a. Given statements: //declare the variables and assign the values int i = 0; int j = 10; int n = 0; //check "i" is less than "j" while(i < j) { //increment the value i++; //decrement the value j--; //increment the value n++; } Trace table for the above loops: i j n 0 10 0 1 9 1 2 8 2 3 7 3 4 6 4 5 5 5 Explanation: Here, the “i” value is “0”, “j” value is “10” and “n” value is “0”. In first iteration of “while” loop checks “i < j” which means “0 < 10”. The condition becomes true, so it increment the “i” value, i = 1, decrement the “j” value, j = 9 and also increment the “n”, n = 1...

Question

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Answer 1

Expert Solution & Answer

Chapter 6, Problem 6RE

Explanation of Solution

a.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 10;

int n = 0;

//check "i" is less than "j"

while(i < j)

{

    //increment the value

    i++;

    //decrement the value

    j--;

    //increment the value

    n++;

}

Trace table for the above loops:

i j n
0 10 0
1 9 1
2 8 2
3 7 3
4 6 4
5 5 5

Explanation:

Here, the “i” value is “0”, “j” value is “10” and “n” value is “0”.

In first iteration of “while” loop checks “i < j” which means “0 < 10”. The condition becomes true, so it increment the “i” value, i = 1, decrement the “j” value, j = 9 and also increment the “n”, n = 1...

Explanation of Solution

b.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 0;

int n = 0;

//check "i" is less than "10"

while(i < 10)

{

    //increment the value

    i++;

    //calculate the "n" value

    n = n + i + j;

    //increment the value

    j++;

}

Trace table for the above loops:

i j n
0 0 1
1 1 4
2 2 9
3 3 16
4 4 25
5 5 36
6 6 49
7 7 64
8 8 81
9 9 100

Explanation:

Here, the “i” value is 0, “j” value is 0 and “n” value is 0.

In first iteration of “while” loop checks “i < 10” which means “0 < 10”. The condition becomes true, so it increment the “i” value, i = 1, calculate “n” value which means (0 + 1 + 0 = 1) and also increment the “j”, j = 1.

In second iteration of “while” loop checks “i < 10” which means “1 < 10”. The condition becomes true, so it increment the “i” value, i = 2, calculate “n” value which means (1+ 2 + 1 = 4) and also increment the “j”, j = 2.

In third iteration of “while” loop checks “i < 10” which means “2 < 10”. The condition becomes true, so it increment the “i” value, i = 3, calculate “n” value which means (4+ 3 + 2 = 9) and also increment the “j”, j = 3.

In fourth iteration of “while” loop checks “i < 10” which means “3 < 10”...

Explanation of Solution

c.

Given statements:

//declare the variables and assign the values

int i = 10;

int j = 0;

int n = 0;

//check "i" is greater than "0"

while(i > 0)

{

    //decrement the value

    i--;

//increment the value

    j++;

    //calculate the "n" value

    n = n + i - j;

}

Trace table for the above loops:

i j n
10 0 8
9 1 14
8 2 18
7 3 20
6 4 20
5 5 18
4 6 14
3 7 8
2 8 0
1 9 -10

Explanation:

Here, the “i” value is 10, “j” value is 0 and “n” value is 0.

In first iteration of “while” loop checks “i > 0” which means “10 > 0”. The condition becomes true, so it decrement the “i” value, i = 9, increment the “j”, j = 1 and calculate “n” value which means (0 + 9 - 1 = 8).

In second iteration of “while” loop checks “i > 0” which means “9 > 0”. The condition becomes true, so it decrement the “i” value, i = 8, increment the “j”, j = 2 and calculate “n” value which means (8 + 8 - 2 = 14).

In third iteration of “while” loop checks “i > 0” which means “8 > 0”. The condition becomes true, so it decrement the “i” value, i = 7, increment the “j”, j = 3 and calculate “n” value which means (14 + 7 - 3 = 18).

In fourth iteration of “while” loop checks “i > 0” which means “7 > 0”...

