Search results
Results From The WOW.Com Content Network
The Shamos–Hoey algorithm [1] applies this principle to solve the line segment intersection detection problem, as stated above, of determining whether or not a set of line segments has an intersection; the Bentley–Ottmann algorithm works by the same principle to list all intersections in logarithmic time per intersection.
LeetCode LLC, doing business as LeetCode, is an online platform for coding interview preparation. The platform provides coding and algorithmic problems intended for users to practice coding . [ 1 ] LeetCode has gained popularity among job seekers in the software industry and coding enthusiasts as a resource for technical interviews and coding ...
The output from the version of the algorithm described by de Berg et al. (2000) consists of the set of intersection points of line segments, labeled by the segments they belong to, rather than the set of pairs of line segments that intersect. A similar approach to degeneracies was used in the LEDA implementation of the Bentley–Ottmann ...
Assume that we want to find intersection of two infinite lines in 2-dimensional space, defined as a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0. We can represent these two lines in line coordinates as U 1 = (a 1, b 1, c 1) and U 2 = (a 2, b 2, c 2). The intersection P′ of two lines is then simply given by [4]
A circular list can be split into two circular lists, in constant time, by giving the addresses of the last node of each piece. The operation consists in swapping the contents of the link fields of those two nodes. Applying the same operation to any two nodes in two distinct lists joins the two list into one.
Several common non-emptiness problems have been shown to be complete for complexity classes ranging from Deterministic Logspace up to PSPACE. [2] The intersection non-emptiness decision problem is concerned with whether the intersection of given languages is non-empty. In particular, the intersection non-emptiness problem is defined as follows.
For example, consider a supermarket with 1000 products and two customers. The basket of the first customer contains salt and pepper and the basket of the second contains salt and sugar. In this scenario, the similarity between the two baskets as measured by the Jaccard index would be 1/3, but the similarity becomes 0.998 using the SMC.
Solution of a travelling salesman problem: the black line shows the shortest possible loop that connects every red dot. In the theory of computational complexity, the travelling salesman problem (TSP) asks the following question: "Given a list of cities and the distances between each pair of cities, what is the shortest possible route that visits each city exactly once and returns to the ...