersection they belong to.
In the end I used this algorithm, inspired by beaker's:
Calculate all the possible dimensions of the grid, given the total number of points
For each possible dimension, fit the points to that dimension and calculate the variance in alignment:
Order the points by x-value
Group the points into columns: the first r points form the first column, where r is the number of rows
Within each column, order the points by y-value to determine which row they're in
For each row/column, calcuate the range of y-values/x-values
The variance in alignment is the maximum range found
Choose the dimension with the least variance in alignment
I wrote this algorithm that accounts for missing coordinates as well as coordinates with errors.
Python Code
# Input [x, y] coordinates of a 'sparse' grid with errors
xys = [[103,101],
[198,103],
[300, 99],
[ 97,205],
[304,202],
[102,295],
[200,303],
[104,405],
[205,394],
[298,401]]
def row_col_avgs(num_list, ratio):
# Finds the average of each row and column. Coordinates are
# assigned to a row and column by specifying an error ratio.
last_num = 0
sum_nums = 0
count_nums = 0
avgs = []
num_list.sort()
for num in num_list:
if num > (1 + ratio) * last_num and count_nums != 0:
avgs.append(int(round(sum_nums/count_nums,0)))
sum_nums = num
count_nums = 1
else:
sum_nums = sum_nums + num
count_nums = count_nums + 1
last_num = num
avgs.append(int(round(sum_nums/count_nums,0)))
return avgs
# Split coordinates into two lists of x's and y's
xs, ys = map(list, zip(*xys))
# Find averages of each row and column within a specified error.
x_avgs = row_col_avgs(xs, 0.1)
y_avgs = row_col_avgs(ys, 0.1)
# Return Completed Averaged Grid
avg_grid = []
for y_avg in y_avgs:
avg_row = []
for x_avg in x_avgs:
avg_row.append([int(x_avg), int(y_avg)])
avg_grid.append(avg_row)
print(avg_grid)
Code Output
[[[102, 101], [201, 101], [301, 101]],
[[102, 204], [201, 204], [301, 204]],
[[102, 299], [201, 299], [301, 299]],
[[102, 400], [201, 400], [301, 400]]]
I am also looking for another solution using linear algebra. See my question here.
Related
In MATLAB, I have a 256x256 RGB image and a 3x3 kernel that passes over it. The 3x3 kernel computes the colour-euclidean distance between every pair combination of the 9 pixels in the kernel, and stores the maximum value in an array. It then moves by 1 pixel and performs the same computation, and so on.
I can easily code the movement of the kernel over the image, as well as the extraction of the RGB values from the pixels in the kernel.
HOWEVER, I do have trouble efficiently computing the colour-euclidean distance operation for every pair combination of pixels.
For example if I had a 3x3 matrix with the following values:
[55 12 5; 77 15 99; 124 87 2]
I need to code a loop such that the 1st element performs an operation with the 2nd,3rd...9th element. Then the 2nd element performs the operation with the 3rd,4th...9th element and so on until finally the 8th element performs the operation with the 9th element. Preferrably, the same pixel combination shouldn't compute again (like if you computed 2nd with 7th, don't compute 7th with 2nd).
Thank you in advance.