Explanation of Solution

d.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 10;

int n = 0;

//check "i" is greater than "0"

while(i != j)

{

    //increment the "i" by 2

    i = i + 2;

//decrement the "j" by 2

    j = j – 2;

    //increment the value

    n++;

}

Trace table for the above loops:

i j n
0 10 0
2 8 1
4 6 2
6 4 3
8 2 4

...

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Check out a sample textbook solution

See solution

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Here are two diagrams. Make them very explicit, similar to Example Diagram 3 (the Architecture of MSCTNN). graph LR subgraph Teacher_Model_B [Teacher Model (Pretrained)] Input_Teacher_B[Input C (Complete Data)] --> Teacher_Encoder_B[Transformer Encoder T] Teacher_Encoder_B --> Teacher_Prediction_B[Teacher Prediction y_T] Teacher_Encoder_B --> Teacher_Features_B[Internal Features F_T] end subgraph Student_B_Model [Student Model B (Handles Missing Labels)] Input_Student_B[Input C (Complete Data)] --> Student_B_Encoder[Transformer Encoder E_B] Student_B_Encoder --> Student_B_Prediction[Student B Prediction y_B] end subgraph Knowledge_Distillation_B [Knowledge Distillation (Student B)] Teacher_Prediction_B -- Logits Distillation Loss (L_logits_B) --> Total_Loss_B Teacher_Features_B -- Feature Alignment Loss (L_feature_B) --> Total_Loss_B Partial_Labels_B[Partial Labels y_p] -- Prediction Loss (L_pred_B) --> Total_Loss_B Total_Loss_B -- Backpropagation -->…

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Answer 2

Expert Solution & Answer

Chapter 6, Problem 6RE

Explanation of Solution

a.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 10;

int n = 0;

//check "i" is less than "j"

while(i < j)

{

    //increment the value

    i++;

    //decrement the value

    j--;

    //increment the value

    n++;

}

Trace table for the above loops:

i j n
0 10 0
1 9 1
2 8 2
3 7 3
4 6 4
5 5 5

Explanation:

Here, the “i” value is “0”, “j” value is “10” and “n” value is “0”.

In first iteration of “while” loop checks “i < j” which means “0 < 10”. The condition becomes true, so it increment the “i” value, i = 1, decrement the “j” value, j = 9 and also increment the “n”, n = 1...

Explanation of Solution

b.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 0;

int n = 0;

//check "i" is less than "10"

while(i < 10)

{

    //increment the value

    i++;

    //calculate the "n" value

    n = n + i + j;

    //increment the value

    j++;

}

Trace table for the above loops:

i j n
0 0 1
1 1 4
2 2 9
3 3 16
4 4 25
5 5 36
6 6 49
7 7 64
8 8 81
9 9 100

Explanation:

Here, the “i” value is 0, “j” value is 0 and “n” value is 0.

In first iteration of “while” loop checks “i < 10” which means “0 < 10”. The condition becomes true, so it increment the “i” value, i = 1, calculate “n” value which means (0 + 1 + 0 = 1) and also increment the “j”, j = 1.

In second iteration of “while” loop checks “i < 10” which means “1 < 10”. The condition becomes true, so it increment the “i” value, i = 2, calculate “n” value which means (1+ 2 + 1 = 4) and also increment the “j”, j = 2.

In third iteration of “while” loop checks “i < 10” which means “2 < 10”. The condition becomes true, so it increment the “i” value, i = 3, calculate “n” value which means (4+ 3 + 2 = 9) and also increment the “j”, j = 3.

In fourth iteration of “while” loop checks “i < 10” which means “3 < 10”...

Explanation of Solution

c.

Given statements:

//declare the variables and assign the values

int i = 10;

int j = 0;

int n = 0;

//check "i" is greater than "0"

while(i > 0)

{

    //decrement the value

    i--;

//increment the value

    j++;

    //calculate the "n" value

    n = n + i - j;

}

Trace table for the above loops:

i j n
10 0 8
9 1 14
8 2 18
7 3 20
6 4 20
5 5 18
4 6 14
3 7 8
2 8 0
1 9 -10

Explanation:

Here, the “i” value is 10, “j” value is 0 and “n” value is 0.

