Lab 5 – Active Control Structures: Loops

PSUC Lab 5 — Iterative Control Structures

Active Logic: Loops (while, for, do-while)

Master loop initialization, continuation conditions, counter updates, prime trial division, base conversions, digit reversal, and Taylor series accumulators.

Loop Constructs Comparison

Loop Construct	Condition Evaluation	Minimum Executions
while Loop	Pre-test (Entry controlled)	0 times (if false initially)
for Loop	Pre-test (init; condition; update)	0 times
do-while Loop	Post-test (Exit controlled)	1 time guaranteed

🎯 Learning Philosophy: Active Logic Building

Test your iterative logic across 3 Interactive Learning Modes:

🧠 Tab 1 (Logic & Hints): Mathematical algorithms & loop invariants.
🧩 Tab 2 (Logic Puzzle): Complete missing loop bounds & accumulator statements!
💻 Tab 3 (C Solution): Verified C source code reference.

Solve loop puzzles before peeking at verified code solutions!

Program 1 of 5

Prime Number Check (Trial Division up to n/2)

Determines if an integer n > 1 has any divisor using a trial division loop running from i = 2 up to i <= n / 2.

Prime Division Algorithm

Base Check: If n <= 1, the number is NOT prime.
Test divisors from i = 2 up to i <= n / 2.
If n % i == 0, a divisor is found → isPrime = false; break;

Interactive Logic Hints

💡 Hint 1: Why break out of the loop immediately upon finding a divisor? ▼

Once a single divisor is found (e.g. 2 divides 10), we know 10 cannot be prime! Continuing the loop would waste CPU cycles!

🧩 Complete Prime Check Loop Logic

bool isPrime = true;
for (i = 2; i n / 2; i++) {
  if (n i == 0) {
    isPrime = false;
    ;
  }
}

Select loop conditions above and click Check!

🔢 Live Prime Tester & Divisor Tracer

Enter Number N:

Result will appear here...

prime_check.cReference C Code

#include <stdio.h>
#include <stdbool.h>

int main() {
    int n, i;
    bool isPrime = true;
    
    printf("Enter number: ");
    scanf("%d", &n);
    
    if (n <= 1) isPrime = false;
    else {
        for (i = 2; i <= n / 2; i++) {
            if (n % i == 0) {
                isPrime = false;
                break;
            }
        }
    }

    if (isPrime) printf("%d IS a Prime Number!\n", n);
    else         printf("%d is NOT a Prime Number.\n", n);
    return 0;
}

Program 2 of 5

Binary to Decimal Conversion

Converts a binary number into decimal using positional weights of powers of 2 (digit × 2ⁱ).

Bit Accumulator Logic

Extract last bit: rem = bin % 10
Accumulate: dec += rem * pow(2, i)
Drop last bit: bin /= 10
Increment power index: i++

Interactive Logic Hints

💡 Hint 1: Why treat binary input as decimal 10s initially? ▼

Using % 10 peels off 0s and 1s right-to-left, allowing us to multiply each bit by 2^i!

🧩 Complete Binary to Decimal Conversion Loop

while (bin != 0) {
  rem = bin 10;
  dec += rem * pow(, i);
  bin /= ;
  i++;
}

Select statements above and click Check!

💻 Binary Positional Visualizer

Binary Input (e.g. 1101):

Result will appear here...

bin_to_dec.cReference C Code

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

int main() {
    long long bin;
    int dec = 0, rem, i = 0;

    printf("Enter binary number: ");
    scanf("%lld", &bin);

    while (bin != 0) {
        rem  = bin % 10;
        dec += rem * pow(2, i);
        bin /= 10;
        i++;
    }

    printf("Decimal Value = %d\n", dec);
    return 0;
}

Program 3 of 5

Reverse Number & Palindrome Check

Reverses integer digits using rev = rev * 10 + rem and compares reversed number with original.

Reversal Accumulator Logic

rem = temp % 10 (Extract digit)
rev = rev * 10 + rem (Shift digits left & add new digit)
temp /= 10 (Drop last digit)
Palindrome check: num == rev

Interactive Logic Hints

💡 Hint 1: Why must we save num in temp before the loop? ▼

Because the loop destroys temp down to 0! We need original num to compare num == rev!

