Can you explain the logic behind question 3 (a) Draw the DFS tree for this graph

Question

Can you explain the logic behind question 3 (a) Draw the DFS tree for this graph, starting from node A. Assume that DFS traverses nodes in (b) Draw the BFS tree for this graph, starting from node A. Assume that BFS traverses nodes in alphabetic order. (That is, if it could go to either B or C, it will always choose B first). alphabetic order. (That is, if it could go to either B or C, it will always choose B first). Give an algorithm to detect whether a given undirected graph G -(V,E) contains a cycle. Prove that your algorithm is correct (i.e., if there is a cycle it will output one and if not it will say there Problem 2: Cycle Detection is no cycle). The running time of your algorithm should be O(VI+IE) Problem 3: Depth First Search ancestor in the DFS-tree as a back edge. Either prove the following statement or provide a counter-example: if G is an undirected, connected graph, then each of its edges is either in the depth-first search tree or is a back edge. (a) During the execution of depth first search, we refer to an edge that connects a vertex to an is called a bridge. Either prove the following statement or provide a counter-example: every bridge e must be an edge in a depth-first search tree of G. (b) Suppose G is a connected undirected graph. An edge whose removal disconnects the graph Problem 4: DFS for DAGs During Spring Break, Emilie's grandmother taught her how make lemon-mering returning, Emilie can remember all the steps she needs to perform, but unfortunately, she ue pie. Upon You do not need to turn in these problems. The goali quiz that will cover the same or similar problems. proble Problem 1: Odd Cycle Detection Given an undirected graph G- (V,E), design au algorithm tha a cycle of odd length. Prove its correctness. Your algorithm sh Problem 2: BFS Shortest Path Suppose that you want you would I

Explanation / Answer

a.that is authentic. suppose it have been now not genuine. then there could be an area in g that is not inside the dfs-tree that is also now not a returned facet. this part connects a node u to a node v. when u was included inside the dfs-tree, we chose now not to include (u, v) in the dfs-tree. the best cause we'd do that became if v had already been included inside the dfs-tree. however v isn't always an ancestor of u inside the dfs-tree, so v ought to be in a few different branch of the dfs-tree. however then,when we protected v inside the dfs-tree (which was before we tested u and u’s friends),we selected not to consist of the edge (v, u). but the simplest reason we might have done this would had been if u had already been in the dfs-tree. so there’s the contradiction

b.remember the threshold e that connects vertices u and v, wherein e is a bridge. this is, if e were removed from the graph g, then we might have a partition of the vertices of g such that no facet crosses the partition.start a depth first seek within the partition that consists of u.subsequently, due to the fact that g is connected, this dfs must attain vertex u. whilst it does, it'll observe all of the vertices adjoining to u and include any who've not yet been “touched” into the dfs-tree. this consists of v. we are assured that v has not but been tested due to the fact that the brink (u, v) is the best aspect that crosses our partition of vertices, and we began our dfs inside the partition that protected u.consequently, the threshold (u, v) need to be included in the dfs, giving us our contradiction

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.