In first iteration of “while” loop checks “i > 0” which means “10 > 0”. The condition becomes true, so it decrement the “i” value, i = 9, increment the “j”, j = 1 and calculate “n” value which means (0 + 9 - 1 = 8).

In second iteration of “while” loop checks “i > 0” which means “9 > 0”. The condition becomes true, so it decrement the “i” value, i = 8, increment the “j”, j = 2 and calculate “n” value which means (8 + 8 - 2 = 14).

In third iteration of “while” loop checks “i > 0” which means “8 > 0”. The condition becomes true, so it decrement the “i” value, i = 7, increment the “j”, j = 3 and calculate “n” value which means (14 + 7 - 3 = 18).

In fourth iteration of “while” loop checks “i > 0” which means “7 > 0”...

Explanation of Solution

d.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 10;

int n = 0;

//check "i" is greater than "0"

while(i != j)

{

    //increment the "i" by 2

    i = i + 2;

//decrement the "j" by 2

    j = j – 2;

    //increment the value

    n++;

}

Trace table for the above loops:

i j n
0 10 0
2 8 1
4 6 2
6 4 3
8 2 4

...

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 3

Expert Solution & Answer

Chapter 6, Problem 6RE

Explanation of Solution

a.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 10;

int n = 0;

//check "i" is less than "j"

while(i < j)

{

    //increment the value

    i++;

    //decrement the value

    j--;

    //increment the value

    n++;

}

Trace table for the above loops:

i j n
0 10 0
1 9 1
2 8 2
3 7 3
4 6 4
5 5 5

Explanation:

Here, the “i” value is “0”, “j” value is “10” and “n” value is “0”.

In first iteration of “while” loop checks “i < j” which means “0 < 10”. The condition becomes true, so it increment the “i” value, i = 1, decrement the “j” value, j = 9 and also increment the “n”, n = 1...

Explanation of Solution

b.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 0;

int n = 0;

//check "i" is less than "10"

while(i < 10)

{

    //increment the value

    i++;

    //calculate the "n" value

    n = n + i + j;

    //increment the value

    j++;

}

Trace table for the above loops:

i j n
0 0 1
1 1 4
2 2 9
3 3 16
4 4 25
5 5 36
6 6 49
7 7 64
8 8 81
9 9 100

Explanation:

Here, the “i” value is 0, “j” value is 0 and “n” value is 0.

In first iteration of “while” loop checks “i < 10” which means “0 < 10”. The condition becomes true, so it increment the “i” value, i = 1, calculate “n” value which means (0 + 1 + 0 = 1) and also increment the “j”, j = 1.

In second iteration of “while” loop checks “i < 10” which means “1 < 10”. The condition becomes true, so it increment the “i” value, i = 2, calculate “n” value which means (1+ 2 + 1 = 4) and also increment the “j”, j = 2.

In third iteration of “while” loop checks “i < 10” which means “2 < 10”. The condition becomes true, so it increment the “i” value, i = 3, calculate “n” value which means (4+ 3 + 2 = 9) and also increment the “j”, j = 3.

In fourth iteration of “while” loop checks “i < 10” which means “3 < 10”...

Explanation of Solution

c.

Given statements:

//declare the variables and assign the values

int i = 10;

int j = 0;

int n = 0;

//check "i" is greater than "0"

while(i > 0)

{

    //decrement the value

    i--;

//increment the value

    j++;

    //calculate the "n" value

    n = n + i - j;

}

Trace table for the above loops:

i j n
10 0 8
9 1 14
8 2 18
7 3 20
6 4 20
5 5 18
4 6 14
3 7 8
2 8 0
1 9 -10

Explanation:

Here, the “i” value is 10, “j” value is 0 and “n” value is 0.

In first iteration of “while” loop checks “i > 0” which means “10 > 0”. The condition becomes true, so it decrement the “i” value, i = 9, increment the “j”, j = 1 and calculate “n” value which means (0 + 9 - 1 = 8).

