giving a cluster is excessively normal. Rather than that, he chose to gift Mila the portion needs his gift to be wonderful, so he chose to pick k -covering sections of the exhibit [1,r1], [12,r2], .. [Ik,rk] uch an extent that: e length of the primary fragment [11,r1]
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- The above order is wrong. Edge (r, y) should come after (t, u) and before (u, v) Edge ry is shown to be a part of the MST but when is it selected? Why is it not shown in the order? There are seven edges in the MST of the diagram. Why are there only 6 edges in the answer of the order in which edges are selected?Correct answer will be upvoted else downvoted. Computer science. It might have been a simple undertaking, yet it worked out that you ought to observe a few guidelines: Before all else, you select any sure integer x. Then, at that point, you do the accompanying activity n times: select two components of cluster with total equivalents x; eliminate them from an and supplant x with limit of that two numbers. For instance, if at first a=[3,5,1,2], you can choose x=6. Then, at that point, you can choose the second and the third components of a with total 5+1=6 and toss them out. After this activity, x equivalents 5 and there are two components in cluster: 3 and 2. You can toss them out on the following activity. Note, that you pick x before the beginning and can't transform it as you need between the activities. Decide how could you act to toss out all components of a. Input The main line contains a solitary integer t (1≤t≤1000) — the number of experiments.…Design an adjacency Matrix of the alphabets of Uzair Bhatti . In accordance with the following conditions: If your name has repeated characters (e.g. character E, 2 times) then you will consider only 1 time. If your Name contains both G and S, then there will be an edge between them and one additional edge from W to each if W is also in your name. If N is the alphabet in your name, it will have an edge to A and S if it available in your name. If P is available then it will have an edge with L if it is available. If there is a blank space in full Name then it will be represented by “_”, and it must have an edge with all alphabets. (Note: Place on in 1 for Edge and 0 for No Edge) Sketch an undirected graph of the above designed adjacency matrix
- Correct answer will be upvoted else downvoted. Computer science. You are given a cluster an of length n. You are approached to deal with q inquiries of the accompanying organization: given integers I and x, duplicate computer based intelligence by x. In the wake of handling each inquiry you really wanted to output the best normal divisor (GCD) of all components of the cluster a. Since the appropriate response can be excessively huge, you are approached to output it modulo 109+7. Input The principal line contains two integers — n and q (1≤n,q≤2⋅105). The subsequent line contains n integers a1,a2,… ,an (1≤ai≤2⋅105) — the components of the cluster a preceding the changes. The following q lines contain inquiries in the accompanying arrangement: each line contains two integers I and x (1≤i≤n, 1≤x≤2⋅105). Output Print q lines: in the wake of handling each inquiry output the GCD of all components modulo 109+7 on a different line.Correct answer will be upvoted else downvoted. Computer science. As per an artwork task b1,b2,… ,bn, the cluster an is parted into two new exhibits a(0) and a(1), where a(0) is the sub-arrangement of all white components in an and a(1) is the sub-grouping of all dark components in a. For instance, assuming a=[1,2,3,4,5,6] and b=[0,1,0,1,0,0], a(0)=[1,3,5,6] and a(1)=[2,4]. The number of portions in a cluster c1,c2,… ,ck, indicated seg(c), is the number of components if we blend all nearby components with a similar worth in c. For instance, the number of fragments in [1,1,2,2,3,3,3,2] is 4, on the grounds that the exhibit will become [1,2,3,2] subsequent to blending adjoining components with a similar worth. Particularly, the number of portions in an unfilled exhibit is 0. Homer needs to find an artistic creation task b, as per which the number of sections in both a(0) and a(1), for example seg(a(0))+seg(a(1)), is pretty much as extensive as could be expected. Track down this…Design an adjacency Matrix of the alphabets of your full name. In accordance with the following conditions: Let my name is Muhammad Adnan If your name has repeated characters (e.g. character E, 2 times) then you will consider only 1 time. If your Name contains both G and S, then there will be an edge between them and one additional edge from W to each if W is also in your name. If N is the alphabet in your name, it will have an edge to A and S if it available in your name. If P is available then it will have an edge with L if it is available. If there is a blank space in full Name then it will be represented by “_”, and it must have an edge with all alphabets. (Note: Place on in 1 for Edge and 0 for No Edge) Sketch an undirected graph of the above designed adjacency matrix.
- Correct answer will be upvoted else downvoted. Computer science. You are given a number n and an exhibit b1,b2,… ,bn+2, acquired by the accompanying calculation: some exhibit a1,a2,… ,a was speculated; cluster a was composed to exhibit b, for example bi=ai (1≤i≤n); The (n+1)- th component of the cluster b is the amount of the numbers in the exhibit a, for example bn+1=a1+a2+… +an; The (n+2)- th component of the cluster b was thought of some number x (1≤x≤109), for example bn+2=x; The cluster b was rearranged. For instance, the cluster b=[2,3,7,12,2] it very well may be acquired in the accompanying ways: a=[2,2,3] and x=12; a=[3,2,7] and x=2. For the given cluster b, find any exhibit a that might have been speculated at first. Input The main line contains a solitary integer t (1≤t≤104). Then, at that point, t experiments follow. The primary line of each experiment contains a solitary integer n (1≤n≤2⋅105). The second column of each experiment…3. Below given a point set in the rectilinear metric (the height/width of any cell=1) where the closest pair of points should be found using divide and conquer. Show - the first partition of the point set (draw a line) -the closest pair in the left part (connect solid), 8en and the right part (connect solid), 6icht - the middle strip (shade) - pairs in the middle strip for which distances should be computed (connect dashed) closest pair in the middle strip (connect solid)Create an SP client that analyses the edge-weighted digraph's edges in relation to a specified pair of vertices, s and t: Create a V-by-V boolean matrix where each item in row v and column w is true for each v and w if the edge v->w is an edge in the edge-weighted digraphs whose weight can be raised without increasing the length of the shortest path from v to w, and false otherwise.
- Correct answer will be upvoted else downvoted. Computer science. You are given a weighted undirected associated diagram comprising of n vertices and m edges. It is ensured that there are no self-circles or various edges in the given diagram. How about we characterize the heaviness of the way comprising of k edges with files e1,e2,… ,ek as ∑i=1kwei−maxi=1kwei+mini=1kwei, where wi — weight of the I-th edge in the chart. Your errand is to track down the base load of the way from the 1-st vertex to the I-th vertex for every I (2≤i≤n). Input The principal line contains two integers n and m (2≤n≤2⋅105; 1≤m≤2⋅105) — the number of vertices and the number of edges in the chart. Following m lines contains three integers vi,ui,wi (1≤vi,ui≤n; 1≤wi≤109; vi≠ui) — endpoints of the I-th edge and its weight separately. Output Print n−1 integers — the base load of the way from 1-st vertex to the I-th vertex for every I (2≤i≤n).Classify the 1’s, 2’s, 3’s for the zip code data in R. (a) Use the k-nearest neighbor classification with k = 1, 3, 5, 7, 15. Report both the training and test errors for each choice. (b) Implement the LDA method and report its training and testing errors. Note: Before carrying out the LDA analysis, consider deleting variable 16 first from the data, since it takes constant values and may cause the singularity of the covariance matrix. In general, a constant variable does not have a discriminating power to separate two classes.What is the property of downward closure? The best way to explain anything is to use an example. This property is maintained by fp-growth algorithm in what way? Explain?