Lab 3 – Active Logic & Control Structures: If Statement

PSUC Lab 3 — Conditional Branching

Active Logic: Decision Control Structures

Master simple if, two-way if-else, multi-way if-else-if ladders, relational operators, and logical operators (&&, ||, !).

C Branching Constructs & Operators

Construct / Operator	Syntax	Key Rule & Trap
Equality Operator	`a == b`	Assignment `a = b` is a bug! Use `==`
Logical AND	`cond1 && cond2`	Both conditions MUST be true
Logical OR	`cond1 \|\| cond2`	At least ONE condition MUST be true
If-Else Ladder	`if ... else if ... else`	First true branch executes; rest skipped
Nested If	`if (c1) { if (c2) ... }`	Evaluates inner condition only if outer is true

🎯 Learning Philosophy: Active Logic Building

Test your decision-making logic across 3 Interactive Learning Modes:

🧠 Tab 1 (Logic & Hints): Decision rules & truth table breakdowns.
🧩 Tab 2 (Logic Puzzle): Fill in missing boolean operators (==, &&, ||) & run live testers.
💻 Tab 3 (C Solution): Reference verified C code solutions.

Use the mode tabs on each slide to solve the branching puzzle before peeking at code!

Program 1 of 6

Odd / Even Check & Multiples Verification

Determine if a number is even (num % 2 == 0) or odd, and verify if integer A is a multiple of integer B (a % b == 0).

Modulus Branching Logic

Even Condition: n % 2 == 0 (Remainder when divided by 2 is 0)
Odd Condition: n % 2 != 0 (Remainder is 1 or -1)
Multiple Condition: a % b == 0 (No remainder when A is divided by B)

Interactive Logic Hints

💡 Hint 1: Why must we write == instead of = in if()? ▼

In C, if (n % 2 = 0) causes a compilation syntax error! = is assignment, == is comparison!

💡 Hint 2: What happens if B = 0 in multiples check? ▼

Division by zero error! Always ensure divisor B is non-zero before evaluating a % b!

🧩 Complete the Even/Odd & Multiples Logic

int num = 24, a = 30, b = 5;
if (num % 2 0) {
printf("EVEN");
}
if (a b 0) {
printf("A is multiple of B");
}

Select operators above and click Check!

🔢 Live Even/Odd & Divisibility Tester

Number N:

Divisor M:

Result will show here...

even_multiples.cReference C Code

#include <stdio.h>

int main() {
    int num, a, b;
    
    printf("Enter number: ");
    scanf("%d", &num);
    
    if (num % 2 == 0) {
        printf("%d is EVEN\n", num);
    } else {
        printf("%d is ODD\n", num);
    }
    
    printf("Enter A and B: ");
    scanf("%d %d", &a, &b);
    
    if (b != 0 && a % b == 0) {
        printf("%d IS a multiple of %d\n", a, b);
    } else {
        printf("%d is NOT a multiple of %d\n", a, b);
    }
    return 0;
}

Program 2 of 6

Leap Year Detection (The Century Rule)

A year is a leap year if it is divisible by 400 OR (divisible by 4 AND NOT divisible by 100).

Leap Year Decision Logic

Rule 1 (Century Years): Year % 400 == 0 (e.g. 2000 is leap)
Rule 2 (Regular Years): Year % 4 == 0 AND Year % 100 != 0 (e.g. 2024 is leap, 1900 is NOT)
Combined: (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0)

Interactive Logic Hints

💡 Hint 1: Why is 1900 NOT a leap year even though 1900 % 4 == 0? ▼

Because century years (ending in 00) must be divisible by 400! 1900 % 100 == 0, but 1900 % 400 != 0.

🧩 Complete the Leap Year Compound Boolean Logic

if ((year % == 0) (year % 4 == 0 year % 100 0)) {
printf("Leap Year");
}

Select operators above and click Check!

📅 Leap Year Rule Evaluator

Enter Year:

Result will show here...

leap_year.cReference C Code

#include <stdio.h>

int main() {
    int year;
    printf("Enter year: ");
    scanf("%d", &year);
    
    if ((year % 400 == 0) || (year % 4 == 0 && year % 100 != 0)) {
        printf("%d IS a Leap Year (366 days)\n", year);
    } else {
        printf("%d is NOT a Leap Year (365 days)\n", year);
    }
    return 0;
}

Program 3 of 6

Roots of a Quadratic Equation (ax² + bx + c = 0)

Classifies discriminant D = b² - 4ac using multi-way if-else-if ladder (D > 0, D == 0, D < 0).

