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Convert the following infix expressions to postfix notations: a. x * (y + z) - ( w + t) b. (x + y) / (z - w) * t c. ((x - y) + (z - w) / t) * u d. x - y / (z + w) * t / u + (v - s)

Short Answer

Expert verified
a: 'x y z + * w t + -'; b: 'x y + z w - / t *'; c: 'x y - z w - t / + u *'; d: 'x y z w + / t * u / - v s - +'.

Step by step solution

01

Understanding Infix to Postfix

Infix notation is the common arithmetic and logical formula notation, where operators are placed between operands. Postfix notation, also known as Reverse Polish Notation (RPN), requires operators to follow their operands. The conversion involves understanding operator precedence and associativity rules, as well as using a stack to store operators when rearranging them according to their precedence.
02

Convert Expression a: x * (y + z) - ( w + t)

1. Process 'x * (y + z)': Push 'x' to the output. Push '*' to the stack. Handle '(y + z)' by processing inside parentheses. Enqueue 'y' and 'z', perform 'y z +' and output the result. Pop '*' and perform 'x y z + *'. 2. Move to '- ( w + t)': Push '-' to the stack temporarily since operators have the same precedence. Process '(w + t)' and output 'w t +'. Finish by outputting '-'. Result: The postfix notation is 'x y z + * w t + -'.
03

Convert Expression b: (x + y) / (z - w) * t

1. Inside parentheses '(x + y)': Process 'x', 'y', and output 'x y +'. 2. Divide by '(z - w)': Process 'z', 'w', and output 'z w -'. 3. Multiply by 't': Push '/' and '*' based on precedence and associativity. 4. Compose 'x y +' and 'z w -' result, divide result, then multiply by 't'. Result: The postfix notation is 'x y + z w - / t *'.
04

Convert Expression c: ((x - y) + (z - w) / t) * u

1. Innermost '(x - y)': Output 'x y -'. 2. Handle '(z - w) / t': Output 'z w - t /'. 3. Add result of previous expressions: Combine 'x y -' and 'z w - t /' to form '(x y - z w - t /) +' resulting in 'x y - z w - t / +'. 4. Finally, multiply by 'u': Move '/ *' operation by operator precedence rules. Result: The postfix notation is 'x y - z w - t / + u *'.
05

Convert Expression d: x - y / (z + w) * t / u + (v - s)

1. Evaluate 'y / (z + w)': Start with 'z', 'w', and output 'z w +', then 'y z w + /'. 2. Multiply by 't': Combine with 't' resulting 'y z w + / t *'. 3. Divide by 'u': Result 'y z w + / t * u /'. 4. Subtract result from 'x': Precedence resolves to 'x y z w + / t * u / -'. 5. Add '(v - s)': Evaluate and append 'v s -' Final expression: 'x y z w + / t * u / - v s - +'.

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Key Concepts

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

Infix to Postfix Conversion
Infix notation is the standard form used in arithmetic expressions where operators sit between the operands, such as in the expression "a + b". However, computers find it easier to understand postfix notation, where operands precede their operators, like "a b +".
To convert an infix expression to postfix, one must account for operator precedence and associativity, which dictate the order in which operations should be performed. Moreover, a stack data structure comes in handy during this conversion process for temporarily holding operators and ensuring they are applied in the correct order.
The process involves scanning the infix expression from left to right and using a stack to ensure operators are appended to the postfix expression only when necessary. This conversion ensures that the resulting postfix expression can be evaluated without needing parentheses to alter precedence.
Operator Precedence
In arithmetic expressions, operators have a hierarchy that dictates their evaluation order, known as operator precedence. For example, in the expression "3 + 5 * 2", the multiplication operator "*" takes precedence over the addition operator "+".
Operator precedence must be considered when converting infix expressions to postfix. If two operators share the same precedence, their associativity rules, either left-to-right or right-to-left, determine the order. For instance, both addition and subtraction share the same precedence and are left-associative, meaning they are evaluated from left to right.
During conversion, when an operator is encountered, the algorithm compares it with the operator on top of the stack. If the current operator has lower precedence than the operator on the stack, the stack operator is added to the postfix expression. Subtle scenarios like these are crucial for achieving the correct postfix notation.
Stack Data Structure
A stack is an essential data structure in the conversion from infix to postfix notation. It follows the Last In, First Out (LIFO) principle, where the last element added is the first one removed. This characteristic makes stacks incredibly useful for managing operators during conversion.
When processing an infix expression, operands are immediately added to the postfix output, while operators are pushed onto the stack. If a new operator with lower or equal precedence than the operator on the stack arises, operators from the stack are popped and added to the postfix expression until the stack is cleared or a lower precedence level is reached.
Parentheses in expressions are utilized to enforce order: for an opening parenthesis "(", push onto the stack, and ""); " signifies popping the stack until its matching "(" is removed.
Utilizing a stack ensures that all operators are applied in the correct order, maintaining the integrity of the mathematical expression during conversion.
Reverse Polish Notation
Reverse Polish Notation (RPN), also known as postfix notation, is a mathematical notation system where operators follow their operands. It eliminates the need for parentheses to denote operation order since the placement uniquely determines it.
RPN is advantageous because it aligns closely with the way stack-based machines evaluate expressions, thus facilitating computational efficiency. Unlike infix notation, where extra syntax (parentheses) is needed to express priority, RPN relies solely on the sequence of operations.
To evaluate an expression in postfix form, you scan it from left to right. When an operand is found, it is pushed onto a stack. When an operator is encountered, the necessary operands are popped from the stack, the operation is performed, and the result is pushed back onto the stack.
Postfix notation greatly simplifies parsing and evaluating arithmetic expressions, particularly for interpreters and compilers. Its efficiency lies in its simplicity and unambiguous nature, making it an excellent choice for calculations in computational systems.

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Most popular questions from this chapter

Add the operation queuecount to the class queueType (the array implementation of queues), which returns the number of elements in the queue. Write the definition of the function template to implement this operation.

Consider the following statements: linkedStackType stack; linkedQueueType queue; int num; Suppose the input is: 48 35 72 88 92 11 10 15 44 52 67 36 Show what is written by the following segment of code: stack.push(0); queue.addQueue(0); cin >> num; while (cin) { switch (num % 3) { case 0: stack.push(num); break; case 1: queue.addQueue(num); break; case 2: if (!stack.isEmptyStack()) { cout << stack.top() << " "; stack.pop(); } else if (!queue.isEmptyQueue()) { cout << queue.front() << " "; queue.deleteQueue(); } else cout << "Stack and queue are empty." << endl; break; } //end switch cin >> num; } //end while cout << endl; cout << "stack: "; while (!stack.isEmptyStack()) { cout << stack.top() << " "; stack.pop(); } cout << endl; cout << "queue: "; while (!queue.isEmptyQueue()) { cout << queue.front() << " "; queue.deleteQueue(); } cout << endl;

Write a function template, reverseQueue, that takes as a parameter a queue object and uses a stack object to reverse the elements of the queue.

Consider the following statements: stackType stack(50); int num; Suppose that the input is: 31 47 86 39 62 71 15 63 Show what is output by the following segment of code: cin >> num; while (cin) { if (!stack.isFullStack()) { if (num % 2 == 0 || num % 3 == 0) stack.push(num); else if (!stack.isEmptyStack()) { cout << stack.top() << " "; stack.pop(); } else stack.push(num + 3); } cin >> num; } cout << endl; cout << "Stack Elements: "; while (!stack.isEmptyStack()) { cout << " " << stack.top(); stack.pop(); } cout << endl;

Describe the two basic operations on a stack.

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