EDIT: My code so far
K=3;
s=1; %If S=0, don't reject, If S=1 Reject first max distance pixel pair
OI=imread('onion.png');
Rch = im2col(OI(:,:,1),[K,K],'sliding')
Gch = im2col(OI(:,:,2),[K,K],'sliding')
Bch = im2col(OI(:,:,3),[K,K],'sliding')
indexes = bsxfun(#gt,(1:K^2)',1:K^2)
a=find(indexes);
[idx1,idx2] = find(indexes);
Rsqdiff = (Rch(idx2,:) - Rch(idx1,:)).^2
Gsqdiff = (Gch(idx2,:) - Gch(idx1,:)).^2
Bsqdiff = (Bch(idx2,:) - Bch(idx1,:)).^2
dists = sqrt(double(Rsqdiff + Gsqdiff + Bsqdiff)) %Distance values for all 36 combinations in 1 column
[maxdist,idx3] = max(dists,[],1) %idx3 is each column's index of max value
if s==0
y = reshape(maxdist,size(OI,1)-K+1,[]) %max value of each column (each column has 36 values)
elseif s==1
[~,I]=max(maxdist);
idx3=idx3(I);
n=size(idx3,2);
for i=1:1:n
idx3(i)=a(idx3(i));
end
[I,J]=ind2sub([K*K K*K],idx3);
for j=1:1:a
[M,N]=ind2sub([K*K K*K],dists(j,:));
M(I,:)=0;
N(:,J)=0;
dists(j,:)=sub2ind; %Incomplete line, don't know what to do here
end
[maxdist,idx3] = max(dists,[],1);
y = reshape(maxdist,size(OI,1)-K+1,[]);
end
If I understood the question correctly, you are looking to form unique pairwise combinations within a sliding 3x3 window, perform euclidean distance calculations consider all three channels, which we are calling as colour-euclidean distances and finally picking out the largest of all distances for each sliding window. So, for a 3x3 window that has 9 elements, you would have 36 unique pairs. If the image size is MxN, because of the sliding nature, you would have (M-3+1)*(N-3+1) = 64516 (for 256x256 case) such sliding windows with 36 pairs each, and therefore the distances array would be 36x64516 sized and the output array of maximum distances would be of size 254x254. The implementation suggested here involves im2col to extract sliding windowed elements as columns, nchoosek to form the pairs and finally performing the square-root of squared differences between three channels of such pairs and would look something like this -
K = 3; %// Kernel size
Rch = im2col(img(:,:,1),[K,K],'sliding')
Gch = im2col(img(:,:,2),[K,K],'sliding')
Bch = im2col(img(:,:,3),[K,K],'sliding')
[idx1,idx2] = find(bsxfun(#gt,(1:K^2)',1:K^2)); %//'
Rsqdiff = (Rch(idx2,:) - Rch(idx1,:)).^2
Gsqdiff = (Gch(idx2,:) - Gch(idx1,:)).^2
Bsqdiff = (Bch(idx2,:) - Bch(idx1,:)).^2
dists = sqrt(Rsqdiff + Gsqdiff + Bsqdiff)
out = reshape(max(dists,[],1),size(img,1)-K+1,[])
Your question is interesting and caught my attention. As far as I understood, you need to calculate euclidean distance between RGB color values of all cells inside 3x3 kernel and to find the largest one. I suggest a possible way to do this by using circshift function and 4D array operations:
Firstly, we pad the input array and create 8 shifted versions of it for each direction:
DIM = 256;
A = zeros(DIM,DIM,3,9);
A(:,:,:,1) = round(255*rand(DIM,DIM,3));%// random 256x256 array (suppose it is your image)
A = padarray(A,[1,1]);%// add zeros on each side of image
%// compute shifted versions of the input array
%// and write them as 4th dimension starting from shifted up clockwise:
A(:,:,:,2) = circshift(A(:,:,:,1),[-1, 0]);
A(:,:,:,3) = circshift(A(:,:,:,1),[-1, 1]);
A(:,:,:,4) = circshift(A(:,:,:,1),[ 0, 1]);
A(:,:,:,5) = circshift(A(:,:,:,1),[ 1, 1]);
A(:,:,:,6) = circshift(A(:,:,:,1),[ 1, 0]);
A(:,:,:,7) = circshift(A(:,:,:,1),[ 1,-1]);
A(:,:,:,8) = circshift(A(:,:,:,1),[ 0,-1]);
A(:,:,:,9) = circshift(A(:,:,:,1),[-1,-1]);
Next, we create an array that calculates the difference for all the possible combinations between all the above arrays:
q = nchoosek(1:9,2);
B = zeros(DIM+2,DIM+2,3,size(q,1));
for i = 1:size(q,1)
B(:,:,:,i) = (A(:,:,:,q(i,1)) - A(:,:,:,q(i,2))).^2;
end
C = sqrt(sum(B,3));
Finally, what we have is all the euclidean distances between all possible pairs within a 3x3 kernel. All we have to do is to extract the maximum values. As far as I understood, you do not consider image edges, so:
C = sqrt(sum(B,3));
D = zeros(DIM-2);
for i = 3:DIM
for j = 3:DIM
temp = C(i-1:i+1,j-1:j+1);
D(i-2,j-2) = max(temp(:));
end
end
D is the 254x254 array with maximum Euclidean distances for A(2:255,2:255), i.e. we exclude image edges.
Hope that helps.
P.S. I am amazed by the shortness of the code provided by #Divakar.
How can I go about trying to order the points of an irregular array from top left to bottom right, such as in the image below?
Methods I've considered are:
calculate the distance of each point from the top left of the image (Pythagoras's theorem) but apply some kind of weighting to the Y coordinate in an attempt to prioritise points on the same 'row' e.g. distance = SQRT((x * x) + (weighting * (y * y)))
sort the points into logical rows, then sort each row.
