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Hunter-Best [27]
4 years ago
5

Write four overloaded methods called randomize. Each method will return a random number based on the parameters that it receives

:
a. randomize() - Returns a random int between min and max inclusive. Must have two int parameters.

b. randomize() - Returns a random int between 0 and max inclusive. Must have one int parameter.

c. randomize() - Returns a random double between min and max inclusive. Must have two double parameters.

d. randomize() - Returns a random double between 0 and max inclusive. Must have one double parameter.
Computers and Technology
2 answers:
Scilla [17]4 years ago
8 0

Answer:

The Java code given below

Explanation:

//import package

import java.util.Scanner;

//Java class

class Lesson_35_Activity {

  // returns random int b/t min and max inclusive; has 2 int parameters

  public static int randomize(int min, int max) {

      int x = (int) (Math.random() * ((max - min) + 1)) + min;

      return x;

 

Yuri [45]4 years ago
6 0

Answer:

The Java code given below

Explanation:

//import package

import java.util.Scanner;

//Java class

class Lesson_35_Activity {

  // returns random int b/t min and max inclusive; has 2 int parameters

  public static int randomize(int min, int max) {

      int x = (int) (Math.random() * ((max - min) + 1)) + min;

      return x;

  }

  // returns random int b/t 0 and max inclusive; has one int parameter

  public static int randomize(int max) {

      int x = (int) (Math.random() * (max + 1));

      return x;

  }

  // returns random double b/t min and max inclusive; two double parameters

  public static double randomize(double min, double max) {

      double x = (double) (Math.random() * ((max - min) + 1)) + min;

      return x;

  }

  // returns random double b/t 0 and max inclusive; has one double parameter.

  public static double randomize(double max) {

      double x = (double) (Math.random() * (max + 1));

      return x;

  }

  // main method, which is entry point of the program

  public static void main(String[] args) {

      // object of Scanner class

      Scanner scan = new Scanner(System.in);

      // asking min and max number

      System.out.print("Enter min number : ");

      int mi = scan.nextInt();// reading min number

      System.out.print("Enter max number : ");

      int ma = scan.nextInt();// reading max number

      // checking number

      if (mi < ma) {

          System.out.println(randomize(mi, ma));// method call

          System.out.println(randomize(0, ma));// method call

          double mii = mi;

          double maa = ma;

          System.out.printf("%.2f", randomize(mii, maa));

          System.out.printf("\n%.2f", randomize(0, maa));

      } else {

          // when minimum number is greater than maximum number then display

          System.out.println("Enter min number " + mi + " should be less than max " + ma);

      }

  }

}

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Musya8 [376]

Answer:

The answer to this question is given below in the explanation section.

Explanation:

The for-loop given in the question is:

for ( j = 0; j < 10; j++ )

{

      appendItem (myList, aNumber); //this loop append a number to a list myList

}

This loop starts from J variable's value zero and when J's value is less than 10, the loop iterate through its body until J's value becomes greater or equal to 10. As J's value exceed nine, the loop will get terminated.

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3 years ago
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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\rightarrowif( 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

\rightarrowvisited[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"

\rightarrowelse

\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
Write a program palindrome.py that prompts for a sequence of words or numbers on a single line and checks if the entries are pal
olya-2409 [2.1K]

Answer:

Explanation:

The program first asks the user for the sequence of words. Then it splits the sequence into an array of words. Then it loops through the array checking each word to see if it is a palindrome. If it is it prints the word, the boolean value, and adds 1 to the palindrome_count variable. Otherwise it prints the word, false, and moves on to the next word in the list. Finally, it prints out the total value of palindrome_count.

word = input("Enter sequence of words: ")

word_list = word.split(' ')

print(word_list)

palindrome_count = 0

for word in word_list:

   print('\n\n')

   reverse = word[::-1]

   if word == reverse:

       print(word, end='\n')

       print(True, end="\n")

       palindrome_count += 1

   else:

       print(word, end='\n')

       print(False, end='\n')

print("\n\nNumber of Palindromes in Sequence: " + str(palindrome_count))

3 0
3 years ago
A user saves a password on a website the user logs into from a desktop. Under which circumstances will the password be saved on
Alja [10]

Answer:

Same browser

Explanation:

A user saves a password on a website the user logs into from a desktop. As long as the the browser used is the same and the user has saved the password on the browser, it does not matter which device the user uses, the password would be saved on the website.

Therefore same browser is important for a website to save the password logged in from any other device.

I hope the answer was helpful.

Thanks for asking.

6 0
3 years ago
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