MIDTERM Preparation. CSS 343: Data Structures Algorithms, and Discrete Math II

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1 MIDTERM Preparation CSS 343: Data Structures Algorithms, and Discrete Math II The midterm is closed-book, closed-notes exam, except for a single sheet of letter-sized, double-sided hand-written or typed notes AND a basic calculator. If you are using typed notes, your font size has to be at least 10. You will submit your notes along with your midterm. You are responsible for creating your own set of notes. Total of 100 points, 120 minutes allowed time. Material covered : Chapter 15-17, 19, 20 in Carrano + lectures + assignments + other readings The actual midterm will have 5-10 questions similar to the questions below. You must attempt these questions yourself in preparing for the exam. Looking at the solutions will give you a false sense of security and will not help you in preparing for the exam. Trees T1. What is the minimum height of a binary tree with 0, 1, 2, 3, 5, 7, 10, 20, 100 nodes? Which of the trees above could be a perfect or full tree? Which of the trees above could be a complete tree? How many nodes does a full binary tree with height 8 have? The total number of nodes in a binary tree is 2 h - 1 Nodes Height (max nodes 1) 2 2 (max nodes 3) (max nodes 7) 7 3 1

10 4 (max nodes 15) 20 5 (max nodes 31) 100 7 (max nodes 127) How about 10,000 nodes? 2 h - 1 >= 10000 2 h >= 9999 h >= log(9999) h >= 13.

2 10 4 (max nodes 15) 20 5 (max nodes 31) (max nodes 127) How about 10,000 nodes? 2 h - 1 >= h >= 9999 h >= log(9999) h >= 13.xx (2 13 is 8192) Minimum height is 14 Prefect or full tree has the maximum nodes possible, so trees with 1, 3 and 7 nodes is full. All of them could be complete, as long as we arrange the leaves properly. When height is 8, gives 255 T2. For the given binary tree below, print the order of nodes that will be visited for preorder, inorder and postorder traversal. Preorder (root, left subtree, right subtree): K, F, C, J, N, L, M, O Inorder (left subtree, root, right subtree): C, F, J, K, L, M, N, O Postorder (left subtree, right subtree, root): C, J, F, M, L, O, N, K T3. Explain the process for removing the node F for the tree above? Find inorder successor, which is J, and swap the node data with J. Since inorder successor was a leaf, we are done. T4. Starting from the initial tree, explain the process for removing K? 2

3 Find inorder successor, which is L, and swap the node data with K. Since the inorder successor has a right child, promote the right child to be the left child of N. T5. Explain tree sort algorithm. What are the different parts, what is its complexity for best-case and worst case. Put all elements from an array into a tree - best case O(n log n), worst case array already sorted O(n 2 ) Inorder traversal to get all items in order - O(n) Total complexity best case O(n log n + n) which is O(n log n), worst case O(n 2 + n) which is O(n 2 ) T6. Create a complete binary search tree with the following elements T7. Given an array representation of a tree with 100 elements, indexes 0 to 99 no free locations. For the item at index 25, what is the array index of its left child, right child and parent? Assuming that the tree has been constructed with only additions where each height was completely filled from left to right before constructing next level. If not, then the nodes can be anywhere and we have to look it up in the table. left = 2 * parentindex + 1 right = 2 * parentindex + 2 parent = (left - 1) / 2 For index 25, parent at 12, left child at 51 and right child at 52 Huffman 3

