S = AU and the intersection of any pair of these sets is the empty set. and an event E. Suppose we have the following results: • P(A) = 1/7 P(B) = 2/7 P(C) = 4/7
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- Write Algorithm for Testing MembershipInput : a group G acting on f~ = { 1,2 ..... n };a permutation g on ~ = { 1,2 ..... n };a base and strong generating set for G;Schreier vectors v (i) ,1 < i < k, for the stabiliser chain;Similarly a non-re, cursiveOutput : a boolean value answer, indicating whether g ~ G;function is_in_group(p : permutation; i : 1..k+l ) : boolean;(* return true if the permutation p is in the group G (i) *)Consider an undirected graph G = (V;E). An independent set is a subset I V such that for any vertices i; j 2 I, there is no edge between i and j in E. A set i is a maximal independent set if no additional vertices of V can be added to I without violating its independence. Note, however, that a maximal independent sent is not necessarily the largest independent set in G. Let (G) denote the size of the largest maximal independent set in G. One way of trying to avoid this dependence on ordering is the use of randomized algorithms. Essentially, by processing the vertices in a random order, you can potentially avoid (with high probability) any particularly bad orderings. So consider the following randomized algorithm for constructing independent sets: @ First, starting with an empty set I, add each vertex of G to I independently with probability p. @ Next, for any edges with both vertices in I, delete one of the two vertices from I (at random). @ Note - in this second step,…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 :
- Correct answer will be upvoted else Multiple Downvoted. Computer science. Polycarp accepts that a bunch of k sections is acceptable in case there is a fragment [li,ri] (1≤i≤k) from the set, with the end goal that it converges each portion from the set (the convergence should be a point or section). For example, a bunch of 3 fragments [[1,4],[2,3],[3,6]] is acceptable, since the portion [2,3] meets each section from the set. Set of 4 fragments [[1,2],[2,3],[3,5],[4,5]] isn't acceptable. Polycarp ponders, what is the base number of sections he has to erase so the remaining portions structure a decent set? Input The primary line contains a solitary integer t (1≤t≤2⋅105) — number of experiments. Then, at that point, t experiments follow. The primary line of each experiment contains a solitary integer n (1≤n≤2⋅105) — the number of portions. This is trailed by n lines depicting the fragments. Each section is portrayed by two integers l and r (1≤l≤r≤109) — coordinates of the…Assuming that the graph G = (V, E) is represented in Adjacency List format, justify in detail the fact Greedy independent sets can be implemented in O(n 2 + m) worst-case running time, where n = |V |, m = |E|. This will require you to take real care in how the adjustment of Adj is done in line 10. The key is to only update/delete what is really necessary for the Algorithm, rather than being concerned with an accurate representation of the residual graph.Consider an undirected graph G = (V;E). An independent set is a subset I V such that for any vertices i; j 2 I, there is no edge between i and j in E. A set i is a maximal independent set if no additional vertices of V can be added to I without violating its independence. Note, however, that a maximal independent sent is not necessarily the largest independent set in G. Let (G) denote the size of the largest maximal independent set in G. Consider the following greedy algorithm for generating maximal independent sets: starting with an empty set I, process the vertices in V one at a time, adding v to I is v is not connected to any vertex already in I. 2) Argue that the output I of this algorithm is a maximal independent set.