In second iteration of “while” loop checks “i > 0” which means “9 > 0”. The condition becomes true, so it decrement the “i” value, i = 8, increment the “j”, j = 2 and calculate “n” value which means (8 + 8 - 2 = 14).

In third iteration of “while” loop checks “i > 0” which means “8 > 0”. The condition becomes true, so it decrement the “i” value, i = 7, increment the “j”, j = 3 and calculate “n” value which means (14 + 7 - 3 = 18).

In fourth iteration of “while” loop checks “i > 0” which means “7 > 0”...

Explanation of Solution

d.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 10;

int n = 0;

//check "i" is greater than "0"

while(i != j)

{

    //increment the "i" by 2

    i = i + 2;

//decrement the "j" by 2

    j = j – 2;

    //increment the value

    n++;

}

Trace table for the above loops:

i j n
0 10 0
2 8 1
4 6 2
6 4 3
8 2 4

...

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 4

Expert Solution & Answer

Chapter 6, Problem 6RE

Explanation of Solution

a.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 10;

int n = 0;

//check "i" is less than "j"

while(i < j)

{

    //increment the value

    i++;

    //decrement the value

    j--;

    //increment the value

    n++;

}

Trace table for the above loops:

i j n
0 10 0
1 9 1
2 8 2
3 7 3
4 6 4
5 5 5

Explanation:

Here, the “i” value is “0”, “j” value is “10” and “n” value is “0”.

In first iteration of “while” loop checks “i < j” which means “0 < 10”. The condition becomes true, so it increment the “i” value, i = 1, decrement the “j” value, j = 9 and also increment the “n”, n = 1...

Explanation of Solution

b.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 0;

int n = 0;

//check "i" is less than "10"

while(i < 10)

{

    //increment the value

    i++;

    //calculate the "n" value

    n = n + i + j;

    //increment the value

    j++;

}

Trace table for the above loops:

i j n
0 0 1
1 1 4
2 2 9
3 3 16
4 4 25
5 5 36
6 6 49
7 7 64
8 8 81
9 9 100

Explanation:

Here, the “i” value is 0, “j” value is 0 and “n” value is 0.

In first iteration of “while” loop checks “i < 10” which means “0 < 10”. The condition becomes true, so it increment the “i” value, i = 1, calculate “n” value which means (0 + 1 + 0 = 1) and also increment the “j”, j = 1.

In second iteration of “while” loop checks “i < 10” which means “1 < 10”. The condition becomes true, so it increment the “i” value, i = 2, calculate “n” value which means (1+ 2 + 1 = 4) and also increment the “j”, j = 2.

In third iteration of “while” loop checks “i < 10” which means “2 < 10”. The condition becomes true, so it increment the “i” value, i = 3, calculate “n” value which means (4+ 3 + 2 = 9) and also increment the “j”, j = 3.

In fourth iteration of “while” loop checks “i < 10” which means “3 < 10”...

Explanation of Solution

c.

Given statements:

//declare the variables and assign the values

int i = 10;

int j = 0;

int n = 0;

//check "i" is greater than "0"

while(i > 0)

{

    //decrement the value

    i--;

//increment the value

    j++;

    //calculate the "n" value

    n = n + i - j;

}

Trace table for the above loops:

i j n
10 0 8
9 1 14
8 2 18
7 3 20
6 4 20
5 5 18
4 6 14
3 7 8
2 8 0
1 9 -10

Explanation:

Here, the “i” value is 10, “j” value is 0 and “n” value is 0.

In first iteration of “while” loop checks “i > 0” which means “10 > 0”. The condition becomes true, so it decrement the “i” value, i = 9, increment the “j”, j = 1 and calculate “n” value which means (0 + 9 - 1 = 8).

In second iteration of “while” loop checks “i > 0” which means “9 > 0”. The condition becomes true, so it decrement the “i” value, i = 8, increment the “j”, j = 2 and calculate “n” value which means (8 + 8 - 2 = 14).

In third iteration of “while” loop checks “i > 0” which means “8 > 0”. The condition becomes true, so it decrement the “i” value, i = 7, increment the “j”, j = 3 and calculate “n” value which means (14 + 7 - 3 = 18).

