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11.14 Tutorial5

intro-to-computer-science-gtiit-tutorial-05-2024.pdf

Assignment Operators

Write the following assignments only using the assignment (=) operator:

  1. ​x += y

    ​x = x + y

  2. ​y -= x

    ​y = y - x

  3. ​x++

    ​x = x + 1

  4. ​z--

    ​z = z - 1

  5. ​a *= 5

    ​a = a * 5

  6. ​x /= 7

    ​x = x / 7

  7. ​b >>= a

    ​b = b >> a

  8. ​p <<= q

    ​p = p << q

  9. ​x &= 1024

    ​x = x & y

  10. ​y |= z

    ​y = y | z

  11. ​x %= 4096

    ​x = x % 4096

  12. ​x ^= 2

    ​x = x ^ 2

For Statement

Manually execute the following program:

2.1

/* 2.1 */
void main() {
    int limit = 5;
    int evenSum = 0;
    int oddSum = 0;
    for (int i = 0; i < limit; i++) {
        if (i % 2 == 0) {
            evenSum += i;
        } else {
            oddSum += i;
        }
    }
}

**Step-by-Step Execution**

1. **Initialization:**

    - ​`limit = 5`
    - ​`evenSum = 0`
    - ​`oddSum = 0`
2. **Loop Execution:**

    - **Iteration 1 (i = 0):**

      - ​`i % 2 == 0` (0 is even)
      - ​`evenSum += i`​ → `evenSum = 0`
      - ​`oddSum = 0`
    - **Iteration 2 (i = 1):**

      - ​`i % 2 != 0` (1 is odd)
      - ​`oddSum += i`​ → `oddSum = 1`
      - ​`evenSum = 0`
    - **Iteration 3 (i = 2):**

      - ​`i % 2 == 0` (2 is even)
      - ​`evenSum += i`​ → `evenSum = 2`
      - ​`oddSum = 1`
    - **Iteration 4 (i = 3):**

      - ​`i % 2 != 0` (3 is odd)
      - ​`oddSum += i`​ → `oddSum = 4`
      - ​`evenSum = 2`
    - **Iteration 5 (i = 4):**

      - ​`i % 2 == 0` (4 is even)
      - ​`evenSum += i`​ → `evenSum = 6`
      - ​`oddSum = 4`
3. **Final Values:**

