I'm looking for an algorithm to find the polygon that surrounds a contiguous grid of squares without holes as shown here:
.
I already have each of the grid squares storing data about the kind of edges with the surrounding area that they are composed of (i.e. top, top-right, top-bottom, no edges, etc.), so I'm thinking that this data could be utilized by the algorithm. If someone could provide some pseudocode for such an algorithm that would also be great.
The input to the algorithm would be a list of data objects, each with a Vector2Int describing the grid positions (note that these are simply positions within a grid, not vertices) as well as an Enum that gives the type of edges that the square has with the surrounding area. The output would be an ordered list of Vector2s describing the vertices of the surrounding polygon, assuming that each grid square is one unit in size.
I have found a similar question in the link below, but I wanted some elaboration on the kind of algorithm that would be specific to my case, especially given the data that I already have stored about the edges. I'd also prefer the algorithm to avoid calculating each of the squares' vertices and running a bunch of straightforward searches to eliminate the shared ones, as I feel that this might be too computationally expensive for my particular application. I just have a suspicion that there has to be a better way.
Outline (circumference) polygon extraction from geometry constructed from equal squares
EDIT: Now I'm beginning to think that some sort of maze walking algorithm might actually be appropriate for my situation. I'm working on a solution that I think will work, but it's very cumbersome to write (involving a tonne of conditional checks against the square edges and the direction of travel around the circumference) and probably isn't as fast as it could be.
I am not sure to understand what your data structure contains, and I assume that you have a list of squares known by the coordinates of some point (corner or center).
Compute the bounding box and create a binary bitmap of the same size. Unless the geometry is really sparse, the area of the bitmap will be of the same order as the number of squares.
For every square, paint the corresponding pixel black. Then use a contouring algorithm. To obtain the outline of the squares, you will need to design a correspondence table between the pixl-to-pixel moves and the outline fragments to be appended.
Came across this post looking for alternatives to my solution. This is what I came up with:
For a cell:
| |
---(0, 0)--------(1, 0)---
| |
| |
| R0C0 |
| |
| |
---(0, 1)--------(1, 1)---
| |
Calculate the borders of each cell as a set of 2 of its corner coordinates:
top: ((c, r), (c, r + 1))
right: ((c, r + 1), (c + 1, r + 1))
bottom: ((c + 1, r + 1), (c + 1, r))
left: ((c + 1, r), (c, r))
Notice how these defined clock-wise, this is important
So for the grid
R0C0 R0C1 R0C2 R0C3
R1C2 R1C3
R2C1 R2C2
you'd get the following edges:
R0C0 (top, bottom, left): (0, 0)-(1, 0), (1, 1)-(0, 1), (0, 1)-(0, 0)
R0C1 (top, bottom): (1, 0)-(2, 0), (2, 1)-(1, 1)
R0C2 (top): (2, 0)-(3, 0)
R0C3 (top, right): (3, 0)-(4, 0), (4, 0)-(4, 1)
R1C2 (left): (2, 2)-(2, 1)
R1C3 (right, bottom): (4, 1)-(4, 2), (4, 2)-(3, 2)
R2C1 (top, bottom, left): (1, 2)-(2, 2), (2, 3)-(1, 3), (1, 3)-(1, 2)
R2C2 (right, bottom): (3, 2)-(3, 3), (3, 3)-(2, 3)
Now it's a question of ordering these in a way that the first coordinate of of one element is the same as second coordinate of its predecessor.
