One way to avoid Runge's problem is to choose non-equally spaced data points. A particularly good set to choose are the so-called Chebyshev nodes. To use these, replace the equidistant points xk produced by xeq with xk = COS 2k+1 2n+2 for k=0,..., n. (3.4) Generate data points using (3.4) and (3.3), then plot the resulting Lagrange interpolating polynomial and, hence, show that using (3.4) solves the Runge problem.
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- • Simplify the Boolean functions by means of the tabulation matching method: a. F(w,x,y,z) = Σm(2,3,12,13,14,15) b. P(m,n,o,p,q,r) =Em(6,9,13,18,19,25,27,29,41,45,57,61) c. P(A,B,C,D,E,F,G) =Σm(20,28,38,39,52,60,102,103)To have random-access lookup, a grid should have a scheme for numbering the tiles.For example, a square grid has rows and columns, which give a natural numberingfor the tiles. Devise schemes for triangular and hexagonal grids. Use the numberingscheme to define a rule for determining the neighbourhood (i.e. adjacent tiles) of agiven tile in the grid. For example, if we have a four-connected square grid, wherethe indices are i for rows and j for columns, the neighbourhood of tile i, j can bedefined asneighbourhood(i, j) = {i ± 1, j,i, j ± 1}def mystery (1st); for idx in range(1, len(1st)); tmp = 1st[idx) idx2 = idx while 1dx2 > 9 and 1st[idx2-11 tmp: 1st[idx2] 1st[10/2 - 11 1dx2 = 1dx2 - 1 1st[idx2] = tmp print(1st) a. If we call this function as follows: mystery(lt) where ist 15, 2, 8, 11, what is printed out t clear about what is printed out, don't make me try to figure it out). b. What does this function do? c. What is the complexity of this function? Oni Ora, On³ Ologinil, Onioginil? Explain your reasoning
- In order to test connectivity, Python code uses the function Connectivity Undirected, which then calls the BFS algorithm, which returns a list of visited vertices and sorts them as shown below. The connectivity is then calculated by comparing this list to every node in the network in the main function. Be aware that in order to match the visited vertices with the sorted array of vertices, we must first sort the visited vertices. However, ordering can be avoided if sets are used in place of lists.Write Algorithm for Interchange Base PointsInput : a group G; a base [131,132 ..... 13k] for G and a strong generating set; an integer j between 1 and k-l;Output : a base B = [131,132 ..... 13j-1, 13L.+1, 13j, 13j+2, 13j+3 ..... 13k] for G; a strong generating set relative to B;How to use python code to plot linear regression 3D graph?x=[22, 25, 34, 41, 51,59]y=[0.65, 0.67,0.65,0.65,0.8,0.7]z=[182, 178, 147, 119, 111,95]A=np.transpose(np.array([[22, 25, 34, 41, 51,59],[0.65, 0.67,0.65,0.65,0.8,0.7],[1,1,1,1,1,1]]))Make x,y,z corresponding poing intercept at one point[eg point m(22,0.65,182)]
- I use this code but I cant get the first graph from the picture, I get second graph. Why?? Code: t=(0:0.5:20); % t is a vector which contains time values from 0 to 20s with a step size of 0.5a=sin(t); % a is a sin(t) functionx=pi/6:15:pi/2; % phase difference φ for 30,45,60,90b=sin(t+x); % b is a sin(t+x) funcion plot(t,a) % plot the sin(t) function with respect to ttitle('sin(t) versus Time') % add title to graphxlabel('time') % add x-label to the graphylabel('sin(t)') % add y-label to the graphlegend('sin(t)') % add legend to the graph figure % to plot multiple linesplot(t,a,t,b,'--') % plot the sint(t) & sin(t+x) function with respect to ttitle('Function versus Time') % add title to the graphxlabel('time') % add x-label to the graphylabel('Function') % add y-label to the graphlegend('sin(t)','sin(t+x)') % add legend to the graph % Create a figure divided into four subplotssubplot(2,2,1) % divides the current figure into an 2-by-2 grid and creates axes in the position…Write Algorithm for Interchange Base PointsInput : a group G;a base [131,132 ..... 13k] for G and a strong generating set;an integer j between 1 and k-l;Output : a base B = [131,132 ..... 13j-1, 13L.+1, 13j, 13j+2, 13j+3 ..... 13k] for G;a strong generating set relative to B;procedure interchange( var B : sequence of points;vat S : set of elements;j : 1..k-1 );(* Given a base B = [131,132 ..... 13~] and a strong generating set S of G,return the base obtained by interchanging 13j and 13j+1,and return a strong generating set relative to the new base. *)To add the two polynomials, we assume that the singly linked lists' nodes are organized in decreasing order of the variable x's exponents.The goal is to generate a new list of nodes reflecting the sum of P1 and P2. This is accomplished by combining the COEFF fields of nodes with comparable powers of variable x in lists P1 and P2, and then adding a new node in the resultant list P1 + P2. The procedure's core is presented below.P1 and P2 are the beginning pointers of the singly linked lists that symbolize polynomials P1 and P2. Furthermore, PTR1 and PTR2 are two temporary pointers that are originally set to P1 and P2, respectively. Procedure code should be written.
- Problem Alice was given a number line containing every positive integer, x, where 1<=x<=n. She was also given m line segments, each having a left endpoint, l (1<=l<=n), a right endpoint, r (1<=r<=n), and an integer weight, w. Her task is to cover all numbers on the given number line using a subset of the line segments, such that the maximum weight among all the line segments in the chosen subset is minimal. In other words, let S=(l1,r1,w1),(l2,r2,w2),....,(lk,rk,wk), represent a set of k line segments chosen by Alice, max(w1,w2,...,wk) should be minimized. All numbers 1,2,....n should be covered by at least one of k the chosen line segments. It is okay for the chosen line segments to overlap. You program should output the minimized maximum weight in the chosen subset that covers all the numbers on the number line, or -1 if it is not possible to cover the number line. Input format The first line of the input contains an integer, n - denoting the range of numbers…Write the algorithm that finds and returns how many paths in k units of length between any given two nodes (source node, destination node; source and target nodes can also be the same) in a non-directional and unweighted line of N nodes represented as a neighborhood matrix. (Assume that each side in the unweighted diagram is one unit long.) Note: By using the problem reduction method of the Transform and Conquer strategy, you have to make the given problem into another problem. Algorithm howManyPath (M [0..N-1] [0..N-1], source, target, k)// Input: NxN neighborhood matrix, source, target nodes, k value.// Ouput: In the given line, there are how many different paths of k units length between the given source and target node.What is the time complexity of functions foo() and bar() respectively in below given code snippets, given that n is the total size of the input and L and R are Vectors def foo(L, R): S = Vector() for x in L: if x in R: S.add_back(x) return S def bar(L, R): R_hash= HashTable() for r in R: R_hash.insert(r, False) S = Vector() for x in L: if R_hash.contains(x): S.add_back(x) return S O(N^2), O(N) O(N), O(N) O(N^2), O(N^2) O(1), O(N)