Given Information : The given code is: // define structure square point_color { // variable declaration int c; int m; int y; int k; }; // declare structure array struct point_color square[16][16]; int i,j; // traverse through the array for(i=0;i<16;i++) { //traverse through elements for(j=0;j<16;j++) { //square all elements of the 2-D array square[j][i].c=0; square[j][i].m=0; square[j][i].y=1; square[j][i].k=0; } } Write hit: If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty. If the state of information is in valid state then it executes a write-through operation. It then updates the memory and block and changes its blocked state to reserved state. Write miss : A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

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Answer 1

Question

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Chapter 6, Problem 6.39HW

A)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[j][i].c=0;

square[j][i].m=0;

square[j][i].y=1;

square[j][i].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

B)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[j][i].c=0;

square[j][i].m=0;

square[j][i].y=1;

square[j][i].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

C)

Program Plan Intro

Given Information:

The given code is:

//define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[j][i].c=0;

square[j][i].m=0;

square[j][i].y=1;

square[j][i].k=0;

}

}

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I need help to solve a simple problem using Grover’s algorithm, where the solution is not necessarily known beforehand. The problem is a 2×2 binary sudoku with two rules: • No column may contain the same value twice. • No row may contain the same value twice. Each square in the sudoku is assigned to a variable as follows: We want to design a quantum circuit that outputs a valid solution to this sudoku. While using Grover’s algorithm for this task is not necessarily practical, the goal is to demonstrate how classical decision problems can be converted into oracles for Grover’s algorithm. Turning the Problem into a Circuit To solve this, an oracle needs to be created that helps identify valid solutions. The first step is to construct a classical function within a quantum circuit that checks whether a given state satisfies the sudoku rules. Since we need to check both columns and rows, there are four conditions to verify: v0 ≠ v1 # Check top row v2 ≠ v3 # Check bottom row…

using r language

I need help to solve a simple problem using Grover’s algorithm, where the solution is not necessarily known beforehand. The problem is a 2×2 binary sudoku with two rules: • No column may contain the same value twice. • No row may contain the same value twice. Each square in the sudoku is assigned to a variable as follows: We want to design a quantum circuit that outputs a valid solution to this sudoku. While using Grover’s algorithm for this task is not necessarily practical, the goal is to demonstrate how classical decision problems can be converted into oracles for Grover’s algorithm. Turning the Problem into a Circuit To solve this, an oracle needs to be created that helps identify valid solutions. The first step is to construct a classical function within a quantum circuit that checks whether a given state satisfies the sudoku rules. Since we need to check both columns and rows, there are four conditions to verify: v0 ≠ v1 # Check top row v2 ≠ v3 # Check bottom row…

Answer 2

Question

100%

Chapter 6, Problem 6.39HW

A)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[j][i].c=0;

square[j][i].m=0;

square[j][i].y=1;

square[j][i].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

B)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[j][i].c=0;

square[j][i].m=0;

square[j][i].y=1;

square[j][i].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

C)

Program Plan Intro

Given Information:

The given code is:

//define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[j][i].c=0;

square[j][i].m=0;

square[j][i].y=1;

square[j][i].k=0;

}

}

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 3

Question

100%

Chapter 6, Problem 6.39HW

A)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[j][i].c=0;

square[j][i].m=0;

square[j][i].y=1;

square[j][i].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

B)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[j][i].c=0;

square[j][i].m=0;

square[j][i].y=1;

square[j][i].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

C)

Program Plan Intro

Given Information:

The given code is:

//define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[j][i].c=0;

square[j][i].m=0;

square[j][i].y=1;

square[j][i].k=0;

}

}

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

100%

Chapter 6, Problem 6.39HW

A)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[j][i].c=0;

square[j][i].m=0;

square[j][i].y=1;

square[j][i].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

B)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[j][i].c=0;

square[j][i].m=0;

square[j][i].y=1;

square[j][i].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

C)

Program Plan Intro

Given Information:

The given code is:

//define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[j][i].c=0;

square[j][i].m=0;

square[j][i].y=1;

square[j][i].k=0;

}

}

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Question

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Answer 6

Question

100%

Answer 7

Chapter 6, Problem 6.39HW

A)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[j][i].c=0;

square[j][i].m=0;

square[j][i].y=1;

square[j][i].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

B)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[j][i].c=0;

square[j][i].m=0;

square[j][i].y=1;

square[j][i].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

C)

Program Plan Intro

Given Information:

The given code is:

//define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[j][i].c=0;

square[j][i].m=0;

square[j][i].y=1;

square[j][i].k=0;

}

}

Answer 8

Chapter 6, Problem 6.39HW

A)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[j][i].c=0;

square[j][i].m=0;

square[j][i].y=1;

square[j][i].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

Answer 9

Chapter 6, Problem 6.39HW

A)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[j][i].c=0;

square[j][i].m=0;

square[j][i].y=1;

square[j][i].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

Answer 10

B)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[j][i].c=0;

square[j][i].m=0;

square[j][i].y=1;

square[j][i].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

Answer 11

B)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[j][i].c=0;

square[j][i].m=0;

square[j][i].y=1;

square[j][i].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

Answer 12

C)

Program Plan Intro

Given Information:

The given code is:

//define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[j][i].c=0;

square[j][i].m=0;

square[j][i].y=1;

square[j][i].k=0;

}

}

Answer 13

C)

Program Plan Intro

Given Information:

The given code is:

//define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[j][i].c=0;

square[j][i].m=0;

square[j][i].y=1;

square[j][i].k=0;

}

}

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

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Chapter 6 Solutions