So, I am currently learning about closest pair algorithms and found this problem in a text book:
Assume you have a sorrted Array of length n that contains one Dimensional points (integers) and are coloured. Now you want to find the closest pair of two points of different colours (either red or blue). How would you do that in O(n)? Since the closest pair of points problem is usually using a divide and conquer algorithm I was wondering whether someone has an idea on how to solve this?
I found a solution using two pointers, but no d&C solution.
You mentioned in a comment that there are only two colours, red and blue.
If we could use a divide-and-conquer approach, it would give a O(logn) solution. I don't think it is possible. However, there is a rather simple O(n) solution. We just have to memorize the value/index of last blue and red samples.
It the last read element is blue (resp. red), we calculate its distance from the last red (resp. blue) element. We then compare the result with the previous minimum distance.
Note: if the number of colours were higher, there is still a simple O(n) approach by using a stack. I originally prepared such a solution, before knowing there are only two colours.
Related
consider an image like this:
by grouping pixels by color into distinct rectangles, different configurations might be achieved, for example:
the goal is to find one of the best configurations, i.e. a configuration which has the least possible number of rectangles (rectangles sizes are not important).
any idea on how to design an efficient algorithm which is able to solve this problem?
EDIT:
i think the best answer is the one by #dshin, as they proved that this problem is a NP-HARD one so there probably isn't any efficient solution that is able to guarantee an optimal result.
other answers provide reasonable compromises to get an acceptable solution, but that won't always be the optimal one.
Each connected colored region is a rectilinear polygon that can be considered independently, and so your problem amounts to solving the minimum rectangle covering for rectilinear polygons. This is a well-studied problem that finds applications in some fields, like VLSI.
For convex rectilinear polygons, there is an algorithm that finds the optimal solution in polynomial time, described in this 1984 thesis.
The non-convex case is NP-hard (reference), so an efficient optimal solution likely does not exist. But there are several algorithms which produce good empirical results. This 1990 publication describes three separate algorithms, each of which are guaranteed to use at most twice as many rectangles as the optimal solution. This 2016 publication describes an algorithm that uses the common IP + LP relaxation technique, which apparently produces better results in real-life problem instances, although lacking in theoretical guarantees. Unfortunately, both publications are behind paywalls, and I haven't been able to find free resources that describe the algorithms.
If you are just looking for something reasonable, and your problem instances are not pathological in nature, then the algorithms described in other answers are probably good enough.
I don't have a proof but my feeling is a greedy approach should solve this problem:
Start on the upper left (or in whichever corner)
Expand rectangle 1px to the right as long as colors match
Expand rectangle 1px to the bottom as long as all colors in that row match
Line by line and column by column, find the next pixel that is not already part of a square (maybe keep track of visited pixels in a second array) and repeat 2 and 3.
You can switch lines and columns and go up and left or whatever instead and end up with different configurations, but from playing this through in my mind I think the number of rectangles should always be the same.
The idea here is based on the following links: Link 1 and Link 2.
In both the cases, the largest possible rectangle is computed within a given polygon/shape. Check both the above links for details.
We can extend the idea above to the problem at hand.
Steps:
Filter the image by color (say red)
Find the largest possible rectangle in the red region. After doing so mask it.
Repeat to find the next biggest rectangle until all the portions in red have been covered.
Repeat the above for every unique color.
Overview:
I have any number of points on an imaginary 2D surface. I also have a grid on the same surface with points at regular intervals along the X and Y access. My task is to map each point to the nearest grid point.
The code is straight forward enough until there are a shortage of grid points. The code I've been developing finds the closest grid point, displaying an already mapped point if the distance will be shorter for the current point.
I then added a second step that compares each mapped point to another and, if swapping the mapping with another point produces a smaller sum of the total mapped distance of both points, I swap them.
This last step seems important as it reduces the number crossed map lines. (This would be used to map points on a plate to a grid on another plate, with pins connecting the two, and lines that don't cross seem to have a higher chance that the pins would not make contact.)
Questions:
Can anyone comment on my thinking that if the image above were truly optimized, (that is, the mapped points--overall--would have the smallest total distance), then none of the lines were cross?
And has anyone seen any existing algorithms to help with this. I've searched but came up with nothing.
The problem could be approached as a variation of the Assignment Problem, with the "agents" being the grid squares and the points being the "tasks", (or vice versa) with the distance between them being the "cost" for that agent-task combination. You could solve with the Hungarian algorithm.
To handle the fact that there are more grid squares than points, find a bounding box for the possible grid squares you want to consider and add dummy points that have a cost of 0 associated with all grid squares.
The Hungarian algorithm is O(n3), perhaps your approach is already good enough.
See also:
How to find the optimal mapping between two sets?
How to optimize assignment of tasks to agents with these constraints?