(0, 0)-(1, 0) (0, 0)-(1, 0)
(1, 1)-(0, 1) (1, 0)-(2, 0)
(0, 1)-(0, 0) (2, 0)-(3, 0)
(1, 0)-(2, 0) (3, 0)-(4, 0)
(2, 1)-(1, 1) (4, 0)-(4, 1)
(2, 0)-(3, 0) (4, 1)-(4, 2)
(3, 0)-(4, 0) (4, 2)-(3, 2)
(4, 0)-(4, 1) => (3, 2)-(3, 3)
(2, 2)-(2, 1) (3, 3)-(2, 3)
(4, 1)-(4, 2) (2, 3)-(1, 3)
(4, 2)-(3, 2) (1, 3)-(1, 2)
(1, 2)-(2, 2) (1, 2)-(2, 2)
(2, 3)-(1, 3) (2, 2)-(2, 1)
(1, 3)-(1, 2) (2, 1)-(1, 1)
(3, 2)-(3, 3) (1, 1)-(0, 1)
(3, 3)-(2, 3) (0, 1)-(0, 0)
Now in the result, let's take only the first coordinate, this is your polygon:
(0, 0)
(1, 0)
(2, 0)
(3, 0)
(4, 0)
(4, 1)
(4, 2)
(3, 2)
(3, 3)
(2, 3)
(1, 3)
(1, 2)
(2, 2)
(2, 1)
(1, 1)
(0, 1)
You can now simplify it by eliminating consecutive points that are on a single line (i.e. in three consecutive points that either have the same x or y coordinate, eliminate the middle one)
(0, 0)
(4, 0)
(4, 2)
(3, 2)
(3, 3)
(1, 3)
(1, 2)
(2, 2)
(2, 1)
(0, 1)
This is now your polygon in clock-wise order:
(0, 0)--------------------------------------(4, 0)
| |
| |
(0, 1)----------------(2, 1) |
| |
| |
(1, 2)-----(2, 2) (3, 2)-----(4, 2)
| |
| |
(1, 3)----------------(3, 3)
This algorithm can be expanded to handle holes as well. You'd just need to account for multiple polygons when ordering the edges. Conveniently, holes will be defined counter-clock-wise, this is handy if you want to draw the result with svg paths or other d2 path algorithms that allow for polygons with overlap.
Related
Given a list of line segments I need to construct a list of polyline while keeping the number of polylines minimum.
The polylines must not visit the same edge more than once.
For example if I was given the 4 edges of a rectangle then one polyline would be sufficient.
If I was given the 6 edges of a rectangle with a cross in the middle then I would need two polyline to cover it.
This problem looks very similar to travelling sales man problem so I am not sure if a solution exists. I can live with a sub optimal solution in that case.
Edit:
Performance is more important than precision for us so ideally we would want to group the ones that almost join (1-2 pixels away from joining) each other into one polyline
Examples:
input for a square:
L(0, 0) - (0, 1), L(0, 1) - (1, 1), L(1, 1) - (1, 0), L(1, 0) - (0, 0)
Expected output: Polyline (0, 0), (0, 1), (1, 1), (1, 0), Close
input for a square with X:
L(0, 0) - (0, 1), L(0, 1) - (1, 1), L(1, 1) - (1, 0), L(1, 0) - (0, 0), L(0, 0) - (1, 1), L(1, 0) - (0, 1)
One possible output: Polyline1 (0, 0), (0, 1), (1, 1), (1, 0), (0, 0), (1, 1) Polyline2 (1, 0), (0, 1)
input for lines close to each other:
L(0, 0) - (1, 0), L(2, 0) - (3, 0)
Ideal output: Polyline (0, 0), (3, 0)
I have two arrays and an empty matrix, I need to perform a function such that the resulting matrix includes every combination of the two arrays.
Unfortunately I cannot run the arrays separately as they are both optional parameters for the function. I thought that the best way to do this was through nested loops but now I am unsure...
I've tried multiplying one of the matrices so that it includes the necessary duplicates, but I struggled with that as the real data is somewhat larger.
I've tried many versions of these nested loops.
a = [ 1 2 3 ]
b = [ 4 5 6 7 ]
ab = zeros(3,4)
for i = 1:length(a)
for j = 1:length(b)
ab[??] = function(x = a[??], y = b[??])
end
end
ab = [1x4 1x5 1x6 1x7, 2x4 2x5 2x6 2x7, 3x4 3x5 3x6 3x7]
Your problem can be solved by broadcasting:
julia> f(x, y) = (x,y) # trivial example
f (generic function with 1 method)
julia> f.([1 2 3]', [4 5 6 7])
3×4 Array{Tuple{Int64,Int64},2}:
(1, 4) (1, 5) (1, 6) (1, 7)
(2, 4) (2, 5) (2, 6) (2, 7)
(3, 4) (3, 5) (3, 6) (3, 7)
The prime in a' transposes a to make the shapes work out correctly.
