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3 years ago

Namespaces are guaranteed to be unique. True False

Computers and Technology

1 answer:

kupik [55]3 years ago

5 0

Answer:

True

Explanation:

Some collection of classes or library functions grouped as one name space. A class which belongs to one namespace is different from the class which belongs to another namespace. we can identify a class uniquely with it's namespace .

for ex:

in c#.net

using system;

System.IO;

here System is the namespace which contains class IO

namespace contains any number of classes . In one namespace we can't define two classes with same Name. We can define two classes with same name in different namespaces

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[1] Please find all the candidate keys and the primary key (or composite primary key) Candidate Key: _______________________ Pri

AVprozaik [17]

Answer:

Check the explanation

Explanation:

1. The atomic attributes can't be a primary key because the values in the respective attributes should be unique.

So, the size of the primary key should be more than one.

In order to find the candidate key, let the functional dependencies be obtained.

The functional dependencies are :

Emp_ID -> Name, DeptID, Marketing, Salary

Name -> Emp_ID

DeptID -> Emp_ID

Marketing -> Emp_ID

Course_ID -> Course Name

Course_Name -> Course_ID

Date_Completed -> Course_Name

Closure of attribute { Emp_ID, Date_Completed } is { Emp_ID, Date_Completed , Name, DeptID, Marketing, Salary, Course_Name, Course_ID}

Closure of attribute { Name , Date_Completed } is { Name, Date_Completed , Emp_ID , DeptID, Marketing, Salary, Course_Name, Course_ID}

Closure of attribute { DeptID, Date_Completed } is { DeptID, Date_Completed , Emp_ID,, Name, , Marketing, Salary, Course_Name, Course_ID}

Closure of attribute { Marketing, Date_Completed } is { Marketing, Date_Completed , Emp_ID,, Name, DeptID , Salary, Course_Name, Course_ID}.

So, the candidate keys are :

{ Emp_ID, Date_Completed }

{ Name , Date_Completed }

{ DeptID, Date_Completed }

{ Marketing, Date_Completed }

Only one candidate key can be a primary key.

So, the primary key chosen be { Emp_ID, Date_Completed }..

The functional dependencies are :

Emp_ID -> Name, DeptID, Marketing, Salary

Name -> Emp_ID

DeptID -> Emp_ID

Marketing -> Emp_ID

Course_ID -> Course Name

Course_Name -> Course_ID

Date_Completed -> Course_Name

For a relation to be in 2NF, there should be no partial dependencies in the set of functional dependencies.

The first F.D. is

Emp_ID -> Name, DeptID, Marketing, Salary

Here, Emp_ID -> Salary ( decomposition rule ). So, a prime key determining a non-prime key is a partial dependency.

So, a separate table should be made for Emp_ID -> Salary.

The tables are R1(Emp_ID, Name, DeptID, Marketing, Course_ID, Course_Name, Date_Completed)

and R2( Emp_ID , Salary)

The following dependencies violate partial dependency as a prime attribute -> prime attribute :

Name -> Emp_ID

DeptID -> Emp_ID

Marketing -> Emp_ID

The following dependencies violate partial dependency as a non-prime attribute -> non-prime attribute :

Course_ID -> Course Name

Course_Name -> Course_ID

So, no separate tables should be made.

The functional dependency Date_Completed -> Course_Name has a partial dependency as a prime attribute determines a non-prime attribute.

So, a separate table is made.

The final relational schemas that follows 2NF are :

R1(Emp_ID, Name, DeptID, Marketing, Course_ID, Course_Name, Date_Completed)

R2( Emp_ID , Salary)

R3 (Date_Completed, Course_Name, Course_ID)

For a relation to be in 3NF, the functional dependencies should not have any transitive dependencies.

The functional dependencies in R1(Emp_ID, Name, DeptID, Marketing, Date_Completed) is :

Emp_ID -> Name, DeptID, Marketing

This violates the transitive property. So, no table is created.