Part of the difficulty is that I do not know how many rows and columns will be present in the image coupled with the irregularity of the array of points. Any advice would be greatly appreciated.
Even though the question is a bit older, I recently had a similar problem when calibrating a camera.
The algorithm is quite simple and based on this paper:
Find the top left point: min(x+y)
Find the top right point: max(x-y)
Create a straight line from the points.
Calculate the distance of all points to the line
If it is smaller than the radius of the circle (or a threshold): point is in the top line.
Otherwise: point is in the rest of the block.
Sort points of the top line by x value and save.
Repeat until there are no points left.
My python implementation looks like this:
#detect the keypoints
detector = cv2.SimpleBlobDetector_create(params)
keypoints = detector.detect(img)
img_with_keypoints = cv2.drawKeypoints(img, keypoints, np.array([]), (0, 0, 255),
cv2.DRAW_MATCHES_FLAGS_DRAW_RICH_KEYPOINTS)
points = []
keypoints_to_search = keypoints[:]
while len(keypoints_to_search) > 0:
a = sorted(keypoints_to_search, key=lambda p: (p.pt[0]) + (p.pt[1]))[0] # find upper left point
b = sorted(keypoints_to_search, key=lambda p: (p.pt[0]) - (p.pt[1]))[-1] # find upper right point
cv2.line(img_with_keypoints, (int(a.pt[0]), int(a.pt[1])), (int(b.pt[0]), int(b.pt[1])), (255, 0, 0), 1)
# convert opencv keypoint to numpy 3d point
a = np.array([a.pt[0], a.pt[1], 0])
b = np.array([b.pt[0], b.pt[1], 0])
row_points = []
remaining_points = []
for k in keypoints_to_search:
p = np.array([k.pt[0], k.pt[1], 0])
d = k.size # diameter of the keypoint (might be a theshold)
dist = np.linalg.norm(np.cross(np.subtract(p, a), np.subtract(b, a))) / np.linalg.norm(b) # distance between keypoint and line a->b
if d/2 > dist:
row_points.append(k)
else:
remaining_points.append(k)
points.extend(sorted(row_points, key=lambda h: h.pt[0]))
keypoints_to_search = remaining_points
Jumping on this old thread because I just dealt with the same thing: sorting a sloppily aligned grid of placed objects by left-to-right, top to bottom location. The drawing at the top in the original post sums it up perfectly, except that this solution supports rows with varying numbers of nodes.
S. Vogt's script above was super helpful (and the script below is entirely based on his/hers), but my conditions are narrower. Vogt's solution accommodates a grid that may be tilted from the horizontal axis. I assume no tilting, so I don't need to compare distances from a potentially tilted top line, but rather from a single point's y value.
Javascript below:
interface Node {x: number; y: number; width:number; height:number;}
const sortedNodes = (nodeArray:Node[]) => {
let sortedNodes:Node[] = []; // this is the return value
let availableNodes = [...nodeArray]; // make copy of input array
while(availableNodes.length > 0){
// find y value of topmost node in availableNodes. (Change this to a reduce if you want.)
let minY = Number.MAX_SAFE_INTEGER;
for (const node of availableNodes){
minY = Math.min(minY, node.y)
}
// find nodes in top row: assume a node is in the top row when its distance from minY
// is less than its height
const topRow:Node[] = [];
const otherRows:Node[] = [];
for (const node of availableNodes){
if (Math.abs(minY - node.y) <= node.height){
topRow.push(node);
} else {
otherRows.push(node);
}
}
topRow.sort((a,b) => a.x - b.x); // we have the top row: sort it by x
sortedNodes = [...sortedNodes,...topRow] // append nodes in row to sorted nodes
availableNodes = [...otherRows] // update available nodes to exclude handled rows
}
return sortedNodes;
};
The above assumes that all node heights are the same. If you have some nodes that are much taller than others, get the value of the minimum node height of all nodes and use it instead of the iterated "node.height" value. I.e., you would change this line of the script above to use the minimum height of all nodes rather that the iterated one.
if (Math.abs(minY - node.y) <= node.height)
I propose the following idea:
1. count the points (p)
2. for each point, round it's x and y coordinates down to some number, like
x = int(x/n)*n, y = int(y/m)*m for some n,m
3. If m,n are too big, the number of counts will drop. Determine m, n iteratively so that the number of points p will just be preserved.
Starting values could be in alignment with max(x) - min(x). For searching employ a binary search. X and Y scaling would be independent of each other.
