Remove Brackets From An Algebraic Expression In C Program

Removing brackets from an algebraic expression is a common string manipulation task in programming. This operation can simplify expressions for display or prepare them for further processing. In this article, you will learn how to implement a C program to effectively remove all types of brackets from a given algebraic expression.

Problem Statement

The core problem is to take an algebraic expression, which may contain various types of brackets (parentheses (), square brackets [], curly braces {}), and produce a new expression where all these brackets have been removed. This is useful for standardizing expressions or when the structural meaning of brackets is no longer required.

Example

Consider the algebraic expression: a+(b-[c*d])/{e-f}. After removing all brackets, the desired output would be: a+b-c*d/e-f.

Background & Knowledge Prerequisites

To understand and implement the solution, a basic understanding of C programming concepts is beneficial:

• Variables and Data Types: Especially char and char arrays (strings).
• Loops: while or for loops for iterating through strings.
• Conditional Statements: if statements for character checking.
• Functions: Defining and calling custom functions.
• String Manipulation: Concepts like null-termination (\0) and character-by-character processing.

Use Cases or Case Studies

Removing brackets from algebraic expressions has several practical applications:

• Simplifying Display: Presenting expressions in a more concise format for users who don't need bracket information.
• Preprocessing for Parsers: As an initial step before feeding an expression to a parser that might have a different way of handling operator precedence or a specific format requirement.
• Data Cleaning: Removing unwanted characters from user input to normalize expressions.
• Text Analysis: When analyzing text containing mathematical or logical expressions, brackets might be considered noise for certain analyses.
• Educational Tools: Creating tools for students to visualize simplified expressions without structural grouping.

Solution Approaches

For the problem of removing brackets, a straightforward and efficient approach involves iterating through the expression character by character and selectively copying only the non-bracket characters to a new string.

Iterative String Traversal

This approach involves traversing the input string from beginning to end. For each character encountered, it checks if the character is one of the bracket types. If it's not a bracket, the character is copied to a new string; otherwise, it's skipped.

One-line summary: Iterate through the source string, copy non-bracket characters to a destination string, and ensure null-termination.

Code example:

// Remove Brackets from Algebraic Expression
#include <stdio.h>
#include <string.h> // Required for strlen, though not strictly used in this solution.

// Function to remove brackets from a string
// Takes the source string and a destination buffer
void removeBrackets(char *source, char *destination) {
    int i = 0; // Index for source string
    int j = 0; // Index for destination string

    // Iterate through the source string until the null terminator is found
    while (source[i] != '\\0') {
        // Check if the current character is NOT any type of bracket
        if (source[i] != '(' && source[i] != ')' &&
            source[i] != '[' && source[i] != ']' &&
            source[i] != '{' && source[i] != '}') {
            // If it's not a bracket, copy it to the destination string
            destination[j] = source[i];
            j++; // Move to the next position in the destination string
        }
        i++; // Always move to the next character in the source string
    }
    // Null-terminate the destination string to make it a valid C string
    destination[j] = '\\0';
}

int main() {
    // Define an example algebraic expression with various brackets
    char expression1[] = "a+(b-[c*d])/{e-f}";
    char result1[100]; // Buffer to store the result, ensuring enough space

    // Call the function to remove brackets
    removeBrackets(expression1, result1);

    // Print the original and modified expressions
    printf("Original expression: %s\\n", expression1);
    printf("Expression without brackets: %s\\n\\n", result1);

    char expression2[] = "((x+y)*z)+{p-q}";
    char result2[100];
    removeBrackets(expression2, result2);
    printf("Original expression: %s\\n", expression2);
    printf("Expression without brackets: %s\\n\\n", result2);

    char expression3[] = "simple+expression*no-brackets";
    char result3[100];
    removeBrackets(expression3, result3);
    printf("Original expression: %s\\n", expression3);
    printf("Expression without brackets: %s\\n", result3);

    return 0;
}

Run

Sample output:

Original expression: a+(b-[c*d])/{e-f}
Expression without brackets: a+b-c*d/e-f

Original expression: ((x+y)*z)+{p-q}
Expression without brackets: xy*z+p-q

Original expression: simple+expression*no-brackets
Expression without brackets: simple+expression*no-brackets

Stepwise explanation for clarity:

1. Include Headers: We include stdio.h for input/output operations and string.h (though strlen isn't explicitly used, it's common for string manipulation).
2. removeBrackets Function:
   • It takes two char * arguments: source (the input string) and destination (a buffer where the result will be stored).

1. main Function:
   • Declares a char array expression1 initialized with the sample algebraic string.

Conclusion

Removing brackets from an algebraic expression is a fundamental string manipulation task in C programming. The iterative string traversal method provides a clear, efficient, and robust way to achieve this by simply copying non-bracket characters to a new string. This technique is valuable for simplifying expressions, preparing data for parsing, or cleaning input.

Summary

• The goal is to eliminate (), [], and {} from an algebraic expression.
• A common approach involves iterating through the input string character by character.
• Only characters that are not brackets are copied to a new string.
• Null-termination (\0) is essential to mark the end of the new string.
• This method is efficient and creates a clean, bracket-free expression.