COS 126 Written Exam 2, Spring 2008
... 8. Theory (14 points). In the blanks, mark each of the statements below as true (T) or false (F). A. _____ Any RE with closure (a "*" or a "+") describes infinitely many strings. B. _____ Any RE without closure describes only finitely many strings. C. _____ Given unlimited memory, TOY can solve any ...
... 8. Theory (14 points). In the blanks, mark each of the statements below as true (T) or false (F). A. _____ Any RE with closure (a "*" or a "+") describes infinitely many strings. B. _____ Any RE without closure describes only finitely many strings. C. _____ Given unlimited memory, TOY can solve any ...
OrderedMap with a BST Data Structure - University of Arizona
... But how did we build this tree? ...
... But how did we build this tree? ...
Tree is a collection of nodes in which there is a root node and all
... operations take Θ(n) worst-case time. The expected height of a randomly built binary search tree is O(lg n),so that basic dynamic -set operations on such a tree take Θ(lg n) time on average. In practice, we can′t always guarantee that binary search trees are built randomly, but there are variations ...
... operations take Θ(n) worst-case time. The expected height of a randomly built binary search tree is O(lg n),so that basic dynamic -set operations on such a tree take Θ(lg n) time on average. In practice, we can′t always guarantee that binary search trees are built randomly, but there are variations ...
Lecture Note 1
... • Suppose k1, k2, …, kn are distinct keys, and that we have a collection T of n records of the form (k1, I1), (k2, I2), …, (kn, In), where Ij is information associated with key kj. Given a particular key value K, the search problem is to locate the record (kj, Ij) in T such that kj = K. ...
... • Suppose k1, k2, …, kn are distinct keys, and that we have a collection T of n records of the form (k1, I1), (k2, I2), …, (kn, In), where Ij is information associated with key kj. Given a particular key value K, the search problem is to locate the record (kj, Ij) in T such that kj = K. ...
Lecture 6 - UCSD CSE
... • We will look at AVL trees first, since the basic techniques there are used in many of the other approaches ...
... • We will look at AVL trees first, since the basic techniques there are used in many of the other approaches ...
Trees
... Siblings: two nodes having the same parent Depth of a node: number of ancestors (excluding the node itself) ...
... Siblings: two nodes having the same parent Depth of a node: number of ancestors (excluding the node itself) ...
Succinct tree representations
... For example, if we can partition the tree into n/k blocks, each of size k, then we can store it using (n/k) lg n + (n/k) k lg k = (n/k) lg n +n lg k bits. A careful two-level `tree covering’ method achieves a space bound of 2n+o(n) bits. ...
... For example, if we can partition the tree into n/k blocks, each of size k, then we can store it using (n/k) lg n + (n/k) k lg k = (n/k) lg n +n lg k bits. A careful two-level `tree covering’ method achieves a space bound of 2n+o(n) bits. ...
Lecture-search
... Use search operation to locate the insertion position or already existing item Use a parent point pointing to the parent of the node currently being examined as descending the tree ...
... Use search operation to locate the insertion position or already existing item Use a parent point pointing to the parent of the node currently being examined as descending the tree ...
Solutions
... that M logM (n) dominates lg (M ) logM (n). Note that in the average case, splits for any reasonably-sized B-tree are rare, so we can amortize the work of splitting over many operations. However, if we’re using a B-tree, it’s because what concerns us the most is the penalty of disk accesses. In that ...
... that M logM (n) dominates lg (M ) logM (n). Note that in the average case, splits for any reasonably-sized B-tree are rare, so we can amortize the work of splitting over many operations. However, if we’re using a B-tree, it’s because what concerns us the most is the penalty of disk accesses. In that ...
CS 61B Data Structures and Programming Methodology
... • What’s the running time for this binary tree? • The running times of find(), insert(), and remove() are all proportional to the depth of the last node encountered, but d = n – 1, so they all run in O(n) worst-case time. ...
... • What’s the running time for this binary tree? • The running times of find(), insert(), and remove() are all proportional to the depth of the last node encountered, but d = n – 1, so they all run in O(n) worst-case time. ...
Persistent Binary Search Trees
... Naive persistence Examples are making a copy of the whole tree after each update, or storing a history of update operations while only retaining the last version of the tree. In general these methods either waste space (first example), or have slow access time to older versions (second example). Par ...
... Naive persistence Examples are making a copy of the whole tree after each update, or storing a history of update operations while only retaining the last version of the tree. In general these methods either waste space (first example), or have slow access time to older versions (second example). Par ...