The functional dependencies in R2 ( Emp_ID , Salary) is :

Emp_ID -> Salary

The functional dependencies in R3 (Date_Completed, Course_Name, Course_ID) are :

Date_Completed -> Course_Name

Course_Name -> Course_ID

Here there is a transitive dependency as a non- prime attribute ( Course_Name ) is determining a non-attribute ( Course_ID ).

So, a separate table is made with the concerned attributes.

The relational schemas which support 3NF re :

R1(Emp_ID, Name, DeptID, Course_ID, Marketing, Date_Completed) with candidate key as Emp_ID.

R2 ( Emp_ID , Salary) with candidate key Emp_ID.

R3 (Date_Completed, Course_Name ) with candidate key Date_Completed.

R4 ( Course_Name, Course_ID ). with candidate keys Course_Name and Course_ID.

6 0

3 years ago

_______ make up the basic structure of a relational database with columns containing field data and rows containing record infor

Klio2033 [76]

QR codes make up the basic structure of a relational database with columns containing field data and rows containing record information.

3 0

3 years ago

Read 2 more answers

Which of the following processes should Angel use to merge cells A1:D4 to create a title?

podryga [215]

Answer:

Highlight cells A1:D4, right-click and select Format Cells, click Alignment, and choose the Merge cells option.

Explanation:

We merge cells with the help of following methods.

Method 1

Highlight cells A1:D4
click on the Home tab
click on the Merge and Center icon in the Alignment group

1st method is not given in options of question.

Method 2

Highlight cells A1:D4
right-click and select Format Cells
click Alignment
choose the Merge cells option

2nd method is matching with option D. So, Option D is the answer.

6 0

3 years ago

Assuming you can use infinite number of CPUS for performance improvements, what would be the maximum speed up when the paralleli

natita [175]

Answer:

correct option is D i.e. 20

Explanation:

Given data:

Parallelizable portion is given as 95%

so remaining will be serial portion i.e. 5%

we know that, for infinite number of cpus, maximum speed is calculated as

$v =\frac{1}{f}$

where f denote serial portion = 5% = 0.05

$v = \frac{1}{0.05}$

v = 20

hence so for serial portion of 5%, maximum speed is 20

correct option is D i.e. 20

3 0

3 years ago

Given a string variable string 's;', implement a tokenizer to identify the unique tokens contained in this string, identify all

Flauer [41]

Answer:

#include <iostream>

#include <string>

#include <vector>

#include <cctype>

#include <algorithm>

using namespace std;

void matrixInit( vector< vector<int> >& matrix, int numRows, int numCols);

struct TokenFreq{

string token;

int freq;

};

string spaces(string& istr, string& newstr);

void printVector(vector<TokenFreq> & tfVec);

void getTokenFreqVec( string& istr, vector<TokenFreq> & tfVecVec);

void selectionSort( vector<TokenFreq> & tokFreqVector );

void insertionSort( vector<TokenFreq> & tokFreqVector );

//main

int main(int argc, const char * argv[]) {

vector<vector<int> > matrix;

matrixInit(matrix, 3, 4);

string ostr1 = " I do keep the main points straight in my head most of the time, and I do know where to find the details when I need them. by Bjarne Stroustrup";

string ostr2 = " ";

string ostr3 = "Your time is limited, so don’t waste it living someone else’s life.Stay Hungry. Stay Foolish.";

//string ostr = ostr2;

string istr;

cout << "Enter the text: ";

getline(cin, istr);

vector<TokenFreq> tfVec;

getTokenFreqVec(istr, tfVec);

selectionSort(tfVec);

insertionSort(tfVec);

return 0;

}

//-----

void selectionSort( vector<TokenFreq> & tokFreqVector ){

TokenFreq temp;

int min;

for( int i = 0; i < int(tokFreqVector.size()-1); i++){

min = i;

for (int j = i+1; j <int(tokFreqVector.size()); j++){

if (tokFreqVector[j].freq < tokFreqVector[min].freq)

min = j;