In natural words this would pin the individual points to grid points by stretching or shrinking the grid distances, until all points have at most one common coordinate (X or Y) but no 2 points overlap. You could call that classifying as well.
I'm not sure how to approach this problem. I'm not sure how complex a task it is. My aim is to have an algorithm that generates any polygon. My only requirement is that the polygon is not complex (i.e. sides do not intersect). I'm using Matlab for doing the maths but anything abstract is welcome.
Any aid/direction?
EDIT:
I was thinking more of code that could generate any polygon even things like this:
I took #MitchWheat and #templatetypedef's idea of sampling points on a circle and took it a bit farther.
In my application I need to be able to control how weird the polygons are, ie start with regular polygons and as I crank up the parameters they get increasingly chaotic. The basic idea is as stated by #templatetypedef; walk around the circle taking a random angular step each time, and at each step put a point at a random radius. In equations I'm generating the angular steps as
where theta_i and r_i give the angle and radius of each point relative to the centre, U(min, max) pulls a random number from a uniform distribution, and N(mu, sigma) pulls a random number from a Gaussian distribution, and clip(x, min, max) thresholds a value into a range. This gives us two really nice parameters to control how wild the polygons are - epsilon which I'll call irregularity controls whether or not the points are uniformly space angularly around the circle, and sigma which I'll call spikeyness which controls how much the points can vary from the circle of radius r_ave. If you set both of these to 0 then you get perfectly regular polygons, if you crank them up then the polygons get crazier.
I whipped this up quickly in python and got stuff like this:
Here's the full python code:
import math, random
from typing import List, Tuple
def generate_polygon(center: Tuple[float, float], avg_radius: float,
irregularity: float, spikiness: float,
num_vertices: int) -> List[Tuple[float, float]]:
"""
Start with the center of the polygon at center, then creates the
polygon by sampling points on a circle around the center.
Random noise is added by varying the angular spacing between
sequential points, and by varying the radial distance of each
point from the centre.
Args:
center (Tuple[float, float]):
a pair representing the center of the circumference used
to generate the polygon.
avg_radius (float):
the average radius (distance of each generated vertex to
the center of the circumference) used to generate points
with a normal distribution.
irregularity (float):
variance of the spacing of the angles between consecutive
vertices.
spikiness (float):
variance of the distance of each vertex to the center of
the circumference.
num_vertices (int):
the number of vertices of the polygon.
Returns:
List[Tuple[float, float]]: list of vertices, in CCW order.
"""
# Parameter check
if irregularity < 0 or irregularity > 1:
raise ValueError("Irregularity must be between 0 and 1.")
if spikiness < 0 or spikiness > 1:
raise ValueError("Spikiness must be between 0 and 1.")
irregularity *= 2 * math.pi / num_vertices
spikiness *= avg_radius
angle_steps = random_angle_steps(num_vertices, irregularity)
# now generate the points
points = []
angle = random.uniform(0, 2 * math.pi)
for i in range(num_vertices):
radius = clip(random.gauss(avg_radius, spikiness), 0, 2 * avg_radius)
point = (center[0] + radius * math.cos(angle),
center[1] + radius * math.sin(angle))
points.append(point)
angle += angle_steps[i]
return points
def random_angle_steps(steps: int, irregularity: float) -> List[float]:
"""Generates the division of a circumference in random angles.
Args:
steps (int):
the number of angles to generate.
irregularity (float):
variance of the spacing of the angles between consecutive vertices.
Returns:
List[float]: the list of the random angles.
"""
# generate n angle steps
angles = []
lower = (2 * math.pi / steps) - irregularity
upper = (2 * math.pi / steps) + irregularity
cumsum = 0
for i in range(steps):
angle = random.uniform(lower, upper)
angles.append(angle)
cumsum += angle
# normalize the steps so that point 0 and point n+1 are the same
cumsum /= (2 * math.pi)
for i in range(steps):
angles[i] /= cumsum
return angles
def clip(value, lower, upper):
"""
Given an interval, values outside the interval are clipped to the interval
edges.
