Randomized Algorithms: Data Structures 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps Randomized Treaps Treap = TREe heAP Or, In Hebrew: ערמץ = ערמה עץ 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps Outline Definition of the Problem Deterministic Data Structures Introducing: Treaps Analyzing Performance Conclusion 11/15/2018 Randomized Algorithms - Treaps

Definition of the Problem Maintain a collection of items. Each of which is given a unique key. Support a group of 7 operations and queries over these items: 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps MakeSet(S) Find(k, S) Insert(k, S) Join(S1, k, S2) 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps Delete(k, S) Paste(S1, S2) Split(k, S) 11/15/2018 Randomized Algorithms - Treaps

Deterministic Data Structures Binary Search Trees A standard solution to the problem. For simplicity we will assume that the trees are endogenous, and thus, also full. 11/15/2018 Randomized Algorithms - Treaps

Binary Search Trees Performance Issues Each of the 7 aforementioned operations could be implemented in o(H). Ideally . Easy to devise an order of insertion in which . 11/15/2018 Randomized Algorithms - Treaps

Binary Search Trees Possible Deterministic Improvements Balancing the bst using rotations at each update operation. Drawback Could render significant overhead at each update operation. 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps Splaying: Move to the root a node accessed by find. Analysis: Guarantees an amortized time bound of o(log n). 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps Drawback: Restructuring of the tree at each find operation. No assurance that a specific action will take a certain amount of time. 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps Introducing: Treaps Consider a binary tree where each node v contains a pair of values (both should be distinct from other nodes) : a key – should suffice the binary search condition. a priority – should suffice the heap condition. 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps For Example: the set could be represented by the valid treap: 7,30 11,27 4,26 2,13 12,22 11/15/2018 Randomized Algorithms - Treaps

Treaps implementation (in brief) All operations are implemented in a similar way to the BST solutions available. When an action violates heap condition, use rotations. 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps How a rotation is made b a a C b A B A B C 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps Note, for future analyzing purposes, that all operations could be based upon find, insert and delete operations. 11/15/2018 Randomized Algorithms - Treaps

On choosing priorities Theorem: Let S= be any set of key-priority pairs such that the keys and the priorities are distinct. Then, there exist a unique treap T(S) for it. 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps Proof: By induction: for n=0, n=1: the theorem holds. 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps for n>1: without the loss of generality, assume that V = (k1,p1) is the node with the highest priority: Assign V to root. recursively apply on both sub trees emerging from splitting by p1. 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps The theorem implies that we may choose any (distinct) priorities to our nodes, ensuring that are we are representing a valid treap. 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps We will choose our priorities in the following manner: Upon insertion of a new key, a priority is randomly and uniformly chosen for it, in such a way that no previous node contains this priority. The priority for a specific element remains constant. The ordering of keys and values is completely uncorrelated. 11/15/2018 Randomized Algorithms - Treaps

Analyzing Performance Mulmuley Games For the sake of time performance, we will present a set of games, called Mulmuley Games. 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps Mulmuley Games Game A Participants: A set of players. Each assigned a distinct numerical key. A set of bystanders. Each assigned a distinct numerical key. 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps The course of the game consists of repeatedly sampling from P union B. Without replacement, until the pool is empty. A random variable V is defined as the number of samples in which a player p is chosen such that p is the largest player chosen to this point. The value of the game Ap is E[V]. 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps Lemma: for all p > 0, Ap = Hp. 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps Mulmuley Games Game D Participants: A set of players. Each assigned a distinct numerical key. A set of bystanders. Each assigned a distinct numerical key. A set of triggers. Each assigned a distinct numerical key. 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps The course of the game consists of repeatedly sampling from P union B union T. Without replacement, until the pool is empty. A random variable V is defined as the number of samples in which a player p is chosen such that p is the largest player chosen to this point. Only after a trigger was chosen. The value of the game is E[V]. 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps Lemma: 11/15/2018 Randomized Algorithms - Treaps

Analyzing Performance We will present and prove several theorems, regarding time bounds of operations on random treaps. For the following, assume that T is a random treap for a set S of size n. 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps Theorem: The expected time for find, insert and delete operations on T is O(log n). 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps Lemma: for an element of rank k (regarding its key), it holds that: The lemma implies that the depth of each element is o(log n) each of the operations ( find, insert and delete) are linear in reagards to the tree depth, it follows that they also are o(log n). 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps Proof of the lemma: let us define: 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps - Ancestors of x 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps it is clear that : depth(x) = And thus E[depth(x) ] = Therfore, we will try and estimate and 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps Let us write S= ordered by their priority. (hence p1 > p2 > … > pn) We will insert the elements into the tree by this order (notice that no rotations are needed) When will an element belong to S- ? 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps In other words, we will be counting right-going arcs in the path between the root and x. Inserting elements larger than x, doesn’t affect the counting An element will belong to q- if it is the largest element of s- inserted yet. 7,30 4,26 11,27 2,23 5,20 12,22 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps Notice that this resembles Game A with P = S- , B = S\S- and thus p = k, following that Similarly it can be shown that Therefore 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps Theorem: The expected number of rotations required during an insert or delete operation is at most 2. We will prove that for each element x defined as before, it holds that the number of rotations is 2. 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps Proof: Let us define: Rx – the right spine of the left subtree of X. Lx – the left spine of the right subtree of X. Observation 1: the number of rotations needed during these operations is |Rx| + |Lx| 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps In a similar to the one taken in the previous theorem, we can show that E[|Rx|] is in fact similar to Game D with p = k-1 , t = 1 , b = n – k Which yeilds that E[|Rx|] = 1 – 1/k. It can also be shown, by the same method that E[|Lx|] = 1 – 1/(n-k+1). Therefore E[|Lx|] + E[|Rx|] <= 2 11/15/2018 Randomized Algorithms - Treaps

Randomized Algorithms - Treaps Theorem: The expected time for join, paste and split involving sets S1 and S2 of sizes n and m respectively, is o(log n + log m). 11/15/2018 Randomized Algorithms - Treaps