4 H1. Given the following distribution of letters, create the Huffman tree. l: 0.03, i: 0.09, e: 0.17, f: 0.14, s: 0.17, b: 0.12, u: 0.13, t: 0.15 What is the average number of bits used to encode a symbol using this encoding? What is the bit representation for the letters: l, i, f, e What is the total number of bits needed to represent the word life? Sort smallest to largest and combine the two smallest in a tree. You should redraw the tree at each step which is not done below. [ The below solution put the heavier nodes on the right, which is not the convention we followed in class. If you were putting the heavier nodes on the left, you'd end up with the mirror image of the below tree. The visualization on puts heavier nodes on the right. I will make sure the question states whether heavier nodes should go on the left or right on the exam. If the question does not state a preference, you can do it either way ] l: 0.03, i: 0.09, b: 0.12, u: 0.13, f: 0.14, t: 0.15, e: 0.17, s: 0.17 Combine l(0.03) i(0.09) to get weight 0.12 b: 0.12, li: 0.12, u: 0.13, f: 0.14, t: 0.15, e: 0.17, s: 0.17 Combine b(0.12) li(0.12) to get weight 0.24 u: 0.13, f: 0.14, t: 0.15, e: 0.17, s: 0.17, bli: 0.24 Combine u(0.13) f(0.14) to get weight 0.27 t: 0.15, e: 0.17, s: 0.17, bli: 0.24, uf:

27, te: 0.32, sbli: 0.41 Combine uf(0.27) te(0.32) to get weight 0.

5 Combine t(0.15) e(0.17) to get weight 0.32 s: 0.17, bli: 0.24, uf: 0.27, te: 0.32 Combine s(0.17) bli(0.24) to get weight 0.41 uf: 0.27, te: 0.32, sbli: 0.41 Combine uf(0.27) te(0.32) to get weight 0.59 sbli: 0.41, ufte: 0.59 Combine sbli(0.41) ufte(0.59) to get weight 1 5

Label the edges with 0s and 1s to get all the encodings s: 00 b 010 l: 0110 i: 0111 u: 100 f: 101 t: 110 e: 111 What is the bit representation for the letters: l, i, f, e Spaces added for

6 Label the edges with 0s and 1s to get all the encodings s: 00 b 010 l: 0110 i: 0111 u: 100 f: 101 t: 110 e: 111 What is the bit representation for the letters: l, i, f, e Spaces added for readability Total: 14 What is the average number of bits used to encode a symbol using this encoding? The average number of bits used to encode a symbol using this encoding is 6

l: 0.03, i: 0.09, b: 0.12, u: 0.13, f: 0.14, t: 0.15, e: 0.17, s: 0.17 s: 00 2 * 0.17 + b 010 3 * 0.12 + l: 0110 4 * 0.03 + i: 0111 4 * 0.09 + u: 100 3 * 0.13 + f: 101 3 * 0.14 + t: 110 3 * 0.

7 l: 0.03, i: 0.09, b: 0.12, u: 0.13, f: 0.14, t: 0.15, e: 0.17, s: 0.17 s: 00 2 * b * l: * i: * u: * f: * t: * e: * 0.17 = = 2.95 Arithmetic Expressions AE1. Draw the corresponding trees representing prefix expressions and evaluate the result 1. + * * / * evaluate right to left when operator found, apply it to the operands and substitute result + * * *

+ 2 3 5 1 + 2 3 4 5 4 625 * / 9 3 + * 2 4 7 6

8 * / * * / * * / * / * AE2. Convert the following infix expression to an algebraic expression tree: (a + b) * (c - d) / (e * f) * g / h Write the preordertraversal corresponding to Polish prefix notation 8

9 / * / * + a b - c d * e f g h Operator Overloading OO1. Assume that imaginary numbers are implemented via the class IN. Write the function signatures for the copy constructor, operator+, operator+= and operator= member functions. IN operator+(const IN &in) const; IN& operator+=(const IN &in); IN& operator=(const IN &in); Priority Queue PQ1. The STL implements priority queue. Given the below definition for a priority queue, write the code to insert 3 objects into the queue, and then remove each item and print it. using IntStr = pair < int, string >; priority_queue<intstr, vector<intstr>, greater<intstr>> pq; pq.push( make_pair(10, "a") ); pq.push( make_pair(30, "b") ); pq.push( make_pair(50, "c") ); while (!pq.empty()) { IntStr pair = pq.top(); pq.pop(); cout << pair.first << " " << pair.second << ", "; } // should print from smallest to largest because of greater // 10 a, 30 b, 50 c, Heap H1. For a maxheap, what is the complexity of the following operations? The heap property must be maintained. 1. Find smallest item 2. Find largest item 9

