All categories

3 years ago

True/False

Computers and Technology

1 answer:

Gnom [1K]3 years ago

8 0

Answer:

True

Explanation:

Since most database applications are Server-Client structured, the User interface sits on the clients and the database sits in the server... when I say database I'm referring to the DBMS (Database Management System). A server is needed to host the DBMS and in turn it stores data from the UI and feeds data back to the UI based on the instruction written by the developer.

You might be interested in

RDBMSs enforce integrity rules automatically. <br> a. True<br> b. False

IRINA_888 [86]

It is true. RDBMS enforce integrity rules automatically. Integrity rules are automatically enforced by RDBMSs.

Integrity rules are automatically enforced by RDBMSs. Because it contains the design choices made regarding tables and their structures, a data dictionary is frequently referred to as "the database designer's database." Any character or symbol intended for mathematical manipulation may be contained in character data.

Data points that are connected to one another are stored and accessible in a relational database, which is a form of database. The relational model, an easy-to-understand method of representing data in tables, is the foundation of RDBMS. Each table row in a relational database is a record with a distinct ID known as the key.

It is simple to determine the associations between data points because the table's columns carry the properties of the data and each record typically has a value for each property.

To know more about RDBMS click on the link:

brainly.com/question/13326182

#SPJ4

8 0

1 year ago

8. Given the array String[] words, which already contains 1 or more values, write a block of code which counts and returns the n

Vikentia [17]

Answer:

Following are the code to this question:

public class Main//defining a class

{

public static void main(String[] arg)//defining main method

{

String[] words={"Key","day", "Know", "kind"};//defining array of String words

int x=0;//defining integer variable for count value

for(int i=0;i<words.length;i++)//defining for loop for count value

{

if(words[i].startsWith("k")||words[i].startsWith("K"))//use if block to check word start from k

x=x+1;//increment the value of x

}

System.out.print("The number of letters which starts from k is: "+ x);//print value with message

}

Output:

The number of letters which starts from k is: 3

Explanation:

In this code, inside the main method an array of String "words" is defined that hold a value, and in the next step an integer variable "x" is defined, which is used to count the letters, which starts from k.

For this, a for loop is used that counts the value and in this an, if block is defined that uses the "startsWith" method to check the value and increment the value of x and at the last, it prints its value.

7 0

3 years ago

To make a duplicate of text, image,chart,graphs etc.

Veronika [31]

Explanation:

You can change the formatting of the rest of the text ( bold, italics, font color or sixe, ect)

5 0

3 years ago

PART 2 - Exercise 2 - Programming Assignment

Oliga [24]

Suppose that a linked list is used as a data structure for the hash table, create a program that includes steps his application counts the number of phrases in documents decided on via way of means of the user #include.

<h3>What is Programming?</h3>

It is the manner of making hard and fast commands that inform a pc the way to carry out a task. Programming may be finished the usage of lots of pc programming languages, inclusive of JavaScript, Python, and C++.

This application counts the quantity of phrases in documents decided on via way of means of the user.
#include
#include
#include the usage of namespace std; //Function 1 int count(string word)go back word.size();
}
//Function 2
int vowel(string word)> filename;
//Open the {input|enter">enter file.
inputfile.open(filename.c_str()); cout<<"nWord listing after format";
cout<<"n____nn"; //If the file succesfully opened, process it.if (inputfile).

Read more about program :

brainly.com/question/1538272

#SPJ1

5 0

2 years ago

There are n poor college students who are renting two houses. For every pair of students pi and pj , the function d(pi , pj ) ou

Nuetrik [128]

Answer:

Here the given problem is modeled as a Graph problem.

Explanation:

Input:- n, k and the function d(pi,pj) which outputs an integer between 1 and n2

Algorithm:-We model each student as a node. So, there would be n nodes. We make a foothold between two nodes u and v (where u and v denote the scholars pu and pv respectively) iff d(pu,pv) > k. Now, Let's call the graph G(V, E) where V is that the vertex set of the graph ( total vertices = n which is that the number of students), and E is that the edge set of the graph ( where two nodes have edges between them if and only the drama between them is bigger than k).

We now need to partition the nodes of the graph into two sets S1 and S2 such each node belongs to precisely one set and there's no edge between the nodes within the same set (if there's a foothold between any two nodes within the same set then meaning that the drama between them exceeds k which isn't allowed). S1 and S2 correspond to the partition of scholars into two buses.

The above formulation is akin to finding out if the graph G(V,E) is a bipartite graph. If the Graph G(V, E) is bipartite then we have a partition of the students into sets such that the total drama <= k else such a partition doesn't exist.

Now, finding whether a graph is bipartite or not is often found using BFS (Breadth First algorithm) in O(V+E) time. Since V = n and E = O(n2) , the worst-case time complexity of the BFS algorithm is O(n2). The pseudo-code is given as

PseudoCode:

// Input = n,k and a function d(pi,pj)

// Edges of a graph are represented as an adjacency list

1. Make V as a vertex set of n nodes.

2. for each vertex u ∈ V

$\rightarrow$ for each vertex v ∈ V

$\rightarrow\rightarrow$ if( d(pu, pj) > k )

$\rightarrow\rightarrow\rightarrow$ add vertex u to Adj[v] // Adj[v] represents adjacency list of v

$\rightarrow\rightarrow\rightarrow$ add vertex v to Adj[u] // Adj[u] represents adjacency list of u

3. bool visited[n] // visited[i] = true if the vertex i has been visited during BFS else false

4. for each vertex u ∈ V

$\rightarrow$ visited[u] = false

5. color[n] // color[i] is binary number used for 2-coloring the graph

6. for each vertex u ∈ V

$\rightarrow$ if ( visited[u] == false)

$\rightarrow\rightarrow$ color[u] = 0;

$\rightarrow\rightarrow$ isbipartite = BFS(G,u,color,visited) // if the vertices reachable from u form a bipartite graph, it returns true

$\rightarrow\rightarrow$ if (isbipartite == false)

$\rightarrow\rightarrow\rightarrow$ print " No solution exists "

$\rightarrow\rightarrow\rightarrow$ exit(0)

7. for each vertex u ∈V

$\rightarrow$ if (color[u] == 0 )

$\rightarrow\rightarrow$ print " Student u is assigned Bus 1"

$\rightarrow$ else

$\rightarrow\rightarrow$ print " Student v is assigned Bus 2"

BFS(G,s,color,visited)

1. color[s] = 0

2. visited[s] = true

3. Q = Ф // Q is a priority Queue

4. Q.push(s)

5. while Q != Ф {

$\rightarrow$ u = Q.pop()

$\rightarrow$ for each vertex v ∈ Adj[u]

$\rightarrow\rightarrow$ if (visited[v] == false)

$\rightarrow\rightarrow\rightarrow$ color[v] = (color[u] + 1) % 2

$\rightarrow\rightarrow\rightarrow$ visited[v] = true

$\rightarrow\rightarrow\rightarrow$ Q.push(v)

$\rightarrow\rightarrow$ else

$\rightarrow\rightarrow\rightarrow$ if (color[u] == color[v])

$\rightarrow\rightarrow\rightarrow\rightarrow$ return false // vertex u and v had been assigned the same color so the graph is not bipartite

}

6. return true

3 0

3 years ago