29 Consider convolving the matrix M with the kernel K with a stride of 1.1 33 4 5 21379.1 51 01М—6 2 7 6 0, K =0 0 01 2 5 415 7 73[1 0 1,9.Consider everything to be 0-indexed. What is the element at position (1, 1) of the resultantmatrix obtained after the convolution?Select one:O 17O 18O 21O 14

Answered: Consider convolving the matrix M with…

Consider convolving the matrix M with the kernel K with a stride of 1. 1 3 4 5 2] 1 0 1 K = | 0 0 0 3 7 9 1 5 M : 2 7 6 0 3 5 4 1 0 1 1 7 7] 2 5

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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If there is a non-singular matrix P such as P-1AP=D, matrix A is called a diagonalizable matrix. A, n x n square matrix is diagonalizable if and only if matrix A has n linearly independent eigenvectors. In this case, the diagonal elements of the diagonal matrix D are the eigenvalues of the matrix A. A=({{1, -1, -1}, {1, 3, 1}, {-3, 1, -1}}) : 1 -1 -1 1 3 1 -3 1 -1 a)Write a program that calculates the eigenvalues and eigenvectors of matrix A using NumPy. b)Write the program that determines whether the D matrix is diagonal by calculating the D matrix, using NumPy. #UsePython
Given the complement of a graph G is a graph G' which contains all the vertices of G, but for each unweighted edge that exists in G, it is not in G', and for each possible edge not in G, it is in G'. What logical operation and operand(s) can be applied to the adjacency matrix of G to produce G'? AND G's adjacency matrix with 0 to produce G' XOR G's adjacency matrix with 0 to produce G' XOR G's adjacency matrix with 1 to produce G' AND G's adjacency matrix with 1 to produce G'
Consider the following pair of adjacency matrices.1. Draw the simple graphs associated with each of the above adjacency matrices.2. Check whether those two simple graphs are isomorphic. Show your work. 1 0 1 1 1 0 0 1 1 1 0 1 1 1 0 0 1 1
Type in Latex **Problem**. Let $$A = \begin{bmatrix} .5 & .2 & .3 \\ .3 & .8 & .3 \\ .2 & 0 & .4 \end{bmatrix}.$$ This matrix is an example of a **stochastic matrix**: its column sums are all equal to 1. The vectors $$\mathbf{v}_1 = \begin{bmatrix} .3 \\ .6 \\ .1 \end{bmatrix}, \mathbf{v}_2 = \begin{bmatrix} 1 \\ -3 \\ 2 \end{bmatrix}, \mathbf{v}_3 = \begin{bmatrix} -1 \\ 0 \\ 1\end{bmatrix}$$ are all eigenvectors of $A$. * Compute $\left[\begin{array}{rrr} 1 & 1 & 1 \end{array}\right]\cdot\mathbf{x}_0$ and deduce that $c_1 = 1$.* Finally, let $\mathbf{x}_k = A^k \mathbf{x}_0$. Show that $\mathbf{x}_k \longrightarrow \mathbf{v}_1$ as $k$ goes to infinity. (The vector $\mathbf{v}_1$ is called a **steady-state vector** for $A.$) **Solution**. To prove that $c_1 = 1$, we first left-multiply both sides of the above equation by $[1 \, 1\, 1]$ and then simplify both sides:$$\begin{aligned}[1 \, 1\, 1]\mathbf{x}_0 &= [1 \, 1\, 1](c_1\mathbf{v}_1 +…
The parts (a) and (b) of this problem are independentof each other.G1 G24 51 236sx yt u v(a) Prove that the graphs G1 and G2 are isomorphic byexhibiting an isomorphism from one to the other byconcrete arguments and verify it by using adjacencymatrices. Please take the ordering of the vertices as1, 2, 3, 4, 5, 6 while forming AG1, adjacency matrix ofG1.Warning: One must stick to the labelings ofthe vertices as they are given, if one changesthe labelings/orderings etc., the solution willnot be taken into account.(b) Consider the complete graph K13 with vertex setV13 = {u1, u2, u3, · · · , u13}.Let H = (V, E) be the simple graph obtained fromK13 by adding a new vertex u, i.e. V = V13 ∪ {u}and deleting the edges {u1, u2} and {u2, u3} andadding the edges {u1, u} and {u, u2} and keepingthe remaining edges same.Determine whether H has an Euler circuit or not,an Euler path or not. One must validate any conclusion by clear arguments.
