Consider the elliptic curve group based on the equation y² = 2³+az+b mod p where a = 2, b=1, and p = 5. This curve contains the point P = (0, 1). The order of this elliptic curve group is the prime number 7, and therefore we can be sure that P is a primitive element. Another element in this group is Q = (0,4). The index of Q with respect to P is the least positive integer d such that Q dP. What is d, the index of Q?
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- ) Let O be the set of odd numbers and O’ = {1, 5, 9, 13, 17, ...} be its subset. Definethe bijections, f and g as:f : O 6 O’, f(d) = 2d - 1, d 0 O.g : ø 6 O, g(n) = 2n + 1, n 0 ø.Using only the concept of function composition, can there be a bijective map from øto O’? If so, compute it. If not, explain in details why not...................................................................................................................................... [2+8]b) A Sesotho word cannot begin with of the following letters of alphabet: D, G, V, W,X, Y and Z.We define the relation: A Sesotho word x is related to another Sesotho word y if xbegins with the same letter as y.Determine whether or not this is an equivalence relation.If it is an equivalence relation then1. Compute C(sekatana)2. How many equivalence classes are there in all, and why?3. What is the partition of the English words under this relation?If it is NOT an equivalence relation then explain in details why it is notOrder the following functions by asymptotics with respect to Ω. That is, find an orderingf1 , f2 , · · ·, of the following functions such that f1 = Ω(f2), f2 = Ω(f3) and so on.n2, (√2)logn , n!, log(n)!, (3/2)n , n3, log2n, loglogn, 4logn, 2n, nlogn, 2logn , 2√2logn , log(n!)A = {0, 1, 2, 3, 4} B = {2, 3, 4, 5}Given the sets A and B, how many constants maps are there from A into B.
- Please answer fast Let (G,g,p) be a cyclic group. Problem P1: Given (g, g^u,g^v), compute g^{u/v} Problem P2: Given (g, g^w, g^z), compute g^{2w/(z+w)} Prove that P1 is reducible to P2.implement Algorithm for Testing MembershipInput : a group G acting on f~ = { 1,2 ..... n };a permutation g of f~ = { 1,2 ..... n };a base and strong ~enerating set for G;Schreier vectors v (i) , 1 < i < k, for the stabiliser chain;Output: 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) *)a. Build an adjacency matrix ? for this map. b. How many paths of length 2 from V5 to V1 exist? c. How many paths of length 3 from V5 to V1 exist?
- give set of R(x): x < x^2 where the domain is ZFor x[5] = {1,2,3,4,5}, use cout << x; in order to show all elements of x?You are given a graph G = (V, E) with positive edge weights, and a minimum spanning tree T = (V, E') with respect to these weights; you may assume G and T are given as adjacency lists. Now suppose the weight of a particular edge e in E is modified from w(e) to a new value w̃(e). You wish to quickly update the minimum spanning tree T to reflect this change, without recomputing the entire tree from scratch. There are four cases. In each case give a linear-time algorithm for updating the tree. Note, you are given the tree T and the edge e = (y, z) whose weight is changed; you are told its old weight w(e) and its new weight w~(e) (which you type in latex by widetilde{w}(e) surrounded by double dollar signs). In each case specify if the tree might change. And if it might change then give an algorithm to find the weight of the possibly new MST (just return the weight or the MST, whatever's easier). You can use the algorithms DFS, Explore, BFS, Dijkstra's, SCC, Topological Sort as…
- 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 :For σarthm := (0, S, +, ·). Prove PA ⊨ ∀x (x ≠ 0 ⟹ ∃y(x = S(y))). In English, this say that every element other than 0 is a successor.Modify the Chebyshev center coding with julia in a simple style using vectors, matrices and for loops # Given matrix A and vector bA = [2 -1 2; -1 2 4; 1 2 -2; -1 0 0; 0 -1 0; 0 0 -1]b = [2; 16; 8; 0; 0; 0] A small sample:Let t_(l),t_(o),t_(m),t_(n),t_(t),t_(s) be starttimes of the associated tasks.Now use the graph to write thedependency constraints:Tasks o,m, and n can't start until task I is finished, and task Itakes 3 days to finish. So the constraints are:t_(l)+3<=t_(o),t_(l)+3<=t_(m),t_(l)+3<=t_(n)Task t can't start until tasks m and n are finished. Therefore:t_(m)+1<=t_(t),t_(n)+2<=t_(t),Task s can't start until tasks o and t are finished. Therefore:t_(o)+3<=t_(s),t_(t)+3<=t_(s)