# Finding the squares in a plane given n points

By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. I tried this by calculating the distances between each 2 pointsthen sort themand look for the squares in the points with four or more equal distances after verifying the points and slopes.

I thought dynamic programming for checking for line segments of equal distances might work S : The squares can be in any manner. They can overlaphave a common side, one square inside another We are interested in a function count i, j that returns, as fast as possible, the number of squares that we can draw by using points i and j. Now, for each pair of points i, jcount i, j will have to iterate H[ d[i][j] ] and count the points in that list that form a square with points i and j.

The basic idea is to group the points by X or Y coordinate, being careful to avoid making groups that are too large. The details are in the paper Finding squares and rectangles in sets of points.

The paper also covers lots of other cases allowing rotated squares, allowing rectangles, and working in higher dimensions. I've paraphrased their 2d axis-aligned square finding algorithm below.

Group the points by their X coordinate. Break any groups with more than sqrt n points apart, and re-group those now-free points by their Y coordinate. This guarantees the groups have at most sqrt n points and guarantees that for each square there's a group that has two of the square's corner points.

For every group gfor every pair of points p,q in gcheck whether the other two points of the two possible squares containing p and q are present. Keep track of how many you find. Watch out for duplicates are the two opposite points also in a group? Why does it work? Well, the only tricky thing is the regrouping.By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service.

Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up. Nothing more to explain. I then have to write the corresponding algorithm.

Thank you. The normal vector of the best-fitting plane is the left singular vector corresponding to the least singular value. So set up matrices like this with all your data:. The theory can be found in many books. Sign up to join this community.

The best answers are voted up and rise to the top. Home Questions Tags Users Unanswered. Asked 8 years, 9 months ago. Active 3 months ago. Viewed 72k times. Siong Thye Goh k 17 17 gold badges 74 74 silver badges bronze badges. G4bri3l G4bri3l 1 1 gold badge 4 4 silver badges 9 9 bronze badges. There are many different measures of how well a plane fits given data, and different measures give rise to different "best" fitting planes.

So you had best tell us what you have in mind as your measure of how well a given plane fits some given data. But I know just a bit. Let's say that the set of Points I have over a already look like a plane, I mean, they are displayed as a plane but not perfectly.Given X i and Y i co-ordinate of N distinct points in the 2-D Plane, the task is to count the number of squares that can be formed from these points that are parallel to both X and Y -axis. Naive Approach: Iterate over all possible combinations of four points and check if a square can be formed that is parallel to both X and Y-axis. The time complexity of this approach would be O N 4.

Efficient Approach: We can observe that for any four points to make a required square, the following conditions must hold true —. Consider any two points on the same horizontal or vertical line from the given ones. Calculate the distance between them. Based on that, form the other two points. Now check if both of the given points are present or not. If so, it ensures that a square is present. Hence, increase the count and proceed further.

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Python3 implementation of the approach. Function that returns the count of.

## Steiner tree problem

Initialize result. Initialize a set to store points. Initialize a map to store the. Store the points in a set.

Store the points in the same vertical line. Check for every two points. Check if other two point are present or not. This code is contributed by AnkitRai Check out this Author's contributed articles. Improved By : AnkitRai Load Comments. We use cookies to ensure you have the best browsing experience on our website.In the first section of this chapter we saw a couple of equations of planes.

However, none of those equations had three variables in them and were really extensions of graphs that we could look at in two dimensions. We would like a more general equation for planes. This vector is called the normal vector.

Also notice that we put the normal vector on the plane, but there is actually no reason to expect this to be the case. We put it here to illustrate the point. It is completely possible that the normal vector does not touch the plane in any way. Recall from the Dot Product section that two orthogonal vectors will have a dot product of zero. In other words. A slightly more useful form of the equations is as follows.

