CSE 100 Practice Final Exam
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1 CSE 100 Practice Final Exam Please note that this practice final only serves as a template and does not exhaustively cover all the topics that are on the actual final. The final exam will cover all the data-structures and algorithms that we have covered during the quarter, including all the readings, class discussions, and the assignments. Be sure that you understand each of the data structures and the algorithms and are able to run the algorithms by hand. 1. [ pts total] A little bit of everything a. [2 pts] Assume that H is the height of a completely full binary search tree that contains N nodes. Which of the following represents the worst case time to insert an element into this tree? Include all relevant bounds, not just the tightest bound. Circle all that apply: A. O(1) B. O(N) C. O(logN) D. O(H) E. O(logH) b. [2 pts] Assume that you have a class in C++ named MyClass. Which of the following statements are true about the following line of code (circle all that apply): MyClass mc; A. This line declares a variable of type MyClass. B. This line creates an object of type MyClass. C. You (the programmer) must call delete to destroy the object created by this line of code D. mc is a pointer type which stores or will store the address of a MyClass object c. [3 pts] Consider this AVL tree (balance factors not shown): For each of the following values, state whether they would cause a single rotation (S), a double rotation (D) or no rotation (N) when inserted into the AVL tree above. Assume that each value would be inserted into the tree above independent of the other values. Write the appropriate letter in the blank next to the value (S, D, or N). 1: 30: 45: 51: 60: 49 d. [2 pts] When is breadth first search guaranteed to find the shortest path through a graph (circle all that apply)? A. In any unweighted graph B. In any directed graph
2 C. In any graph with no cycles D. In any fully connected graph E. Never e. [5 pts] Consider the following treap, in which letters represent keys (using alphabetical ordering) and numbers represent priorities. We are midway through the insertion of the key F into this treap. G 50 C 35 B 24 F 45 A 21 E 33 What was the last (previous) action performed during the insert? A. A left rotation with E and F B. A right rotation with E and F C. A left rotation with F and C D. A right rotation with F and C E. F was inserted directly into its current location (no rotations performed yet) After the insert is complete, what will be the children of the following nodes? If a node has no child in that position, write NONE. Node G: left: right: Node C: left: right: Node F: left: right: Node B: left: right:
3 Name: PID: 2. More on Trees and treaps a. In the space below, draw a legal AVL tree with 4 nodes that is currently in balance but that will require a double rotation to insert at least one value that is NOT currently in the tree. You will be asked what this value is in part b, and you will be asked to do the double rotation in part c. However, you will get full credit in part a for drawing any legal AVL tree with 4 nodes. Be sure to clearly annotate each node with its balance factor. b. What value could you insert that would cause a double rotation? c. Insert this value and then draw the resulting tree below. d. Now, take your tree from part a (your original AVL tree, before you inserted the value from part b), and convert it into a treap with the same structure using the same values as keys, and any legal priority values that lead to the same structure. Draw the treap below. e. Finally, imagine that you will now insert the key from part b into this treap, using a priority that is greater than any other priority currently in the treap. Which of the following best describes why you do or do not need to perform a double rotation when you first insert this key, priority pair into the treap? A. You do need a double rotation, because if you use only single rotations the treap will become unbalanced. B. You do need a double rotation, because if you use only single rotations the treap will lose the BST property on the priorities. C. You do not need a double rotation, because a double rotation would violate the BST property on the keys.
