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Explain what this code is doing

I'm a newbie in java. I was going through some tutorials and came across this code I was not able to understand the code. Please explain what it means.
class Randoms
{
public static void main(String[] args)
{
Random rand = new Random();
int freq[] = new int[7];
for(int roll = 1; roll < 10; roll++)
{
(++freq[1 + rand.nextInt(6)]);
}
...

Line by line:
Random rand = new Random(); create new instance of the Random object, this is responsible for the creation of random numbers.
int[] freq = new int[7]; create a new int array that can store 7 elements, with indices from 0...6. It is worth noting that in Java, the ints stored in the array are initialized to 0. (This is not true for all languages, an example being C, as in C the int arrays initially store memory junk data, and must be explicitly initialized to zero).
for(int roll = 1; roll < 10; roll++) this rolls 9 times (because 1...9, but it's better practice to go from 0)
(++freq[1 + rand.nextInt(6)]); this line is something that you shouldn't ever do in this sort of fashion, because it's a monstrosity as you can see.
Do something like this:
for(int roll = 0; roll < 9; roll++)
{
int randomNumber = rand.nextInt(6); //number between 0...5
int index = 1 + randomNumber; //number between 1...6
freq[index]++; //increment the number specified by the index by 1
//nearly equivalent to freq[index] += 1;
}
So basically it randomizes the number of 9 dice throws, and stores the dice throw count (or so it calls it, frequency) in the array.
Thus, it's simulating 9 dice throws (numbers from 1...6), and each time it "rolls" a particular number, it increases the number stored in the array at that specific location.
So in the end, if you say:
for(int i = 1; i <= 6; i++)
{
System.out.println("Thrown " + freq[i] + " times of number " + i);
}
Then it will be clearly visible what's happened.

(++freq[1 + rand.nextInt(6)]); // this line of code.
The above line of code is pre-incrementing the value of freq[] array at the specified position,i.e., 1+rand.nextInt(6) --- referred value is ++freq[some-position to be evaluated] which we will evaluate below.
This rand.nextInt(6) will generate an integer number lesser than 6 and greater than 0,as it is a pre-defined method of Random Class ,randomly.We can't predict it.
And,then say number generated is 4. SO, 1+rand.nextInt(6)=5.
Hence,your code would simplify to (++freq[1 + rand.nextInt(6)]) OR `(++freq[5]).
So,simplification of this code will be equivalent to a number which equals 1 more than 6th element of array freq[].
// as freq[5] is the 6th element of the array freq[].
Also,there are some other points which SIR David Wallace suggested me to include which I would like to explain a bit more.It goes below :-
++a here ++ is called pre-increment operator and it increases the value of a by 1. There also exists an altered reverse version of it.
a++ here this ++ is called post-increment operator and it also increases the value of a by 1.But,WAIT,you might have thought that there aren't differences,but there are.
For the differences potion,I'd like to suggest to have a reading of What is the difference between pre-increment and post-increment in the cycle (for/while)?, though it is questioned in sense of C++,the same is in Java too!

// Create a new Random Object, this helps you generate random numbers
Random rand = new Random();
// create a integer array with 7 integers
int freq[] = new int[7];
// loop 9 times
for(int roll = 1; roll < 10; roll++)
{
// rand.nextInt(6) generates a number between 0 and 5 (<6). add one to it
// ++ adds one to the integer in the array that is at the index of 1-6.
(++freq[1 + rand.nextInt(6)]);
}
Some strange things about this code:
Roll loop starts at 1 then goes to 10 so at first glance it would seem to loop 10 times but actually runs 9 times.
The ++ inside the loop would generally be located on the right and could lead to some confusion among newer programmers.
freq[1 + rand.nextInt(6)] causes freq[0] to never be used.

At first a new object of the Random-Class and an array with 7 elements are created. Each element of the Array has the value 0. Inside the for-loop you randomly pick element 2 to 7 of the Array and increase its current value by 1. This is done 9 times.
Your code will never pick the first element of the Array which has the index 0.
I would rewrite the code to make it more clear:
Random rand = new Random();
int freq[] = new int[6];
int randomIndex = 0;
for(int roll = 0; roll < 9; ++roll)
{
randomIndex = rand.nextInt(6);
freq[randomIndex] = freq[randomIndex] + 1;
}
This code has not been tested, but it should basicly do the same.