Start with the first form of the vector equation and write down a vector for the difference. This second form is often how we are given equations of planes. Notice that if we are given the equation of a plane in this form we can quickly get a normal vector for the plane. A normal vector is. We need to find a normal vector. Recall however, that we saw how to do this in the Cross Product section. These two vectors will lie completely in the plane since we formed them from points that were in the plane. Notice as well that there are many possible vectors to use here, we just chose two of the possibilities.

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Now, we know that the cross product of two vectors will be orthogonal to both of these vectors. Since both of these are in the plane any vector that is orthogonal to both of these will also be orthogonal to the plane.

Therefore, we can use the cross product as the normal vector. This is not as difficult a problem as it may at first appear to be.

We can pick off a vector that is normal to the plane. We can also get a vector that is parallel to the line.

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Now, if these two vectors are parallel then the line and the plane will be orthogonal. If you think about it this makes some sense. So, if the two vectors are parallel the line and plane will be orthogonal. If the line is parallel to the plane then any vector parallel to the line will be orthogonal to the normal vector of the plane. Notes Quick Nav Download. You appear to be on a device with a "narrow" screen width i.By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service.

Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. This question was asked to me in an online test. There are N points given in a Cartesian plane.

An integer K will be given. The aim is to find the area of the square minimum enclosing at least K points. The sides of square should be parallel to the axis.

The vertices of the square should be integers. Any points lying on the sides are not considered to be inside the square. One approach for general solution would be to find all possible combination of K points among N points and applying the above method for all combinations but that is not good. Please advise. The square's bottom left point can be anywhere within the bounding square. Constraints like minimum size square required to accommodate at least k points sqrt k are considered. If a square encloses at least k points and area is lesser than current min area, then update area. It can be seen that we always can move the square so that it would have points on the left and bottom edges.

We'll iterate through all combinations of left and bottom edges of the square. Then we will need to find upper or right edge of the square. For every point we can determine at what edge it would lie. For example if point. It will be the smallest square enclosing at least K points with specified left bottom corner. The previously best known algorithm for this problem takes O k 2n log n time and uses O kn space.

Asked 5 years, 3 months ago. Active 2 years, 10 months ago. Viewed 9k times. Ouney Ouney 1, 8 8 silver badges 21 21 bronze badges. You could take a statistical approach and calculate the standard deviation of each of the points along the x and y axis, then remove the first N - K points with highest standard deviation.

Do you need to find the square enclosing at least K points or exactly K points? I really do not remember the question, but it should be at least. Will change the question as well. Active Oldest Votes.Plane equation given three points Calculator. Calculates the plane equation given three points. Customer Voice. Plane equation given three points. Thank you for your questionnaire. Sending completion. To improve this 'Plane equation given three points Calculator', please fill in questionnaire.

Male or Female? Bug report Click here to report questionnaire. Text bug Please enter information such as wrong and correct texts Your feedback and comments may be posted as customer voice. The hyperlink to [Plane equation given three points] Plane equation given three points Calculator. Related Calculator. Shortest distance between two lines Plane equation given three points Volume of a tetrahedron and a parallelepiped Shortest distance between a point and a plane Cartesian to Spherical coordinates Cartesian to Cylindrical coordinates Spherical to Cartesian coordinates Spherical to Cylindrical coordinates Cylindrical to Cartesian coordinates Cylindrical to Spherical coordinates New coordinates by 3D rotation of points.

Point A. Point B.

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Point C. Disp-Num 5 10 30 50 The Steiner tree problemor minimum Steiner tree problemnamed after Jakob Steineris an umbrella term for a class of problems in combinatorial optimization. While Steiner tree problems may be formulated in a number of settings, they all require an optimal interconnect for a given set of objects and a predefined objective function.

One well-known variant, which is often used synonymously with the term Steiner tree problem, is the Steiner tree problem in graphs. Given an undirected graph with non-negative edge weights and a subset of vertices, usually referred to as terminals, the Steiner tree problem in graphs requires a tree of minimum weight that contains all terminals but may include additional vertices.

