To generate random integer numbers between specific range

[Mudit Agarwal]

I have to write a code to generate set of arrays in which each array should contain distinct five integers arranged in ascending order and range in which random number can be generated is between 1-52 both inclusive.....but i have written code below for just generating five random integers and it is not working first two numbers are always 0....can any body tell me whats the heck!!??

import java.util.*;
import java.math.*;
public class CollTest2{
	public static void main(String...args){
		Random poke=new Random();
		int[] A=new int[5];
		int[]B=new int[5];
	
			B=A;
		for(int i=0;i<5;i++)
		{
			A[i]=1+poke.nextInt(52-1);
			Arrays.sort(A);
		}
		for(int n : A)
			System.out.println(n);		
	}	
}

[Pat Farrell]

Random number generators in nearly all languages default to using the same seed, which makes then produce the same sequence of results. If you want to get different numbers each time you run, use something that varies, like the current time, as an argument to set the seed. Its usually easier to debug your code when you have the same sequence over and over -- once its working, you then set a non-default seed.

[Campbell Ritchie]

Welcome to the Ranch

I have added code tags, which you should always use, to your post, and you can see how much better it looks I also removed some blank lines, because they don't always make it easier to read.

I ran your code, and removed lines 2, 7, 8 and 9, which don't do anything. I think the problem has nothing to do with the randomness of the numbers. You will occasionally get 0 0 0 .... printed out.

java CollTest2
0
0
7
22
30
java CollTest2
0
0
7
31
47
java CollTest2
0
0
14
35
43
etc...

I suggest you go through the flow of control like this, on a sheet of paper.

first random number is 28

array is now[28,0,0,0,0]

The next thing I write down will explain why the first number is always 0

As I said, follow the flow control carefully.

[Mudit Agarwal]

hey Campbell thanks for posting on my query will definitely go through flow again..and will post the problem if it occurs again

yea actually i was writing code for something else ..code i posted is a very small part of that program,complete question i am mentioning below

Question -1- write a program (any language you want) to loop through all possible poker hands once each. Here, a “poker hand” is a list of 5 distinct numbers from 1..52, e.g. {2, 3, 7, 13, 40}. Your program should be able to generate each such array, with
the 5 numbers in ascending order.....

[Richard Tookey]

Mudit Agarwal wrote:hey Campbell thanks for posting on my query will definitely go through flow again..and will post the problem if it occurs again

yea actually i was writing code for something else ..code i posted is a very small part of that program,complete question i am mentioning below

Question -1- write a program (any language you want) to loop through all possible poker hands once each. Here, a “poker hand” is a list of 5 distinct numbers from 1..52, e.g. {2, 3, 7, 13, 40}. Your program should be able to generate each such array, with
the 5 numbers in ascending order.....

This does not require any random numbers; it requires that you pick all combination of 5 from 52 using an ordered approach. This can be done either by using 5 'for' loops or by recursion to a depth of 5 and no actual sorting is required since both approaches automatically sort the output.

[Piet Souris]

hi Mudit,

hope you figured out why you always get two zeroes.

If I look at your original listing, I wonder: how are you going to make sure you have 5 distinct values?
And a randomizer will not help you to list all possible hands (which are over 300M).

Greetings,
Piet

[Mudit Agarwal]

Hey Piet ,
yes i figured out why i was getting two zeroes at the starting it got resolved.

I think my approach to the problem is not correct..and yes randomizer is not always giving 5 distinct number and also i was thinking how i will loop through all the sets of possible poker hands.... is it possible can you help with the algorithm of the problem.

Regards

Mudit

[Richard Tookey]

Mudit Agarwal wrote:. is it possible can you help with the algorithm of the problem.t

It is almost trivial ! Run a loop with an index (say a) from 1 to (52-4). Inside that run a loop with an index (say b) from ((a+1) to (52-3). Inside that run a loop with an index (say c) ...

Get the idea?

Inside the inner most loop the loop variables are the values you want !

[Piet Souris]

hi Mudit,

What Richard writes is what I would have written. What I always do with questions like these, is to picture a very
simplified verios, write it out and see if I see some patterns arising.

So, lets take a sorted set of 3 out of a collection of 5.

Now, if you write out all possibilities, in an orderly way, you get the following possible outcomes:

{1, 2, 3}

(1, 2, 4}

(1, 2, 5}

{1,3, 4}

{1, 3, 5}

{1, 4, 5}

{2, 3, 4}

{2, 3, 5}

{2, 4, 5}

{3, 4, 5}

Ten in all. Do you see a pattern here? It is exacly what Richard described.

Now, the general formula for a sorted set of K diifrent elements out of a set of N different elements, is

N * (N-1) * ... * (N - K + 1) / [ 1 * 2 * ... * K ] usually written as

( N )
( K )

In the example from 3 out of 5, you get 5 * 4 * 3 / (1 * 2 * 3) = 10

and for the poker hands: 5 out of 52, or ...

Indeed, it is almost trivial (but only if you are almost a mathematician!)

Greetings,
Piet

[Mudit Agarwal]

Thanks Piet and Richard for nice explanation to the problem i got my code working...now i have to make few amendments in my base code to solve more quetions related to the base code. Here i am posting my code and the questions that i am trying to solve based on it.

public class Poker {
	public static void main(String...args){
		int i,j=0,k=0,l=0,m=0;
		long count=0;
		int eq=0;
		int sum=0;
		for(i=1;i<=(52-4);i++)
		{
			for(j=i+1;j<=(52-3);j++)
			{
				for(k=j+1;k<=(52-2);k++)
				{
					for(l=k+1;l<=(52-1);l++)
					{
						for(m=l+1;m<=52;m++)
						{
							int C=0;
							int []A={i,j,k,l,m};
							for(int n=0;n<A.length;n++)
							{
								
							System.out.print(A[n]+" ");
							 eq=eq+A[n];
							
							 C++;
							 if(C==5)
							 {
								 System.out.println();
							     C=0;
							 }
							 
							 
							}
							if(eq==130)
								sum=sum+1;
							  count++;
							  
						      
						}
						
						
					
			}
		}
	}
}
  System.out.println("number of lines=" + count);	
  System.out.println("number of poker hands whose numbers add up to 130 =" + sum);	
  
}
}

1.Print each poker hand on its own line, and count the number of unique lines you have....(This also helps verifying your code)
2. Check if there are any lines that are duplicates
3. Count the number of poker hands whose numbers add up to 130.
3. Count the number of poker hands whose numbers add up to a multiple of 16.
4. Count the number of poker hands whose numbers multiply together to form a product that is a multiple of 45 but not a multiple of 27.
5. Instead of simply looping through every poker hand, you should be able to input any number between 1 and total number of lines, and your program should output a unique poker hand associated with that number.

[Piet Souris]

hi Mudit,

now, questions 1 and 2 are simple to answer. The method used to generate all the hands garantee the uniqueness of each hand.
The 'count' variable is the total number. You have that already,

Also, you have the number of hands for which the sum of the values equals 130.
In the same fashion you can check if that value is a multiple of 16, and also if the product
is a multiple of 45 but not a multiple of 27.

Nastier is the last question. The way you generate the hands now makes that you loose
the hand after each iteration. Also, if you look at the code,you will probably admit that
it looks a bit messy.

To tackle both problems, I would suggest to define a 'Hand' class, something simple like

class Hand {
   private int[] card;
   private int sumOfCards;
   private int productOfCards;

   Hand(int first, int second, int third, int fourth, int fifth) {
      card = new int[] {first, second, third, fourth, fifth};
      sumOfCards = first + // et cetera
      productOfCards = first * second * ...  // et cetera
   }

   public int getSum() { return sumOfCards;}
   public int getProduct() {return productOfCards; }

   @Override
   public String toString() { return ... }

}  // end of class

and instead of your int[]A, you simply create a new Hand.

Now, I don't know if you are allowed to use an ArrayList, but if so,
create an ArrayList<Hand>, and add each Hand to that list.

If not, I suggest you create an array of about 3.000.000 entries,
like Hand[] hands = new Hand[3000000],
and when you have created a new Hand, you add it to the array by
hand[count] = Hand_just_created.

Having produced a complete array of Hands, you can do, glass of bear
next to the pc, big sigar in the mouth, huge grin on the face, perform the questions, like:

for(Hand hand : hands) {
   int sum = hand.getSum();
   int prod = hand.getProduct;
   if (sum is multiple of 16 or prod is multiple of whatever) {...}
}

and (in pseudo code)

input a number between 0 and <count>: x
println("Hand is: " + hands[x])

Greetz,
Piet

Edit: by the way, almost forgot to tell: did you reset the variable 'eq' to 0, in ech iteration?

Edit 2: added a 'toString' method to the Hand class

[nithatufro tufro]

Hey! I was doing this task with modern java and I was able to code a very simple function generateRandomInteger following this tutorial to generate random integers with java 1.7 or later.

Hope this information helps to anyone. If you don't want to follow the link and just use the function:

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

    Random generator = new Random(System.currentTimeMillis()); // see comments!
    int randomNum = rand.nextInt((max - min) + 1) + min;

    return rand;
}

[Carey Brown]

Pat Farrell wrote:Random number generators in nearly all languages default to using the same seed, which makes then produce the same sequence of results. If you want to get different numbers each time you run, use something that varies, like the current time, as an argument to set the seed. Its usually easier to debug your code when you have the same sequence over and over -- once its working, you then set a non-default seed.

From the documentation for Random:

public Random()
Creates a new random number generator. This constructor sets the seed of the random number generator to a value very likely to be distinct from any other invocation of this constructor.

So, using this constructor will start with a random seed, not a default seed. If you want to control the seed being used you need to use the Random(seed) constructor.

[Carey Brown]

nithatufro tufro wrote:Hey! I was doing this task with modern java and I was able to code a very simple function generateRandomInteger following this tutorial to generate random integers with java 1.7 or later.

Hope this information helps to anyone. If you don't want to follow the link and just use the function:

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

    Random generator = new Random(System.currentTimeMillis()); // see comments!
    int randomNum = rand.nextInt((max - min) + 1) + min;

    return rand;
}

See prior post. Using currentTimeMillis() is not as random as what you'd get using the empty constructor, Random().

[Campbell Ritchie]

nithatufro tufro wrote:. . . this tutorial. . . .

That tutorial is not very helpful; as well as what Carey says, it doesn't explain the potential problems with the different techniques, nor give you a recommendation about which is best. It also for some peculiar reason uses thread local random rather than java.util‍.Random. Yoiu will find a thorough discussion here, but of course all those discussions are out of date now the Random#ints() method and simlar methods are available.

And welcome to the Ranch