CODEWITHSUNDEEP
javaarraysloops
BeEasyImNoob
BeEasyImNoob

Reputation: 21

Shuffling a deck of cards until getting a specific number

I'm still new at this and I am having a hard time in this project. I need to shuffle the cards and get the first four cards and computing the sum of the four cards. I've done this task so far. Now my problem is I need to repeat this method until I can get the sum total of 24. I'm having a hard time on how to loop the method. Here's is the code I've done so far:

public class cards
{
    public static void main(String[] args)
    {
       int[] deck = new int[52];
       String[] suits = {"Spades", "Hearts", "Diamonds", "Clubs"};
       String[] ranks = {"Ace", "2", "3", "4", "5", "6", "7", "8", "9", "10", "Jack", "Queen", "King"};
       int rankedVal[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13};
       int tot = 0;
            if (tot != 24){  
       //Initialize cards
    tot = 0;
   for (int i = 0; i < deck.length; i++)
        deck[i] = i;



   //Shuffle cards    
    for (int i = 0; i < deck.length; i++)
   {
        int index = (int)(Math.random() * deck.length);
        int temp = deck[i];
        deck[i] = deck[index];
        deck[index] = temp;
   }

    //Display first four                
    for (int i = 0; i < 4; i++)
   {
        String suit = suits[deck[i] / 13];
        String rank = ranks[deck[i] % 13];
        System.out.println("Card number " + deck[i] + ": " + rank + " of " + suit);
   }  

     //find the total 
           for (int i = 0; i<4;i++){
        tot = tot + rankedVal[deck[i] % 13];
    }


 System.out.println("Total Hand = " + tot);
}
}

}

Upvotes: 2

Views: 286

Answers (3)

Anshul
Anshul

Reputation: 730

That's why you should learn how to use methods and subroutines. And if you are using Java then definitely you should first learn a bit about Object Oriented Programming. If you don't know about OOP, I wouldn't recommend Java as a beginner language. Go for C or Python maybe.

public class cards
{
    int[] deck = new int[52];
    String[] suits = {"Spades", "Hearts", "Diamonds", "Clubs"};
    String[] ranks = {"Ace", "2", "3", "4", "5", "6", "7", "8", "9", "10", "Jack", "Queen", "King"};
    int rankedVal[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13};

    void init() {
       // initialise logic
       for (int i = 0; i < deck.length; i++)
           deck[i] = i;
    }

    void shuffleCards() {
       for (int i = 0; i < deck.length; i++) {
        // Shuffle logic
       }
    }

    void displayFirstFour() {
       // display logic
    }

    int sumFirstFour() {
      int sum = 0;
      // logic to sum first four cards
      return sum;
    }

    public static void main(String[] args)
    {
       init();
       int total = 0;
       // LOOP. NOT 'if' . 'if' simply tests condition once and moves on
       // loop tests condition continually until true
       while(total != 24) {
         //Shuffle cards    
         shuffleCards();  
         //Display first four                
         displayFirstFour();
         total = sumFirstFour();
       }
       System.out.println("Total Hand = " + total);
    }
}

Upvotes: 1

Michel Keijzers
Michel Keijzers

Reputation: 15367

Without making a spoiler, you could use a condition in a while loop until 24 is reached, like:

do
{
    // Get 4 cards and add sum.
 } while (sum < 24);

Also you need to check you didn't run out of cards, so an improvement would be:

int cardsUsed = 0
do
{
   // Get 4 cards and add sum.
   cardsUsed += 4;
} ((sum < 24) or (cards_used >= 52));

Upvotes: 1

user3079666
user3079666

Reputation: 1165

I will use a little more generic code as different representations suit different programmers, but this should work with anything:

int sum = 0;
While (sum != 24)
{
    Shuffle();
    sum = deck[0].value + deck[2].value + deck[3].value + deck[4].value;
}

It would be better if you made deck an array of cards and exchanged between pairs when shuffling, and cards having an integer value within them.. Initialize once and then it's all swapping the objects. It's Java, isn't it?

Upvotes: 1

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