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Represent the expression as a binary tree and write the prefix and postfix forms of the expression. $$ (A+B) *(C-D) $$

Short Answer

Expert verified
The prefix form of the expression \((A+B) *(C-D)\) when represented as a binary tree is '*+AB-CD', and the postfix form is 'AB+CD-*'

Step by step solution

01

Understanding the expression

The given expression is \((A+B) *(C-D)\). According to the orders of operation in arithmetic, addition/subtraction gets done last. So, '+' and '-' would become children nodes of '*'. The binary tree will have three levels, with '*' at the root, '+' and '-' at the second level, and the individual letters at the third.
02

Constructing the binary tree

Begin by placing the '*' at the root of the tree. Place '+' and '-' on the second level, '+' to the left of '*' and '-' to the right. Afterwards, place 'A' and 'B' as left and right children of '+', respectively. Similarly, place 'C' and 'D' as left and right children of '-', respectively. Now, the binary tree for the expression \((A+B) *(C-D)\) has been created.
03

Writing the Prefix notation

Prefix notation involves writing down the root, then the left and right subtree. In prefix notation, the operator comes before its operands. Therefore, the prefix form of the expression \((A+B) *(C-D)\) would be '*+AB-CD'.
04

Writing the Postfix notation

Postfix notation, the operators are written after their operands. Therefore, the operator '*' would come at the end. The postfix form of \((A+B) *(C-D)\) would be 'AB+CD-*'.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expression Trees
An expression tree is a binary tree representation of an arithmetic expression. It's particularly useful for visually understanding and evaluating expressions. In our example, the expression \((A+B)*(C-D)\) transforms into a binary tree composed of nodes and edges.

In this tree, each node represents either an operator (like '*', '+', or '-') or an operand (like 'A', 'B', 'C', or 'D'). The operators are typically non-leaf nodes, while operands are leaf nodes.
  • Root Node: This is the highest node in the tree, representing the main operation, which is multiplication (*) in our case.
  • Sub-nodes or Children: The other operations, addition (+) and subtraction (-), become sub-nodes to the multiplication operator.
  • Leaf Nodes: These are the operands, situated at the end of the branches, consisting of 'A', 'B', 'C', and 'D'.
By creating an expression tree, one effectively outlines the hierarchical structure of operations dictated by the original expression.
Prefix Notation
Prefix notation, also known as Polish notation, rearranges an arithmetic expression such that each operator precedes its operands. This style of notation eliminates the need for parentheses and clarifies the order of operations.

To transform an expression like \((A+B)*(C-D)\) into prefix notation, you follow these steps:

1. Start from the root of the expression tree, which is the multiplication (*).
2. Recursively visit each left subtree before writing down the operator and operands.
  • So, from the root '*', move to the left child '+' and then to its children 'A' and 'B'.
  • The prefix form writes the operator '+', followed by operands 'A' and 'B'.
  • Move to the right subtree with '-' and do the same for 'C' and 'D'.
Thus, the prefix form becomes '*+AB-CD', reflecting the deep hierarchical structure of your operations. This systematic approach translates the operation flow from top-down in the tree to a linear prefix expression.
Postfix Notation
Postfix notation, or Reverse Polish Notation (RPN), places every operator after its operands. This method efficiently deals with ambiguous expressions without the need for parentheses.

After constructing the expression tree for \((A+B)*(C-D)\), you can determine its postfix notation by reading the tree nodes in a specific sequence:

1. Start from the leftmost leaf node and proceed left to right.
2. Handle operators after traversing their operands.
  • In our tree, collect the operands 'A' and 'B', then write their operator '+'.
  • Next, collect 'C' and 'D', then write their operator '-'.
  • Finally, write the root operator '*' at the end.
This traversal results in 'AB+CD-*'. Postfix notation simplifies both the evaluation of expressions and certain kinds of arithmetic processing acting directly on this output sequence.
Arithmetic Operations
Arithmetic operations form the backbone of mathematical expressions and are composed of operators and operands. In our discussion, the primary operations are addition, subtraction, and multiplication.

Here, the expression \((A+B)*(C-D)\) can be broken down into:
  • Addition ('+'): The operation here is among the operands 'A' and 'B'.
  • Subtraction ('-'): Within its brackets, this operation occurs between 'C' and 'D'.
  • Multiplication ('*'): This global operator combines the results of the two internal operations.
Each of these operations follows specific precedence rules, dictating the sequence in which they should be performed. Multiplication has the highest precedence, making it the root in our binary expression tree. The association of these operations in binary trees highlights their order, crucial for accurate computation.

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Most popular questions from this chapter

What can you say about a vertex in a rooted tree that has no ancestors?

What is wrong with the following "proof" that the greedy algorithm is optimal for all amounts of postage for the denominations \(1,5,\) and \(6 ?\) We will prove that for all \(i \geq 1,\) the greedy algorithm is optimal for all amounts of postage \(n \leq 6 i\). The Basis Step is \(i=1,\) which is true by inspection. For the Inductive Step, assume that the greedy algorithm is optimal for all amounts of postage \(n \leq 6 i\). We must show that the greedy algorithm is optimal for all amounts of postage \(n \leq 6(i+1)\). We may assume that \(n>6 i\). Now \(n-6 \leq 6 i\), so by the inductive assumption, the greedy algorithm is optimal for \(n-6 .\) Suppose that the greedy algorithm chooses \(k\) stamps for \(n-6 .\) In determining the postage for the amount \(n\), the greedy algorithm will first choose a 6 -cent stamp and then stamps for \(n-6\) for a total of \(k+1\) stamps. These \(k+1\) stamps must be optimal or otherwise \(n-6\) would use less than \(k\) stamps. The Inductive Step is complete.

Refer to tournament sort. Tournament Sort. We are given a sequence \(s_{1}, \ldots, s_{2^{k}}\) to sort in nondecreasing order. We will build a binary tree with terminal vertices labeled \(s_{1}, \ldots, s_{2^{k}} .\) An example is shown. Working left to right, create a parent for each pair and label it with the maximum of the children. Continue in this way until you reach the root. At this point, the largest value, \(m\), has been found. To find the second-largest value, first pick a value vless than all the items in the sequence. Replace the terminal vertex w containing \(m\) with \(v\). Relabel the vertices by following the path from w to the root, as shown. At this point, the secondlargest value is found. Continue until the sequence is ordered. Show that any algorithm that finds the largest value among \(n\) items requires at least \(n-1\) comparisons.

Refer to the following situation. Suppose that we have stamps of various denominations and that we want to choose the minimum number of stamps to make a given amount of postage. Consider a greedy algorithm that selects stamps by choosing as many of the largest denomination as possible, then as many of the second-largest denomination as possible, and so on. Find positive integers \(a_{1}\) and \(a_{2}\) such that \(a_{1}>2 a_{2}>1, a_{2}\) does not divide \(a_{1},\) and the algorithm, with available denominations \(1, a_{1}, a_{2},\) does not always produce the fewest number of stamps to make a given amount of postage.

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