Discriminant Conditions

disc > 0: Real & Distinct roots: (-b ± √D) / (2a)
disc == 0: Real & Equal roots: -b / (2a)
disc < 0: Complex Imaginary roots: -b/(2a) ± i √(-D)/(2a)

Interactive Logic Hints

💡 Hint 1: What happens if a = 0? ▼

If a = 0, division by 2*a causes undefined division by zero! (It becomes a linear equation, not quadratic).

🧩 Complete Discriminant Ladder Conditions

disc = b*b - 4*a*c;
if (disc 0) {
  printf("Real & Distinct");
} else if (disc 0) {
  printf("Real & Equal");
} else {
  printf("Imaginary");
}

Select operators above and click Check!

🧮 Interactive Quadratic Solver

a:

b:

c:

Result will show here...

quadratic_roots.cReference C Code

#include <stdio.h>
#include <math.h>

int main() {
    double a, b, c, disc, r1, r2, realPart, imagPart;
    printf("Enter coefficients a, b, c: ");
    scanf("%lf %lf %lf", &a, &b, &c);

    disc = b*b - 4*a*c;

    if (disc > 0) {
        r1 = (-b + sqrt(disc)) / (2*a);
        r2 = (-b - sqrt(disc)) / (2*a);
        printf("Real & Distinct: %.2lf, %.2lf\n", r1, r2);
    } else if (disc == 0) {
        r1 = -b / (2*a);
        printf("Real & Equal: %.2lf\n", r1);
    } else {
        realPart = -b / (2*a);
        imagPart = sqrt(-disc) / (2*a);
        printf("Imaginary: %.2lf ± %.2lfi\n", realPart, imagPart);
    }
    return 0;
}

Program 4 of 6

Grades Ladder & Triangle Inequality Validation

Assigns letter grades using if-else-if ladder and validates triangle inequality (a + b > c && a + c > b && b + c > a).

Triangle Inequality Theorem

The sum of any two sides MUST be strictly greater than the third side!
Condition: (a + b > c) && (a + c > b) && (b + c > a)
If valid, semi-perimeter s = (a+b+c)/2.0 & Heron's Area sqrt(s*(s-a)*(s-b)*(s-c))

Interactive Logic Hints

💡 Hint 1: Why must we use && instead of comma or separate ifs? ▼

All 3 side conditions MUST hold simultaneously for a valid 2D triangle. Logical AND && enforces all 3!

🧩 Complete Triangle Inequality & Grade Logic

if (marks >= 90) printf("A+");
else if (marks >= 80) printf("A");

if (a + b c a + c > b && b + c > a) {
s = (a + b + c) / 2.0;
area = sqrt(s * (s-a) * (s-b) * (s-c));
}

Select operators above and click Check!

📊 Grade & Triangle Tester

Marks (0-100):

Side a:

Side b:

Side c:

Result will appear here...

grade_triangle.cReference C Code

#include <stdio.h>
#include <math.h>

int main() {
    float marks, a, b, c, s, area;
    
    printf("Enter marks: ");
    scanf("%f", &marks);
    if (marks >= 90)      printf("Grade: A+\n");
    else if (marks >= 80) printf("Grade: A\n");
    else if (marks >= 70) printf("Grade: B\n");
    else if (marks >= 60) printf("Grade: C\n");
    else                  printf("Grade: F\n");

    printf("Enter sides a b c: ");
    scanf("%f %f %f", &a, &b, &c);
    if (a+b > c && a+c > b && b+c > a) {
        s = (a + b + c) / 2.0;
        area = sqrt(s * (s-a) * (s-b) * (s-c));
        printf("Valid Triangle! Area = %.2f\n", area);
    } else {
        printf("Invalid Triangle!\n");
    }
    return 0;
}

Program 5 of 6

Days in a Month Calculator (Leap Year Aware)

Groups months with 31 days, 30 days, and handles February (28 vs 29 days based on leap year).

Month Days Rules

31 Days: Month 1, 3, 5, 7, 8, 10, 12 (Use OR ||)
30 Days: Month 4, 6, 9, 11
February (Month 2): 29 days if leap year, 28 days otherwise!

Interactive Logic Hints

💡 Hint 1: Why use || (OR) operator for month grouping? ▼

A month can only be ONE number at a time (e.g. month == 1 OR month == 3). Using && would be impossible!