But note that a = [ 1 2 3 ] constructs a 1×3 Array{Int64,2}, which is a matrix. For a vector (what you probably call "array"), use commas: a = [ 1, 2, 3 ] etc. If you have your data in that form, you have to transpose the other way round:
julia> f.([1,2,3], [4,5,6,7]')
3×4 Array{Tuple{Int64,Int64},2}:
(1, 4) (1, 5) (1, 6) (1, 7)
(2, 4) (2, 5) (2, 6) (2, 7)
(3, 4) (3, 5) (3, 6) (3, 7)
BTW, this is called an "outer product" (for f = *), or a generalization of it. And if f is an operator ∘, you can use dotted infix broadcasting: a' ∘. b.
Isn't that just
a'.*b
?
Oh, now I have to write some more characters to get past the minimum acceptable answer length but I don't really have anything to add, I hope the code is self-explanatory.
Also a list comprehension:
julia> a = [1,2,3];
julia> b = [4,5,6,7];
julia> ab = [(x,y) for x in a, y in b]
3×4 Array{Tuple{Int64,Int64},2}:
(1, 4) (1, 5) (1, 6) (1, 7)
(2, 4) (2, 5) (2, 6) (2, 7)
(3, 4) (3, 5) (3, 6) (3, 7)
So I'm trying to sort data in this format...
[((0, 4), 3), ((4, 0), 3), ((1, 6), 1), ((3, 2), 3), ((0, 5), 1)...
Ascending by key and then descending by value. I'm able to achieve this via...
test = test.sortBy(lambda x: (x[0], -x[1]))
which would give me based on shortened version above...
[((0, 4), 3), ((0, 5), 1), ((1, 6), 1), ((3, 2), 3), ((4, 0), 3)...
The problem I'm having is that after the sorting I no longer want the value but do need to retain the sort after grouping the data. So...
test = test.map(lambda x: (x[0][0],x[0][1]))
Gives me...
[(0, 4), (0, 5), (1, 6), (3, 2), (4, 0)...
Which is still in the order I need it but I need the elements to be grouped up by key. I then use this command...
test = test.groupByKey().map(lambda x: (x[0], list(x[1])))
But in the process I lose the sorting. Is there any way retain?
I managed to retain the order by changing the format of the tuple...
test = test.map(lambda x: (x[0][0],(x[0][1],x[1]))
test = test.groupByKey().map(lambda x: (x[0], sorted(list(x[1]), key=lambda x: (x[0],-x[1]))))
[(0, [(4, 3), (5, 1)] ...
which leaves me with the value (2nd element in the tuple) that I want to get rid of but took care of that too...
test = test.map(lambda x: (x[0], [e[0] for e in x[1]]))
Feels a bit hacky but not sure how else it could be done.
Say, we have an N-dimensional grid and some point X in it with coordinates (x1, x2, ..., xN).
For simplicity we can assume that the grid is unbounded.
Let there be a radius R and a sphere of this radius with center in X, that is the set of all points in grid such that their manhattan distance from X is equal to R.
I suspect that their will be 2*N*R such points.
My question is: how do I enumerate them in efficient and simple way? By "enumerate" I mean the algorithm, which, given N, X and R will produce the list of points which form this sphere (where point is the list of it's coordinates).
UPDATE: Initially I called the metric I used "Hamming distance" by mistake. My apologies to all who answered the question. Thanks to Steve Jessop for pointing this out.
Consider the minimal axis-aligned hypercube that bounds the hypersphere and write a procedure to enumerate the grid points inside the hypercube.
Then you only need a simple filter function that allows you to discard the points that are on the cube but not in the hypersphere.
This is a simple and efficient solution for small dimensions. For instance, for 2D, 20% of the points enumerated for the bounding square are discarded; for 6D, almost 90% of the hypercube points are discarded.
For higher dimensions, you will have to use a more complex approach: loop over every dimension (you may need to use a recursive function if the number of dimensions is variable). For every loop you will have to adjust the minimal and maximal values depending on the values of the already calculated grid components. Well, try doing it for 2D, enumerating the points of a circle and once you understand it, generalizing the procedure to higher dimensions would be pretty simple.
update: errh, wait a minute, you want to use the Manhattan distance. Calling the cross polytope "sphere" may be correct but I found it quite confusing! In any case you can use the same approach.
