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4 years ago

Modern database tools support ________________, which entails separating data from the programs that manipulate data.

1 answer:

8 0

The correct answer: Data independence

Data independence is the type of data transparency that matters for a centralised DBMS. It refers to the immunity of user applications to changes made in the definition and organization of data. Physical data independence deals with hiding the details of the storage structure from user applications.

The logical structure of the data is known as the 'schema definition'. In general, if a user application operates on a subset of the attributes of a relation, it should not be affected later when new attributes are added to the same relation. Logical data independence indicates that the conceptual schema can be changed without affecting the existing schemas.

The physical structure of the data is referred to as "physical data description". Physical data independence deals with hiding the details of the storage structure from user applications. The application should not be involved with these issues since, conceptually, there is no difference in the operations carried out against the data.

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3 years ago

g Write a program to sort an array of 100,000 random elements using quicksort as follows: Sort the arrays using pivot as the mid

shtirl [24]

Answer:

header.h->function bodies and header files.

#include<iostream>

#include<cstdlib>

#include<ctime>

using namespace std;

/* Partitioning the array on the basis of piv value. */

int Partition(int a[], int bot, int top,string opt)

{

int piv, ind=bot, i, swp;

/*Finding the piv value according to string opt*/

if(opt=="Type1 sort" || opt=="Type3 sort")

{

piv=(top+bot)/2;

}

else if(opt=="Type2 sort" || opt=="Type4 sort")

{

piv=(top+bot)/2;

if((a[top]>=a[piv] && a[top]<=a[bot]) || (a[top]>=a[bot] && a[top]<=a[piv]))

piv=top;

else if((a[bot]>=a[piv] && a[bot]<=a[top]) || (a[bot]>=a[top] && a[bot]<=a[piv]))

piv=bot;

}

swp=a[piv];

a[piv]=a[top];

a[top]=swp;

piv=top;

/*Getting ind of the piv.*/

for(i=bot; i < top; i++)

{

if(a[i] < a[piv])

{

swp=a[i];

a[i]=a[ind];

a[ind]=swp;

ind++;

}

swp=a[piv];

a[piv]=a[ind];

a[ind]=swp;

return ind;

}

void QuickSort(int a[], int bot, int top, string opt)

{

int pindex;

if((opt=="Type3 sort" || opt=="Type4 sort") && top-bot<19)

{

/*then insertion sort*/

int swp,ind;

for(int i=bot+1;i<=top;i++){

swp=a[i];

ind=i;

for(int j=i-1;j>=bot;j--){

if(swp<a[j]){

a[j+1]=a[j];

ind=j;

}

else

break;

}

a[ind]=swp;

}

else if(bot < top)

{

/* Partitioning the array*/

pindex =Partition(a, bot, top,opt);

/* Recursively implementing QuickSort.*/

QuickSort(a, bot, pindex-1,opt);

QuickSort(a, pindex+1, top,opt);

}

return ;

}

main.cpp->main driver file

#include "header.h"

int main()

{

int n=100000, i;

/*creating randomized array of 100000 numbers between 0 and 100001*/

int arr[n];

int b[n];

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

arr[i]=(rand() % 100000) + 1;

clock_t t1,t2;

t1=clock();

QuickSort(arr, 0, n-1,"Type1 sort");

t2=clock();

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

arr[i]=b[i];

cout<<"Quick sort time, with pivot middle element:"<<(t2 - t1)*1000/ ( CLOCKS_PER_SEC );

cout<<"\n";

t1=clock();

QuickSort(arr, 0, n-1,"Type2 sort");

t2=clock();

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

arr[i]=b[i];

cout<<"Quick sort time, with pivot median element:"<<(t2 - t1)*1000/ ( CLOCKS_PER_SEC );

cout<<"\n";

t1=clock();

QuickSort(arr, 0, n-1,"Type3 sort");

t2=clock();

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

arr[i]=b[i];

cout<<"Quick sort time and insertion sort time, with pivot middle element:"<<(t2 - t1)*1000/ ( CLOCKS_PER_SEC );

cout<<"\n";

t1=clock();

QuickSort(arr, 0, n-1,"Type4 sort");

t2=clock();

cout<<"Quick sort time and insertion sort time, with pivot median element:"<<(t2 - t1)*1000/ ( CLOCKS_PER_SEC );

return 0;

}

Explanation:

Change the value of n in the main file for different array size. Output is in the same format as mentioned, time is shown in milliseconds.

7 0

3 years ago

Computer system with a 32-bit logical address and 4k-byte page size. assume that each entry of a page table consists of 4bytes.

Annette [7]

The number of bits in the physical address is 26 bits. The number of entries in a page table is $\mathbf{2^{20}}$ entries. The size of the page table in a one-level paging scheme is 4MB.

<h3>
What is paging in Operating System?</h3>

Paging is a storage method used in operating systems to recover activities as pages from secondary storage and place them in primary memory. The basic purpose of pagination is to separate each procedure into pages.

We are given the following parameters:

32-bit logical address
Page size = 4 KB = $\mathbf{2^{12} \ bytes }$
Size of each Page table entry = 4 bytes

Suppose the system supports physical memory size = 64 MB = $\mathbf{2^{26} \ bytes}$

Thus, the number of bits in the physical address is computed as:

= $\mathbf{log_2 \{Physical-memory-size\}}$

$=\mathbf{log_2(2^{26})}}$

= 26 bits

The number of entries in a page table = logical address space size/page size

The number of entries in a page table $\mathbf{= \dfrac{2^{32}}{2^{12}}}$

$\mathbf{=2^{20}}$ entries

In a one-level paging scheme, the size of the table is:

= entire no. of page entries × page table size

= $\mathbf{2^{20}\times 4 \ bytes}$

= 4 MB

suppose that this system supports up to 2^30 bytes of physical memory.

The size of the page table will be the same as 4 MB due to the fact that the number of entries, as well as, the page table entry size is the same.

Since the size of the page table surpasses that of a single page. A page cannot include a whole page table. Therefore, the page table must be broken into parts to fit onto numerous pages, and an additional level of the page table is required to access this page table.

This is called the Multi-Level Paging system.

Therefore, we can conclude that the number of bits in the physical address is 26 bits, the number of entries in a page table is $\mathbf{2^{20}}$ entries, and the size of the page table in a one-level paging scheme is 4MB.

Learn more about Paging in Operating System here:

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4 0

2 years ago

Which of the following would be the most appropriate way to address and greet your teacher, Mr. Joseph Herman, in an email? Hi J

Valentin [98]

To determine what would be the most appropriate way to address and greet your teacher during an email, we should eliminate some greetings, such as:
"Yo wassup?", "How u doin?", or any other grammatical and socially inappropriate errors.
Let's look at our first option.
"Hi Joseph, How u doin??". This is incorrect as it is not appropriate to address anyone in such a manner and with grammatical errors.
Let's look at our second option.
"Dear Joseph Herman, how are you doing!!". This was on the right path, but didn't end well. The ending of the message, "how are you doing!!" is incorrect punctuation, and has too much excitement.
How about our third option?
"Dear Mr. Herman, I hope you're doing fine.". This is a great email. It has perfect punctuation, grammar, and is appropriate.
What about our fourth?
"Dear Joseph, i hope you are doing great.". This is a good email, but has incorrect punctuation.
Your answer is C.) Dear Mr. Herman, I hope you're doing fine.

8 0

3 years ago

Need help with this

andrezito [222]

Answer:

13:a. 15:c. 14:Unknown answer

6 0

3 years ago