Given Information : The definition for the code is mentioned below: //Traverse through the grid for (i =0; i < 16; i++) { for (j = 0; j < 16; j++) { //add values of x into grid total_x += grid[i][j].x; //add values of y into grid total_y += grid[i][j].y; } }

Question

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Answer 1

Question

Chapter 6.5, Problem 6.20PP

A)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

for (i =0; i < 16; i++)

{

for (j = 0; j < 16; j++)

{

//add values of x into grid

total_x += grid[i][j].x;

//add values of y into grid

total_y += grid[i][j].y;

}

}

B)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

for (i =0; i < 16; i++)

{

for (j = 0; j < 16; j++)

{

//add values of x into grid

total_x += grid[i][j].x;

//add values of y into grid

total_y += grid[i][j].y;

}

}

C)

Explanation of Solution

Miss rate:

The cache can only hold half of the elements in the array, so that means that a read to grid[8][0] will evict the block that was loaded when we read grid[0][0]. Since this block also contained grid[0][1], the first read of grid[0][1] will be a miss.

Hence, each iteration will have one hit and one miss.

This means one will have 128 hits and 128 misses.

Hence,

miss rate = ( number of miss/total reads

D)

Explanation of Solution

New Miss Rate:

If the cache were twice as big them also no change will be made to the mi...

Expert Solution & Answer

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Check out a sample textbook solution

See solution

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7. See the code below and solve the following. public class Test { public static void main(String[] args) { int result = 0; } result = fn(2,3); System.out.println("The result is: + result); // fn(x, 1) = x // fn(x, y) = fn(x, y-1) + 2, when y>1 public static int fn(int x, int y) { if (x <= 1) return x; else return fn(x, y-1) + 2; } } 7-1. This program has a bug that leads to infinite recursion. Modify fn(int x, int y) method to fix the problem. (2 point) 7-2. Manually trace the recursive call, fn(2,3) and show the output (step by step). (2 point) 7-3. Can you identify the Base Case in recursive method fn(int x, int y)? (1 point)

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Write a program that reads a list of integers from input and determines if the list is a palindrome (values are identical from first to last and last to first). The input begins with an integer indicating the length of the list that follows. Assume the list will contain a maximum of 20 integers. Output "yes" if the list is a palindrome and "no" otherwise. The output ends with a newline. Hints: - use a for loop to populate the array based on the specified size (the first number entered) - use a for loop to check first value with last value, second value with second from end, etc. - if the values do not match, set a Boolean variable to flag which statement to output (yes or no) Ex: If the input is (remember to include spaces between the numbers): 6 1 5 9 9 5 1 the output is: yes Ex: If the input is: 5 1 2 3 4 5 the output is: C++ coding

Answer 2

Question

Chapter 6.5, Problem 6.20PP

A)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

for (i =0; i < 16; i++)

{

for (j = 0; j < 16; j++)

{

//add values of x into grid

total_x += grid[i][j].x;

//add values of y into grid

total_y += grid[i][j].y;

}

}

B)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

for (i =0; i < 16; i++)

{

for (j = 0; j < 16; j++)

{

//add values of x into grid

total_x += grid[i][j].x;

//add values of y into grid

total_y += grid[i][j].y;

}

}

C)

Explanation of Solution

Miss rate:

The cache can only hold half of the elements in the array, so that means that a read to grid[8][0] will evict the block that was loaded when we read grid[0][0]. Since this block also contained grid[0][1], the first read of grid[0][1] will be a miss.

Hence, each iteration will have one hit and one miss.

This means one will have 128 hits and 128 misses.

Hence,

miss rate = ( number of miss/total reads

D)

Explanation of Solution

New Miss Rate:

If the cache were twice as big them also no change will be made to the mi...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 6.5, Problem 6.20PP

A)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

for (i =0; i < 16; i++)

{

for (j = 0; j < 16; j++)

{

//add values of x into grid

total_x += grid[i][j].x;

//add values of y into grid

total_y += grid[i][j].y;

}

}

B)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

for (i =0; i < 16; i++)

{

for (j = 0; j < 16; j++)

{

//add values of x into grid

total_x += grid[i][j].x;

//add values of y into grid

total_y += grid[i][j].y;

}

}

C)

Explanation of Solution

Miss rate:

The cache can only hold half of the elements in the array, so that means that a read to grid[8][0] will evict the block that was loaded when we read grid[0][0]. Since this block also contained grid[0][1], the first read of grid[0][1] will be a miss.

