What is the name of the output devices of an computer?

The IT person most involved with system development is the _________

laila [671]

Answer:

douch

Explanation:

3 0

3 years ago

To achieve the maximum confidentiality and integrity found in a completely secure information system would require that the syst

fgiga [73]

Answer:

The answer is "Option a".

Explanation:

Privacy is a set of rules, which limiting access to the data, and integrity ensures impartiality, accuracy, and reliability is assured to ensure, which authorized persons can have secure access to the data.

The system needs no access for everyone to achieve its secrecy and integrity contained in a completely safe data system, that's why the given statement is "true".

5 0

3 years ago

2. Write a program with a function that accepts a string as an argument and returns a copy of the string with the first characte

Oduvanchick [21]

Answer:

Following are the code to this question:

def capital(val):#defining a method that takes string value as parameter

x=True#defining boolean variable

r=''#defining string variable

for i in val:#defining loop to convert the first character into upper case

if i.isalpha() and x:#defining condition to check input value is string

r=r+i.upper()#change value into uppercase and hold value in r variable

x=False#assign value false in boolean variable

elif i=='.' or i=='?' or i=='!':#check symbols

r=r+i#add value in r variable

x=True#assign value True in boolean variable

else:

r=r+i#add all value in r variable

return r#return r variable value

val=input()#input value in val variable

print(capital(val))#called the function which print the return value

Output:

please find the attachment.

Explanation:

In the above python program, a method "capital" is declared, which accepts a "val" a string value in its parameter, and inside the method one boolean "x" and one string "r" variable is used, in which r stores return value.
In the next step, for loop is declared, inside the loop, the conditional statement is used, in if the block it checks string value and converts the first character into upper case and assigns value false in the boolean variable.
In the next step, elif block is defined that adds value in r variable and at the last, it will return function value, at the last step "val" variable is declared that input value from the user and pass into the method and print its return value.

4 0

2 years ago

PGP encryption can be performed from the command line as well. What is the PGP command line syntax to encrypt the my-message.txt

Mrac [35]

Answer:

pgp --encrypt "my-message.txt" --recipient (Sean) --output ("secret-message.txt.gpg")

Explanation:

PGP encryption from the command line can be done using the following format:

pgp --encrypt (input) --recipient (user) --output (output file)

Therefore, PGP command line syntax to encrypt the my-message.txt file for a specific user (Sean) and save the output as secret-message.txt.gpg is:

pgp --encrypt "my-message.txt" --recipient (Sean) --output ("secret-message.txt.gpg")

6 0

3 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