Question

can you please solve this CORRECTLY?

Exercise 4 - Shortest path (25 pts) Using Dijkstra's algorithm, find the shortest path from A to E in the following weighted graph: a- Once done, indicate the sequence (min distance, previous node) for nodes D and E. (15pts) b- Below is a high-level code for Dijkstra's algorithm. The variables used in the code are self-explanatory. Clearly explain why its running time (when we use a min-heap to store the values min distance of the nodes) is O(VI+ |E| LogVI). (10 pts) Dijkstra - High-level algorithm Initialize the min distance of all the nodes of the graph G -(V, E) Store the nodes in a list L While L is not empty: Select the node u that has the lowest min distance value Mark and remove u from L For all the (unmarked) nodes v that are adjacent to u if (min distance(u) + w y)<min distance()) min distance(v) - min distance(u) + (y) previous.node(v) - u Hint: one of the above instructions in the for loop require more than constant time.

Answer 1

Given graph is Moco, coe want to find the shortest path from "A" to s! For this we want to assume that Source rede "A"=o and Remaining nodes aro Equal to infinity (60)". Ao como 6 [31 ao & 8 6 we get these values from the one formula the New destination value = rniniumfold destination, Marked value t edge coeight). SILN

3. Shortest path From N to ' Ps. A-CB. E- I Now, we want to find the sequence (min distance, Previous node) for nodes "" and "E". For node "D" From D-A From D - B frono D- C D- A9 D-C-A => 8] D-B. A 10 B=74), D-E-B =>9. 10-c-By D- c. 75] D-Buch Danc-712 D-B-C-A -> 9 79 D-A-B 715 | D-E-B-A ->15 DEB-cell From D-> E L from all this, Sequence ( min distance, previous noda DEI = D-E >3] verter Previous Node. D--B-E ->10 DALA DC-B-E=> 13 D-B D DOD TO DELE 13

3. Shortest path From N to ' Ps. A-CB. E- I Now, we want to find the sequence (min distance, Previous node) for nodes "" and "E". For node "D" From D-A From D - B frono D- C D- A9 D-C-A => 8] D-B. A 10 B=74), D-E-B =>9. 10-c-By D- c. 75] D-Buch Danc-712 D-B-C-A -> 9 79 D-A-B 715 | D-E-B-A ->15 DEB-cell From D-> E L from all this, Sequence ( min distance, previous noda DEI = D-E >3] verter Previous Node. D--B-E ->10 DALA DC-B-E=> 13 D-B D DOD TO DELE 13

For Node "E" = E YA ' From E B From E- C E-B-A 712 | E->B-] E-D-c=8 ] E-B-D-A719 E70- 87 E-D-B-C 9 E-O-C-B => 10 E-D-a A => E-D-A-<=>15 E-D-A-B => 18 E-B-Arc=715 E-D-B-C-A =Y12 E-B-C-A => 111 From all these, Sequence (min distance, From E-70 l previous node is (E-D3 E-B-D10 E-B-C-D=> 13 E.B.LA--D=> 211 Vertex Mindist previous I node Е

For Node "E" = E YA ' From E B From E- C E-B-A 712 | E->B-] E-D-c=8 ] E-B-D-A719 E70- 87 E-D-B-C 9 E-O-C-B => 10 E-D-a A => E-D-A-<=>15 E-D-A-B => 18 E-B-Arc=715 E-D-B-C-A =Y12 E-B-C-A => 111 From all these, Sequence (min distance, From E-70 l previous node is (E-D3 E-B-D10 E-B-C-D=> 13 E.B.LA--D=> 211 Vertex Mindist previous I node Е

For the given dijkstra's algorithm code, the running time of the is O(lvl + 161 cogvt) Becances, the main lexp Exerutes, one venter is Every time edracted from the quente. Assuming that there are V vertices in the graph, the queue may contain ofv} vertices. Each Extract - min operation takes (loov) time assuming the heap implementation of priority queures so the total time required to Greate the main loop itself is O(V log v), in addition, ane meest consider the time spent in the function expand, which applies the function handle.edge for each outgoing Edge. Because Expand is only called once pe vertex, handle edge is only called once per edge 31 maight call insert (V), but there can be at most of loan Seact calls during the entire Execution so that total cost of that case arm i

For the given dijkstra's algorithm code, the running time of the is O(lvl + 161 cogvt) Becances, the main lexp Exerutes, one venter is Every time edracted from the quente. Assuming that there are V vertices in the graph, the queue may contain ofv} vertices. Each Extract - min operation takes (loov) time assuming the heap implementation of priority queures so the total time required to Greate the main loop itself is O(V log v), in addition, ane meest consider the time spent in the function expand, which applies the function handle.edge for each outgoing Edge. Because Expand is only called once pe vertex, handle edge is only called once per edge 31 maight call insert (V), but there can be at most of loan Seact calls during the entire Execution so that total cost of that case arm i

atroost olv logo) The other cose arm may be called O(E) Himes, however, and Cach ciell to increase priority tokes Olloqv) time with the heap eroplementation : the total Bur time is O(vlog v I loqu (0 Oliviel log lvl), which is O cr log w because V is ole) assuming a connected graph