Given the adjacency matrix of an undirected simple graph G = (V, E) mapped in a natural fashion onto a mesh of size n2, in Θ(n) time a directed breadth-first spanning forest T = (V, A) can be created. As a byproduct, the undirected breadth first spanning forest edge set EA can also be created, where EA consists of the edges of A and the edges of A directed in the opposite direction.give proof of theorem.
Consider now the interchange of 1 and 8, so that the new base will be[9, 8, 1, 10, 2, 12].We require the generators for ~(3)=G98. The orbit of 8 under G (2) =G 9 is{ 8,10,11,13,14,16,21 }, and so we calculate that' I A (3) I = 7. Initially,A= {1}, F= {2..7}, andT= {$3,S5,$6}.Choosing T = 2 from F givesg I = s7 = (1,2,3,4,6,5,7).Then 8 gl-t = 8, sog2 is the identity, and we add s7 to T. Thus,A = {1..7}, F= empty, and T= {s3,ss,s6, sT}.Since A has the correct size, we are finished.The new strong generating set is the same as the old. By checking not only S (/+2) but alsoS (/) for elements that fix 13j+1, we could have saved ourselves the trouble of duplicatingpermutations already in S. This simple improvement is used in implementations, We leave itas an exercise for the reader to make the necessary modifications to Algorithm 3.Analysis of Interchanging Base Points
Let G = (V, E) be a connected, undirected graph, and let s be a fixed vertex in G. Let TB be the spanning tree of G found by a bread first search starting from s, and similarly TD the spanning tree found by depth first search, also starting at s. (As in problem 1, these trees are just sets of edges; the order in which they were traversed by the respective algorithms is irrelevant.) Using facts, prove that TB = TD if and only if TB = TD = E, i.e., G is itself a tree.
Algorithm for Normalizing GeneratorInput: a permutation group H with base B=[~I ,~2, -.-., 13k],strong generating set T, Schreier vectors v t~ , and basic orbits A (i) ;a permutation y normalizing H;Output: a base for K = <H,y> which may extend B;a sequence of elements [g 1 ,g 2 ..... gr] such that, for each j,(a) T L.) {gj,gj+l ..... gr } is a strong generating set of <H, gj,gj+l ..... gr >,(b) gj ~ <H, gj+l,gj§ ..... gr>,(c) gi normalizes <H,gj§ ,g/+2 ..... gr>, and(d) gp ~ <H, gj+l,gj+2 ..... gr>, for some prime pj;
Let C be an (n, k) linear code over F such that the minimum weightof any nonzero code word is 2r + 1. Show that not every vector ofweight t + 1 in Fn can occur as a coset leader.
Given the adjacency matrix of an undirected simple graph G = (V, E) mapped in a naturalfashion onto a mesh of size n2, in Θ(n) time a directed breadth-first spanning forest T = (V, A) can becreated. As a byproduct, the undirected breadth-first spanning forest edge set EA can also be created,where EA consists of the edges of A and the edges of A directed in the opposite direction.
for G such that1. cq,0c2 ..... cx r e A, and2. G c~l,0h ..... ct, is the pointwise stabilizer of A in G.Applying the base change algorithm if necessary, we may assume a strong generating set S ofU relative to B is known.Let us return to the above example where G is the symmetries of the square acting on pairs ofpoints and A is A 1 , the set of edges of the square. The points (x I =1 and cz2=3 form a base forG, so G1,3 = G1,3,4,6 = < identity > (and s=0). Hence, ~1 =1 and ~2=2 form a base for imrThe stabiliser G 1 is generated by b• so {a,b,bxa} is a strong generating set of Grelative to the chosen base. Furthermore, the stabiliser of 1 in imr is generated by bxa=(2,3).Hence, the set of images { -d, b, bxa } = { (1,3,4,2), (1,2)(3,4), (2,3) } is a strong generatingset of im(p relative to the base [1,2]. The kernel of the homomorphism is the trivial subgroup,< identity > perform each of the basic tasks :