"""
return min(upper, max(value, lower))
#MateuszKonieczny here is code to create an image of a polygon from a list of vertices.
vertices = generate_polygon(center=(250, 250),
avg_radius=100,
irregularity=0.35,
spikiness=0.2,
num_vertices=16)
black = (0, 0, 0)
white = (255, 255, 255)
img = Image.new('RGB', (500, 500), white)
im_px_access = img.load()
draw = ImageDraw.Draw(img)
# either use .polygon(), if you want to fill the area with a solid colour
draw.polygon(vertices, outline=black, fill=white)
# or .line() if you want to control the line thickness, or use both methods together!
draw.line(vertices + [vertices[0]], width=2, fill=black)
img.show()
# now you can save the image (img), or do whatever else you want with it.
There's a neat way to do what you want by taking advantage of the MATLAB classes DelaunayTri and TriRep and the various methods they employ for handling triangular meshes. The code below follows these steps to create an arbitrary simple polygon:
Generate a number of random points equal to the desired number of sides plus a fudge factor. The fudge factor ensures that, regardless of the result of the triangulation, we should have enough facets to be able to trim the triangular mesh down to a polygon with the desired number of sides.
Create a Delaunay triangulation of the points, resulting in a convex polygon that is constructed from a series of triangular facets.
If the boundary of the triangulation has more edges than desired, pick a random triangular facet on the edge that has a unique vertex (i.e. the triangle only shares one edge with the rest of the triangulation). Removing this triangular facet will reduce the number of boundary edges.
If the boundary of the triangulation has fewer edges than desired, or the previous step was unable to find a triangle to remove, pick a random triangular facet on the edge that has only one of its edges on the triangulation boundary. Removing this triangular facet will increase the number of boundary edges.
If no triangular facets can be found matching the above criteria, post a warning that a polygon with the desired number of sides couldn't be found and return the x and y coordinates of the current triangulation boundary. Otherwise, keep removing triangular facets until the desired number of edges is met, then return the x and y coordinates of triangulation boundary.
Here's the resulting function:
function [x, y, dt] = simple_polygon(numSides)
if numSides < 3
x = [];
y = [];
dt = DelaunayTri();
return
end
oldState = warning('off', 'MATLAB:TriRep:PtsNotInTriWarnId');
fudge = ceil(numSides/10);
x = rand(numSides+fudge, 1);
y = rand(numSides+fudge, 1);
dt = DelaunayTri(x, y);
boundaryEdges = freeBoundary(dt);
numEdges = size(boundaryEdges, 1);
while numEdges ~= numSides
if numEdges > numSides
triIndex = vertexAttachments(dt, boundaryEdges(:,1));
triIndex = triIndex(randperm(numel(triIndex)));
keep = (cellfun('size', triIndex, 2) ~= 1);
end
if (numEdges < numSides) || all(keep)
triIndex = edgeAttachments(dt, boundaryEdges);
triIndex = triIndex(randperm(numel(triIndex)));
triPoints = dt([triIndex{:}], :);
keep = all(ismember(triPoints, boundaryEdges(:,1)), 2);
end
if all(keep)
warning('Couldn''t achieve desired number of sides!');
break
end
triPoints = dt.Triangulation;
triPoints(triIndex{find(~keep, 1)}, :) = [];
dt = TriRep(triPoints, x, y);
boundaryEdges = freeBoundary(dt);
numEdges = size(boundaryEdges, 1);
end
boundaryEdges = [boundaryEdges(:,1); boundaryEdges(1,1)];
x = dt.X(boundaryEdges, 1);
y = dt.X(boundaryEdges, 2);
warning(oldState);
end
And here are some sample results:
The generated polygons could be either convex or concave, but for larger numbers of desired sides they will almost certainly be concave. The polygons are also generated from points randomly generated within a unit square, so polygons with larger numbers of sides will generally look like they have a "squarish" boundary (such as the lower right example above with the 50-sided polygon). To modify this general bounding shape, you can change the way the initial x and y points are randomly chosen (i.e. from a Gaussian distribution, etc.).
For a convex 2D polygon (totally off the top of my head):
Generate a random radius, R
Generate N random points on the circumference of a circle of Radius R
Move around the circle and draw straight lines between adjacent points on the circle.
As #templatetypedef and #MitchWheat said, it is easy to do so by generating N random angles and radii. It is important to sort the angles, otherwise it will not be a simple polygon. Note that I am using a neat trick to draw closed curves - I described it in here. By the way, the polygons might be concave.
Note that all of these polygons will be star shaped. Generating a more general polygon is not a simple problem at all.
