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Gre4nikov [31]
2 years ago
10

In cyber security, one of the best ways to protect a computer or network is with a strategy called defense in depth. This strate

gy means that there are multiple defenses put in place. If one fails, it is likely another will catch the problem. Which method of reducing complexity is this concept most closely aligned with
Computers and Technology
2 answers:
emmainna [20.7K]2 years ago
5 0

This idea is most closely related to the Layering in terms of how to reduce complexity.

<h3>Describe cybersecurity?</h3>

The term "cybersecurity" refers to the safeguarding of computer systems and networks against the leakage, theft, or destruction of data as well as the interruption of the services they offer.

Defense in depth is a technique used in cybersecurity that is one of the greatest ways to safeguard a computer or network. If one fails, it's likely that another will identify the issue. The approach to simplifying that is most in line with is called layering.

Thus, This idea is most closely related to the Layering in terms

For further details about cybersecurity, click here:

brainly.com/question/13016321

#SPJ1

galina1969 [7]2 years ago
3 0

The method of reducing complexity is this concept most closely aligned with the Layering.

<h3>What is cyber security?</h3>

Cyber security refers to the protection of computer systems and networks from information disclosure, theft, or damage as well as the disruption of the services they provide.

In cyber security, one of the best ways to protect a computer or network is with a strategy called defense in depth. If one fails, it is likely another that will catch the problem. Layering refers to the method of reducing complexity that is most closely aligned with.

Learn more about cyber security here:

brainly.com/question/17175779

#SPJ1

You might be interested in
Write a program that reads numbers from a file (or allow user to input data) and creates an ordered binary tree. The program sho
IgorLugansk [536]

Answer:

See explaination

Explanation:

include<bits/stdc++.h>

using namespace std;

typedef struct Node

{

int data;

struct Node *left,*right;

}Node;

bool search(Node *root,int data)

{

if(root==NULL)

return false;

if(root->data==data)

return true;

queue<Node*> q;

q.push(root);

while(!q.empty())

{

Node *temp=q.front();

q.pop();

if(temp->data==data)

return true;

if(temp->left)

q.push(temp->left);

if(temp->right)

q.push(temp->right);

}

return false;

}

Node *insert(Node *root,int data)

{

if(root==NULL)

{

Node *temp=new Node();

temp->data=data;

temp->left=NULL;

temp->right=NULL;

return temp;

}

if(data < root->data)

root->left=insert(root->left,data);

if(data>root->data)

root->right=insert(root->right,data);

return root;

}

Node *get_smallest_element_right_subtree(Node *root)

{

while(root && root->left!=NULL)

root=root->left;

return root;

}

Node *delete_node(Node *root,int data)

{

if(root==NULL)

return root;

if(data < root->data)

root->left=delete_node(root->left,data);

else if(data > root->data)

root->right=delete_node(root->right,data);

else

{

if(root->left==NULL) //If right only presents means - delete the curr node and return right node

{

Node *temp=root->right;

free(root);

return temp;

}

else if(root->right==NULL) //If left only presents means - delete the curr node and return let node

{

Node *temp=root->left;

free(root);

return temp;

}

else

{

Node *temp=get_smallest_element_right_subtree(root->right);

root->data=temp->data;

root->right=delete_node(root->right,temp->data);

}

return root;

}

}

void inorder(Node *root)

{

if(root!=NULL)

{

inorder(root->left);

cout<<root->data<<" ";

inorder(root->right);

}

}

void postorder(Node *root)

{

if(root!=NULL)

{

inorder(root->left);

inorder(root->right);

cout<<root->data<<" ";

}

}

void preorder(Node *root)

{

if(root!=NULL)

{

cout<<root->data<<" ";

inorder(root->left);

inorder(root->right);

}

}

int main()

{

fstream f;

string filename;

cout<<"\n\n1 - Input through File ";

cout<<"\n\n2 - Input through your Hand";

int h;

cout<<"\n\n\nEnter Your Choice : ";

cin>>h;

Node *root=NULL; // Tree Declaration

if(h==1)

{

cout<<"\n\nEnter the Input File Name : ";

cin>>filename;

f.open(filename.c_str());

if(!f)

cout<<"\n\nError in Opening a file !";

else

{

cout<<"\n\nFile is Being Read ........";

string num;

int value;

int node=0;

while(f>> num)