Hence, each iteration will have one hit and one miss.

This means one will have 128 hits and 128 misses.

Hence,

miss rate = ( number of miss/total reads

D)

Explanation of Solution

New Miss Rate:

If the cache were twice as big them also no change will be made to the mi...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 6.5, Problem 6.20PP

A)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

for (i =0; i < 16; i++)

{

for (j = 0; j < 16; j++)

{

//add values of x into grid

total_x += grid[i][j].x;

//add values of y into grid

total_y += grid[i][j].y;

}

}

B)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

for (i =0; i < 16; i++)

{

for (j = 0; j < 16; j++)

{

//add values of x into grid

total_x += grid[i][j].x;

//add values of y into grid

total_y += grid[i][j].y;

}

}

C)

Explanation of Solution

Miss rate:

The cache can only hold half of the elements in the array, so that means that a read to grid[8][0] will evict the block that was loaded when we read grid[0][0]. Since this block also contained grid[0][1], the first read of grid[0][1] will be a miss.

Hence, each iteration will have one hit and one miss.

This means one will have 128 hits and 128 misses.

Hence,

miss rate = ( number of miss/total reads

D)

Explanation of Solution

New Miss Rate:

If the cache were twice as big them also no change will be made to the mi...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 6.5, Problem 6.20PP

A)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

for (i =0; i < 16; i++)

{

for (j = 0; j < 16; j++)

{

//add values of x into grid

total_x += grid[i][j].x;

//add values of y into grid

total_y += grid[i][j].y;

}

}

B)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

for (i =0; i < 16; i++)

{

for (j = 0; j < 16; j++)

{

//add values of x into grid

total_x += grid[i][j].x;

//add values of y into grid

total_y += grid[i][j].y;

}

}

C)

Explanation of Solution

Miss rate:

The cache can only hold half of the elements in the array, so that means that a read to grid[8][0] will evict the block that was loaded when we read grid[0][0]. Since this block also contained grid[0][1], the first read of grid[0][1] will be a miss.

Hence, each iteration will have one hit and one miss.

This means one will have 128 hits and 128 misses.

Hence,

miss rate = ( number of miss/total reads

D)

Explanation of Solution

New Miss Rate:

If the cache were twice as big them also no change will be made to the mi...

Answer 6

Chapter 6.5, Problem 6.20PP

A)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

for (i =0; i < 16; i++)

{

for (j = 0; j < 16; j++)

{

//add values of x into grid

total_x += grid[i][j].x;

//add values of y into grid

total_y += grid[i][j].y;

}

}

Answer 7

Chapter 6.5, Problem 6.20PP

A)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

for (i =0; i < 16; i++)

{

for (j = 0; j < 16; j++)

{

//add values of x into grid

total_x += grid[i][j].x;

//add values of y into grid

total_y += grid[i][j].y;

}

}

Answer 8

B)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

for (i =0; i < 16; i++)

{

for (j = 0; j < 16; j++)

{

//add values of x into grid

total_x += grid[i][j].x;

//add values of y into grid

total_y += grid[i][j].y;

}

}

Answer 9

B)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

for (i =0; i < 16; i++)

{

for (j = 0; j < 16; j++)

{

//add values of x into grid

total_x += grid[i][j].x;

//add values of y into grid

total_y += grid[i][j].y;

}

}

Answer 10

C)

Explanation of Solution

Miss rate:

The cache can only hold half of the elements in the array, so that means that a read to grid[8][0] will evict the block that was loaded when we read grid[0][0]. Since this block also contained grid[0][1], the first read of grid[0][1] will be a miss.

Hence, each iteration will have one hit and one miss.

This means one will have 128 hits and 128 misses.

