We Keep Coding

java random number generator and their distribution

I am relying on Java's Random class, specifically on the nextInt method to generate N random numbers. I do not know what N will be ahead of time, it is decided on the fly.
One of the requirements of my todo list is to have the random numbers be representative of the distribution.
For example, if N=100, in the range from 1-100 there should be 10 (approximately) numbers between 1-10, 20 numbers between 1-20 etc.
But N can potentially grow on the fly from 100 to 100,000 and as such the distribution of the generated randoms should adjust on the fly to represent 100,000 generated numbers between 1-100
I'm not sure if this is possible, hope it makes sense what I am trying to achieve.

You appear to be describing a uniform distribution.
Looking at the Javadoc of Random.nextInt(int):
Returns a pseudorandom, uniformly distributed int value between 0 (inclusive) and the specified value (exclusive)
So, just use Random.nextInt, passing N as the parameter, and add 1 to the result to get it 1 to N instead of 0 to (N-1).

Generate even numbers 1-4 using Math.random()?

I would like to generate the numbers 1-4 (whole integers) using Math.random. I have only succeeded in getting doubles or large doubles, and cannot figure out how to set a limit on the minimum and maximum.
Math.random(); = something between 0-1 as a double?
I have seen some people suggest something like this: num = Math.random() * 60 + 25; but have no idea what that does, or how it works.
I am not sure if this is a true question, and feel free to let me know if I should delete it.
Edit: Is there a way to not get the numbers to repeat, yet still be random every time the program is run?

int rand = (Math.random() * 4) + 1;

Math.Random is redundant here, use the Random class.
Random rand = new Random();
rand.nextInt(4)+1; //starts at 0, so add 1
Import this class by:
import java.util.*; or import java.util.Random;

the random number in math gives you a decimal number between zero and one.
you need to tell it to be within a certain range.
something like:
(4*Math.random())+1 should give you between 1-4 I think. correct me if I am wrong anyone.

Random rand = new Random();
System.out.println(rand.nextInt(4) + 1); // we add 1 because it starts with 0

If you really have to use Math.random you need to multiply (and add).
It's quite basic math, Math.random()
Returns a double value with a positive sign, greater than or equal to 0.0 and less than 1.0.
So multiplying it with X will give a number greater than or equal to 0.0 and less than X. Cast that to an int to get rid of decimals and if you only want even numbers you can do a few things, the easiest probably being int even = (notSureIfEven >> 1) << 1;. [I'm kind of assuming that with 'even' numbers you actually meant 'whole' numbers though, in which case you can ignore that.] Then if you don't want the range to be 0->X but Y->X you just add Y to your outcome (make sure to subtract Y from X before the multiplication or your range will be Y->X+Y).
To not generate the same number twice you can do different things. One way is to store all the numbers you generated so far in a List and then when you generate a new number, check if the list contains that number already, if so generate a new one until you got one that isn't in the list (and then when you do obviously add that to the list). Another way could be to preload all numbers it could generate into a list and then remove a random number out of that list.
Both ways probably won't scale very well to really large ranges of numbers though. The first one since it might get in a very long loop trying to find a number it hadn't generated yet, the second one because you'll have to create a really large list at the start.
I'm not sure if there's something you could do in the case of a really large range.

How to get the last digits of 2 power

I’m working on a problem where its demanded to get the last 100 digits of 2^1001. The solution must be in java and without using BigInteger, only with int or Long. For the moment I think to create a class that handle 100 digits. So my question is, if there is another way the handle the overflow using int or long to get a number with 100 digits.
Thanks for all.

EDITED: I was off by a couple powers of 10 in my modulo operator (oops)
The last 100 digits of 2^1001 is the number 2^1001 (mod 10^100).
Note that 2^1001 (mod 10^100) = 2*(2^1000(mod 10^100)) (mod 10^100).
Look into properties of modulo: http://www.math.okstate.edu/~wrightd/crypt/lecnotes/node17.html
This is 99% math problem, 1% programming problem. :)
With this approach, though, you wouldn't be able to use just an integer, as 10^100 won't fit in an int or long.
However, you could use this to find, say, the last 10 digits of 2^1001, then use a separate routine to find the next 10 digits, etc, where each routine uses this feature...

The fastest way to do it (coding wise) is to create an array of 100 integers and then have a function which starts from the ones place and doubles every digit. Make sure to take the modulus of 10 and carry over 1s if needed. For the 100th digit, simply eliminate the carry over. Do it 1001 times and you're done.

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;}
}
}

Generate random integers in java

How to generate random integers but making sure that they don't ever repeat?
For now I use :
Random randomGenerator = new Random();
randomGenerator.nextInt(100);
EDIT I
I'm looking for most efficient way, or least bad
EDIT II
Range is not important

ArrayList<Integer> list = new ArrayList<Integer>(100);
for(int i = 0; i < 100; i++)
{
list.add(i);
}
Collections.shuffle(list);
Now, list contains the numbers 0 through 99, but in a random order.