- The off-line minimum problem maintains a dynamic set T of elements from the domain {1, 2,...,n}under the operations INSERT and EXTRACT-MIN. A sequence S of n INSERT and m EXTRACT-MIN calls are given, where each key in {1, 2,...,n} is inserted exactly once. Let a sequence S berepresented by I1 , E, I2, E, ... , E, Im+1 , where each Ij stands for a subsequence (possibly empty) ofINSERT and each E stands for a single EXTRACT-MIN. Let Kj be the set of keys initially obtainedfrom insertions in Ij. The algorithm to build an array extracted[1..m], where for i = 1, 2, ..., m,extracted[i] is the key returned by the ith EXTRACT-MIN call is given below: Off-Line-Minimum(m, n)for i = 1 to n determine j such that i ∈ Kj if j ≠ m + 1 extracted[j] = i let L be the smallest value greater than j for which KL exists KL = KL U Kj, Kjreturn extracted (1) Given the operation sequence 9, 4, E, 6, 2, E, E, 5, 8, E, 1, 7, E, E, 3; where eachnumber stands for its insertion. Draw a…The off-line minimum problem maintains a dynamic set T of elements from the domain {1, 2,...,n}under the operations INSERT and EXTRACT-MIN. A sequence S of n INSERT and m EXTRACT-MIN calls are given, where each key in {1, 2,...,n} is inserted exactly once. Let a sequence S berepresented by I1 , E, I2, E, ... , E, Im+1 , where each Ij stands for a subsequence (possibly empty) ofINSERT and each E stands for a single EXTRACT-MIN. Let Kj be the set of keys initially obtainedfrom insertions in Ij. The algorithm to build an array extracted[1..m], where for i = 1, 2, ..., m,extracted[i] is the key returned by the ith EXTRACT-MIN call is given below: Off-Line-Minimum(m, n)for i = 1 to n determine j such that i ∈ Kj if j ≠ m + 1 extracted[j] = i let L be the smallest value greater than j for which KL exists KL = KL U Kj, destoying Kjreturn extracted Given the operation sequence 9, 4, E, 6, 2, E, E, 5, 8, E, 1, 7, E, E, 3; where eachnumber stands for its insertion.…The off-line minimum problem maintains a dynamic set T of elements from the domain {1, 2,...,n}under the operations INSERT and EXTRACT-MIN. A sequence S of n INSERT and m EXTRACT-MIN calls are given, where each key in {1, 2,...,n} is inserted exactly once. Let a sequence S berepresented by I1 , E, I2, E, ... , E, Im+1 , where each Ij stands for a subsequence (possibly empty) ofINSERT and each E stands for a single EXTRACT-MIN. Let Kj be the set of keys initially obtainedfrom insertions in Ij. The algorithm to build an array extracted[1..m], where for i = 1, 2, ..., m,extracted[i] is the key returned by the ith EXTRACT-MIN call is given below: Off-Line-Minimum(m, n)for i = 1 to n determine j such that i ∈ ?? if j ≠ m + 1 extracted[j] = i let L be the smallest value greater than j for which KL exists KL = KL U Kj, destroying ????return extracted (1) Given the operation sequence 9, 4, E, 6, 2, E, E, 5, 8, E, 1, 7, E, E, 3; where eachnumber stands for its…
- The off-line minimum problem maintains a dynamic set T of elements from the domain {1, 2,...,n}under the operations INSERT and EXTRACT-MIN. A sequence S of n INSERT and m EXTRACT-MIN calls are given, where each key in {1, 2,...,n} is inserted exactly once. Let a sequence S berepresented by I1 , E, I2, E, ... , E, Im+1 , where each Ij stands for a subsequence (possibly empty) ofINSERT and each E stands for a single EXTRACT-MIN. Let Kj be the set of keys initially obtainedfrom insertions in Ij. The algorithm to build an array extracted[1..m], where for i = 1, 2, ..., m,extracted[i] is the key returned by the ith EXTRACT-MIN call is given below: Off-Line-Minimum(m, n)for i = 1 to n determine j such that i ∈ Kj if j ≠ m + 1 extracted[j] = i let L be the smallest value greater than j for which KL exists KL = KL U Kj, destoying Kjreturn extracted Given the operation sequence 9, 4, E, 6, 2, E, E, 5, 8, E, 1, 7, E, E, 3; where eachnumber stands for its insertion.…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.Given an abstraction of the City Tube Map where each node represents the city’s attraction, plan the trip for the tourist to visit. Start from node 1 that is their hotel, a group of tourists wants to visit all places. However, due to the time limitation, they can only visit each attraction once. To find the best order of places to visit, genetic algorithm can be used to derive the best possible solutions.