If I understand your main concern correctly, minimising total length of line segments, the algorithm you used does not find the best mapping and it is clear in your image. e.g. when two line segments cross each other, simple mathematic says that if you rearrange their endpoints such that they do not cross, it provides a better total sum. You can use this simple approach (rearranging crossed items) to get better approximation to the optimum, you should apply swapping for more somehow many iterations.
In the following picture you can see why crossing has longer length than non crossing (first question) and also why by swapping once there still exists crossing edges (second question and w.r.t. Comments), I just drew one sample, in fact one may need many iterations of swapping to get non crossed result.
This is a heuristic algorithm certainly not optimum but I expect to be very good and efficient and simple to implement.
I'm looking for a general algorithm for creating an evenly spaced grid, and I've been surprised how difficult it is to find!
Is this a well solved problem whose name I don't know?
Or is this an unsolved problem that is best done by self organising map?
More specifically, I'm attempting to make a grid on a 2D Cartesian plane in which the Euclidean distance between each point and 4 bounding lines (or "walls" to make a bounding box) are equal or nearly equal.
For a square number, this is as simple as making a grid with sqrt(n) rows and sqrt(n) columns with equal spacing positioned in the center of the bounding box. For 5 points, the pattern would presumably either be circular or 4 points with a point in the middle.
I didn't find a very good solution, so I've sadly left the problem alone and settled with a quick function that produces the following grid:
There is no simple general solution to this problem. A self-organizing map is probably one of the best choices.
Another way to approach this problem is to imagine the points as particles that repel each others and that are also repelled by the walls. As an initial arrangement, you could already evenly distribute the points up to the next smaller square number - for this you already have a solution. Then randomly add the remaining points.
Iteratively modify the locations to minimize the energy function based on the total force between the particles and walls. The result will of course depend on the force law, i.e. how the force depends on the distance.
To solve this, you can use numerical methods like FEM.
A simplified and less efficient method that is based on the same principle is to first set up an estimated minimal distance, based on the square number case which you can calculate. Then iterate through all points a number of times and for each one calculate the distance to its closest neighbor. If this is smaller than the estimated distance, move your point into the opposite direction by a certain fraction of the difference.
This method will generally not lead to a stable minimum but should find an acceptable solution after a number ot iterations. You will have to experiment with the stepsize and the number of iterations.
To summarize, you have three options:
FEM method: Efficient but difficult to implement
Self organizing map: Slightly less efficient, medium complexity of implementation.
Iteration described in last section: Less efficient but easy to implement.
Unfortunately your problem is still not very clearly specified. You say you want the points to be "equidistant" yet in your example, some pairs of points are far apart (eg top left and bottom right) and the points are all different distances from the walls.
Perhaps you want the points to have equal minimum distance? In which case a simple solution is to draw a cross shape, with one point in the centre and the remainder forming a vertical and horizontal crossed line. The gap between the walls and the points, and the points in the lines can all be equal and this can work with any number of points.
I believe I understand the algorithm quite clearly, except for the step where you look to see if there's any points that are close by looking across the division and create a strip where points within the strip are candidates.
But then the algorithm states to sort the points by their y coordinates and then check each other point in the strip to find if there is a smaller distance than the one previously found. It basically sounds like you brute force within the strip.
For example, here's what Introduction to Algorithms states:
So it seems you just take each point and compare it against all the others to find the closest points? Why is it necessary to sort by y value then? You already have them sorted by x, why not brute force with that?
You don't brute force compare against all points in Y' but only against the one next to p. If that one is already too far away you can just stop, because all other points will be even further away. You only continue evaluating the next closest neighbor if the last one was still within your search distance.
The text explains it in the As we will see shortly section.
Sorting is an optimization here that allows you to iterate nearest neighbors in O(1) after paying the sorting costs of O(n log n) once.
I was asked this question in Yahoo for machine learning profile. Given a set of points (x,y) coordinates I was asked to find points with lowest distance in O(n) or O(log n )time.
Obviously I was able to come up with O(n^2) time but was no way near getting the better algorithm. Even though the problem statement was screaming for Divide and Conquer I just could not come up with the reasoning for the merge step. I also googled for this question on the internet and found that It is actually very popular but I still could not get hold of the reasoning of the merge step.
Can anyone help me out with this?
Input: (x1,y1),(x2,y2),(x3,y3),(x4,y4),(x5,y5)
The problem can be solved in O(n log n) time using the recursive divide and conquer approach, e.g., as follows:
1.Sort points according to their x-coordinates.
2.Split the set of points into two equal-sized subsets by a vertical line x=xmid.
3.Solve the problem recursively in the left and right subsets. This yields the left-side and right-side minimum distances dLmin and dRmin, respectively.
4.Find the minimal distance dLRmin among the pair of points in which one point lies on the left of the dividing vertical and the second point lies to the right.
5.The final answer is the minimum among dLmin, dRmin, and dLRmin.
http://en.wikipedia.org/wiki/Closest_pair_of_points