Just to give you a taste of the problem - check out
http://www.cosy.sbg.ac.at/~held/projects/rpg/rpg.html
and http://compgeom.cs.uiuc.edu/~jeffe/open/randompoly.html.
function CreateRandomPoly()
figure();
colors = {'r','g','b','k'};
for i=1:5
[x,y]=CreatePoly();
c = colors{ mod(i-1,numel(colors))+1};
plotc(x,y,c);
hold on;
end
end
function [x,y]=CreatePoly()
numOfPoints = randi(30);
theta = randi(360,[1 numOfPoints]);
theta = theta * pi / 180;
theta = sort(theta);
rho = randi(200,size(theta));
[x,y] = pol2cart(theta,rho);
xCenter = randi([-1000 1000]);
yCenter = randi([-1000 1000]);
x = x + xCenter;
y = y + yCenter;
end
function plotc(x,y,varargin)
x = [x(:) ; x(1)];
y = [y(:) ; y(1)];
plot(x,y,varargin{:})
end
Here is a working port for Matlab of Mike Ounsworth solution. I did not optimized it for matlab. I might update the solution later for that.
function [points] = generatePolygon(ctrX, ctrY, aveRadius, irregularity, spikeyness, numVerts)
%{
Start with the centre of the polygon at ctrX, ctrY,
then creates the polygon by sampling points on a circle around the centre.
Randon noise is added by varying the angular spacing between sequential points,
and by varying the radial distance of each point from the centre.
Params:
ctrX, ctrY - coordinates of the "centre" of the polygon
aveRadius - in px, the average radius of this polygon, this roughly controls how large the polygon is, really only useful for order of magnitude.
irregularity - [0,1] indicating how much variance there is in the angular spacing of vertices. [0,1] will map to [0, 2pi/numberOfVerts]
spikeyness - [0,1] indicating how much variance there is in each vertex from the circle of radius aveRadius. [0,1] will map to [0, aveRadius]
numVerts - self-explanatory
Returns a list of vertices, in CCW order.
Website: https://stackoverflow.com/questions/8997099/algorithm-to-generate-random-2d-polygon
%}
irregularity = clip( irregularity, 0,1 ) * 2*pi/ numVerts;
spikeyness = clip( spikeyness, 0,1 ) * aveRadius;
% generate n angle steps
angleSteps = [];
lower = (2*pi / numVerts) - irregularity;
upper = (2*pi / numVerts) + irregularity;
sum = 0;
for i =1:numVerts
tmp = unifrnd(lower, upper);
angleSteps(i) = tmp;
sum = sum + tmp;
end
% normalize the steps so that point 0 and point n+1 are the same
k = sum / (2*pi);
for i =1:numVerts
angleSteps(i) = angleSteps(i) / k;
end
% now generate the points
points = [];
angle = unifrnd(0, 2*pi);
for i =1:numVerts
r_i = clip( normrnd(aveRadius, spikeyness), 0, 2*aveRadius);
x = ctrX + r_i* cos(angle);
y = ctrY + r_i* sin(angle);
points(i,:)= [(x),(y)];
angle = angle + angleSteps(i);
end
end
function value = clip(x, min, max)
if( min > max ); value = x; return; end
if( x < min ) ; value = min; return; end
if( x > max ) ; value = max; return; end
value = x;
end
I am trying to come up with a dynamic programming algorithm that finds the largest sub matrix within a matrix that consists of the same number:
example:
{5 5 8}
{5 5 7}
{3 4 1}
Answer : 4 elements due to the matrix
5 5
5 5
This is a question I already answered here (and here, modified version). In both cases the algorithm was applied to binary case (zeros and ones), but the modification for arbitrary numbers is quite easy (but sorry, I keep the images for the binary version of the problem). You can do this very efficiently by two pass linear O(n) time algorithm - n being number of elements. However, this is not a dynamic programming - I think using dynamic programming here would be clumsy and inefficient in the end, because of the difficulties with problem decomposition, as the OP mentioned - unless its a homework - but in that case you can try to impress by this algorithm :-) as there's obviously no faster solution than O(n).
Algorithm (pictures depict binary case):
Say you want to find largest rectangle of free (white) elements.
Here follows the two pass linear O(n) time algorithm (n being number of elemets):
1) in a first pass, go by columns, from bottom to top, and for each element, denote the number of consecutive elements available up to this one:
repeat, until:
Pictures depict the binary case. In case of arbitrary numbers you hold 2 matrices - first with the original numbers and second with the auxiliary numbers that are filled in the image above. You have to check the original matrix and if you find a number different from the previous one, you just start the numbering (in the auxiliary matrix) again from 1.