If you only want to enumerate the points on the hyper-surface of the cross polytope, well, the solution is also very similar, you have to loop over every dimension with appropriate limits. For instance:
for (i = 0; i <= n; i++)
for (j = 0; j + i <= n; j++)
...
for (l = 0; l + ...+ j + i <= n; l++) {
m = n - l - ... - j - i;
printf(pat, i, j, ..., l, m);
}
For every point generated that way, then you will have to consider all the variations resulting of negating any of the components to cover all the faces and then displace them with the vector X.
update
Perl implementation for the case where X = 0:
#!/usr/bin/perl
use strict;
use warnings;
sub enumerate {
my ($d, $r) = #_;
if ($d == 1) {
return ($r ? ([-$r], [$r]) : [0])
}
else {
my #r;
for my $i (0..$r) {
for my $s (enumerate($d - 1, $r - $i)) {
for my $j ($i ? (-$i, $i) : 0) {
push #r, [#$s, $j]
}
}
}
return #r;
}
}
#ARGV == 2 or die "Usage:\n $0 dimension radius\n\n";
my ($d, $r) = #ARGV;
my #r = enumerate($d, $r);
print "[", join(',', #$_), "]\n" for #r;
Input: radius R, dimension D
Generate all integer partitions of R with cardinality ≤ D
For each partition, permute it without repetition
For each permutation, twiddle all the signs
For example, code in python:
from itertools import *
# we have to write this function ourselves because python doesn't have it...
def partitions(n, maxSize):
if n==0:
yield []
else:
for p in partitions(n-1, maxSize):
if len(p)<maxSize:
yield [1] + p
if p and (len(p)<2 or p[1]>p[0]):
yield [ p[0]+1 ] + p[1:]
# MAIN CODE
def points(R, D):
for part in partitions(R,D): # e.g. 4->[3,1]
part = part + [0]*(D-len(part)) # e.g. [3,1]->[3,1,0] (padding)
for perm in set(permutations(part)): # e.g. [1,3,0], [1,0,3], ...
for point in product(*[ # e.g. [1,3,0], [-1,3,0], [1,-3,0], [-...
([-x,x] if x!=0 else [0]) for x in perm
]):
yield point
Demo for radius=4, dimension=3:
>>> result = list( points(4,3) )
>>> result
[(-1, -2, -1), (-1, -2, 1), (-1, 2, -1), (-1, 2, 1), (1, -2, -1), (1, -2, 1), (1, 2, -1), (1, 2, 1), (-2, -1, -1), (-2, -1, 1), (-2, 1, -1), (-2, 1, 1), (2, -1, -1), (2, -1, 1), (2, 1, -1), (2, 1, 1), (-1, -1, -2), (-1, -1, 2), (-1, 1, -2), (-1, 1, 2), (1, -1, -2), (1, -1, 2), (1, 1, -2), (1, 1, 2), (0, -2, -2), (0, -2, 2), (0, 2, -2), (0, 2, 2), (-2, 0, -2), (-2, 0, 2), (2, 0, -2), (2, 0, 2), (-2, -2, 0), (-2, 2, 0), (2, -2, 0), (2, 2, 0), (-1, 0, -3), (-1, 0, 3), (1, 0, -3), (1, 0, 3), (-3, -1, 0), (-3, 1, 0), (3, -1, 0), (3, 1, 0), (0, -1, -3), (0, -1, 3), (0, 1, -3), (0, 1, 3), (-1, -3, 0), (-1, 3, 0), (1, -3, 0), (1, 3, 0), (-3, 0, -1), (-3, 0, 1), (3, 0, -1), (3, 0, 1), (0, -3, -1), (0, -3, 1), (0, 3, -1), (0, 3, 1), (0, -4, 0), (0, 4, 0), (0, 0, -4), (0, 0, 4), (-4, 0, 0), (4, 0, 0)]
>>> len(result)
66
(Above I used set(permutations(...)) to get permutations without repetition, which is not efficient in general, but it might not matter here due to the nature of the points. And if efficiency mattered, you could write your own recursive function in your language of choice.)
This method is efficient because it does not scale with the hypervolume, but just scales with the hypersurface, which is what you're trying to enumerate (might not matter much except for very large radii: e.g. will save you roughly a factor of 100x speed if your radius is 100).
You can work your way recursively from the center, counting zero distance once and working on symmetries. This Python implementation works on the lower-dimension "stem" vector and realizes one 1-dimensional slice at a time. One might also do the reverse, but it would imply iterating on the partial hyperspheres. While mathematically the same, the efficiency of both approaches is heavily language-dependent.