If what you want is a pseudo-random non-repeating sequence of numbers then you should look at a linear feedback shift register. It will produce all the numbers between 0 and a given power of 2 without ever repeating. You can easily limit it to N by picking the nearest larger power of 2 and discarding all results over N. It doesn't have the memory constraints the other colleciton based solutions here have.
You can find java implementations here

How to generate random integers but making sure that they don't ever repeat?
First, I'd just like to point out that the constraint that the numbers don't repeat makes them non-random by definition.
I think that what you really need is a randomly generated permutation of the numbers in some range; e.g. 0 to 99. Even then, once you have used all numbers in the range, a repeat is unavoidable.
Obviously, you can increase the size of your range so that you can get a larger number without any repeats. But when you do this you run into the problem that your generator needs to remember all previously generated numbers. For large N that takes a lot of memory.
The alternative to remembering lots of numbers is to use a pseudo-random number generator with a long cycle length, and return the entire state of the generator as the "random" number. That guarantees no repeated numbers ... until the generator cycles.
(This answer is probably way beyond what the OP is interested in ... but someone might find it useful.)

If you have a very large range of integers (>>100), then you could put the generated integers into a hash table. When generating new random numbers, keep generating until you get a number which isn't in your hash table.

Depending on the application, you could also generate a strictly increasing sequence, i.e. start with a seed and add a random number within a range to it, then re-use that result as the seed for the next number. You can set how guessable it is by adjusting the range, balancing this with how many numbers you will need (if you made incremental steps of up to e.g., 1,000, you're not going to exhaust a 64-bit unsigned integer very quickly, for example).
Of course, this is pretty bad if you're trying to create some kind of unguessable number in the cryptographic sense, however having a non-repeating sequence would probably provide a reasonably effective attack on any cypher based on it, so I'm hoping you're not employing this in any kind of security context.
That said, this solution is not prone to timing attacks, which some of the others suggested are.

Matthew Flaschen has the solution that will work for small numbers. If your range is really big, it could be better to keep track of used numbers using some sort of Set:
Set usedNumbers = new HashSet();
Random randomGenerator = new Random();
int currentNumber;
while(IStillWantMoreNumbers) {
do {
currentNumber = randomGenerator.nextInt(100000);
} while (usedNumbers.contains(currentNumber));
}
You'll have to be careful with this though, because as the proportion of "used" numbers increases, the amount of time this function takes will increase exponentially. It's really only a good idea if your range is much larger than the amount of numbers you need to generate.

Since I can't comment on the earlier answers above due to not having enough reputation (which seems backwards... shouldn't I be able to comment on others' answers, but not provide my own answers?... anyway...), I'd like to mention that there is a major flaw with relying on Collections.shuffle() which has little to do with the memory constraints of your collection:
Collections.shuffle() uses a Random object, which in Java uses a 48-bit seed. This means there are 281,474,976,710,656 possible seed values. That seems like a lot. But consider if you want to use this method to shuffle a 52-card deck. A 52-card deck has 52! (over 8*10^67 possible configurations). Since you'll always get the same shuffled results if you use the same seed, you can see that the possible configurations of a 52-card deck that Collections.shuffle() can produce is but a small fraction of all the possible configurations.
In fact, Collections.shuffle() is not a good solution for shuffling any collection over 16 elements. A 17-element collection has 17! or 355,687,428,096,000 configurations, meaning 74,212,451,385,344 configurations will never be the outcome of Collections.shuffle() for a 17-element list.
Depending on your needs, this can be extremely important. Poor choice of shuffle/randomization techniques can leave your software vulnerable to attack. For instance, if you used Collections.shuffle() or a similar algorithm to implement a commercial poker server, your shuffling would be biased and a savvy computer-assisted player could use that knowledge to their benefit, as it skews the odds.

If you want 256 random numbers between 0 and 255, generate one random byte, then XOR a counter with it.
byte randomSeed = rng.nextInt(255);
for (int i = 0; i < 256; i++) {
byte randomResult = randomSeed ^ (byte) i;
<< Do something with randomResult >>
}
Works for any power of 2.

If the Range of values is not finite, then you can create an object which uses a List to keep track of the Ranges of used Integers. Each time a new random integer is needed, one would be generated and checked against the used ranges. If the integer is unused, then it would add that integer as a new used Range, add it to an existing used Range, or merge two Ranges as appropriate.
But you probably really want Matthew Flaschen's solution.

Linear Congruential Generator can be used to generate a cycle with different random numbers (full cycle).