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3 years ago

Below is a list of 32-bit memory address references, given as word addresses.

2 answers:

5 0

For a direct mapped cache the general rule is: first figure out the bits of the offset (the right-most bits of the address), then figure out the bits of the index (the next-to right-most address bits), and then the tag is everything left over (on the left side).

One way to think of a direct mapped cache is as a table with rows and columns. The index tells you what row to look at, then you compare the tag for that row, and if it matches, the offsettells you which column to use. (Note that the order you use the parts: index/tag/offset, is different than the order in which you figure out which bits are which: offset/index/tag.)

So in part (a) The block size is 1 word, so you need 0 offset bits (because 20=120=1). You have 16 blocks, so you need 4 index bits to give 16 different indices (because 24=1624=16). That leaves you with the remaining 28 bits for the tag. You seem to have gotten this mostly right (except for the rows for "180" and "43" where you seem to have missed a few bits, and the row for "181" where you interchanged some bits when converting to binary, I think). You are correct that everything is a miss.

For part (b) The block size is 2 words, so you need 1 offset bit (because 21=221=2). You have 8 blocks, so you need 3 index bits to give 8 different row indices (because 23=823=8). That leaves you with the remaining 28 bits for the tag. Again you got it mostly right except for the rows for "180" and "43" and "181". (Which then will change some of the hits and misses.)

koban [17]3 years ago

5 0

i don't know this answer

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Implement a C program to solve the 15-puzzle problem using the A* search algorithm.

fomenos

Answer:

#include<stdio.h>

#include<conio.h>

int m=0,n=4;

int cal(int temp[10][10],int t[10][10])

{

int i,j,m=0;

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

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

{

if(temp[i][j]!=t[i][j])

m++;

}

return m;

}

int check(int a[10][10],int t[10][10])

{

int i,j,f=1;

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

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

if(a[i][j]!=t[i][j])

f=0;

return f;

}

void main()

{

int p,i,j,n=4,a[10][10],t[10][10],temp[10][10],r[10][10];

int m=0,x=0,y=0,d=1000,dmin=0,l=0;

clrscr();

printf("\nEnter the matrix to be solved,space with zero :\n");

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

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

scanf("%d",&a[i][j]);

printf("\nEnter the target matrix,space with zero :\n");

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

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

scanf("%d",&t[i][j]);

printf("\nEntered Matrix is :\n");

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

{

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

printf("%d\t",a[i][j]);

printf("\n");

}

printf("\nTarget Matrix is :\n");

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

{

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

printf("%d\t",t[i][j]);

printf("\n");

}

while(!(check(a,t)))

{

l++;

d=1000;

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

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

{

if(a[i][j]==0)

{

x=i;

y=j;

}

//To move upwards

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

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

temp[i][j]=a[i][j];

if(x!=0)

{

p=temp[x][y];

temp[x][y]=temp[x-1][y];

temp[x-1][y]=p;

}

m=cal(temp,t);

dmin=l+m;

if(dmin < d)

{

d=dmin;

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

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

r[i][j]=temp[i][j];

}

//To move downwards

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

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

temp[i][j]=a[i][j];

if(x!=n-1)

{

p=temp[x][y];

temp[x][y]=temp[x+1][y];

temp[x+1][y]=p;

}

m=cal(temp,t);

dmin=l+m;

if(dmin < d)

{

d=dmin;

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

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

r[i][j]=temp[i][j];

}

//To move right side

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

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

temp[i][j]=a[i][j];

if(y!=n-1)

{

p=temp[x][y];

temp[x][y]=temp[x][y+1];

temp[x][y+1]=p;

}

m=cal(temp,t);

dmin=l+m;

if(dmin < d)

{

d=dmin;

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

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

r[i][j]=temp[i][j];

}

//To move left

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

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

temp[i][j]=a[i][j];

if(y!=0)

{

p=temp[x][y];

temp[x][y]=temp[x][y-1];

temp[x][y-1]=p;

}

m=cal(temp,t);

dmin=l+m;

if(dmin < d)

{

d=dmin;

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

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

r[i][j]=temp[i][j];

}

printf("\nCalculated Intermediate Matrix Value :\n");

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

{

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

printf("%d\t",r[i][j]);

printf("\n");

}

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

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

{

a[i][j]=r[i][j];

temp[i][j]=0;

}

printf("Minimum cost : %d\n",d);

}

getch();

}

Explanation:

7 0

3 years ago

How do I create a videogame

mote1985 [20]

Answer:

-Learn how to code

-Make Character

-Design

-Go to classes

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-Play video games and see the scripts

-Decide what type of video game your making(horror, family game, battle game, etc.)

Explanation:

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3 years ago

How to connect local account to domain account?

muminat

Join them to the domain
Login with their domain credentials, logout
Login as local administrator (not the old account, not the new one, a 3rd local admin)
Right click My Computer and select properties
Select advanced system settings
Go to the Advanced tab
Click settings under user profiles
Select their old profile you want and click Copy to
Browse to the location of their new profile and overwrite it

If the Copy button is grayed out then reboot and repeat steps two to nine.

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4 years ago

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S_A_V [24]

Answer: Option [C] : The longest time path through the network.

Explanation:

The critical path of a network is the longest time path through the network.

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Reptile [31]

P {
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font-size: 16px;
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4 years ago