Simplify if condition

[Tiberius Marius]

I m trying to simplify an if condition as much as possible and i m curious if it can be made more short then i managed to .The problem is simple , a integer number with 3 digits is generated and the user is asked to input another 3 digits integer number ... One of the conditions that need to be verified is if all the digits in each number match but not exactly ( aka 123 and 132 ) .Using only an if statement to verify this what is the most short condition that can be checked to identify when the digits in a 3 digits number match the digits of another 3 digits integer number (not interested if they match exactly) .

The simplest condition i managed to find is :

else if (guessDigit1 + guessDigit2 + guessDigit3
        == lotteryDigit1 + lotteryDigit2 + lotteryDigit3 
   && (guessDigit1 == lotteryDigit1 || guessDigit1
        == lotteryDigit2 || guessDigit1 == lotteryDigit3)
   && (guessDigit2 == lotteryDigit1 || guessDigit2
        == lotteryDigit2 || guessDigit2 == lotteryDigit3)) {
      System.out.println("Match all digits: you win $3,000");}

Is there a better(shorter ) condition that can be used ?

[Rico Felix]

This sort of problem can be made much simpler using some logic and the Java API

Here is one proposed way you can accomplish such a task:

import java.util.Arrays;
import java.util.Random;

public class Lottery {

	public static void main(String[] args) {

		// Number generator
		Random randomNumber = new Random();

		// Generate a number and transform it into a character array
		char[] winningNumbers = Integer.toString(randomNumber.nextInt(50))
				.toCharArray();

		// Get user input and transform it into a character array
		char[] userNumbers = System.console().readLine("Enter a number: ")
				.toCharArray();

		// Sort both arrays
		Arrays.sort(winningNumbers);
		Arrays.sort(userNumbers);

		// Check for match
		if (Arrays.equals(winningNumbers, userNumbers)) {
			System.out.println("Match all digits: you win $3,000");
		} else {
			System.out.println("Sorry try again");
		}
	}
}

[Tiberius Marius]

Rico Felix wrote:This sort of problem can be made much simpler using some logic and the Java API

Here is one proposed way you can accomplish such a task:

import java.util.Arrays;
import java.util.Random;

public class Lottery {

	public static void main(String[] args) {

		// Number generator
		Random randomNumber = new Random();

		// Generate a number and transform it into a character array
		char[] winningNumbers = Integer.toString(randomNumber.nextInt(50))
				.toCharArray();

		// Get user input and transform it into a character array
		char[] userNumbers = System.console().readLine("Enter a number: ")
				.toCharArray();

		// Sort both arrays
		Arrays.sort(winningNumbers);
		Arrays.sort(userNumbers);

		// Check for match
		if (Arrays.equals(winningNumbers, userNumbers)) {
			System.out.println("Match all digits: you win $3,000");
		} else {
			System.out.println("Sorry try again");
		}
	}
}

I m using a book to learn and in that book this problem is meant to be resolved using only an if statement ( even for statement / char data type ware not introduced yet and i dont know allot about java arrays at my current study level) . Can you look for a shorter condition inside a simple if statement ?

[Rico Felix]

Oh I see ... let me think

[Paul Ngom]

try this

String lottery=Integer.toString(lotteryDigit1)
   + Integer.toString(lotteryDigit2)
   + Integer.toString(lotteryDigit3);

else if(lottery.contains(Integer.toString(guessDigit1))
   && lottery.contains(Integer.toString(guessDigit2))
   && lottery.contains(Integer.toString(guessDigit3))){
System.out.println("Match all digits: you win $3,000");}  

But i wonder if you will find it shorter

[Tiberius Marius]

No strings

[Tony Docherty]

It doesn't work anyway - if the guess is 222 and the lottery number is 123 it will return true.

A similar problem occurs in your solution Tiberius.

[Paul Ngom]

You are perfectly right Tony. I think i needed to check the other way to like:

String guess=Integer.toString(guessDigit1)
   + Integer.toString(guessDigit2) + Integer.toString(guessDigit3);
String lottery=Integer.toString(lotteryDigit1)
   + Integer.toString(lotteryDigit2) + Integer.toString(lotteryDigit3);

else if(lottery.contains(Integer.toString(guessDigit1))
 && lottery.contains(Integer.toString(guessDigit2))
 && lottery.contains(Integer.toString(guessDigit3))
 && guess.contains(Integer.toString(lotteryDigit1))
 && guess.contains(Integer.toString(lotteryDigit2))
 && guess.contains(Integer.toString(lotteryDigit3))){
System.out.println("Match all digits: you win $3,000");}  

which makes it too long :-)

Is it correct now?