🧩 Complete Reversal & Palindrome Logic

temp = num;
while (temp > 0) {
  rem = temp % 10;
  rev = rev * rem;
  temp /= 10;
}
if (num rev) printf("Palindrome");

Select statements above and click Check!

🔄 Reversal & Palindrome Simulator

Enter Number N:

Result will appear here...

reverse_num.cReference C Code

#include <stdio.h>

int main() {
    int num, temp, rev = 0, rem;
    printf("Enter number: ");
    scanf("%d", &num);

    temp = num;
    while (temp > 0) {
        rem = temp % 10;
        rev = rev * 10 + rem;
        temp /= 10;
    }

    printf("Reversed = %d\n", rev);
    if (num == rev) printf("It IS a Palindrome!\n");
    else            printf("NOT a Palindrome.\n");
    return 0;
}

Program 4 of 5

Armstrong Number Verification

An n-digit number is an Armstrong number if the sum of its digits raised to the power n equals the number itself (for example: 153 = 1³ + 5³ + 3³).

2-Pass Loop Algorithm

Pass 1: Count total digits d using digits++; temp /= 10;
Pass 2: Accumulate sum += pow(digit, d) for each digit.
Check: sum == num

Interactive Logic Hints

💡 Hint 1: What are classic 3-digit Armstrong numbers? ▼

153, 370, 371, 407! (e.g. 3^3 + 7^3 + 0^3 = 27 + 343 + 0 = 370).

🧩 Complete Armstrong Power Sum Logic

while (temp > 0) { digits++; temp /= 10; }

temp = num;
while (temp > 0) {
  rem = temp % 10;
  sum += (rem, );
  temp /= 10;
}

Select functions above and click Check!

✨ Armstrong Power Sum Visualizer

Enter Number N:

Result will appear here...

armstrong.cReference C Code

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

int main() {
    int num, temp, digits = 0, rem, sum = 0;
    printf("Enter number: ");
    scanf("%d", &num);

    temp = num;
    while (temp > 0) { digits++; temp /= 10; }

    temp = num;
    while (temp > 0) {
        rem  = temp % 10;
        sum += pow(rem, digits);
        temp /= 10;
    }

    if (sum == num) printf("%d IS an Armstrong Number!\n", num);
    else            printf("%d is NOT an Armstrong Number.\n", num);
    return 0;
}

Program 5 of 5

Sum of Sine Series sin(x)

Computes Taylor expansion sin(x) = x - x³/3! + x⁵/5! - … up to N terms using term recurrence term = -term * x*x / ((2i)*(2i+1)).

Sine Recurrence Formula

Convert degrees to radians: xrad = xdeg * (3.14159 / 180.0)
First term: term = xrad; sum = term;
Next term recurrence: term = -term * xrad * xrad / ((2*i)*(2*i+1))

Interactive Logic Hints

💡 Hint 1: Why use term recurrence instead of calculating factorials separately? ▼

Factorials like 15! overflow integer limits! Recurrence updating avoids factorial overflow completely!

🧩 Complete Sine Series Recurrence Logic

xrad = xdeg * (3.14159 / );
term = xrad; sum = term;
for (i = 1; i < n; i++) {
term = * xrad * xrad / ((2*i) * (2*i + 1));
sum += term;
}

Select recurrence factors above and click Check!

📈 Sine Taylor Series Visualizer

Angle x (Degrees):

Terms N:

Series sum result...

sin_series.cReference C Code

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

int main() {
    int n, i;
    float xdeg, xrad, sum = 0, term;
    printf("Enter x in degrees & terms N: ");
    scanf("%f %d", &xdeg, &n);

    xrad = xdeg * (3.14159 / 180.0);
    term = xrad;
    sum  = term;

    for (i = 1; i < n; i++) {
        term = -term * xrad * xrad / ((2*i) * (2*i + 1));
        sum += term;
    }

    printf("sin(%.2f°) = %.6f\n", xdeg, sum);
    return 0;
}

Knowledge Check

Test Your Loop Logic & Series Mastery

Question 1 of 5

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