{

value=stoi(num);

root=insert(root,value);

node++;

}

cout<<"\n\nTree has been successfully created with : "<<node<<" Nodes"<<endl;

}

}

if(h==2)

{

int y;

cout<<"\n\nEnter the Total No of Input :";

cin>>y;

int i=1,g;

while(i!=y+1)

{

cout<<"\n\nEnter Input "<<i<<" : ";

cin>>g;

root=insert(root,g);

i++;

}

cout<<"\n\nTree has been successfully created with : "<<y<<" Nodes"<<endl;

}

if(h>=3)

{

cout<<"\n\nInvalid Choice !!! ";

return 0;

}

int n=0;

while(n!=6)

{

cout<<"\n\n\n1 - Insert Element";

cout<<"\n\n2 - Remove Element";

cout<<"\n\n3 - Inorder (LNR) Display ";

cout<<"\n\n4 - Pre (NLR) Order Display";

cout<<"\n\n5 - Post (LRN) Order Display";

cout<<"\n\n6 - Quit";

cout<<"\n\nEnter Your Choice : ";

cin>>n;

switch(n)

{

case 1:

{

int k;

cout<<"\n\nEnter Element to insert : ";cin>>k;

root=insert(root,k);

cout<<"\n\nElement Sucessfully Inserted !!!!!";

break;

}

case 2:

{

int k;

cout<<"\n\nEnter Element to Remove : ";

cin>>k;

if(search(root,k))

{

root=delete_node(root,k);

cout<<"\n\nValue Successfully Deleted !!!";

}

else

cout<<"\n\n!!!!!!!!!!!!!!!!!!!! No Such Element !!!!!!!!!!!!!!!!!!!!!!";

break;

}

case 3:

{

cout<<"\n\nThe Elements (LNR) are : ";

inorder(root);

break;

}

case 4:

{

cout<<"\n\nThe Elements (NLR) are : ";

preorder(root);

break;

}

case 5:

{

cout<<"\n\nThe Elements (LRN) are : ";

postorder(root);

break;

}

case 6:

{

break;

}

}

}

cout<<"\n\nBye!!!! See You !!!"<<endl;

7 0
4 years ago
Question 2 of 10
NeTakaya

Answer:

D.

Because you said it was.

3 0
2 years ago
A technician is using a network-attached desktop computer with a Type 2 hypervisor to run two VMs. One of the VMs is infected wi
lianna [129]

Answer:

Place both VMs on virtual NICs that are isolated from the company LAN.

Explanation:

From the scenario above, for the technician to best protect the company's LAN while performing the above mentioned tests, it us best to place both VMs on virtual NICs that are isolated from the company LAN.

Cheers

4 0
4 years ago
Create an Entity-Relationship Diagram with the following requirements.
melamori03 [73]

Answer:

idk

Explanation:

6 0
3 years ago
Write a program with a function that accepts a string as an argument and returns the number of uppercase, lowercase, vowel, cons
diamong [38]

Answer:

Explanation:

The following code is written in Python. It is a function called checkString that takes in a string as an argument and loops through each char in that string and checking to see if it is lowercase, uppercase, vowel, consonant, or punctuations. It adds 1 to the correct variable. At the end of the loop it prints out all of the variables. The picture below shows a test output with the string "Brainly, Question."

def checkString(word):

   uppercase = 0

   lowercase = 0

   vowel = 0

   consonants = 0

   punctuation = 0

   vowelArray = ['a', 'e', 'i', 'o','u', 'y' ]

   consonantArray = ['b', 'c', 'd', 'f', 'g', 'h', 'j', 'k', 'l', 'm', 'n', 'p', 'q', 'r', 's', 't', 'v', 'w', 'x', 'y', 'z']

   punctuationArray = ['.', '!', '?', ',', ';', '(', ')']

   for char in word:

       if char.isupper():

           uppercase += 1

       else:

           lowercase += 1

       if char.lower() in vowelArray:

           vowel += 1

       elif char.lower() in consonantArray:

           consonants += 1

       if char in punctuationArray:

           punctuation += 1

   print('Uppercase: ' + str(uppercase))

   print('Lowercase: ' + str(lowercase))

   print('Vowels: ' + str(vowel))

   print('Consonants: ' + str(consonants))

   print('Punctuations: ' + str(punctuation))

8 0
3 years ago
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