CS 314H Algorithms and Data Structures Fall 2012 Programming Assignment #6 Treaps Due November 11/14/16, 2012

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1 CS H Algorithms and Data Structures Fall 202 Programming Assignment # Treaps Due November //, 202 In this assignment ou will work in pairs to implement a map (associative lookup) using a data structure called a treap, which is a combination of a tree and a heap. Your ke challenge in this assignment will be to carefull and thoroughl test our data structure, so ou will also be asked to design a testing program for our code. For this assignment, ou should not discuss testing strategies with other teams. You will again be writing Peer Reviews, with the following schedule. Treaps Program and Report Due :00pm Sunda November Peer Reviews Due :00pm Wednesda November Testing Report and Test Code Due :00pm Frida November A treap is a binar search tree that uses randomiation to produce balanced trees. In addition to holding a ke-value pair (a map entr), each node of a treap holds a randoml chosen priorit value, such that the priorit values satisf the heap propert: Each node other than the root has a priorit that is at least as large as its parent s priorit. An eample treap is shown in Figure, where the kes are shown at the top of each node and the priorities are shown at the bottom of each node. Notice that the kes obe the binar search tree (BST) propert and the priorities obe the heap propert. Because the kes obe the BST propert, a lookup operation can be performed just as with an BST. However, the insert and remove operations are more comple Figure : A treap for a map with ke set {,,,,,, }. For each node, the ke is shown in the top half, while the priorit is shown in the bottom half. The priorit values are chosen at random, with smaller numbers indicating higher priorit. To insert a new node with ke k, we first perform the insertion at the appropriate leaf position according to the BST propert, eactl as in a binar search tree (See Figure 2). The node is assigned a randoml chosen priorit p, and because s parent ma have priorit greater than p, the heap propert ma be violated. To restore the heap propert, we perform a rotation, making the parent of, as shown in Figure 2(b). Specificall, if is the left child of, then we rotate right around, and if is the right child of, then we rotate left around. Node now has a new parent, and the heap propert ma still be violated, requiring another rotation. In general, the heap propert is restored b rotating the new node up the treap as long as it has a parent with higher priorit. Figure 2 shows an insertion requiring 2 rotations. To remove a node, we reverse the insertion. We rotate down the treap until it becomes a leaf, and then we simpl clip it off. At each step, the decision to rotate left or right is governed b the relative priorit of the children.

2 (a) (b) (c) Figure 2: Inserting new node into a treap. (a) The new node, with ke k= and priorit p=2, is added as a leaf according to the BST propert. The heap propert with respect to s parent is violated. (b) The situation after a right rotation around ; the heap propert with respect to s new parent is violated. (c) After a left rotation around, the heap propert is restored. The child with the higher priorit should become the new parent. Thus, if s left child has higher priorit than s right child, then we rotate right around. Conversel, if s right child has smaller priorit than s left child, then we rotate left around. Figure illustrates a removal requiring 2 rotations. This removal reverses the insertion of Figure 2. All three map operations lookup, insert, and remove run in time O(h), where h is the height of the treap. It is not hard to show that a treap with n nodes has epected height Θ(log n). Note that the root of a treap is determined b the randoml chosen priorities. The node with the highest priorit (smallest ke value) is the root. Thus, the root node is equall likel to contain an of the map entries, regardless of the order in which the entries are inserted or removed. Consequentl, we epect that half of the entries will be in the left subtreap and the other half in the right subtreap. The analsis of treap height is therefore similar to the analsis of recursion depth in quicksort. 2 Your Assignment Implement a map using a treap. In particular, ou should implement the following interface. Your treap should store entries with kes that are Comparable objects and values that are Objects. The lookup(k) method should return null if no entr with ke k is in the map. public interface Treap { Object lookup(comparable ke); void insert(comparable ke, Object value); void insert(comparable ke, Object value, int priorit); Object remove(comparable ke); Treap [] split(comparable ke); void join(treap t); void printtreap(printwriter o); void resetiterator(); Comparable netke(); double balancefactor() throws OperationNotSupportedEception; void meld(treap t) throws OperationNotSupportedEception; void difference(treap t) throws OperationNotSupportedEception; } A more detailed description of the interface is in Treap.java. Implement our treap-based map in TreapMap.java 2