[Winston Gutkowski]

Paul Ngom wrote:which makes it too long :-)

YES.

Guys (specifically Paul and Tiberius): DontWriteLongLines. They make your thread very hard to read.

And that applies whether you write them yourself OR include them in a quote.

I've broken up all of yours THIS TIME. Any more, and you're on your own.

Winston

[Paul Ngom]

DontWriteLongLines

Ok, Winston. I will try to stick to the rule next time.
Regards

[Winston Gutkowski]

Tiberius Marius wrote:Using only an if statement to verify this what is the most short condition that can be checked to identify when the digits in a 3 digits number match the digits of another 3 digits integer number (not interested if they match exactly) .

This is a simple permutation problem, and doesn't jive with your other statement:
"I m trying to simplify an if condition as much as possible"

If you're NOT allowed to use anything other than an if statement involving the three separate digits, then you can't simplify it. The permutations are what they are, even if digits are duplicated (and checking for that, while it might make the logic work faster, is also likely to make it bigger).

There are, however, several ways of simplifying it if you CAN use other logic. My suggestion:
1. Put the digits in Sets (java.util.Set).
2. Check the lengths of those sets. If they're different, the digits can't possibly match.
3. (if needed) Put the digits in Lists (java.util.List), sort them, and check if they are identical. Hopefully, that one is self-explanatory.

Winston


[Edit] Actually, there is ONE check you could do (although it's an addition to all the others): check if the numbers that the digits come from are the same before you embark on your "digit permutation odyssey".

[Paul Ngom]

There are, however, several ways of simplifying it if you CAN use other logic. My suggestion:
1. Put the digits in Sets (java.util.Set).
2. Check the lengths of those sets. If they're different, the digits can't possibly match.
3. Put the digits in Lists (java.util.List), sort them, and check if they are identical. Hopefully, that one is self-explanatory.

Winston, i am waiting to hear from Tiberius. I am almost sure he wouldn't like to use Sets ot Lists as he said that he has has just begun the language. By the way, could you post some code that achieves what you had said?

[Tony Docherty]

You could simply do:

    int[]lottery = new int[]{3,2,1};
    int[]guess = new int[]{1,2,3};

    Set<Integer> lotterySet = new TreeSet<Integer>();
    Set<Integer> guessSet = new TreeSet<Integer>();

    for ( int i = 0; i < lottery.length; i++ )
        {
        lotterySet.add(lottery[i]);
        guessSet.add(guess[i]);
        }

    System.out.println(lotterySet.equals(guessSet));

[Tiberius Marius]

To avoid any more confusion i m going to post the whole exercise .Also remember that while it can be solved with strings/arrays or other objects the problem is in the third chapter and strings/ characters / objects / the for statement has not been introduced . What was introduced is the if statement and all the logical operators . My level of knowledge ends at objects (no knowledge of arrays /stacks because i didnt get that far yet.


*3.15 (from the book Introduction to Java Programming, Tenth Edition by Daniel Liang)
(Game: lottery) Revise Listing 3.8, Lottery.java, to generate a lottery of a three-digit number. The program prompts the user to enter a three-digit number and determines whether the user wins according to the following rules:

1. If the user input matches the lottery number in the exact order, the award is $10,000.

2. If all digits in the user input match all digits in the lottery number, the award is $3,000.

3. If one digit in the user input matches a digit in the lottery number, the award is $1,000

This is Lottery.java :

import java.util.Scanner;

public class Lottery {
  public static void main(String[] args) {
    // Generate a lottery
    int lottery = (int)(Math.random() * 100);

    // Prompt the user to enter a guess
    Scanner input = new Scanner(System.in);
    System.out.print("Enter your lottery pick (two digits): ");
    int guess = input.nextInt();

    // Get digits from lottery
    int lotteryDigit1 = lottery / 10;
    int lotteryDigit2 = lottery % 10;

    // Get digits from guess
    int guessDigit1 = guess / 10;
    int guessDigit2 = guess % 10;

    System.out.println("The lottery number is " + lottery);

    // Check the guess
    if (guess == lottery)
      System.out.println("Exact match: you win $10,000");
    else if (guessDigit2 == lotteryDigit1
          && guessDigit1 == lotteryDigit2)
      System.out.println("Match all digits: you win $3,000");
    else if (guessDigit1 == lotteryDigit1 
          || guessDigit1 == lotteryDigit2 
          || guessDigit2 == lotteryDigit1 
          || guessDigit2 == lotteryDigit2)
      System.out.println("Match one digit: you win $1,000");
    else
      System.out.println("Sorry, no match"); 
  }
}

