**Binary Trees and Prefix Free Codes**To explore the concept of optimal prefix-free codes using binary trees, we need to establish why a non-full binary tree cannot represent such a code optimally.**Definition:**A *full binary tree* is defined as a tree where every non-leaf node has exactly two children.**Problem Statement:**Demonstrate that if a binary tree is not full, it cannot correlate with an optimal prefix-free code. **Steps for Proof:**1. **Consider a Prefix-Free Code (C):** - Use a corresponding binary tree, T, that has at least one node with a single child.2. **Transformation of Tree (T to T'):** - Show that tree T can be modified into another binary tree, T′. - Tree T′ should be structured such that the new code, C′, derived from it, has a reduced average length compared to C. 3. **Reasoning:** - Demonstrate why C′, the code from T′, is more efficient than C. - A smaller average length indicates better codification efficiency, confirming that C was suboptimal.The key aspect of the proof involves transforming the binary tree to decrease the code's average length, proving that having nodes with single children does not lead to optimal coding efficiency. This demonstrates why a full binary tree must be used for optimal prefix-free codes.

We need to prove that a full binary tree that is not full cannot correspond to an optimal prefix…

Prove that a binary tree that is not full cannot correspond to an optimal prefix free code. Note: a full binary tree is a tree in which any non-leaf node has 2 children (see page 430). Your proof (which is very simple) should have the following structure: Consider a prefix-free code C, whose corresponding binary tree T has some node with only one child; show that you can transform T into another binary tree T', whose corresponding code C' has smaller average length and so is better than C (which means that C is not optimal). In your proof you need to indicate the transformation from T into T' and explain why the code C' is better than C.

Prove that a binary tree that is not full cannot correspond to an optimal prefix free code. Note: a full binary tree is a tree in which any non-leaf node has 2 children (see page 430). Your proof (which is very simple) should have the following structure: Consider a prefix-free code C, whose corresponding binary tree T has some node with only one child; show that you can transform T into another binary tree T', whose corresponding code C' has smaller average length and so is better than C (which means that C is not optimal). In your proof you need to indicate the transformation from T into T' and explain why the code C' is better than C.

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Similar questions

Match the term with each definition: In a binary tree, the left child (if any) of a node plus its descendants The number of nodes on the longest path from the root to a leaf A node on a path from a node to a leaf Set of nodes, T, such that either T is empty or T is partitioned into three disjoint sets (a single node r, called the root, and two (possibly empty) sets that are binary trees, called left and right subtrees of T). A binary tree such that for any given data item, x, in the tree, every node in the left subtree of x contains only items less than x. Every node in the right subtree of x contain only items greater than x. Nodes which have the same parent A node on the path from the root to a node A node directly "above" a node in the tree The node in a tree that has no parent If the node is the root, then the value is one. Otherwise, it's the value of one greater…
A binary tree is \emph{full} if every non-leaf node has exactly two children. For context, recall that we saw in lecture that a binary tree of height $h$ can have at most $2^{h+1}-1$ nodes and at most $2^h$ leaves, and that it achieves these maxima when it is \emph{complete}, meaning that it is full and all leaves are at the same distance from the root. Find $\nu(h)$, the \emph{minimum} number of leaves that a full binary tree of height $h$ can have, and prove your answer using ordinary induction on $h$. Note that tree of height of 0 is a single (leaf) node. \textit{Hint 1: try a few simple cases ($h = 0, 1, 2, 3, \dots$) and see if you can guess what $\nu(h)$ is.}
Prove that any binary tree of height h (where the empty tree is height 0, and a tree witha single node is height 1) has between h and 2h − 1 nodes, inclusive. A binary tree is onein which every node has at most three edges (at most one to the ’parent’ and two to the’children.’)
No node in a binary tree may produce more than two children. Exhibit that in any binary tree, the number of nodes producing two offspring is exactly one less than the number of leaves.
Problem 3 Recall from lecture that a binary tree T is either: 1. null (the empty tree); or 2. a root noder with two subtrees T₁ and TR, both of which are binary trees. Consider the following algorithm numLeaves (T) that takes as input a binary tree T and returns the number of leaves in the tree. numLeaves (T): 1: if T == null then return 0 2: 3: else 4: 5: 6: 7: 8: TL, TR the left and right subtrees of T, respectively if T₁ == null and TR == null then return 1 else return numLeaves (TL) + numLeaves (TR) Prove by structural induction that for every binary tree T, numLeaves (T) returns the number of leaves of T.