10 3. Remove smallest item 4. Remove largest item 5. Inserting an item that is smaller than all items in the tree 6. Inserting an item that is larger than all items in the tree Assuming the heap is implemented as an array 1. Find smallest item - last element O(n) - finding a small item is easy but smallest requires visiting all leaves which is about n/2 2. Find largest item - first element O(1) 3. Remove smallest item - finding smallest is O(n), removing it is O(1), but then we need to make the heap structure again by swapping it with the last item in the heap and calling heapify which is O(log n). Total O(n) 4. Remove largest item - swap largest with smallest and trickle down, O(log n) 5. Inserting an item that is smaller than all items in the tree - insert at last free spot, no bubble up required, O(1) 6. Inserting an item that is larger than all items in the tree - insert at last free spot, bubble up all the way to the root, O(log n) H2. Draw the maxheap tree structure and the array representation that results from the following order of insertions: 9, 15, 6, 1, 5, 7, 3, 50 (fixed: removed 2, 2019/02/06) 9 Insert 15 and bubble up

11 Insert 7 and bubble up

12 Insert 50 and bubble up H3: Remove 50 and then 15 from the heap created above. Draw the tree structure and the corresponding array representation Cannot remove 15, must remove 50 first and then 15 12

Swap 1 and 50, delete the node that now has 50 Trickle down 1: Swap with 15, Swap with 9 15 9 7 1 5 6 3 Swap 15 with 3, delete the node that now has 15 Trickle down 3: Swap with 9, Swap with 5 9 5 7

13 Swap 1 and 50, delete the node that now has 50 Trickle down 1: Swap with 15, Swap with Swap 15 with 3, delete the node that now has 15 Trickle down 3: Swap with 9, Swap with H4: Given the array below, show the steps to convert the array into a maxheap. Redraw the array every time a number changes or two number are swapped. Write out what is happening in each step

14 size is 9 Call heaprebuild starting from (size - 2) / 2 to 0, that is from 3 to 0. Index 0 is the first node that has a child Calling heaprebuild on index 3, corresponding to 40 Bigger child, 200, bis promoted replacing Calling heaprebuild on index 2, corresponding to 5 Bigger child, 100, is promoted replacing

Calling heaprebuild on index 1, corresponding to 20 Bigger child, 200, is promoted replacing 20 heaprebuild calls itself recursively since 20 now has a child of 40 which is bigger Bigger child, 40,

15 Calling heaprebuild on index 1, corresponding to 20 Bigger child, 200, is promoted replacing 20 heaprebuild calls itself recursively since 20 now has a child of 40 which is bigger Bigger child, 40, is promoted replacing Calling heaprebuild on index 0, corresponding to 10 Bigger child, 200, is promoted replacing 10 heaprebuild calls itself recursively since 10 now has a child of 40 and 50 which are bigger Bigger child, 50, is promoted replacing

16 The array now represents heap. H5: Using the heap array above, show the steps to sort the array using heap sort. Redraw the array every time a number changes or two number are swapped. Write out what is happening in each step. The sorted section of the array is currently empty Swap index-0 with index-8 Trickle 20 down, swap with 100 Sorted section is just index Swap index-0 with index-7 Trickle 3 down, swap with 50, swap with 40 Sorted section is [7-8] Swap index-0 with index-6 16

Trickle 5 down, swap with 40, swap with 10 Sorted section is [6-8] 40 10 20 3 5 6 50 100 200 Swap index-0 with index-5 Trickle 6 down, swap with 20 Sorted section is [5-8] 20 10 6 3 5 40 50 100 200