4 D. You do not need a double rotation because single rotations are sufficient to get the node to bubble up to its proper location in the treap 3. Huffman coding and running time Consider the following symbols with the given frequency distribution: Symbol Frequency A 0.46 M 0.18 L 0.15 Y 0.05 P 0.16 a. [6 pts] Is the tree below a correct Huffman Code tree for these symbols with this distribution (circle one, below)? You should NOT consider it a problem that the frequencies are not explicitly marked. Assume they are present (calculated based on the table to the left), but not shown. Y 0 Yes No 1 1 Y 0 P Y 0 A M P L Y b. [3 pts] Regardless of whether or not the tree is correct, is there more than one correct tree for these symbols with this distribution? You should consider two trees different if they lead to different codes for the symbols. Yes No c. [20 pts, 4 pts each blank] In this part, you will analyze the running time for each stage of a naïve implementation of Dijkstra s algorithm on a graph G(V, E). If an edge exists between any two vertices, v, w, the weight of that edge is denoted by l vw 1. Initialize the distance of all nodes to Inf: L(u) =Inf, for all u in V 2. Set the distance of source, s to zero: L(s) =0 3. Initialize the set of vertices explored, X={s} 4. Iterate until X = V Among all crossing edges (v, w), with v in X and w in V-X, pick (v*, w*) such that (v*, w*) =arg min L(v) + l vw v,w Include w* in X: X= X U {w*}
5 Name: PID: What is the overall running time of the above algorithm 4. Graphs, C++ a. [13 pts] In this part, you will run Dijkstra s algorithm on graph shown below, with the source node V0. The data structure for each node is shown below. You will modify values in it. See below. V0: dist=0 prev= -1 done= f adj: (V1,1), (V2,6), (V3,3) V1: dist= prev= -1 done= f adj: (V2,4) V2: dist= prev= -1 done= f adj: V3: dist= prev= -1 done= f adj: (V2,1) The boxes below represent the priority queue which is used in the algorithm. Each set of boxes represents the priority queue after the first pair has been dequeued and its associated node has been fully expanded (i.e. its neighbors, along with their distances, have been added to the priority queue as appropriate). You will fill these in (see below for more instructions). V0 3 V3 6 1 V2 (V0, 0) (initial priority queue) 1 4 V1 (priority queue after expanding V0)
6 Simulate Dijkstra s algorithm by (1) modifying the values in the dist, prev, and done fields as the algorithm runs, and (2) showing the values stored in the priority queue after each new node is expanded. For the data structures at the top, you should cross off old values as you replace them. For the priority queue, you should show the contents of the whole queue in the next set of boxes after you have fully expanded/explored from the node that you just removed from the front of the queue in the previous step. The initial priority queue and initial values are already filled in for you. You may stop the process when each node has been marked as done. You may not need all of the queues. If you need more, just draw them in below the final array, to the side, or on the scratch paper. b. [13 pts] Consider the classes below which represent a graph node (vertex) and a graph. Complete the code to perform breadth first search on the graph from the source node with index source. After BFS is called on the graph, all of the Vertex objects pointed to by nodes should have their members set as follows: visited should be true if this node is reachable from the source node, false otherwise prev should store the integer index of the node (vertex) from which this node (vertex) was first reached Here are some helpful C++ commands and information of relevance here. You may not use all of these methods. You can access an element at position i in vector v using v[i] For an example of how to create and work with an iterator over a vector object, see the code provided to you at the start of BFS. Useful methods in class queue (Please do not use any other methods in the queue class. You do not need any others.): o q.front() accesses (returns) the next element in queue q (but not remove it) o q.pop()removes the next element (from the front) in queue q (but not return it) o q.push( d ) places the element d into the queue q (at the back). o q.empty() returns true if the queue q is empty, and false if there is at least one element in it #include <iostream> #include <queue> #include <vector> class Vertex { public: vector<int> adj; this vertex bool visited; int index; int prev; was found // Indexes of vertexes directly connected to // Has this vertex been visited? // The vertex s index // The index of the vertex from which this vertex Vertex(int ind, vector<int> alist ) :
7 Name: PID: }; index(ind), adj(alist), prev(-1), visited(false) {} class Graph { public: vector<vertex*> nodes; // Pointers to vertexes in the graph. You can access a pointer to Vertex at index i using nodes[i]. Graph( vector<vertex*> thenodes ) : nodes(thenodes) {} // Continued on next page, where you will add your code
8 void BFS( int source ) { vector<vertex*>::iterator it = nodes.begin(); for ( ; it!= nodes.end(); ++it ) { // initialize the graph (*it)->visited = false; (*it)->prev = -1; } queue<vertex*> toexplore; // Your code here (see previous page) } };
9 Name: PID:
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