2) in a second pass you go by rows, holding data structure of potential rectangles, i.e. the rectangles containing current position somewhere at the top edge. See the following picture (current position is red, 3 potential rectangles - purple - height 1, green - height 2 and yellow - height 3):
For each rectangle we keep its height k and its left edge. In other words we keep track of the sums of consecutive numbers that were >= k (i.e. potential rectangles of height k). This data structure can be represented by an array with double linked list linking occupied items, and the array size would be limited by the matrix height.
Pseudocode of 2nd pass (non-binary version with arbitrary numbers):
var m[] // original matrix
var aux[] // auxiliary matrix filled in the 1st pass
var rect[] // array of potential rectangles, indexed by their height
// the occupied items are also linked in double linked list,
// ordered by height
foreach row = 1..N // go by rows
foreach col = 1..M
if (col > 1 AND m[row, col] != m[row, col - 1]) // new number
close_potential_rectangles_higher_than(0); // close all rectangles
height = aux[row, col] // maximal height possible at current position
if (!rect[height]) { // rectangle with height does not exist
create rect[height] // open new rectangle
if (rect[height].next) // rectangle with nearest higher height
// if it exists, start from its left edge
rect[height].left_col = rect[height].next.left_col
else
rect[height].left_col = col;
}
close_potential_rectangles_higher_than(height)
end for // end row
close_potential_rectangles_higher_than(0);
// end of row -> close all rect., supposing col is M+1 now!
end for // end matrix
The function for closing rectangles:
function close_potential_rectangles_higher_than(height)
close_r = rectangle with highest height (last item in dll)
while (close_r.height > height) { // higher? close it
area = close_r.height * (col - close_r.left_col)
if (area > max_area) { // we have maximal rectangle!
max_area = area
max_topleft = [row, close_r.left_col]
max_bottomright = [row + height - 1, col - 1]
}
close_r = close_r.prev
// remove the rectangle close_r from the double linked list
}
end function
This way you can also get all maximum rectangles. So in the end you get:
And what the complexity will be? You see that the function close_potential_rectangles_higher_than is O(1) per closed rectangle. Because for each field we create 1 potential rectangle at the maximum, the total number of potential rectangles ever present in particular row is never higher than the length of the row. Therefore, complexity of this function is O(1) amortized!
So the whole complexity is O(n) where n is number of matrix elements.
A dynamic solution:
Define a new matrix A wich will store in A[i,j] two values: the width and the height of the largest submatrix with the left upper corner at i,j, fill this matrix starting from the bottom right corner, by rows bottom to top. You'll find four cases:
case 1: none of the right or bottom neighbour elements in the original matrix are equal to the current one, i.e: M[i,j] != M[i+1,j] and M[i,j] != M[i,j+1] being M the original matrix, in this case, the value of A[i,j] is 1x1
case 2: the neighbour element to the right is equal to the current one but the bottom one is different, the value of A[i,j].width is A[i+1,j].width+1 and A[i,j].height=1
case 3: the neighbour element to the bottom is equal but the right one is different, A[i,j].width=1, A[i,j].height=A[i,j+1].height+1
case 4: both neighbours are equal: A[i,j].width = min(A[i+1,j].width+1,A[i,j+1].width) and A[i,j].height = min(A[i,j+1]+1,A[i+1,j])
the size of the largest matrix that has the upper left corner at i,j is A[i,j].width*A[i,j].height so you can update the max value found while calculating the A[i,j]
the bottom row and the rightmost column elements are treated as if their neighbours to the bottom and to the right respectively are different
in your example, the resulting matrix A would be:
{2:2 1:2 1:1}
{2:1 1:1 1:1}
{1:1 1:1 1:1}
being w:h width:height
Modification to the above answer:
Define a new matrix A wich will store in A[i,j] two values: the width and the height of the largest submatrix with the left upper corner at i,j, fill this matrix starting from the bottom right corner, by rows bottom to top. You'll find four cases:
case 1: none of the right or bottom neighbour elements in the original matrix are equal to the current one, i.e: M[i,j] != M[i+1,j] and M[i,j] != M[i,j+1] being M the original matrix, in this case, the value of A[i,j] is 1x1
case 2: the neighbour element to the right is equal to the current one but the bottom one is different, the value of A[i,j].width is A[i+1,j].width+1 and A[i,j].height=1
case 3: the neighbour element to the bottom is equal but the right one is different, A[i,j].width=1, A[i,j].height=A[i,j+1].height+1
case 4: both neighbours are equal:
Three rectangles are considered:
1. A[i,j].width=A[i,j+1].width+1; A[i,j].height=1;
A[i,j].height=A[i+1,j].height+1; a[i,j].width=1;
A[i,j].width = min(A[i+1,j].width+1,A[i,j+1].width) and A[i,j].height = min(A[i,j+1]+1,A[i+1,j])
The one with the max area in the above three cases will be considered to represent the rectangle at this position.