17 Trickle 5 down, swap with 40, swap with 10 Sorted section is [6-8] Swap index-0 with index-5 Trickle 6 down, swap with 20 Sorted section is [5-8] Swap index-0 with index-4 Trickle 5 down, swap with 10 Sorted section is [4-8] Swap index-0 with index-3 Trickle 3 down, swap with 6 Sorted section is [3-8]

Swap index-0 with index-2 Trickle 3 down, swap with 5 Sorted section is [2-8] 5 3 6 10 20 40 50 100 200 Swap index-0 with index-1 Sorted section is [1-8] 3 5 6 10 20 40 50 100 200 Swap index-0 with

18 Swap index-0 with index-2 Trickle 3 down, swap with 5 Sorted section is [2-8] Swap index-0 with index-1 Sorted section is [1-8] Swap index-0 with index-0 Sorted section is [0-8] Graphs G1. For the graph below, list the order of nodes that will be visited by depth-first and breadth-first search algorithms, starting from 7. If there are any ties on what edge to process, break the ties using alphabetical/numerical order. 18

For the graph given below, write out the topological order of the vertices and describe the algorithm used.

19 DFS: 7, 3, 1, 2, 5, 10, 6, 4, 8, 9, 13, 14, 11, 12 BFS: 7, 3, 4, 9, 11, 12, 1, 6, 8, 13, 2, 10, 14, 5 // Sanity check for BFS // Distance 0 edge: 7 // Distance 1 edge: 3, 4, 9, 11, 12 // Distance 2 edges: 1, 6, 8, 13 // Distance 3 edges: 2, 8, 10, 14 // Distance 4 edges: 5 G2. For the graph given below, write out the topological order of the vertices and describe the algorithm used. Remove vertex without any successors, add that to the topological list and then repeat. Not a unique solution K A O C N D L B (fixed 2019/02/06) G3. For the graph given below, use Prim's algorithm to create the minimum spanning tree and indicate the total length. Use 3 as your starting vertex. 19

20 The edge between 4-8 is not labelled, assuming 0 weight for that edge (fixed 2019/02/06) Edge Weight Total Weight: 21 20

21 G4. For the graph given below, find the shortest path from S to all the vertices using Dijkstra's algorithm. At each step, you must show what vertices are in the vertex set and what the current weight to each vertex is. Assume the graph below: vertex vertexse t A B C D E F G H S S INF INF INF INF INF A S,A 7 B S,A,B C E F S,A,B,C S,A,B,C, E S,A,B,C,E,F 7 9 G SABCEFG 8 H SABCEFGH Final Distances

22 G5. Given the graph below, assume the Dobreta-Craiova road has been removed. List the order of cities that will be visited by greedy search algorithm when starting from Timisiora and going to Bucharest. Dobreta-Craiova road has been removed. Greedy search will choose the city that has the shortest straight-line distance. Using Best-first search algorithm with f(u) set to straight line distance Add initial vertex to PQ OPEN while PQ is not empty w = top of queue if w is targetvertex, return add w to CLOSED list for each vertex u adjacent to w if u is not in CLOSED calculate f(u), push u to PQ OPEN with cost f(u) SORTED OPEN LIST : Timisoara(329) Next city: Timisoara Closed: Timisoara 22