If you know beforehand the cardinality of the target space, I would recommend to write an iterative implementation.
The following enumerates the points on a R=16 hyper-LEGO block in six dimensions in about 200 ms on my laptop. Of course, performance rapidly decreases with more dimensions or larger spheres.
def lapp(lst, el):
lst2 = list(lst)
lst2.append(el)
return lst2
def hypersphere(n, r, stem = [ ]):
mystem = lapp(stem, 0)
if 1 == n:
ret = [ mystem ]
for d in range(1, r+1):
ret.append(lapp(stem, d))
ret.append(lapp(stem, -d))
else:
ret = hypersphere(n-1, r, mystem)
for d in range(1, r+1):
mystem[-1] = d
ret.extend(hypersphere(n-1, r-d, mystem))
mystem[-1] = -d
ret.extend(hypersphere(n-1, r-d, mystem))
return ret
(This implementation assumes the hypersphere is centered in the origin. It would be easier to translate all points afterwards than carrying along the coordinates of the center).
I'm working on a program for class that involves solving the Chinese Postman problem. Our assignment only requires us to write a program to solve it for a hard-coded graph but I'm attempting to solve it for the general case on my own.
The part that is giving me trouble is generating the partitions of pairings for the odd vertices.
For example, if I had the following labeled odd verticies in a graph:
1 2 3 4 5 6
I need to find all the possible pairings / partitions I can make with these vertices.
I've figured out I'll have i paritions given:
n = num of odd verticies
k = n / 2
i = ((2k)(2k-1)(2k-2)...(k+1))/2^n
So, given the 6 odd verticies above, we will know that we need to generate i = 15 partitions.
The 15 partions would look like:
1 2 3 4 5 6
1 2 3 5 4 6
1 2 3 6 4 5
...
1 6 ...
Then, for each partition, I take each pair and find the shortest distance between them and sum them for that partition. The partition with the total smallest distance between its pairs is selected, and I then double all the edges between the shortest path between the odd vertices (found in the selected partition).
These represent the edges the postman will have to walk twice.
At first I thought I had worked out an appropriate algorithm for generating these partitions:
Start with all the odd verticies sorted in increasing order
12 34 56
Select the pair behind the pair that currently has the max vertice
12 [34] 56
Increase the second digit in this pair by 1. Leave everything to the
left of the selected pair the same,
and make everything to the right of
the selected pair the remaining
numbers in the set, sorted in
increasing order.
12 35 46
Repeat
However, this is flawed. For example, I realized that when I reach to the end and the select pair is at the left most position (ie):
[16] .. ..
The algorithm I worked out will stop in this case, and not generate the rest of the pairs that begin [16], because there is no pair to the left of it to alter.
So, it is back to the drawing board.
Does anyone who has studied this problem before have any tips that can help point me in the right direction for generating these partitions?
You can construct the partitions using a recursive algorithm.
Take the lowest node, in this case node 1. This must be paired with one of the other unpaired nodes (2 to 6). For each of these nodes, create with match 1, then find all of the pairs of the remaining 4 elements using the same algorithm on the remaining four elements.
In Python:
def get_pairs(s):
if not s: yield []
else:
i = min(s)
for j in s - set([i]):
for r in get_pairs(s - set([i, j])):
yield [(i, j)] + r
for x in get_pairs(set([1,2,3,4,5,6])):
print x
This generates the following solutions:
[(1, 2), (3, 4), (5, 6)]
[(1, 2), (3, 5), (4, 6)]
[(1, 2), (3, 6), (4, 5)]
[(1, 3), (2, 4), (5, 6)]
[(1, 3), (2, 5), (4, 6)]
[(1, 3), (2, 6), (4, 5)]
[(1, 4), (2, 3), (5, 6)]
[(1, 4), (2, 5), (3, 6)]
[(1, 4), (2, 6), (3, 5)]
[(1, 5), (2, 3), (4, 6)]
[(1, 5), (2, 4), (3, 6)]
[(1, 5), (2, 6), (3, 4)]
[(1, 6), (2, 3), (4, 5)]
[(1, 6), (2, 4), (3, 5)]
[(1, 6), (2, 5), (3, 4)]