How do I get a code to generate 4 different random numbers from 1 through 56?

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This is a followup to my previous question. Using this code, I'd like to know how to generate 4 different random numbers from 1-56 for the yes case, same from 1-51 for the no case, and a loop back to the start for if neither case is used.

import java.util.Scanner;

public class RandomPerkSelector {

public static void main(String [] args){
Scanner userInputReader = new Scanner(System.in);
System.out.println("Are you playing as a survivor?");
while(true){
String userInput = userInputReader.nextLine();
if(userInput.equals("Yes")){
   //"Yes" case
   //generate your numbers for "Yes"
   break;
}else if(userInput.equals("No")){
   //"No" case
   //generate your numbers for "No"
   break;
}else{
   System.out.println("This is a yes or no question.");
   continue; 
   }
  }
 }
}

Im not sure what your asking but this should make 4 random numbers up to 56.

Random rand = new Random();
int[] nums = new int[4];
for (int i = 0; i < 4; i++) {
    int[i] = rand.nextint(57);
}

New Perspectives Computer Concepts 2016 Enhanced, Comprehensive, How do I get a code to generate 4 different random numbers from 1 through 56? random number generator how to generate random numbers in java within range How to generate many random various numbers? We need an array of random various numbers between x[0] and x[n] like shuffling the cards. For example: 8, 10, 5, 4, 7, 6, 3, 1, 9, 2 (between 1 and 10)

You can create a method to avoid duplicating code.

    private static int obtainRandomNumberInRange(int min, int max) {
        if (min >= max) {
            throw new IllegalArgumentException("Min in greater than Max value");
        }

        Random r = new Random();
        return r.nextInt((max - min) + 1) + min;
    }

This, will return a int value between min and max. You can take this method and call it the times you want.

For example, If you are generating numbers for yes:

List<Integer> numbers = new ArrayList<Integer>();
//This will call the function 4 times, and get numbers between 100 and 200. 
IntStream.range(0, 4).forEach(x-> numbers.add(obtainRandomNumberInRange(100, 200)));

Now the numbers have 4 random numbers between 100 and 200.

Or you can use a for statement:

for (int i = 1; i <= 4; i++) {
       numbers.add(obtainRandomNumberInRange(100, 200));
}

Random Number Generator, number. generator? Most programming languages provide a random number generator to create the random numbers that are needed for certain programs, Enter the following lines of code: import random x = random 5: # Get random number in range 1 through 56. n = random.randint(1,56) print(n) count = count + 1  Hello, To begin with, i'm new to programming so try to keep it simple We're making a mastermind game at school, and i've done some of it already, but i need help with how to generate 4 unique random numbers, put into an array.

i do not know what you are going to do but you can use this code

int n = 2 * (Integer.MAX_VALUE / 3);
int low = 0;
for (int i = 0; i < 4; i++)
if (random(n) < n/2) {
    low++;
 System.out.println(low);
 }

by this method you will never get repeated number i hope you will understand.

Generate Random Numbers using Python, By default the tool is generating a random number from 1 to 10. number sequences (default 1); Select if you want unique random numbers in a This tool can be used for any games, lottery numbers, bets where you need random numbers. channel to find the best coupon codes and deals for your online shopping! Store 1 to 100 numbers in an Array. Generate random number between 1 to 100 as position and return array[position-1] to get the value . Once you use a number in array, mark the value as -1 ( No need to maintain another array to check if this number is already used) If value in array is -1, get the random number again to fetch new location in array.

How to easily Generate Random Numbers in Java, This random module contains pseudo-random number generators for various call the random() method to generate a real (float) number between 0 and 1. 36​, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57,  When you generate random numbers it's often the case that each generated number number must be unique. A good example is picking lottery numbers. A good example is picking lottery numbers. Each number picked randomly from a range (e.g., 1 to 40) must be unique, otherwise, the lottery draw would be invalid.

COMP101, random(). You can use the following loop to generate them- public class DemoRandom{ public static void main(String[] args) { for(int xCount  A random number generator, like the ones above, is a device that can generate one or many random numbers within a defined scope. Random number generators can be hardware based or pseudo-random number generators. Hardware based random-number generators can involve the use of a dice, a coin for flipping, or many other devices.

Random Number Generators, Therefore if a suitable predefined method exist in the Java SDK Main method */ public static void main(String[] args) { int ranNum Table 4: Source code for random number application class (version2) 92 96 68 44 56 97 74 12 86 54 26 66 90 7 51 10 88 37 84 81 1  Most programming languages have inbuilt code for generating pseudo-random numbers. Some of them can give you random numbers from a distribution like (binomial, normal, poisson) or you can transform uniformly distributed random numbers using a func

Comments
  • I'm not sure if this code will loop yet or not with an answer outside of the "yes" or "no" boundaries. - Don't guess if it works. test the code!!! I'm sure you can run your code ant type yes/no/maybe just like any of us can.
  • The problem is that I honestly have no idea how to do the random number generation and have tried to find an answer and failed.
  • Okay, so if I can still put in input after I put in an answer different from "yes" or "no", is that considered it looping? Sorry, I'm actually quite new to java.
  • You tell me. Only you know what the requirement is. Does it do what YOU want? I would expect if the person enters an invalid value you would display a message and then prompt for the input again.
  • Well, it lets me still put in the input, so I guess?
  • When you post code you should then 1) select the code you posted 2) click on the "{}" button so the code will retain its formatting. Also this code won't work, since you have a return statement when you do the first iteration of the loop. If you want 4 number you would add them to an Array or List.
  • (1+) for a simple loop.
  • When I input this code, it tells me that it's missing ] and ; on the 4th line even though it looks right, so I can't test it.
  • @SolFox. so fix the problem. Take the time to understand the suggestion. The OP made a typo. The code you find here is not tested and is used to give you an example. You need to specify the variable name.
  • Okay, so by specify that you meant to put something in such as int PerkNumber; right?
  • Seriously? the OP doesn't understand the basics of using if statements and a while loop and you are trying to solve the problem with a stream. Am I the only person who thinks people should learn the basics first before moving on to more advance features of the language?
  • @camickr "Am I the only person who thinks people should learn the basics.." Nope. I SMH at the Swing (about the only tag I explicitly follow these days) questions where the OP does not understand simple if / else statements or looping structures, or solving (or at least investigating) basic compilation errors.