23 Add to OPEN: Arad(366), Lugoj(244) SORTED OPEN LIST : Lugoj(244), Arad(366) Next city: Lugoj Closed: Lugoj Add to OPEN: Mehadia(241) -- Timisoara is closed, so not added SORTED OPEN LIST : Mehadia(241), Arad(366) Next city: Mehadia Closed: Mehadia Add to OPEN: Dobreta(242) -- Lugoj is closed, so not added SORTED OPEN LIST : Dobreta(242), Arad(366) Next city: Dobreta Closed: Dobreta Add to OPEN: none - road to Craiova removed, Mehadia closed SORTED OPEN LIST : Arad(366) Next city: Arad Closed: Arad Add to OPEN: Zerind(374), Sibiu(253) -- Timisoara is closed, so not added SORTED OPEN LIST : Sibiu(253), Zerind(374) Next city: Sibiu Closed: Sibiu Add to OPEN: Rimnicu Vilcea(193), Fagaras(178), Oradea(380) SORTED OPEN LIST : Fagaras(178), Rimnicu Vilcea(193), Zerind(374), Oradea(380) Next city: Fagaras Close: Fagaras Add to OPEN: Bucharest(0) SORTED OPEN LIST : Bucharest(0), Rimnicu Vilcea(193), Zerind(374), Oradea(380) Next city: Bucharest Found destination - not the shortest path, but good enough The path found was Timisoara-Arad-Sibiu-Fagaras-Bucharest. The shorter path is Timisoara-Arad-Sibiu-Rimnicu-Pitesti-Bucharest G6. Given the graph above, make the following changes 23

24 1. Pitesti-Bucharest road is now 200 instead of The straight line distance from Pitesti to Bucharest is 198 instead of 98 Starting from Arad, list the order of the that will be explored by the A* search algorithm A* search will choose the city based on priority. Using Best-first search algorithm with f(u) set to distance to get to the city PLUS estimated distance from that city to destination Add initial vertex to PQ OPEN while PQ is not empty w = top of queue if w is targetvertex, return add w to CLOSED list for each vertex u adjacent to w if u is not in CLOSED calculate f(u), push u to PQ OPEN with cost f(u) SORTED OPEN LIST : Arad(329=0+329) Next city: Arad, distance traveled 0 Closed: Arad Add to OPEN: Zerind(449= ), Sibiu(393= ), Timisoara(447= ) SORTED OPEN LIST : Sibiu(393), Timisoara(447), Zerind(449) Next city: Sibiu,, distance traveled 140 Closed: Sibiu Add to OPEN: Oradea(671= ), Fagaras(417= ), Rimnicu(413= ) SORTED OPEN LIST : Rimnicu(413), Fagaras(417), Timisoara(447), Zerind(449), Oradea(671) Next city: Rimnicu, distance traveled 220 Closed: Rimnicu Add to OPEN: Craiova(526= ), Pitesti(515= ) Pitesti road estime 198 instead of 98, so Pitesti has lower priority SORTED OPEN LIST : Fagaras(417), Timisoara(447), Zerind(449), Bucharest(450), Pitesti(515), Craiova(526), Oradea(671) Next city: Fagaras, distance traveled 239 Closed: Fagaras 24

Add to OPEN: Bucharest(450=239+211+0) SORTED OPEN LIST : Timisoara(447), Zerind(449), Bucharest(450), Pitesti(515), Craiova(526),

.. Next city: Bucharest Target city found, distance traveled 450 A* does find the shortest path Arad-Sibiu-Fagaras-Bucharest, but explores

25 Add to OPEN: Bucharest(450= ) SORTED OPEN LIST : Timisoara(447), Zerind(449), Bucharest(450), Pitesti(515), Craiova(526), Oradea(671) Next city: Timisoara... Next city: Zerind... Next city: Bucharest Target city found, distance traveled 450 A* does find the shortest path Arad-Sibiu-Fagaras-Bucharest, but explores far away options like Zerind because the estimate is always underestimating the required distance. Balanced Trees B1. For the 2-3 given below, insert the following elements one by one and draw the tree after each insertion:

26 B2. For the tree given below, insert the following elements one by one and draw the tree after each insertion. Use preemptive split if needed:

27 B3. For the AVL tree given below, insert the following elements one by one and draw the tree after each insertion

28 General S1. Who is Mr Landau? How many hands did he shake? What was served at the dinner party? 28

29 29

Introduction to Artificial Intelligence. Informed Search

Introduction to Artificial Intelligence Informed Search Bernhard Beckert UNIVERSITÄT KOBLENZ-LANDAU Winter Term 2004/2005 B. Beckert: KI für IM p.1 Outline Best-first search A search Heuristics B. Beckert: