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math question about random (x) and random() % x - Java [duplicate]

This question already has answers here:
Why do people say there is modulo bias when using a random number generator?
(10 answers)
Closed 2 years ago.
so my question is at Java but it can be in any programming language.
there is this declaration :
Random rnd = new Random();
We want to get a random number at range 0 to x
I want to know if there is any mathematical difference between the following:
rnd.nextInt() % x;
and
rnd.nextInt(x)
The main question is, are one of these solutions more random than the other? Is one solution more appropriate or "correct" than the other? If they are equal I will be happy to see the mathematics proof for it

Welcome to "mathematical insight" with "MS Paint".
So, from a statistical standpoint, it would depend on the distribution of the numbers being generated. First of all, we'll treat the probability of any one number coming up as an independant event (aka discarding the seed, which RNG, etc). Following that, a modulus simply takes a range of numbers (e.g. a from N, where 0<=a<N), and subdivides them based on the divisor (the x in a % x). While the numbers are technically from a discrete population (integers), the range of integers for a probability mass function would be so large that it'd end up looking like a continuous graph anyhow. So let's consider a graph of the probability distribution function for a range of numbers:
If your random number generator doesn't generate with a uniform distribution across the range of numbers (aka, any number is as likely to come up as another number), then modulo would (potentially) be breaking up the results of a non-uniform distribution. When you consider the individual integers in those ranges as discrete (and individual) outcomes, the probability of any number i (0 <= i < x) being the result is the multiplication of the individual probabilities (i_1 * i_2 * ... * i_(N/x)). To think of it another way, if we overlaid the subdivisions of the ranges, it's plain to see that in non-symmetric distributions, it's much more likely that a modulo would not result in equally likely outcomes:
Remember, the likelihood of an outcome i in the graph above would be achieved through multiplying the likelihood of the individuals numbers (i_1, ..., i_(N/x)) in the range N that could result in i. For further clarity, if your range N doesn't evenly divide by the modular divisor x, there will always be some amount of numbers N % x that will have 1 addditional integer that could produce their result. This means that most modulus divisors that aren't a power of 2 (and similarly, ranges that are not a multiple of their divisor) could be skewed towards their lower results, regardless of having a uniform distribution:
So to summarize the point, Random#nextInt(int bound) takes all of these things (and more!) into consideration, and will consistently produce an outcome with uniform probability across the range of bound. Random#nextInt() % bound is only a halfway step that works in some specific scenarios. To your teacher's point, I would argue it's more likely you'll see some specific subset of numbers when using the modulus approach, not less.

new Random(x) just creates the Random object with the given seed, it does not ifself yield a random value.
I presume you are asking what the difference is between nextInt() % x and nextInt(x).
The difference is as follows.
nextInt(x)
nextInt(x) yields a random number n where 0 ≤ n < x, evenly distributed.
nextInt() % x
nextInt() % x yields a random number in the full integer range1, and then applies modulo x. The full integer range includes negative numbers, so the result could also be a negative number. With other words, the range is −x < n < x.
Furthermore, the distribution is not even in by far the most cases. nextInt() has 232 possibilities, but, for simplicity's sake, let's assume it has 24 = 16 possibilities, and we choose x not to be 16 or greater. Let's assume that x is 10.
All possibilities are 0, 1, 2, …, 14, 15, 16. After applying the modulo 10, the results are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5. That means that some numbers have a greater likelihood to occur than others. That also means that the change of some numbers occurring twice has increased.
As we see, nextInt() % x has two problems:
Range is not as required.
Uneven distribution.
So you should definetely use nextInt(int bound) here. If the requirement is get only unique numbers, you must exclude the numbers already drawn from the number generator. See also Generating Unique Random Numbers in Java.
1 According to the Javadoc.

I want to generate random numbers within given range in java

For that, I got the solution but I can't understand it:
I have used below function to generate random numbers within the given range but what is the meaning of below function?
static int randomRangeInNumber(int min, int max) {
Random r=new Random();
return r.nextInt((max-min)+1)+min;
}
What i have returned i didnt get
Please help making me understand its meaning
any help will be appreciated

nextInt is normally exclusive of the top value, so add 1 to make it
inclusive
So, in your case, we'll say the max is 10 and min is 5.
nextInt(n) will give you a random number between 0 and n - 1.
So adding the +1 will let you include the value of n.
nextInt(max-min) therefore gives you a random number between (in this case) 0 and 5 - 1 (and include the 5 because of the +1)
Then adding the min again, means it will give you the above random number, but add 5 to it, so the result will be a random number between 5 and 10.

The documentation for nextInt(int bound) says
Returns a pseudorandom, uniformly distributed int value between 0 (inclusive) and the specified value (exclusive), drawn from this random number generator's sequence.
So your nextInt call is
taking the return value, which is in the range [0..bound] (where the bound is (max-min)+1) and then
scaling it into the range [min..max] with +min.
return r.nextInt((max-min)+1)+min;
\__________1_________/\_2_/

Is it possible to generate random negative number range in Java?

I have seen MANY posts on here in regard to generating random numbers with a specific range in JAVA (especially this one). However, I have not found one that describes how to generate a random number between a negative MAX and a negative MIN. Is this possible in Java?
For example, if I want to generate a random number that is between (-20) and (-10). Using something like the below will only result in a JAVA Exception that screams about n having to be positive:
int magicNumber=(random.nextInt(-20)-10);

Just generate a random number between 10 and 20 and then negate it.

Another option would be to generate a random number between 0 and 10 and then subtract 20, if that feels less like a work-around to you.

I'm not entirely sure what you want, but ThreadLocalRandom has a method which accepts a range, which can also have negative values:
ThreadLocalRandom.current().nextInt(-20, -10 + 1)
There is no practical difference to just negating the result of a positive random though.

What is the randomness of Java.nextFloat()

Specifically, if used in the form of:
Random.nextFloat() * N;
can I expect a highly randomized distribution of values from 0 to N?
Would it be better to do something like this?
Random.nextInt(N) * Random.nextFloat();

A single random number from a good generator--and java.util.Random is a good one--will be evenly distributed across the range... it will have a mean and median value of 0.5*N. 1/4 of the numbers will be less than 0.25*N and 1/4 of the numbers will be larger than 0.75*N, etc.
If you then multiply this by another random number generator (whose mean value is 0.5), you will end up with a random number with a mean value of 0.25*N and a median value of 0.187*N... So half your numbers are less than 0.187*N! 1/4 of the numbers will be under .0677*N! And only 1/4 of the numbers will be over 0.382*N. (Numbers obtained experimentally by looking at 1,000,000 random numbers generated as the product of two other random numbers, and analyzing them.)
This is probably not what you want.

At first, Random in Java doesn't contain rand() method. See docs. I think you thought about Random.next() method.
Due to your question, documentation says that nextFloat() is implemented like this:
public float nextFloat() {
return next(24) / ((float)(1 << 24));
}
So you don't need to use anything else.

Random#nextFloat() will give you an evenly distributed number between 0 and 1.
If you take an even distribution and multiply it by N, you scale the distribution up evenly. So you get a random number between 0 and N evenly distributed.
If you multiply this by a random number between 0 and N, then you'll get an uneven distribution. If multiplying by N gives you an even distribution between 0 and N, then multiplying by a number between 0 and N, must give you an answer that is less or equal to if you just multiplied by N. So your numbers on average are smaller.

How to be sure that random numbers are unique and not duplicated?

I have a simple code which generates random numbers
SecureRandom random = new SecureRandom();
...
public int getRandomNumber(int maxValue) {
return random.nextInt(maxValue);
}
The method above is called about 10 times (not in a loop). I want to ensure that all the numbers are unique (assuming that maxValue > 1000).
Can I be sure that I will get unique numbers every time I call it? If not, how can I fix it?
EDIT: I may have said it vaguely. I wanted to avoid manual checks if I really got unique numbers so I was wondering if there is a better solution.

There are different ways of achieving this and which is more appropriate will depend on how many numbers you need to pick from how many.
If you are selecting a small number of random numbers from a large range of potential numbers, then you're probably best just storing previously chosen numbers in a set and "manually" checking for duplicates. Most of the time, you won't actually get a duplicate and the test will have practically zero cost in practical terms. It might sound inelegant, but it's not actually as bad as it sounds.
Some underlying random number generation algorithms don't produce duplicates at their "raw" level. So for example, an algorithm called a XORShift generator can effectively produce all of the numbers within a certain range, shuffled without duplicates. So you basically choose a random starting point in the sequence then just generate the next n numbers and you know there won't be duplicates. But you can't arbitrarily choose "max" in this case: it has to be the natural maximum of the generator in question.
If the range of possible numbers is small-ish but the number of numbers you need to pick is within a couple of orders of magnitude of that range, then you could treat this as a random selection problem. For example, to choose 100,000 numbers within the range 10,000,000 without duplicates, I can do this:
Let m be the number of random numbers I've chosen so far
For i = 1 to 10,000,000
Generate a random (floating point) number, r, in the range 0-1
If (r < (100,000-m)/(10,000,000-i)), then add i to the list and increment m
Shuffle the list, then pick numbers sequentially from the list as required
But obviously, there's only much point in choosing the latter option if you need to pick some reasonably large proportion of the overall range of numbers. For choosing 10 numbers in the range 1 to a billion, you would be generating a billion random numbers when by just checking for duplicates as you go, you'd be very unlikely to actually get a duplicate and would only have ended up generating 10 random numbers.

A random sequence does not mean that all values are unique. The sequence 1,1,1,1 is exactly as likely as the sequence 712,4,22,424.
In other words, if you want to be guaranteed a sequence of unique numbers, generate 10 of them at once, check for the uniqueness condition of your choice and store them, then pick a number from that list instead of generating a random number in your 10 places.

Every time you call Random#nextInt(int) you will get
a pseudorandom, uniformly distributed int value between 0 (inclusive)
and the specified value (exclusive).
If you want x unique numbers, keep getting new numbers until you have that many, then select your "random" number from that list. However, since you are filtering the numbers generated, they won't truly be random anymore.

For such a small number of possible values, a trivial implementation would be to put your 1000 integers in a list, and have a loop which, at each iteration, generates a random number between 0 and list.size(), pick the number stored at this index, and remove it from the list.

This is code is very efficient with the CPU at the cost of memory. Each potiental value cost sizeof(int) * maxValue. An unsigned integer will work up to 65535 as a max. long can be used at the cost of a lot of memory 2000 bytes for 1000 values of 16 bit integers.
The whole purpose of the array is to say have you used this value before or not 1 = yes
'anything else = no
'The while loop will keep generating random numbers until a unique value is found.
'after a good random value is found it marks it as used and then returns it.
'Be careful of the scope of variable a as if it goes out of scope your array could erased.
' I have used this in c and it works.
' may take a bit of brushing up to get it working in Java.
unsigned int a(1000);
public int getRandomNumber(int maxValue) {
unsigned int rand;
while(a(rand)==1) {
rand=random.nextInt(maxValue);
if (a(rand)!=1) { a(rand)=1; return rand;}
}
}