In fourth iteration of “while” loop checks “i > 0” which means “7 > 0”...

Explanation of Solution

d.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 10;

int n = 0;

//check "i" is greater than "0"

while(i != j)

{

    //increment the "i" by 2

    i = i + 2;

//decrement the "j" by 2

    j = j – 2;

    //increment the value

    n++;

}

Trace table for the above loops:

i j n
0 10 0
2 8 1
4 6 2
6 4 3
8 2 4

...

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

Chapter 6, Problem 6RE

Explanation of Solution

a.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 10;

int n = 0;

//check "i" is less than "j"

while(i < j)

{

    //increment the value

    i++;

    //decrement the value

    j--;

    //increment the value

    n++;

}

Trace table for the above loops:

i j n
0 10 0
1 9 1
2 8 2
3 7 3
4 6 4
5 5 5

Explanation:

Here, the “i” value is “0”, “j” value is “10” and “n” value is “0”.

In first iteration of “while” loop checks “i < j” which means “0 < 10”. The condition becomes true, so it increment the “i” value, i = 1, decrement the “j” value, j = 9 and also increment the “n”, n = 1...

Explanation of Solution

b.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 0;

int n = 0;

//check "i" is less than "10"

while(i < 10)

{

    //increment the value

    i++;

    //calculate the "n" value

    n = n + i + j;

    //increment the value

    j++;

}

Trace table for the above loops:

i j n
0 0 1
1 1 4
2 2 9
3 3 16
4 4 25
5 5 36
6 6 49
7 7 64
8 8 81
9 9 100

Explanation:

Here, the “i” value is 0, “j” value is 0 and “n” value is 0.

In first iteration of “while” loop checks “i < 10” which means “0 < 10”. The condition becomes true, so it increment the “i” value, i = 1, calculate “n” value which means (0 + 1 + 0 = 1) and also increment the “j”, j = 1.

In second iteration of “while” loop checks “i < 10” which means “1 < 10”. The condition becomes true, so it increment the “i” value, i = 2, calculate “n” value which means (1+ 2 + 1 = 4) and also increment the “j”, j = 2.

In third iteration of “while” loop checks “i < 10” which means “2 < 10”. The condition becomes true, so it increment the “i” value, i = 3, calculate “n” value which means (4+ 3 + 2 = 9) and also increment the “j”, j = 3.

In fourth iteration of “while” loop checks “i < 10” which means “3 < 10”...

Explanation of Solution

c.

Given statements:

//declare the variables and assign the values

int i = 10;

int j = 0;

int n = 0;

//check "i" is greater than "0"

while(i > 0)

{

    //decrement the value

    i--;

//increment the value

    j++;

    //calculate the "n" value

    n = n + i - j;

}

Trace table for the above loops:

i j n
10 0 8
9 1 14
8 2 18
7 3 20
6 4 20
5 5 18
4 6 14
3 7 8
2 8 0
1 9 -10

Explanation:

Here, the “i” value is 10, “j” value is 0 and “n” value is 0.

In first iteration of “while” loop checks “i > 0” which means “10 > 0”. The condition becomes true, so it decrement the “i” value, i = 9, increment the “j”, j = 1 and calculate “n” value which means (0 + 9 - 1 = 8).

In second iteration of “while” loop checks “i > 0” which means “9 > 0”. The condition becomes true, so it decrement the “i” value, i = 8, increment the “j”, j = 2 and calculate “n” value which means (8 + 8 - 2 = 14).

In third iteration of “while” loop checks “i > 0” which means “8 > 0”. The condition becomes true, so it decrement the “i” value, i = 7, increment the “j”, j = 3 and calculate “n” value which means (14 + 7 - 3 = 18).

In fourth iteration of “while” loop checks “i > 0” which means “7 > 0”...

Explanation of Solution

d.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 10;

int n = 0;

//check "i" is greater than "0"

while(i != j)

{

    //increment the "i" by 2

    i = i + 2;

//decrement the "j" by 2

    j = j – 2;

    //increment the value

    n++;

}

Trace table for the above loops:

i j n
0 10 0
2 8 1
4 6 2
6 4 3
8 2 4

...