Generating 1000 random numbers between 13 and 100

I'm trying to generate 1000 random numbers between 13 and 100. So far it's only generating 75% of what I want repeatedly a thousand times. Here's what I have so far:
Random rand = new Random();
for (int j = 0; j < 1000; j++)
{
int pick = rand.nextInt((87) + 13);
pick++;
}
Why isn't it working?

Pay attention to nextInt() covering the 0 inclusively and the specified value exclusively! So it has to be rand.nextInt(88) to make the highest int generated be 87. Here is what you want:
Random rand = new Random();
for (int j = 0; j<1000; j++)
{
int pick = rand.nextInt(88)+13;
}

rand.nextInt(88) + 13; should give you numbers between 13 and 100, and you just put it in your loop.

The line :
So far its only generating 75% of what i want repeatedly a thousand times
Really doesn't add up to me. It might be a seeding issue you're having though. Make sure to always re-seed the random number, using time.
But I agree with Abdul , you need to take the +13 out of that parenthesis:
rand.nextInt(87) + 13;
Because rand.nextInt((87) + 13) is the same as rand.nextInt((67) + 23) as rand.nextInt((1) + 99)
But if you want more "true" randomness, look into something called buzzhash (though that is for hashing ; yet may be modded for number)

I use this in my codes:
public static int randomInteger(int min, int max)
{
java.security.SecureRandom rand = new java.security.SecureRandom();
//get bounded [0, max) from nextInt()
int randomNum = rand.nextInt(max) + min;
return randomNum;
}
value = randomInteger(13,100); //13..99
value = randomInteger(13,101); //13..100

Generate random numbers increasing by multiples of N in Java

I need to write a short randomizer which generates a random number between 1 to N where the random number is increasing by multiples of M.
For example: generate numbers between 1 and (N=30) increasing by multiples of (M=5). The only possible generated numbers can be then: 1,5,10,15,20,25 and 30. Hope you get what I mean :)
Normally if you use new Random().nextInt(30)+1, you get numbers increasing by multiples of 1 (1,2,3,4,5,6,7,etc.). That is not what I want.
Any help, links, or directions are very appreciated?
EDIT:
That the sequence of generated random numbers could include 1 not zero is one of the requirements of the method. Precisely, the sequence always starts with the lower bound (min). In the example the lower bound is 1 and therefore there can't be zero in the sequence. It's odd I know, but those are the requirements I have to follow ;)

You could multiply the random value by 5. Since you want to have 1 instead of 0 just look for that value and alter it specifically:
static final Random random = new Random(System.currentTimeMillis());
public int random(int range, int multiple) {
int value = random.nextInt(range / multiple) * multiple;
return value == 0 ? 1 : value;
}

I think that your example is a little bit flawed: the sequence should be (0, 5, 10, 15, 20, 25, 30)
If I'm correct, you can then use:
new Random().nextInt(7) * 5; // will generate a number between 0 and 7 not included then multiply by 5 which should give what you want
If you want to start by 1, then just add it to the previous statement

Try this..
public static void main(String[] args){
int N = 50;
int M = 5;
System.out.println(getRandomNumber(N, M));
}
static int getRandomNumber(int n, int m){
Random r = new Random();
int rnd = r.nextInt(n);
return rnd%m == 0?rnd:getRandomNumber(n, m);
}

int n=Random.range(1,10);
n=n*100;

How to get a 50/50 chance in random generator

I am trying to get a 50/50 chance of get either 1 or 2 in a random generator.
For example:
Random random = new Random();
int num = random.nextInt(2)+1;
This code will output either a 1 or 2.
Let's say I run it in a loop:
for ( int i = 0; i < 100; i++ ) {
int num = random.nextInt(2)+1 ;
}
How can I make the generator make an equal number for 1 and 2 in this case?
So I want this loop to generate 50 times of number 1 and 50 times of number 2.

One way: fill an ArrayList<Integer> with fifty 1's and fifty 2's and then call Collection.shuffle(...) on it.