🧩 Complete Month Grouping Conditions

if (m == 1 m == 3 || m == 5 || m == 7 || m == 8 || m == 10 || m == 12) {
  printf("31 Days");
} else if (m == 4 || m == 6 || m == 9 || m == 11) {
  printf("30 Days");
} else if (m 2) {
  // February leap check...
}

Select operators above and click Check!

📆 Month Days Calculator

Month:

Year:

Result will show here...

month_days.cReference C Code

#include <stdio.h>

int main() {
    int month, year;
    printf("Enter month (1-12) & year: ");
    scanf("%d %d", &month, &year);
    
    if (month == 1 || month == 3 || month == 5 || month == 7 ||
        month == 8 || month == 10 || month == 12) {
        printf("31 Days\n");
    } else if (month == 4 || month == 6 || month == 9 || month == 11) {
        printf("30 Days\n");
    } else if (month == 2) {
        if ((year%400==0) || (year%4==0 && year%100!=0))
            printf("29 Days (Leap Year)\n");
        else
            printf("28 Days\n");
    } else {
        printf("Invalid Month Number!\n");
    }
    return 0;
}

Program 6 of 6

Character Classifier (Vowel, Consonant, Digit, Special)

Classifies input character using vowel checks and <ctype.h> helper functions (isalpha, isdigit).

Classification Rules

Header required: #include <ctype.h>
Vowels: ch=='a'||ch=='e'||ch=='i'||ch=='o'||ch=='u' (also uppercase A,E,I,O,U)
Alphabet Check: isalpha(ch)
Digit Check: isdigit(ch)

Interactive Logic Hints

💡 Hint 1: Why space before %c in scanf(" %c", &ch)? ▼

The leading space in " %c" instructs scanf to skip any leftover newline or whitespace characters!

🧩 Complete Character Classifier Logic

#include <>

if (ch=='a'||ch=='e'||ch=='i'||ch=='o'||ch=='u') printf("Vowel");
else if ((ch)) printf("Consonant");
else if ((ch)) printf("Digit");

Select functions above and click Check!

🔤 Interactive Character Classifier

Enter Character:

Result will appear here...

char_classifier.cReference C Code

#include <stdio.h>
#include <ctype.h>

int main() {
    char ch;
    printf("Enter a character: ");
    scanf(" %c", &ch);

    if (ch=='a'||ch=='e'||ch=='i'||ch=='o'||ch=='u'||
        ch=='A'||ch=='E'||ch=='I'||ch=='O'||ch=='U') {
        printf("'%c' is a VOWEL\n", ch);
    } else if (isalpha(ch)) {
        printf("'%c' is a CONSONANT\n", ch);
    } else if (isdigit(ch)) {
        printf("'%c' is a DIGIT\n", ch);
    } else {
        printf("'%c' is a SPECIAL SYMBOL\n", ch);
    }
    return 0;
}

Knowledge Check

Test Your Control Structures Mastery

Question 1 of 5

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Active Logic: Decision Control Structures

C Branching Constructs & Operators

🎯 Learning Philosophy: Active Logic Building

Odd / Even Check & Multiples Verification

Modulus Branching Logic

Interactive Logic Hints

🧩 Complete the Even/Odd & Multiples Logic

🔢 Live Even/Odd & Divisibility Tester

Leap Year Detection (The Century Rule)

Leap Year Decision Logic

Interactive Logic Hints

🧩 Complete the Leap Year Compound Boolean Logic

📅 Leap Year Rule Evaluator

Roots of a Quadratic Equation (ax2 + bx + c = 0)

Discriminant Conditions

Interactive Logic Hints

🧩 Complete Discriminant Ladder Conditions

🧮 Interactive Quadratic Solver

Grades Ladder & Triangle Inequality Validation

Triangle Inequality Theorem

Interactive Logic Hints

🧩 Complete Triangle Inequality & Grade Logic

📊 Grade & Triangle Tester

Days in a Month Calculator (Leap Year Aware)

Month Days Rules

Interactive Logic Hints

🧩 Complete Month Grouping Conditions

📆 Month Days Calculator

Character Classifier (Vowel, Consonant, Digit, Special)

Classification Rules

Interactive Logic Hints

🧩 Complete Character Classifier Logic

🔤 Interactive Character Classifier

Test Your Control Structures Mastery

Roots of a Quadratic Equation (ax² + bx + c = 0)