Discrete mathematics. Let G = (V, E) be a simple graph4 with n = |V| vertices, and let A be its adjacency matrix of dimension n × n. We want to count the L-cycles : such a cycle, denoted by C = u0u1 · · · uL with  uL = u0 contains L distinct vertices u0, . . . , uL-1 et L edges E(C) = {uiui+1 | 0 ≤ i ≤ L − 1} ⊆ E. Two cycles are distinct if the edge sets are different : C = C' if and only if E(C) = E(C'). We define the matrices D, T, Q, the powers of A by matrix multiplication : D = A · A = A2,   T = A · D = A3,   Q = A · T = A4. Consider the values on the diagonals. Prove that the number of 3-cycles N3 in the whole graph is N3 = 1/6 ∑ u∈V   Tu,u

We are given an undirected connected graph G = (V, E) and vertices s and t.Initially, there is a robot at position s and we want to move this robot to position t by moving it along theedges of the graph; at any time step, we can move the robot to one of the neighboring vertices and the robotwill reach that vertex in the next time step.However, we have a problem: at every time step, a subset of vertices of this graph undergo maintenance andif the robot is on one of these vertices at this time step, it will be destroyed (!). Luckily, we are given theschedule of the maintenance for the next T time steps in an array M [1 : T ], where each M [i] is a linked-listof the vertices that undergo maintenance at time step i.Design an algorithm that finds a route for the robot to go from s to t in at most T seconds so that at notime i, the robot is on one of the maintained vertices, or output that this is not possible. The runtime ofyour algorithm should ideally be O((n + m) ·T ) but you will…

One can manually count path lengths in a graph using adjacency matrices. Using the simple example below, produces the following adjacency matrix: A B A 1 1 B 1 0 This matrix means that given two vertices A and B in the graph above, there is a connection from A back to itself, and a two-way connection from A to B. To count the number of paths of length one, or direct connections in the graph, all one must do is count the number of 1s in the graph, three in this case, represented in letter notation as AA, AB, and BA. AA means that the connection starts and ends at A, AB means it starts at A and ends at B, and so on. However, counting the number of two-hop paths is a little more involved. The possibilities are AAA, ABA, and BAB, AAB, and BAA, making a total of five 2-hop paths. The 3-hop paths starting from A would be AAAA, AAAB, AABA, ABAA, and ABAB. Starting from B, the 3-hop paths are BAAA, BAAB, and BABA. Altogether, that would be eight 3-hop paths within this graph. Write a program…

Consider eight points on the Cartesian two-dimensional xx-yy plane.   For each pair of vertices uu and vv, the weight of edge uvuv is the Euclidean (Pythagorean) distance between those two points. For example, dist(a,h) = \sqrt{4^2 + 1^2} = \sqrt{17}dist(a,h)=42+12​=17​ and dist(a,b) = \sqrt{2^2 + 0^2} = 2dist(a,b)=22+02​=2.   Using the algorithm of your choice, determine one possible minimum-weight spanning tree and compute its total distance, rounding your answer to one decimal place. Clearly show your steps.