}

if(min!=i)

{

temp = tokFreqVector[i];

tokFreqVector[i] = tokFreqVector[min];

tokFreqVector[min] = temp;

}

cout<<"-----selectionSort------------------"<<endl;

printVector(tokFreqVector);

}

void insertionSort( vector<TokenFreq> & tokFreqVector ){

TokenFreq temp;

int max;

for( int i = 0; i < int(tokFreqVector.size()-1); i++){

max = i;

for (int j = i+1; j < int(tokFreqVector.size()); j++){

if (tokFreqVector[j].freq > tokFreqVector[max].freq)

max = j;

}

if(max!=i)

{

temp = tokFreqVector[i];

tokFreqVector[i] = tokFreqVector[max];

tokFreqVector[max] = temp;

}

cout<<"-----insertionsort------------------"<<endl;

printVector(tokFreqVector);

}

string spaces(string& istr, string & newstr){

int j;

//Check double space and create new string with lowercase

for (int i = 0; i < int(istr.length()); ++i){

istr[i] = tolower(istr[i]);

if (isspace(istr[i])){

//check double space again

j = i+1;

while(true){

if (istr[j] == '\342') istr[j]=' ';

if (istr[j] == '\200') istr[j]=' ';

if (istr[j] == '\224') istr[j]=' ';

if(isspace(istr[j])) j++;

else break;

}

i=j-1;

newstr = newstr + istr[i];

}else

newstr = newstr + istr[i];

}

return newstr;

}

void getTokenFreqVec( string& istr, vector<TokenFreq> & tfVec){

string newstr;

int curr_beg = 0;

int curr_end = 0;

string word;

bool check = false;

int index=0;

//call func spaces to get new string

newstr = spaces(istr, newstr);

//cout << newstr << endl;

//cout << endl;

//cout << newstr << "||" << endl;

string tempstr = newstr;

if(newstr != " "){

tfVec.push_back(TokenFreq());

}

for (int i = 0; i < int(newstr.length()); ++i){

if(isspace(newstr[i])){

check = false;

curr_end = i;

word = tempstr.substr(curr_beg, curr_end - curr_beg);

curr_beg = i + 1;

for (int j = 0; j < int(tfVec.size()); j++){

if (tfVec[j].token == word){

tfVec[j].freq++;

check = true;

break;

}

}//end-for;

if (check == false && i!=(int(newstr.length())-1)){

//cout <<"**"<< word <<"**"<<word.length()<< endl;

tfVec[index].token=word;

tfVec[index].freq=1;

index++;

tfVec.push_back(TokenFreq());

}

word = "";

//add last value

}

}//end-for

if(!isspace(newstr[int(newstr.length())-1])){

word = tempstr.substr(curr_beg, curr_end - (int(newstr.length())-1));

tfVec[index].token=word;

tfVec[index].freq=1;

//tfVec.push_back(TokenFreq());

}

printVector(tfVec);

}

void printVector(vector<TokenFreq> & tfVec){

cout << "size= {" << tfVec.size() << endl;

for (int i = 0; i < int(tfVec.size()); i++){

cout << " [" << i << "] = (token = \""<< tfVec[i].token <<"\", freq = " << tfVec[i].freq << ")" << endl;

//cout << tfVec[i].token<<endl;;

}

cout << endl;

}

void matrixInit( vector< vector<int> >& matrix, int numRows, int numCols){

int i;

int j;

matrix.resize(numRows, vector<int>(numCols) );

for(i = 0; i< numRows; i++ ){

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

matrix[i][j] = i*j;

}

cout << "size of matrix is: " << numRows << "x" << numCols << endl;

for (i = 0; i < numRows; i++){

for (j = 0; j < numCols; j++){

cout<< "matrix[" << i << "][" << i << "]=" << matrix[i][j] << endl;

}

cout << endl;

}

Explanation:

3 0

3 years ago