    - ​`evenSum = 6`
    - ​`oddSum = 4`

### Program 2.2 Execution

```c
void main() {
    int limit = 10;
    int sum = 0;
    for (int i = 0; i < limit; i += 2) {
        if (i % 2 == 0) {
            continue;
        }
        sum += i;
    }
}

  1. Initialization:

    • ​limit = 10
    • ​sum = 0

    • Loop Execution:

    • The loop increments i​ by 2 each time, starting from i = 0.

    • Iteration 1 (i = 0):

    • ​i % 2 == 0 (true)

    • ​continue​ is executed, skipping sum += i
    • ​sum = 0

    • Iteration 2 (i = 2):

    • ​i % 2 == 0 (true)

    • ​continue​ is executed, skipping sum += i
    • ​sum = 0

    • Iteration 3 (i = 4):

    • ​i % 2 == 0 (true)

    • ​continue​ is executed, skipping sum += i
    • ​sum = 0

    • Iteration 4 (i = 6):

    • ​i % 2 == 0 (true)

    • ​continue​ is executed, skipping sum += i
    • ​sum = 0

    • Iteration 5 (i = 8):

    • ​i % 2 == 0 (true)

    • ​continue​ is executed, skipping sum += i
    • ​sum = 0

    • Final Value:

    • ​sum = 0

Explanation: Since i​ is incremented by 2 and starts from 0, it is always even, so the continue​ statement is executed in every iteration, and sum is never modified.

Program 2.3 Execution

void main() {
    int limit = 0;
    int product = 0;
    for (int i = 6; i > limit; i--) {
        product += i;
        if (product <= i * 2) {
            continue;
        }
        product *= 2;
    }
}

  1. Initialization:

    • ​limit = 0
    • ​product = 0

    • Loop Execution:

    • The loop decrements i​ by 1 each time, starting from i = 6.

    • Iteration 1 (i = 6):

    • ​product += i​ → product = 0 + 6 = 6

    • ​product <= i * 2​ → 6 <= 6 * 2 (true)
    • ​continue​ is executed, skipping product *= 2
    • ​product = 6

    • Iteration 2 (i = 5):

    • ​product += i​ → product = 6 + 5 = 11

    • ​product <= i * 2​ → 11 <= 5 * 2 (true)
    • ​continue​ is executed, skipping product *= 2
    • ​product = 11

    • Iteration 3 (i = 4):

    • ​product += i​ → product = 11 + 4 = 15

    • ​product <= i * 2​ → 15 <= 4 * 2 (false)
    • ​product *= 2​ → product = 15 * 2 = 30

    • Iteration 4 (i = 3):

    • ​product += i​ → product = 30 + 3 = 33

    • ​product <= i * 2​ → 33 <= 3 * 2 (false)
    • ​product *= 2​ → product = 33 * 2 = 66

    • Iteration 5 (i = 2):

    • ​product += i​ → product = 66 + 2 = 68

    • ​product <= i * 2​ → 68 <= 2 * 2 (false)
    • ​product *= 2​ → product = 68 * 2 = 136

    • Iteration 6 (i = 1):

    • ​product += i​ → product = 136 + 1 = 137

    • ​product <= i * 2​ → 137 <= 1 * 2 (false)
    • ​product *= 2​ → product = 137 * 2 = 274

    • Final Value:

    • ​product = 274

Linear Accumulation with For

Translate the following summations to C programs with for statements:

a. \(\sum_{i=0}^{1024} i\)

#include <stdio.h>

int main () {
    int sum = 0;
    for (int i = 0; i <= 1024; i++) {
        sum = sum + i;
    }
    printf("%d", sum);
    return 0;
}

b. \(\sum_{i=1}^{10} \frac{36}{i}\)

#include <stdio.h>

int main () {
    float sum = 0;
    for (int i = 1; i <= 36; i++) {
        sum = sum + (36 / i);
    }
    printf("%f", sum);
    return 0;
}

c. \(\sum_{i=0}^{10} i^i\)

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

int main () {
    double sum = 0;
    for (double i = 0; i <= 10; i++) {
        sum = sum + pow(i, i);
    }
    printf("%lld", (long long int)sum);
    return 0;
}

d. \(\sum_{i=0}^{10} \frac{i}{2}\)

#include <stdio.h>

int main () {
    float sum = 0;
    for (int i = 0; i <= 10; i++) {
        sum = sum + (i / 2);
    }
    printf("%f", sum);
    return 0;
}

e. \(\sum_{i=1}^{100} i \cdot \sum_{k=1}^{i} k \cdot \sum_{j=1}^{k} j\)

#define Oo0Oo0oO int
#define o0O0OoOo long
#define O0oO0o0O for
#define oOo0O0o0 main 
#define OO0oO0Oo printf
#define o0O0oO0o return

#include <stdio.h> 
Oo0Oo0oO oOo0O0o0(){o0O0OoOo o0O0OoOo Oo0Oo0oO O0o0O0o0 = 0;O0oO0o0O
(Oo0Oo0oO o0O0o0OO = 1; o0O0o0OO <=
100; o0O0o0OO++) {O0oO0o0O (Oo0Oo0oO 
Oo0Oo0Oo = 1; Oo0Oo0Oo <= o0O0o0OO; 
Oo0Oo0Oo++) {O0oO0o0O (Oo0Oo0oO oO0oO0oO = 1; oO0oO0oO 
<= Oo0Oo0Oo; oO0oO0oO++) {O0o0O0o0 += 
o0O0o0OO * Oo0Oo0Oo * 
oO0oO0oO;}}}OO0oO0Oo(
"%lld", O0o0O0o0);o0O0oO0o 0;
}

#include <stdio.h>

int main () {
    long long int sum=0;
    for (int i = 1; i <= 100; i++) {
        for (int k = 1; k <= i; k++) {
            for (int j = 1; j <= k; j++) {
                sum += i * k * j;
            }
        }
    }
    printf("%lld", sum);
    return 0;
}

f. \(\frac{\sum_{i=0}^{10} 2 \cdot i}{\sum_{i=0}^{5} 4 \cdot i}\)

#include <stdio.h>

int main () {
    int sum1 = 0, sum2 = 0;
    for (int i = 0; i <= 10; i++) {
        sum1 = sum1 + 2 * i;
    }
    for (int j = 0; j <= 5; j++) {
        sum2 = sum2 + 4 * j;
    }
    // pay attention to the type of variable!
    // if the result may be float, then the variable should have at least one float
    // or this may casue integer division, then leading to loss of the fractional part.
    float result = (float)sum1 / (float)sum2;
    printf("%f", result);
    return 0;
}

4. Comma operator

Rewrite the following C programs without using the comma operator and correctly indenting them:

image

#include <stdio.h>

int main() {
    int i = 1, number = 1000;

    while (i < number) {
        if (number % i == 0) {
            printf("%d\n", i);
        }
        i++;
    }

    return 0;
}

#include <stdio.h>

int main() {
    int a, b;
    int number = 10;

    for (a = 1; a < number; a++) {
        for (b = 1; b < number; b++) {
            printf("%d^2 + %d^2 = %d\n", a, b, (a * a) + (b * b));
        }
    }

    return 0;
}

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