Answer 8

Chapter 6, Problem 6RE

Explanation of Solution

a.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 10;

int n = 0;

//check "i" is less than "j"

while(i < j)

{

    //increment the value

    i++;

    //decrement the value

    j--;

    //increment the value

    n++;

}

Trace table for the above loops:

i j n
0 10 0
1 9 1
2 8 2
3 7 3
4 6 4
5 5 5

Explanation:

Here, the “i” value is “0”, “j” value is “10” and “n” value is “0”.

In first iteration of “while” loop checks “i < j” which means “0 < 10”. The condition becomes true, so it increment the “i” value, i = 1, decrement the “j” value, j = 9 and also increment the “n”, n = 1...

Answer 9

Chapter 6, Problem 6RE

Explanation of Solution

a.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 10;

int n = 0;

//check "i" is less than "j"

while(i < j)

{

    //increment the value

    i++;

    //decrement the value

    j--;

    //increment the value

    n++;

}

Trace table for the above loops:

i j n
0 10 0
1 9 1
2 8 2
3 7 3
4 6 4
5 5 5

Explanation:

Here, the “i” value is “0”, “j” value is “10” and “n” value is “0”.

In first iteration of “while” loop checks “i < j” which means “0 < 10”. The condition becomes true, so it increment the “i” value, i = 1, decrement the “j” value, j = 9 and also increment the “n”, n = 1...

Answer 10

Explanation of Solution

b.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 0;

int n = 0;

//check "i" is less than "10"

while(i < 10)

{

    //increment the value

    i++;

    //calculate the "n" value

    n = n + i + j;

    //increment the value

    j++;

}

Trace table for the above loops:

i j n
0 0 1
1 1 4
2 2 9
3 3 16
4 4 25
5 5 36
6 6 49
7 7 64
8 8 81
9 9 100

Explanation:

Here, the “i” value is 0, “j” value is 0 and “n” value is 0.

In first iteration of “while” loop checks “i < 10” which means “0 < 10”. The condition becomes true, so it increment the “i” value, i = 1, calculate “n” value which means (0 + 1 + 0 = 1) and also increment the “j”, j = 1.

In second iteration of “while” loop checks “i < 10” which means “1 < 10”. The condition becomes true, so it increment the “i” value, i = 2, calculate “n” value which means (1+ 2 + 1 = 4) and also increment the “j”, j = 2.

In third iteration of “while” loop checks “i < 10” which means “2 < 10”. The condition becomes true, so it increment the “i” value, i = 3, calculate “n” value which means (4+ 3 + 2 = 9) and also increment the “j”, j = 3.

In fourth iteration of “while” loop checks “i < 10” which means “3 < 10”...

Answer 11

Explanation of Solution

b.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 0;

int n = 0;

//check "i" is less than "10"

while(i < 10)

{

    //increment the value

    i++;

    //calculate the "n" value

    n = n + i + j;

    //increment the value

    j++;

}

Trace table for the above loops:

i j n
0 0 1
1 1 4
2 2 9
3 3 16
4 4 25
5 5 36
6 6 49
7 7 64
8 8 81
9 9 100

Explanation:

Here, the “i” value is 0, “j” value is 0 and “n” value is 0.

In first iteration of “while” loop checks “i < 10” which means “0 < 10”. The condition becomes true, so it increment the “i” value, i = 1, calculate “n” value which means (0 + 1 + 0 = 1) and also increment the “j”, j = 1.

In second iteration of “while” loop checks “i < 10” which means “1 < 10”. The condition becomes true, so it increment the “i” value, i = 2, calculate “n” value which means (1+ 2 + 1 = 4) and also increment the “j”, j = 2.

In third iteration of “while” loop checks “i < 10” which means “2 < 10”. The condition becomes true, so it increment the “i” value, i = 3, calculate “n” value which means (4+ 3 + 2 = 9) and also increment the “j”, j = 3.

In fourth iteration of “while” loop checks “i < 10” which means “3 < 10”...

Answer 12

Explanation of Solution

c.

