Let R = (J,K, L, M, N, P) and F = {J → KL, L → J, MNP → K, KP – M, LJ → N}, give a canonical cover F. for F.
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- Consider the Omega network and Butterfly network from p nodes in the leftmost column to p nodes in the rightmost column for some p=2^k. The Omega network is defined in Chapter 2 of the text book such that Si is connected to element S j if j=2i for or j=2i+1-p for See Chapter 2 in text book for its definition. The Butterfly network is an interconnection network composed of log p levels (as the omega network). In a Butterfly network, each switching node i at a level l is connected to the identically numbered element at level l + 1 and to a switching node whose number differs from itself only at the lth most significant bit. Therefore, switching node Si is connected to element S j at level l if j = i or j . Prove that for each node Si in the leftmost column and a node Sj in the rightmost column, there is a path from Si to Sj in the Omega network. Prove that for each node Si in the leftmost and a node Sj in the rightmost, there is a path from Si to Sj in the Butterfly network.Suppose there is undirected graph F with nonnegative edge weights we ≥ 0. You have also calculated the minimum spanning tree of F and also the shortest paths to all nodes from a particular node p ∈ V . Now, suppose that each edge weight is increased by 1, so the new weights are we′ = we + 1. (a) Will there be a change of the minimum spanning tree? Provide an example where it does or prove that it cannot change. (b) (3 points) Will the shortest paths from p change? Provide an example where it does or prove that it cannot change.6. Consider a binary classification problem using 1-nearest neighbors with the Euclidean distance metric. We have N 1-dimensional training points x(1), x(2), . . . x(N ) and corresponding labelsy(1), y(2), . . . y(N ) with x(i ) ∈ R and y(i ) ∈ {0, 1}. Assume the points x(1), x(2), . . . x(N ) are in ascending order by value. If there are ties during the 1-NN algorithm, we break ties by choosing the labelcorresponding to the x(i ) with lower value.
- The set of letters S consists of 30 As, 6 Bs, 24 Cs, 15 Ds, and 15 Es. The set of letters T consists of 25 As, 20 Bs, 15 Cs, 5 Ds and 35 Es. What is the mutual index of coincidence between sets S and T? Leave your answer as a fraction in lowest termsQuestion 1: In graph theory, a graph X is a "complement" of a graph F if which of the following is true? Select one: a. If X is isomorph to F, then X is a complement of F. b. If X has half of the vertices of F (or if F has half of the vertices of X) then X is a complement of F. c. If X has the same vertex set as F, and as its edges ONLY all possible edges NOT contained in F, then X is a complement of F. d. If X is NOT isomorph to F, then X is a complement of F. Question 2: Which statement is NOT true about Merge Sort Algorithm: Select one: a. Merge Sort time complexity for worst case scenarios is: O(n log n) b. Merge Sort is a quadratic sorting algorithm c. Merge Sort key disadvantage is space overhead as compared to Bubble Sort, Selection Sort and Insertion Sort. d. Merge Sort adopts recursive approachThe Barabasi and Albert model ´• A discrete time network evolution process,relating the graph G(t + 1) to G(t).• Start at t=0 with a single isolated node.• At each discrete time step, a new node arrives.• This new node makes m edges to already existing nodes.(Why m edges? i.e., what happens if m = 1?)• The likelihood of a new edge to connect to an existing node jis proportional to the degree of node j, denoted dj.• We are interested in the limit of large graph size, n → ∞.Visualizing a PA graph (m = 1) at n = 5000Probabilistic treatment (kinetic theory)• Start at t = 0 with one isolated node (or a small core set).– At time t the total number of nodes added n = t.– At time t the total number of edges added is mt.• Let dj(t) denote the degree of node j at time t.• Probability an edge added at t + 1 connects to node j:P r(t + 1 → j) = dj(t)/Pjdj(t).• Normalization constant easy (but time dependent):Pjdj(t) = 2mt(Each node 1 through t, contributes m edges.)(Each edge augments the degree of…
- Suppose that the road network is defined by the undirected graph, where the vertices represent cities and edges represent the road between two cities. The Department of Highways (DOH) decides to install the cameras to detect the bad driver. To reduce the cost, the cameras are strategically installed in the cities that a driver must pass through in order to go from one city to another city. For example, if there are two cities A and B such that the path that goes from A to B and the path that goes from B to A must pass the city C, then C must install the camera. Suppose that there are m cities and n roads. Write an O (m + n) to list all cities where cameras should be installed.Consider F = {BC -> D, B -> E, CE -> D, E -> CA, BF -> G} and R(A,B,C,D,E,F,G) 1.Find a minimal cover for F. 2.Find all many keys for R.Let A, B, C, D be the vertices of a square with side length 100. If we want to create a minimum-weight spanning tree to connect these four vertices, clearly this spanning tree would have total weight 300 (e.g. we can connect AB, BC, and CD). But what if we are able to add extra vertices inside the square, and use these additional vertices in constructing our spanning tree? Would the minimum-weight spanning tree have total weight less than 300? And if so, where should these additional vertices be placed to minimize the total weight? Let G be a graph with the vertices A, B, C, D, and possibly one or more additional vertices that can be placed anywhere you want on the (two-dimensional) plane containing the four vertices of the square. Determine the smallest total weight for the minimum-weight spanning tree of G. Round your answer to the nearest integer. Note: I encourage you to add n additional points (for n=1, 2, 3) to your graph and see if you can figure out where these point(s) need to…
- Let A, B, C, D be the vertices of a square with side length 100. If we want to create a minimum-weight spanning tree to connect these four vertices, clearly this spanning tree would have total weight 300 (e.g. we can connect AB, BC, and CD). But what if we are able to add extra vertices inside the square, and use these additional vertices in constructing our spanning tree? Would the minimum-weight spanning tree have total weight less than 300? And if so, where should these additional vertices be placed to minimize the total weight? Let G be a graph with the vertices A, B, C, D, and possibly one or more additional vertices that can be placed anywhere you want on the (two-dimensional) plane containing the four vertices of the square. Determine the smallest total weight for the minimum-weight spanning tree of G. Round your answer to the nearest integer. Attention: Please don't just copy these two following answers, which are not correct at all. Thank you.…Consider a Diffie-Hellman scheme with a common prime q = 17 and a primitive root α = 3. a) If user A has a private key XA=4, what is A’s public key, YA? b) A sends YA to B. If B has a private key XB=6, what is the shared secret key, K that B can calculate and share with A? c) If B computes YB and sends it to A, what is the shared secret Key, K computed by A?Consider the following knowledge base Prove that Q is true with: 1. P → Q 2. L ∧ M → P 3. B ∧ L → M 4. A ∧ P → L 5. A ∧ B → L 6. A 7. B Forward-Chaining Backward-Chaining Resolution Prove t → s: 1. p → q 2. [q ∧ r] → s 3. [t ∧ u] → r 4. u → w 5. t → y 6. y → u 7. r → p 8. p → m i. Express in clause form ii. Forward-Chaining iii. Backward-Chaining iv. Resolution