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vladimir2022 [97]
2 years ago
8

Algorithmic Complexity: what is the asymptotic complexity (Big-O) of each code section? Identify the critical section of each.\

Line 1: for (int i=0; i<532; i++) { f(n) = O( ) Line 2: for (int j=1; j
Computers and Technology
1 answer:
AleksandrR [38]2 years ago
4 0

Answer:

Check the explanation

Explanation:

1) f(n) = O( 1 ), since the loops runs a constant number of times independent of any input size

there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.

2) f(n) = O( log n! ), the outer loop runs for n times, and the inner loop runs log k times when i = k,ie the total number of print will be – log 1 + log2 +log3 +log4+…...+ log n = log (1 . 2 . 3 . 4 . ……. . n ) =log n!

there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.

Note : Log (m *n) = Log m + Log n : this is property of logarithm

3) f(n) = O( n^2 ), since both outer and inner loop runs n times hence , the total iterations of print statement will be : n +n+n+…+n

for n times, this makes the complexity – n * n = n2

there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.

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What are common names for some primary discrete components used on circuit boards?
Alenkinab [10]
Transistor, resistor, capacitor, inductor, integrated circuit
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2 years ago
If you were investigating login issues on a Windows computer, which portion of the Event Viewer logs would be a good place to st
USPshnik [31]

Answer:

If we are investigating login issues then we have to start with 'security logs' in 'windows logs' portion of Event viewer.

Explanation:

Much information about login issues is contained in log files which are related to security because it is mostly security issue. Some it is also better to start with 'system logs' portion of windows logs portion of Event viewer when there may be system problems instead of security issues. But in most cases there is security issues so 'security logs' is better option overall

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2 years ago
Describe a strategy for avoiding nested conditionals. Give your own example of a nested conditional that can be modified to beco
harina [27]

Answer:

One of the strategies to avoid nested conditional is to use logical expressions such as the use of AND & operator.

One strategy is to use an  interface class with a method. That method can be created to be used for a common functionality or purpose. This is also called strategy design pattern. You can move the chunk of conditional statement to that method. Then each class can implement that interface class and use that shared method according to their own required task by creating objects of sub classes and call that common method for any such object. This is called polymorphism.

Explanation:

Nested conditionals refers to the use of if or else if statement inside another if or else if statement or you can simply say a condition inside another condition. For example:

if( condition1) {  

//executes when condition1 evaluates to true

 if(condition2) {

//executes when condition1  and condition2 evaluate to true

 }  else if(condition3) {

 //when condition1 is true and condition3 is true

} else {

 //condition1  is true but neither condition2 nor conditions3 are true

}  }

The deeply nested conditionals make the program difficult to understand or read if the nested conditionals are not indented properly. Also the debugging gets difficult when the program has a lot of nested conditionals.

So in order to avoid nested conditionals some strategies are used such as using a switch statement.

Here i will give an example of one of the strategies i have mentioned in the answer.

Using Logical Expressions:

A strategy to avoid nested conditionals is to use logical expressions with logical operators such as AND operator. The above described example of nested conditionals can be written as:

if(condition1 && condition2){  //this executes only when both condition1 and condition2 are true

} else if(condition1 && condition3) {

this executes only when both condition1 and condition3 are true

} else if(condition1 ){

//condition1  is true but neither condtion2 nor condtion3 are true  }

This can further be modified to one conditional as:

if(!condition3){

// when  condition1 and condition2 are true

}

else

// condition3 is true

Now lets take a simple example of deciding to go to school or not based on some conditions.

if (temperature< 40){

  if (busArrived=="yes")   {

      if (!sick)       {

          if (homework=="done")           {

              printf("Go to school.");

          }

      }     

  }

}

This uses nested conditionals. This can be changed to a single conditional using AND logical operator.

if ((temperature <40) && (busArrived=="yes") &&

(!sick) && (homework=="done"))

{    cout<<"Go to school."; }

6 0
2 years ago
I am doing keyboarding keyboarding is very boring and yeah
Bingel [31]

Answer:

yes I agree with you

Explanation:

I have been keyboarding for the past 7 hours.

7 0
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Read 2 more answers
This question involves the creation of user names for an online system. A user name is created based on a user’s first and last
Evgen [1.6K]

Answer:

See explaination

Explanation:

import java.util.*;

class UserName{

ArrayList<String> possibleNames;

UserName(String firstName, String lastName){

if(this.isValidName(firstName) && this.isValidName(lastName)){

possibleNames = new ArrayList<String>();

for(int i=1;i<firstName.length()+1;i++){

possibleNames.add(lastName+firstName.substring(0,i));

}

}else{

System.out.println("firstName and lastName must contain letters only.");

}

}

public boolean isUsed(String name, String[] arr){

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

if(name.equals(arr[i]))

return true;

}

return false;

}

public void setAvailableUserNames(String[] usedNames){

String[] names = new String[this.possibleNames.size()];

names = this.possibleNames.toArray(names);

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

if(isUsed(usedNames[i],names)){

int index = this.possibleNames.indexOf(usedNames[i]);

this.possibleNames.remove(index);

names = new String[this.possibleNames.size()];

names = this.possibleNames.toArray(names);

}

}

}

public boolean isValidName(String str){

if(str.length()==0) return false;

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

if(str.charAt(i)<'a'||str.charAt(i)>'z' && (str.charAt(i)<'A' || str.charAt(i)>'Z'))

return false;

}

return true;

}

public static void main(String[] args) {

UserName person1 = new UserName("john","smith");

System.out.println(person1.possibleNames);

String[] used = {"harta","hartm","harty"};

UserName person2 = new UserName("mary","hart");

System.out.println("possibleNames before removing: "+person2.possibleNames);

person2.setAvailableUserNames(used);

System.out.println("possibleNames after removing: "+person2.possibleNames);

}

}

8 0
2 years ago
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