The size of the largest matrix that has the upper left corner at i,j is A[i,j].width*A[i,j].height so you can update the max value found while calculating the A[i,j]
the bottom row and the rightmost column elements are treated as if their neighbours to the bottom and to the right respectively are different.
This question is a duplicate. I have tried to flag it as a duplicate. Here is a Python solution, which also returns the position and shape of the largest rectangular submatrix:
#!/usr/bin/env python3
import numpy
s = '''5 5 8
5 5 7
3 4 1'''
nrows = 3
ncols = 3
skip_not = 5
area_max = (0, [])
a = numpy.fromstring(s, dtype=int, sep=' ').reshape(nrows, ncols)
w = numpy.zeros(dtype=int, shape=a.shape)
h = numpy.zeros(dtype=int, shape=a.shape)
for r in range(nrows):
for c in range(ncols):
if not a[r][c] == skip_not:
continue
if r == 0:
h[r][c] = 1
else:
h[r][c] = h[r-1][c]+1
if c == 0:
w[r][c] = 1
else:
w[r][c] = w[r][c-1]+1
minw = w[r][c]
for dh in range(h[r][c]):
minw = min(minw, w[r-dh][c])
area = (dh+1)*minw
if area > area_max[0]:
area_max = (area, [(r, c, dh+1, minw)])
print('area', area_max[0])
for t in area_max[1]:
print('coord and shape', t)
Output:
area 4
coord and shape (1, 1, 2, 2)
I have a N-dimensional vector, X and 'n' equidistant points along each dimension and a parameter 'delta'. I need a way to find the total of n^N vectors enclosed by the Hypercube defined with the vector X at the center and each side of Hypercube being of size 2*delta.
For example:
Consider a case of N=3, so we have a Cube of size (2*delta) enclosing the point X.
------------\
|\--------|--\
| | X | |
----------- |
\ |_2*del___\|
Along each dimension I have 'n' points. So, I have a total of n^3 vectors around X. I need to find all the vectors. Is there any standard algorithm/method for the same? If you have done anything similar, please suggest.
If the problem is not clear, let me know.
This is what I was looking at: Considering one dimension, length of a side is 2*delta and I have n divisions. So, each sub-division is of size (2*delta/n). So I just move to the origin that is (x-delta) (since x is the mid point of the side) and obtain the 'n' points by {(x-delta) + 1*(2*delta/n),(x-delta) + 2*(2*delta/n)....+ (x-delta) + 1*(n*delta/n) } . I do this for all the N-dimensions and then take a permutation of the co-ordinates. That way I have all the points.
(I would like to close this)
If i understand your problem correctly, you have an axis-aligned hypercube centred around a point X, and you have subdivided the interior of this hypercube into a regular lattice where the lattice points and spacing are in the coordinate system of the hypercube. All you have to do is let X = 0, find the vectors to each of the lattice points, and then go back and translate them by X.
Edit: let me add an example
let x = (5,5,5), delta = 1 and n = 3
then, moving x to the origin, your lattice points are (-1, -1, -1), (0, -1, -1), (1, -1, -1) and so on for a total of 27. translating back, we have (4, 4, 4), (5, 4, 4), (6, 4, 4) and so on.
Ok, I didn't fully understand your question. There are total of 2^(N-1)*N "lines" about a point in an N-dimensional hypercube.
If you just want to create n points on lines which look like the axis, but translated at a distance of delta from the origin, here's some (poorly written, for clarity) MATLAB code:
n = 10;
delta = 10;
N = 3;
step = (2*delta)/(n-1);
P = zeros(n,N,N);
X = [20 30 25];
for line_dim = 1:N
for point = 1:n
for point_dim = 1:N
if(point_dim ~= line_dim)
P(point,point_dim,line_dim) = X(point_dim)-delta;
else
P(point,point_dim,line_dim) = X(point_dim)-delta+step*(point-1);
end
end
end
end
The code's for a cube, but it should work for any N. All I've done is:
Draw those n equidistant points on the axes.
Translate the axes by (X-delta)
Display:
% Display stuff
PP = reshape(permute(P,[1 3 2]),[n*N N]);
plot3(X(1),X(2),X(3),'r*',PP(:,1),PP(:,2),PP(:,3),'.')
axis([0 Inf 0 Inf 0 Inf]);
grid on;