on the first iteration- I want to create a random number from 0..10
This is how I've implement it-
int generator= (int)(Math.random() * (10));
On the first iteration I want to create another random number, but without one of the values. For example I would like the random number without the number 4. only 1,2,3,5,6,7,8,9,0
How can I remove one of the values from the numbers 0..10 when I generate a random number?
You can also maintain a List of all the numbers. Then use Collections.shuffle to shuffle the list and get the first element. And remove it.
List<Integer> list = new ArrayList<Integer>();
list.add(1);
list.add(4);
list.add(2);
list.add(3);
System.out.println("Original List: " + list);
Collections.shuffle(list);
System.out.println("Shuffled List: " + list);
int number1 = list.remove(0); // Remove the first value
System.out.println("Number removed: " + number1);
Collections.shuffle(list);
System.out.println("Shuffled List: " + list);
number1 = list.remove(0);
System.out.println("Number removed: " + number1);
OUTPUT: -
Original List: [1, 4, 2, 3]
Shuffled List: [4, 3, 1, 2]
Number removed: 4
Shuffled List: [1, 3, 2]
Number removed: 1
A better way would be to generate a random number from 0 to the size of the list, and get the value of that index and remove it.
int size = list.size();
Random random = new Random();
for (int i = 0; i < size; i++) {
int index = random.nextInt(list.size());
int number1 = list.remove(index);
System.out.println(number1);
}
NOTE: -
If you want to exclude a fixed set of numbers from being generated, then you can have those numbers in a list. And then remove all elements of that list from your list of numbers. And then use any of the approach above.
List on a 10-elements issue and shuffling on each iteration. "Terribly inefficient" would be an understatement, no offence intended. ___ Also, the OP never really specified removal of elements in the first place. Actually, the fact he removes the 4 before the first iteration indicates he's simply wanting an exception in the randomizer's range of output values.
int generator= (int)(Math.random() * (10));
if (generator >= 4) generator++;
This will include the number 10 as a possible outcome, if you do not want this, multiply by 9 instead of 10.
5, while others have a 1/10 chance.
You could store the values in a list, and generate a random index to take from that list.
private Random random = new Random();
private ArrayList<Integer> values = new ArrayList<Integer>();
public static int nextInt() {
if (values.size() > 0) {
return values.remove(random.nextInt(values.size()));
} else {
throw new RuntimeException("no values possible");
}
}
Random.nextInt(10)==number generate again
public int generateExcept(int number) {
Random random = new Random();
int generated = 0;
while ((generated = random.nextInt(11)) == number) {
}
return generated;
}
As quoted by OP
On the first iteration I want to create another random number, but without one of the values
If I understand correctly This is single operation not repetitive operation.
Without Regeneration
public static int generateExcept(int number, int max) {
Random random = new Random();
int generated = 0;
if ((generated = random.nextInt(max - 1)) == number) {
generated++;
}
return generated;
}
Random is always generated max-1 because we want to increment by one
nextInt(11) for 0 to 10 - although it is not clear if the OP wants 10 to be included or not in the range of possible outcome...random.nextInt(11). It's not so much an issue with one value excluded, but it's extremely non-scalable (in the case of excluding several values).
There are two basic efficient approaches:
1 - Generate an array of valid values (0 to 10 except 4), and use the randomizer to pick an index on that array. An example follows:
Random random = new Random();
int[] unwanted = {4};
int[] values = new int[10-unwanted.length];
for(int i=0; i<10-unwanted.length; i++){
for(int curr : unwanted){
if(i == curr) continue;
}
values[i] = i;
}
int result = values[random.nextInt(10-unwanted.length)];
2 - Follow Simon André's approach.
Generate an array of valid values (0 to 10 except 4) how will you define generation except 4 its just an example from OPint[] values = {0,1,2,3,5,6,7,8,9,10} or adaptive: int[] unwanted = {4}; int[] values = new int[10-unwanted.length]; for(int i=0; i<10-unwanted.length; i++){for(int curr:unwanted){if(i==curr) continue;} values[i] = i;}What you could do (perhaps there is a cleaner way of handling this) is to check the 2nd generated number to see if it matches the first, and if it does, just have it run the random num. generator again. A simple if...else loop should take care of it.
int excluded[] = { 1,4,3};
Random r = new Random();
int generator;
do {
generator= r.nextInt(10);
} while (! Arrays.asList(excluded).contains(generator) )
Random.nextInt(10)
4. -1
Random.nextInt(10)==4generate again4several times before anything else.