50/50 is quite easy with Random.nextBoolean()
private final Random random = new Random();
private int next() {
if (random.nextBoolean()) {
return 1;
} else {
return 2;
}
}
Test Run:
final ListMultimap<Integer, Integer> histogram = LinkedListMultimap.create(2);
for (int i = 0; i < 10000; i++) {
nal Integer result = Integer.valueOf(next());
histogram.put(result, result);
}
for (final Integer key : histogram.keySet()) {
System.out.println(key + ": " + histogram.get(key).size());
}
Result:
1: 5056
2: 4944

You can't achieve this with random. If you need exactly 50 1s and 50 2s, you should try something like this:
int[] array = new int[100];
for (int i = 0; i < 50; ++i)
array[i] = 1;
for (int i = 50; i < 100; ++i)
array[i] = 2;
shuffle(array); // implement shuffling algorithm or use an already existing one

EDIT:
I understand that if you are looking to accomplish exactly 50-50 results, then my answer was not accurate. You should use a pre-filled collection, since it is impossible to achive that using any kind of randomness. This considered, my answer is still valid for the title of the question, so, this is it:
Well, you do not need the rnd generator to do this.
Comming from javascript, I would go with a single liner:
return Math.random() > 0.5 ? 1: 2;
Explanation: Math.random() returns a number between 0(inclusive) and 1(exclusive), so, we just examine weather is larger than 0.5 (middle value). In theory there is a 50% change that does.
For a more generic use, you can just replace 1:2 to true:false

You can adjust the probability along the way so that the probability of getting a one decreases as you get more ones. This way you don't always have a 50% chance of getting a one, but you can get the result you expected (exactly 50 ones):
int onesLeft = 50;
for(int i=0;i<100;i++) {
int totalLeft = 100 - i;
// we need a probability of onesLeft out of (totalLeft)
int r = random.nextInt(totalLeft);
int num;
if(r < onesLeft) {
num = 1;
onesLeft --;
} else {
num = 2;
}
}
This has an advantage over shuffling because it generates numbers incrementally so it desn't need memory to store the numbers.

You have already successfully created a random generator that returns 1 or 2 with equal probability.
As (many) other's have mentioned, your next request, to force an exact 50/50 distributions in 100 trials, does not fall in line with random number generation. As shown in https://math.stackexchange.com/questions/12348/probability-of-getting-50-heads-from-tossing-a-coin-100-times, the realistic expectation of that occurring is only around 8%. So even while you might expect 50 of each, that exact outcome is actually rather rare.
The Law of Large Numbers states that you should close in on expected value as your number of trials increases.
So for your actual question: How can I make the generator make an equal number for 1 and 2 in this case?
The best (humorous) answer I can come up with is: "Run it in an infinite loop."

What is the best way to create a random value?

I am triying to create random value for my game to show enemies on screen. BUt it some times shows 2 together some times 3 ...I want to ask that which is the best formul for creating random value.. This is my so far random value
random = 1 * (int) (Math.random() * 100);

"BUt it some times shows 2 together some times 3"
Given perfectly random numbers... In every 100 random values from 0 to 99, you'll find an average of 1.0 doubles. A triple will occur on average once for every 10,000 values. Given 10 million random numbers, java.util.Random yeilds the following results on my machine:
Doubles: 99873
Triples: 985
Double Rate: 1 in 100
Triple Rate: 1 in 10152
Source code:
import static java.lang.System.*;
import java.util.Random;
public class Sandbox {
public static final int NUM_ITERATIONS = 10000000;
public static void main(String[] args) {
Random rand = new Random();
int cur;
int last = -1;
int secondLast = -2;
int nDoubles = 0;
int nTriples = 0;
for (int i = 0; i < NUM_ITERATIONS; i++) {
cur = rand.nextInt(100);
if (cur == last) {
nDoubles++;
if (cur == secondLast) nTriples++;
}
secondLast = last;
last = cur;
}
out.println("Doubles: " + nDoubles);
out.println("Triples: " + nTriples);
out.println();
out.println("Double Rate: 1 in " + Math.round(1.0 * NUM_ITERATIONS / nDoubles));
out.println("Triple Rate: 1 in " + Math.round(1.0 * NUM_ITERATIONS / nTriples));
exit(0);
}
}

Actually the creation of genuinely random random numbers is a complex game in its own right. The Wikipedia article on this subject will give you an insight into the complexity that lies therein. Simple approximations such as those outlined above are probably sufficient for game purposes but will, it should be noted, be inclined to be 'streaky' from time to time.

You can use java.util.Random:
Random random = new Random(); // uses System.nanoTime() as seed
int enemies = random.nextInt(100);
Anyway, your approach is also fine, as it is in fact equivalent (behind the scene) with the above.
You can print a sequence of 100 random numbers generated your way and see for yourself that there isn't a problem.

What you use if perfectly fine.
In case you want something simplier you might like to use Random class like this:
Random generator = new Random(seed);
int number = generator.nextInt(100);

...and sometimes 77, sometimes 23, etc, as expected in a uniform distribution?
If you would like a normal distribution instead for your "enemies", so that extremes are less likely, it seems to be there in Java.