CODEWITHSUNDEEP
java
KAKAK
KAKAK

Reputation: 899

Please help me understand this piece of code

I have a java project done for school. This is a piece of code which i am having a hard time to understand its logic. Please shed some light on it.

for(int i = 0; i< leftbut.length; i++){
          int randomNumber =(int)(Math.random()*leftbut.length);
             tempNum = leftbut[randomNumber];
            leftbut[randomNumber] = leftbut[i];
            leftbut[i]=tempNum;      

     }

The leftbut in this case is actually an array of 9 buttons. This code is supposed to shuffle the 9 buttons in different positions. I just cant understand how this code works.

Upvotes: 0

Views: 84

Answers (3)

amit
amit

Reputation: 178411

The code generate a random permutation of the original array.

However, note that this is biased - it does not generate all permutations in uniform distribution. This thread discusses what is the affect of this bias.

To overcome this issue - you might want to have a look on fisher yates shuffle (which main difference is, to generate a random number in range [i,n) in each iteration instead of in range [0,n).)

EDIT:
You might better understand it if you encapsulate the assignments in a method:

private static void swap(int[] array, int i, int j) { 
       tempNum = array[j];
       array[j] = array[i];
       array[i]=tempNum; 
}

Now, the code will be simpler to follow:

for(int i = 0; i< leftbut.length; i++) {
          //choose a random index in the array
          int randomNumber =(int)(Math.random()*leftbut.length);
          //swap the element in index i with the random element chosen in the array
          swap(leftbut, i, randomNumber);
}

The idea is you swap() each element in the array with a random index. This random index is chosen randomly from the array, and its index is denoted as randomNumber.
Since you only swap() items around, you can easily prove that the output array is a permutation of the original.

Upvotes: 2

lelloman
lelloman

Reputation: 14173

for(int i = 0; i< leftbut.length; i++){

Is a loop, it inizialize variable i to 0, and increment it each loop by 1

int randomNumber =(int)(Math.random()*leftbut.length);

declare the integer variable randomNumber and assign a random value in range 0 - length of array

 tempNum = leftbut[randomNumber];         
 leftbut[randomNumber] = leftbut[i];
 leftbut[i]=tempNum;

this actually invert the 2 buttons position in the array, the value i become the random one and viceversa

Upvotes: 1

Damian Jeżewski
Damian Jeżewski

Reputation: 1246

It just 9 times randomly swaps to elements of leftbut array.

Upvotes: 1

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