Question

Question 3. Below is the result of the 1st and 2nd iteration of the Bellman-Ford single source shortest path algorithm starting at node A A B C D E B 2 000 0-14 E 0000 DO (D Please note the above table does not contain the pi or previous node values. Please provide the changes to the tables that occure during the third iteration only for distance(shortest path estimation) when processing only the edges: edges (D,C), (B,C),(D,B), (B,D) (B,E) and (E, D) in this order. Show your work. Question 4. a.)Given the graph below from part 4.b. you are using the priority queue prims algorithm. Let the priority queue currently contains nodes A, B. What is the value of node you usextract_min(Q). b.) Run 1 iteration of Prims while loop, showing the priority queue. 5 А. E 1 10 7 2 B с D 3 4 c.) What if you read on the news prioriy queue was improved such that both update_key, and extract min became O(log(n)) how would Prims, Dijkstras and Kurskals change. Please explain for each

Answer 1

Ques 3.

Ques3. Bellman - ford has Single algorithm the following source shortest path following steps . each One- One 1) 2) Relan Relax edge edges all the where (3-1) times number vertices زين we better shorter then the contains If even after (V-1) relaxations get fath graph negative weight cycles the graphe given in the question, we get following the table : A с E 8 Null distance O Null i=o Hull qull parent ою Null А i=2 iterations Null А Null 4 A 2 B 2 B 2 B B 1 i=3 B -2 E -2 Null А B i=4 The last iteration the shortest paths has as in some values. the (1-1)th iteration, of the . There is no negative weight cycle in cycle in the graph.

Ques 4. (a),(b)

(c)

Ques 4. For the iterations giren graph, Prim's algorithm are summarised below: A с E Null Mull Null AK lowp 8C 7 A 3 B ор Null 10 А 10 A 4 с Nall 5 А 5 А 5 c. 4 2 IT Every time, extract-mind function. The minimum node is then, is extracted corre selected. using corresponding to that minimum So, if if priority queue already has A, B, then next node that will be extracted C. we get the Finally spanning following s minimum tree A E 2 B D 3 4

update_key o (loge (ar)) extract - min o (log? (m) Dijkstras Algorithm Time complenity n times extract_min() + build time to priority queue + scan adjacency list + E times update key 2 n to log? (n)) + 0(m) + 0(E) Ex nt n- no. 8 nodes E no. of edges Prins Algorithm Time complexity n times extract min time + build to scan priority queue + adjacency lict E times update key = O(log?)) + O(n) + 2x0 (BE + E E (loge (n)) because it is undirected graph o (n ++) x (m)

(c)