Further well-known variants are the Euclidean Steiner tree problem and the rectilinear minimum Steiner tree problem. The Steiner tree problem in graphs can be seen as a generalization of two other famous combinatorial optimization problems: the non-negative shortest path problem and the minimum spanning tree problem.

If a Steiner tree problem in graphs contains exactly two terminals, it reduces to finding the shortest path. If, on the other hand, all vertices are terminals, the Steiner tree problem in graphs is equivalent to the minimum spanning tree.

### Count of Squares that are parallel to the coordinate axis from the given set of N points

However, while both the non-negative shortest path and the minimum spanning tree problem are solvable in polynomial time, the decision variant of the Steiner tree problem in graphs is NP-complete which implies that the optimization variant is NP-hard ; in fact, the decision variant was among Karp's original 21 NP-complete problems.

The Steiner tree problem in graphs has applications in circuit layout or network design. However, practical applications usually require variations, giving rise to a multitude of Steiner tree problem variants. Most versions of the Steiner tree problem are NP-hardbut some restricted cases can be solved in polynomial time.

Despite the pessimistic worst-case complexity, several Steiner tree problem variants, including the Steiner tree problem in graphs and the rectilinear Steiner tree problem, can be solved efficiently in practice, even for large-scale real-world problems. The original problem was stated in the form that has become known as the Euclidean Steiner tree problem or geometric Steiner tree problem : Given N points in the planethe goal is to connect them by lines of minimum total length in such a way that any two points may be interconnected by line segments either directly or via other points and line segments.

It may be shown that the connecting line segments do not intersect each other except at the endpoints and form a tree, hence the name of the problem.

This paper was long overlooked, but it already contains "virtually all general properties of Steiner trees" later attributed to other researchers, including the generalization of the problem from the plane to higher dimensions.

For the Euclidean Steiner problem, points added to the graph Steiner points must have a degree of three, and the three edges incident to such a point must form three degree angles see Fermat point.

For general Nthe Euclidean Steiner tree problem is NP-hardand hence it is not known whether an optimal solution can be found by using a polynomial-time algorithm. The rectilinear Steiner tree problem is a variant of the geometric Steiner tree problem in the plane, in which the Euclidean distance is replaced with the rectilinear distance.

The problem arises in the physical design of electronic design automation. In VLSI circuitswire routing is carried out by wires that are often constrained by design rules to run only in vertical and horizontal directions, so the rectilinear Steiner tree problem can be used to model the routing of nets with more than two terminals.

Steiner trees have been extensively studied in the context of weighted graphs. The prototype is, arguably, the Steiner tree problem in graphs. A Steiner tree is a tree in G that spans S. There are two versions of the problem: in the optimization problem associated with Steiner trees, the task is to find a minimum-weight Steiner tree; in the decision problem the edge weights are integers and the task is to determine whether a Steiner tree exists whose total weight does not exceed a predefined natural number k.

The decision problem is one of Karp's 21 NP-complete problems ; hence the optimization problem is NP-hard.

Find the Equation of the Plane Given a Point and Perpendicular Line

Put otherwise, the edge weights satisfy the triangle inequality. This variant is known as the metric Steiner tree problem. Given an instance of the non-metric Steiner tree problem, we can transform it in polynomial time into an equivalent instance of the metric Steiner tree problem; the transformation preserves the approximation factor.

In a special case of the graph problem, the Steiner tree problem for quasi-bipartite graphsS is required to include at least one endpoint of every edge in G. The Steiner tree problem has also been investigated in higher dimensions and on various surfaces. Algorithms to find the Steiner minimal tree have been found on the sphere, torus, projective planewide and narrow cones, and others.

Other generalizations of the Steiner tree problem are the k -edge-connected Steiner network problem and the k -vertex-connected Steiner network problemwhere the goal is to find a k -edge-connected graph or a k -vertex-connected graph rather than any connected graph.