[Tiberius Marius]

The correct solution(at list i think so ) that can verify the rule 2 is :

if (guess == lottery)
      System.out.println("Exact match: you win $10,000");
    
    else if ((guessDigit1 != lotteryDigit1 && guessDigit1 != lotteryDigit2 
            && guessDigit1 != lotteryDigit3)|| (guessDigit2 != lotteryDigit1 && 
            guessDigit2 != lotteryDigit2 && guessDigit2 != lotteryDigit3) ||
            (guessDigit3 != lotteryDigit1 && guessDigit3 != lotteryDigit2 &&
            guessDigit3 != lotteryDigit3)){
        
      System.out.println("Match all digits: you win $3,000");}

If you tell me it cannot be a more simple condition in the if statement i believe you !

[Campbell Ritchie]

The existence of so many != operators entails a simpler solution in my opinion.
You are doing it wrongly by trying to solve the problem in code. What you should do is solve the problem on a sheet of paper. Write down exactly how you would solve it on paper, and then when you have got it nice and simple, you can turn it into code easily.

[Paul Ngom]

And i will say that it is better that your if statement reads:

    else if (
        (guessDigit1 == lotteryDigit1 || guessDigit1 == lotteryDigit2 || guessDigit1 == lotteryDigit3)
       && (guessDigit2 == lotteryDigit1 || guessDigit2 == lotteryDigit2 || guessDigit2 == lotteryDigit3)
       && (guessDigit3 == lotteryDigit1 || guessDigit3 == lotteryDigit2 || guessDigit3 == lotteryDigit3)){  
          
      System.out.println("Match all digits: you win $3,000");} 

to avoid confusion

[Tiberius Marius]

Paul Ngom wrote:And i will say that it is better that your if statement reads:

    else if (
        (guessDigit1 == lotteryDigit1 || guessDigit1 == lotteryDigit2 || guessDigit1 == lotteryDigit3)
       && (guessDigit2 == lotteryDigit1 || guessDigit2 == lotteryDigit2 || guessDigit2 == lotteryDigit3)
       && (guessDigit3 == lotteryDigit1 || guessDigit3 == lotteryDigit2 || guessDigit3 == lotteryDigit3)){  
          
      System.out.println("Match all digits: you win $3,000");} 

to avoid confusion

Think it is incorrect in the sense 111 and 100 will pass the condition ...

[Paul Ngom]

Think it is incorrect in the sense 111 and 100 will pass the condition

You are right, i am making the same mistake as previously. What about this:

  else if (
	(guessDigit1 == lotteryDigit1 || guessDigit1 == lotteryDigit2 || guessDigit1 == lotteryDigit3)
       && (guessDigit2 == lotteryDigit1 || guessDigit2 == lotteryDigit2 || guessDigit2 == lotteryDigit3)
       && (guessDigit3 == lotteryDigit1 || guessDigit3 == lotteryDigit2 || guessDigit3 == lotteryDigit3)
 
	&& (lotteryDigit1 == guessDigit1 || lotteryDigit1 == guessDigit2 || lotteryDigit1 == guessDigit3)
       && (lotteryDigit2 == guessDigit1 || lotteryDigit2 == guessDigit2 || lotteryDigit2 == guessDigit3)
       && (lotteryDigit3 == guessDigit1 || lotteryDigit3 == guessDigit2 || lotteryDigit3 == guessDigit3)){  
          
      System.out.println("Match all digits: you win $3,000");}  

[Tiberius Marius]

This I think works but the one posted earlier by me is simpler i think ( the one with != )

[Winston Gutkowski]

Paul Ngom wrote:And i will say that it is better that your if statement reads:

Paul,

After all my work, and my post (and your reply), you post ANOTHER piece of code that is FAR TOO LONG.

Please edit it, because this thread is now (again) unreadable for me.
(Unlike you, I don't have 20:10 vision )

Winston

[Edit] And I notice that it has now been quoted by both of you.
If you simply want to swap replies, carry on; if you want to get answers from the people here:
EDIT YOUR POSTS AND BREAK UP THOSE LINES (and now you'll both have to do it).

[Paul Ngom]

This I think works but the one posted earlier by me is simpler i think ( the one with != )

I think the one using != is not correct. You are also doing the same mistake.
You need to apply a ! to your if condition (like if(!expression)) for it to work.

[Tiberius Marius]

Yes , your correct . Forgot to negate the whole condition .

[Campbell Ritchie]

Pancil and paper. When you have got it worked out on paper, then you can easily translate it into code.