Given statements:

//declare the variables and assign the values

int i = 10;

int j = 0;

int n = 0;

//check "i" is greater than "0"

while(i > 0)

{

    //decrement the value

    i--;

//increment the value

    j++;

    //calculate the "n" value

    n = n + i - j;

}

Trace table for the above loops:

i j n
10 0 8
9 1 14
8 2 18
7 3 20
6 4 20
5 5 18
4 6 14
3 7 8
2 8 0
1 9 -10

Explanation:

Here, the “i” value is 10, “j” value is 0 and “n” value is 0.

In first iteration of “while” loop checks “i > 0” which means “10 > 0”. The condition becomes true, so it decrement the “i” value, i = 9, increment the “j”, j = 1 and calculate “n” value which means (0 + 9 - 1 = 8).

In second iteration of “while” loop checks “i > 0” which means “9 > 0”. The condition becomes true, so it decrement the “i” value, i = 8, increment the “j”, j = 2 and calculate “n” value which means (8 + 8 - 2 = 14).

In third iteration of “while” loop checks “i > 0” which means “8 > 0”. The condition becomes true, so it decrement the “i” value, i = 7, increment the “j”, j = 3 and calculate “n” value which means (14 + 7 - 3 = 18).

In fourth iteration of “while” loop checks “i > 0” which means “7 > 0”...

Answer 13

Explanation of Solution

c.

Given statements:

//declare the variables and assign the values

int i = 10;

int j = 0;

int n = 0;

//check "i" is greater than "0"

while(i > 0)

{

    //decrement the value

    i--;

//increment the value

    j++;

    //calculate the "n" value

    n = n + i - j;

}

Trace table for the above loops:

i j n
10 0 8
9 1 14
8 2 18
7 3 20
6 4 20
5 5 18
4 6 14
3 7 8
2 8 0
1 9 -10

Explanation:

Here, the “i” value is 10, “j” value is 0 and “n” value is 0.

In first iteration of “while” loop checks “i > 0” which means “10 > 0”. The condition becomes true, so it decrement the “i” value, i = 9, increment the “j”, j = 1 and calculate “n” value which means (0 + 9 - 1 = 8).

In second iteration of “while” loop checks “i > 0” which means “9 > 0”. The condition becomes true, so it decrement the “i” value, i = 8, increment the “j”, j = 2 and calculate “n” value which means (8 + 8 - 2 = 14).

In third iteration of “while” loop checks “i > 0” which means “8 > 0”. The condition becomes true, so it decrement the “i” value, i = 7, increment the “j”, j = 3 and calculate “n” value which means (14 + 7 - 3 = 18).

In fourth iteration of “while” loop checks “i > 0” which means “7 > 0”...

Answer 14

Explanation of Solution

d.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 10;

int n = 0;

//check "i" is greater than "0"

while(i != j)

{

    //increment the "i" by 2

    i = i + 2;

//decrement the "j" by 2

    j = j – 2;

    //increment the value

    n++;

}

Trace table for the above loops:

i j n
0 10 0
2 8 1
4 6 2
6 4 3
8 2 4

...

Answer 15

Explanation of Solution

d.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 10;

int n = 0;

//check "i" is greater than "0"

while(i != j)

{

    //increment the "i" by 2

    i = i + 2;

//decrement the "j" by 2

    j = j – 2;

    //increment the value

    n++;

}

Trace table for the above loops:

i j n
0 10 0
2 8 1
4 6 2
6 4 3
8 2 4

...

Answer 16

Expert Solution & Answer

Answer 17

Expert Solution & Answer

Answer 18

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Answer 19

Ch. 6.1 - Prob. 1SC Ch. 6.1 - Prob. 2SC Ch. 6.1 - Prob. 3SC Ch. 6.1 - Prob. 4SC Ch. 6.1 - What does the following loop print? int n =...Ch. 6.2 - Hand-trace the following code, showing the value...Ch. 6.2 - Prob. 7SC Ch. 6.2 - Prob. 8SC Ch. 6.2 - Prob. 9SC Ch. 6.2 - Prob. 10SC

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