Logical problems in programming are those that primarily involve decision-making based on a set of conditions and rules. Solving them requires a systematic approach to ensure all possible scenarios are handled correctly. The algorithm for a logical problem is essentially a structured map of these decisions.

Systematic Steps for Algorithm Design

A reliable method for designing any algorithm, especially for logical problems, involves four key steps:

1.   Understand the Problem: Clearly define the inputs, the required outputs, and the rules or conditions that connect them.

2.   Devise a Plan: Outline the sequence of logical checks needed to get from the input to the output.

3.   Formulate the Algorithm: Translate the plan into a precise, step-by-step format, such as pseudocode.

4.   Test and Refine: Trace the algorithm with sample data to verify its correctness for different scenarios.

1. Basic Example: Voter Eligibility ✅

·         Problem Statement: Determine if a person is eligible to vote based on their age. The voting age is 18.

·         Step 1: Understand the Problem

o    Input: A person's age (an integer).

o    Output: A message: "Eligible to Vote" or "Not Eligible to Vote".

o    Rule: If age is 18 or greater, the person is eligible.

·         Step 2: Devise a Plan The plan involves a single conditional check. We will compare the input age to the number 18. If it's greater than or equal to 18, we select one outcome; otherwise, we select the other.

·         Step 3: Formulate the Algorithm (Pseudocode)

START

  READ age

  IF age >= 18 THEN

    PRINT "Eligible to Vote"

  ELSE

    PRINT "Not Eligible to Vote"

  ENDIF

END

·         Step 4: Test and Refine

o    If age = 25, the condition 25 >= 18 is true, so the output is "Eligible to Vote". Correct.

o    If age = 17, the condition 17 >= 18 is false, so the output is "Not Eligible to Vote". Correct.

2. Moderate Example: Assigning a Student's Grade 📜

·         Problem Statement: Assign a letter grade (A, B, C, D, F) based on a student's score.

·         Step 1: Understand the Problem

o    Input: A student's score (e.g., from 0 to 100).

o    Output: A single letter grade.

o    Rules:

§  A: 90-100

§  B: 80-89

§  C: 70-79

§  D: 60-69

§  F: Below 60

·         Step 2: Devise a Plan We need a chain of conditional checks. We'll start from the highest grade and move down. We check if the score meets the 'A' grade criteria. If not, we then check for 'B', and so on. This structure is known as an if-else-if ladder.

·         Step 3: Formulate the Algorithm (Pseudocode)

START

  READ score

  IF score >= 90 THEN

    PRINT "Grade is A"

  ELSE IF score >= 80 THEN

    PRINT "Grade is B"

  ELSE IF score >= 70 THEN

    PRINT "Grade is C"

  ELSE IF score >= 60 THEN

    PRINT "Grade is D"

  ELSE

    PRINT "Grade is F"

  ENDIF

END

·         Step 4: Test and Refine

o    If score = 85, the first condition (85 >= 90) is false. The second condition (85 >= 80) is true, so the output is "Grade is B". The algorithm stops checking further. Correct.

o    If score = 50, all IF and ELSE IF conditions are false, leading to the final ELSE block. The output is "Grade is F". Correct.

3. Advanced Example: The FizzBuzz Problem

·         Problem Statement: For a given number, print "Fizz" if it's divisible by 3, "Buzz" if it's divisible by 5, and "FizzBuzz" if it's divisible by both 3 and 5. Otherwise, print the number itself.

·         Step 1: Understand the Problem

o    Input: An integer number.

o    Output: "FizzBuzz", "Fizz", "Buzz", or the original number.

o    Rules: The core logic relies on divisibility, checked using the modulo operator (%).

·         Step 2: Devise a Plan The order of checks is critical. The most specific condition (divisible by both 3 and 5) must be checked first. If we check for divisibility by 3 first, a number like 15 would incorrectly result in "Fizz" and the algorithm would stop, never reaching the "FizzBuzz" check.

·         Step 3: Formulate the Algorithm (Pseudocode)

START

  READ number

  // Check the most specific condition first

  IF (number % 3 == 0) AND (number % 5 == 0) THEN

    PRINT "FizzBuzz"

  // Check the next conditions

  ELSE IF (number % 3 == 0) THEN

    PRINT "Fizz"

  ELSE IF (number % 5 == 0) THEN

    PRINT "Buzz"

  // The default case if no other conditions are met

  ELSE

    PRINT number

  ENDIF

END

·         Step 4: Test and Refine

o    If number = 15, 15 % 3 == 0 AND 15 % 5 == 0 is true. Output: "FizzBuzz". Correct.

o    If number = 9, the first condition is false. The second (9 % 3 == 0) is true. Output: "Fizz". Correct.

o    If number = 10, the first two conditions are false. The third (10 % 5 == 0) is true. Output: "Buzz". Correct.

o    If number = 7, all conditions are false. The ELSE block is executed. Output: 7. Correct.