Hence,

miss rate = ( number of miss/total reads

Answer 11

C)

Explanation of Solution

Miss rate:

The cache can only hold half of the elements in the array, so that means that a read to grid[8][0] will evict the block that was loaded when we read grid[0][0]. Since this block also contained grid[0][1], the first read of grid[0][1] will be a miss.

Hence, each iteration will have one hit and one miss.

This means one will have 128 hits and 128 misses.

Hence,

miss rate = ( number of miss/total reads

Answer 12

D)

Explanation of Solution

New Miss Rate:

If the cache were twice as big them also no change will be made to the mi...

Answer 13

D)

Explanation of Solution

New Miss Rate:

If the cache were twice as big them also no change will be made to the mi...

Answer 14

Expert Solution & Answer

Answer 15

Expert Solution & Answer

Answer 16

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Check out a sample textbook solution

See solution

Answer 17

Ch. 6.1 - Prob. 6.1PP Ch. 6.1 - Prob. 6.2PP Ch. 6.1 - Prob. 6.3PP Ch. 6.1 - Prob. 6.4PP Ch. 6.1 - Prob. 6.5PP Ch. 6.1 - Prob. 6.6PP Ch. 6.2 - Prob. 6.7PP Ch. 6.2 - Prob. 6.8PP Ch. 6.4 - Prob. 6.9PP Ch. 6.4 - Prob. 6.10PP

Given Information : The definition for the code is mentioned below: //Traverse through the grid for (i =0; i < 16; i++) { for (j = 0; j < 16; j++) { //add values of x into grid total_x += grid[i][j].x; //add values of y into grid total_y += grid[i][j].y; } }

Solution Summary: The author assumes that sizeof(int)=4 Grid begins at memory address 0. The cache is empty and only memory accesses are to the array grid entries.

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

for (i =0; i < 16; i++)

{

for (j = 0; j < 16; j++)

{

//add values of x into grid

total_x += grid[i][j].x;

//add values of y into grid

total_y += grid[i][j].y;

}

}

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

for (i =0; i < 16; i++)

{

for (j = 0; j < 16; j++)

{

//add values of x into grid

total_x += grid[i][j].x;

//add values of y into grid

total_y += grid[i][j].y;

}

}

Explanation of Solution

Explanation of Solution

New Miss Rate:

If the cache were twice as big them also no change will be made to the mi...

Want to see the full answer?

Chapter 6 Solutions

Given Information : The definition for the code is mentioned below: //Traverse through the grid for (i =0; i &lt; 16; i++) { for (j = 0; j &lt; 16; j++) { //add values of x into grid total_x += grid[i][j].x; //add values of y into grid total_y += grid[i][j].y; } }

Solution Summary: The author assumes that sizeof(int)=4 Grid begins at memory address 0. The cache is empty and only memory accesses are to the array grid entries.

Given Information: The definition for the code is mentioned below: //Traverse through the grid for (i =0; i < 16; i++) { for (j = 0; j < 16; j++) { //add values of x into grid total_x += grid[i][j].x; //add values of y into grid total_y += grid[i][j].y; } }

Given Information: The definition for the code is mentioned below: //Traverse through the grid for (i =0; i < 16; i++) { for (j = 0; j < 16; j++) { //add values of x into grid total_x += grid[i][j].x; //add values of y into grid total_y += grid[i][j].y; } }

Explanation of Solution

Explanation of Solution

New Miss Rate: If the cache were twice as big them also no change will be made to the mi...

Want to see the full answer?

Chapter 6 Solutions

Given Information : The definition for the code is mentioned below: //Traverse through the grid for (i =0; i < 16; i++) { for (j = 0; j < 16; j++) { //add values of x into grid total_x += grid[i][j].x; //add values of y into grid total_y += grid[i][j].y; } }

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

for (i =0; i < 16; i++)

{

for (j = 0; j < 16; j++)

{

//add values of x into grid

total_x += grid[i][j].x;

//add values of y into grid

total_y += grid[i][j].y;

}

}

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

for (i =0; i < 16; i++)

{

for (j = 0; j < 16; j++)

{

//add values of x into grid

total_x += grid[i][j].x;

//add values of y into grid

total_y += grid[i][j].y;

}

}

New Miss Rate:

If the cache were twice as big them also no change will be made to the mi...