Data Structures
... Binary Tree A binary tree is a tree such that each node has at most two children. A binary search tree is a binary tree that has its data values arranged as follows: each node in the tree contains a value that is larger than each value in its left subtree and smaller than each value in its right su ...
... Binary Tree A binary tree is a tree such that each node has at most two children. A binary search tree is a binary tree that has its data values arranged as follows: each node in the tree contains a value that is larger than each value in its left subtree and smaller than each value in its right su ...
Transform-and-conquer
... Instance simplification - Presorting Solve a problem’s instance by transforming it into another simpler/easier instance of the same problem Presorting Many problems involving lists are easier when list is sorted. searching computing the median (selection problem) checking if all elements are ...
... Instance simplification - Presorting Solve a problem’s instance by transforming it into another simpler/easier instance of the same problem Presorting Many problems involving lists are easier when list is sorted. searching computing the median (selection problem) checking if all elements are ...
(6-up)
... concatenate together to give a suffix of s – all suffixes are represented by some path – the leaf of the path is labeled with the index of the first character of the suffix in s ...
... concatenate together to give a suffix of s – all suffixes are represented by some path – the leaf of the path is labeled with the index of the first character of the suffix in s ...
Proofs, Recursion and Analysis of Algorithms
... Because a tree is also a graph, representations for graphs in general can also be used for trees. Binary trees, however, have special characteristics that we want to capture in the representation, namely, the identity of the left and right child. The equivalent of an adjacency matrix is a twocolumn ...
... Because a tree is also a graph, representations for graphs in general can also be used for trees. Binary trees, however, have special characteristics that we want to capture in the representation, namely, the identity of the left and right child. The equivalent of an adjacency matrix is a twocolumn ...
Midterm 2 Exam Study Guide
... binary tree are NOT allowed)? Why did we need KeyedItem? (What possible violation of the ADT does this solve.) The BST Search Algorithm: be able to simulate it. Insertion and how this uses the search algorithm. Deletion: how it uses the search algorithm; handling the three cases. Quantitative facts ...
... binary tree are NOT allowed)? Why did we need KeyedItem? (What possible violation of the ADT does this solve.) The BST Search Algorithm: be able to simulate it. Insertion and how this uses the search algorithm. Deletion: how it uses the search algorithm; handling the three cases. Quantitative facts ...
tree
... children at every node is specified. If each node must have a specific number of children appearing in a specific order, then we have an M-ary tree The simplest type of M-ary tree is the binary tree ...
... children at every node is specified. If each node must have a specific number of children appearing in a specific order, then we have an M-ary tree The simplest type of M-ary tree is the binary tree ...
COMP 261 Lecture 14
... • For each node, split other nodes into – Subtree nodes – Non-subtree nodes • Whether there is an alternative path (edges in the graph but not in the tree) from each subtree node to the non-subtree nodes? ...
... • For each node, split other nodes into – Subtree nodes – Non-subtree nodes • Whether there is an alternative path (edges in the graph but not in the tree) from each subtree node to the non-subtree nodes? ...
Complete Binary Trees
... Binary trees contain nodes. Each node may have a left child and a right child. If you start from any node and move upward, you will eventually reach the root. Every node except the root has one parent. The root has no parent. Complete binary trees require the nodes to fill in each level from ...
... Binary trees contain nodes. Each node may have a left child and a right child. If you start from any node and move upward, you will eventually reach the root. Every node except the root has one parent. The root has no parent. Complete binary trees require the nodes to fill in each level from ...
Trees
... Lecture Notes 15 - Trees CSS 501 – Data Structures and Object-Oriented Programming – Professor Clark F. Olson Reading: ...
... Lecture Notes 15 - Trees CSS 501 – Data Structures and Object-Oriented Programming – Professor Clark F. Olson Reading: ...
Binary search tree
In computer science, binary search trees (BST), sometimes called ordered or sorted binary trees, are a particular type of containers: data structures that store ""items"" (such as numbers, names and etc.) in memory. They allow fast lookup, addition and removal of items, and can be used to implement either dynamic sets of items, or lookup tables that allow finding an item by its key (e.g., finding the phone number of a person by name).Binary search trees keep their keys in sorted order, so that lookup and other operations can use the principle of binary search: when looking for a key in a tree (or a place to insert a new key), they traverse the tree from root to leaf, making comparisons to keys stored in the nodes of the tree and deciding, based on the comparison, to continue searching in the left or right subtrees. On average, this means that each comparison allows the operations to skip about half of the tree, so that each lookup, insertion or deletion takes time proportional to the logarithm of the number of items stored in the tree. This is much better than the linear time required to find items by key in an (unsorted) array, but slower than the corresponding operations on hash tables.They are a special case of the more general B-tree with order equal to two.