3 (a) (b) (c) Figure : Removing a node from a treap. (a) Node has two children, of which the left child has smaller priorit. (b) After a right rotation around, node now has onl one child,. (c) After a left rotation around, node is now a leaf and can be clipped off like an ecessivel long toenail. The insert method. The remove method. Insertion into the treap should be implemented as outlined earlier. Removal from the treap should be implemented as outlined earlier. The split method. A treap T can be split, using a ke k, to produce two treaps, T and T 2, such that T contains all of the entries in T with ke less than k, and T 2 contains all of the entries in T with ke greater than or equal to k. To perform the split, we insert into T a new entr with ke k and priorit p = 0, forming a new treap T. (We assume that 0 is the smallest possible priorit value.) Because has the smallest possible priorit, is the root of T, so the split has been accomplished with T being the left subtreap and T 2 being the right subtreap. You should not lose an value associated with k if k is alread in the treap, although it is ok if ou destro the old treap. The join method. The inverse of a split is join, in which two treaps, T and T 2, with all kes in T being smaller than all kes in T 2, are merged to form a new treap T. To perform the join, we create a new treap T with an arbitrar new root node and with T and T 2 as the left and right subtreaps. We then remove from T to form the joined result T. Split and join both take time O(h), where h is the height of the T (the input to split or the result of join). The epected height is Θ(log n), where n is the sie of T, so split and join both run in O(log n) epected time. More interestingl, split and join can be used as subroutines to meld two treaps or take the difference between two treaps. You ma implement these for additional karma. Testing Since the treap in this assignment is not part of a larger application, ou will not be able to use or test our treap without writing our own test program. Write a program (in TreapTest.java) to test our treap for correctness. Your test program should work with an implementation of the Treap interface. It should not find an errors in a correct implementation. For errors that it does find, it should produce output that would be useful to a human. You are not required to test the portions that ou do not implement ourself, but ou should test everthing that ou do implement.

4 Karma Three of the operations in the interface (balancefactor(), meld() and difference()) are optional. Implement them for etra karma. If ou do not implement them, throw an OperationNotSupportedEception. Meld A meld takes two treaps, T and T 2 and merges them into a new treap T, much like the Vulcan mind meld for which it is named. Unlike a join, a meld does not require an relationship between the kes in T and T 2. Meld is a naturall recursive procedure and should be able to meld two treaps of sie n and m (m n) in O(m log(n/m)) time. Describe how ou meld treaps and how our algorithm meets the specified asmptotic time bound..2 Difference The difference between two treaps, T and T 2, is a treap T containing the kes of T with an kes in T 2 removed. The difference can also be computed recursivel and also runs in O(m log(n/m)) time. Describe how ou take a difference and how our algorithm satisfies this time bound.. Diagnosing Problems Through Testing Tpicall, the goal of a test program is to identif bugs. With some additional work, ou can attempt to diagnose common problems b using the observed behavior of the program. For eample, if the iterator misses one ke, it is likel that the missing ke is the first or last ke added. A test program can attempt to verif this hpothesis and provide a suggestion to the user. Can ou use our test program to provide assistance in finding common mistakes?. Balance statistics It would be useful to know how balanced or imbalanced our treap is. The balance factor is the ratio between the height of the treap and the minimum possible height. A perfectl balanced treap will have a balance factor of.0. Include observations on how well the treap seems to keep itself balanced in our report. What to turn in Ecept for the specific names of the files, the directions for what to turn in are the same as for Assignment, but we repeat them below for our convenience.. Deadline Your code and assignment report are due, as normal. If ou turn in our code late, our reviewers will have the choice of whether or not to review our code, so it is in our best interest to not take an late das for this assignment. Regardless of whether ou have submitted our assignment et or not, on Sunda night, ou will be assigned several projects to review. Turn in TreapMap.java and TreapTest.java and an other files necessar for our implementation. Be sure to comment an non-obvious portions of our assignment. Please do not use packages. When ou use a package, our Peer Reviewer, who onl has access to our btecode, cannot easil run our code. Make sure that our code compiles and that all classes have the correct names and are in the correct files. If ou do not submit our solution on time, ou will lose points for being late, and our reviewers get to choose whether or not the will review our solution. Since the reviews can help ou test our program to improve our final submission, it s in our best interest to submit our solution on time..2 Deadline 2 Your peer reviews are due. If ou turn these in late, our grade will be penalied. Not reall.

5 . Deadline Your revised code, testing report, and reviewer grades are due. You have the option to fi our code based on the feedback ou received, if ou have time. The testing report should cover all insight into the review process, what ou did, how ou did it, what ou learned, what ou learned from the reviews of our own code, what this revealed about our code, and what changes ou made/would make given more time. Acknowledgments. We thank Bobb Blumofe, now at Akamai, for the original version of this assignment, and Walter Chang for his subsequent modifications.