How to generate a sequence of random numbers

v.k.

Re: How to generate a sequence of random numbers

10 months ago

If they can repeat:

int[] numbers = new int[6];
void setup()
{
for (int i = 0; i< numbers.length; i++)
{
numbers[i] = (int)random(1, 61);
}
println(numbers);// before sort
numbers = sort(numbers);
println(numbers);//sorted
}

If you need unique numbers, I think, the way to go is:

- fill an array with numbers from 1 to 60 in order;

- shuffle the array

- pick six indexes (the first six for instance)

edited: see codes and links below.

:P

command_z

Re: How to generate a sequence of random numbers

10 months ago

Thanks. The numbers cannot repeat. I'll look into the second example you posted. Thank you!!!

v.k.

Re: How to generate a sequence of random numbers

10 months ago

Well sorry for this messy example (removed), forget it. Here a much better one by benja in another post.

Integer[] numbers = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12};
void setup(){
println(numbers);
}
void draw(){}
void mouseReleased(){
Collections.shuffle(Arrays.asList(numbers));
println(numbers);
}
and even other:
int[] randomarray = new int[18];
for(int t=0; t<randomarray.length; t++){
randomarray[t]=t;
}

and even from other post by ceeszz

int[] randomarray = new int[18];
for(int t=0; t<randomarray.length; t++){
randomarray[t]=t;
}
for (int k=0; k < randomarray.length; k++) {
// Goal: swap the value at pos k with a rnd value at pos x.
// save current value from pos/index k into temp
int temp = randomarray[k];
// make rnd index x
int x = (int)random(0, randomarray.length);
// overwrite value at current pos k with value at rnd index x
randomarray[k]=randomarray[x];
// finish swapping by giving the old value at pos k to the
// pos x.
randomarray[x]=temp;
}
println(randomarray);

command_z

Re: How to generate a sequence of random numbers

10 months ago

I should have mentioned that I need the random numbers back in numerical order.

I'm looking for something that can produce results from low to high like this:

04 20 33 39 54 56

Thanks for your help.

command_z

Re: How to generate a sequence of random numbers

10 months ago

What about something like this?

void setup(){
}
void draw(){}
void mousePressed(){
int one = int(random(1, 22));
int two = int(random(one+1, 35));
int three = int(random(two+1, 47));
int four = int(random(three+1, 53));
int five = int(random(four+1, 60));
}

PhiLho

Re: How to generate a sequence of random numbers

10 months ago

Use the first code of v.k. Then look at the result, looping over the array. If two successive numbers are equal, draw a new random number for the second one. At the end, sort again and check again, until it is OK.

command_z

Re: How to generate a sequence of random numbers

10 months ago

This is great, I never knew about sort().

Is there a more efficient way to write the code below? How can I get this into a for loop?

Thanks everyone for the help!

int[] numbers = new int[6];
void setup()
{
}
void draw(){
background(255);
}
void mouseReleased(){
for (int i = 0; i< numbers.length; i++)
{
numbers[i] = (int)random(1, 61);
}
numbers = sort(numbers);
if (numbers[0] == numbers[1]){
background(255,0,0);
numbers[0] = (int)random(1, 61);
}
if (numbers[1] == numbers[2]){
background(255,0,0);
numbers[1] = (int)random(1, 61);
}
if (numbers[2] == numbers[3]){
background(255,0,0);
numbers[2] = (int)random(1, 61);
}
if (numbers[3] == numbers[4]){
background(255,0,0);
numbers[3] = (int)random(1, 61);
}
if (numbers[4] == numbers[5]){
background(255,0,0);
numbers[4] = (int)random(1, 61);
}
//println(numbers);// before sort
numbers = sort(numbers);
if (numbers[0] == numbers[1]){
background(255,0,0);
numbers[0] = (int)random(1, 61);
}
if (numbers[1] == numbers[2]){
background(255,0,0);
numbers[1] = (int)random(1, 61);
}
if (numbers[2] == numbers[3]){
background(255,0,0);
numbers[2] = (int)random(1, 61);
}
if (numbers[3] == numbers[4]){
background(255,0,0);
numbers[3] = (int)random(1, 61);
}
if (numbers[4] == numbers[5]){
background(255,0,0);
numbers[4] = (int)random(1, 61);
}
numbers = sort(numbers);
println(numbers);//sorted
println(" ");
}

v.k.

Re: How to generate a sequence of random numbers

10 months ago

HI, I made a try using recursion, it is kind of working. But I thought that passing an Array to a function it would be a reference and the actual array passed would be modified, but this is not happening. So as i made a void method, the array in setup still holding the original numbers. The question is: why?

code with void method:

int[] numbers = new int[6];
int range = 6;
public void setup()
{
for (int i = 0; i< numbers.length; i++)
{
numbers[i] = (int)random(1, range+1);
}
println(numbers);// before sort
testRepeated(numbers, range);
print("in setup: \n");
//array not modified here... ??
//i though it was a reference when passed to function
println(numbers);
}
public void testRepeated(int[] n, int range)
{
//sort
n = sort(n);
// to hold new number for testing and getting a different one
int newNumber = n[0];
//debug here we got the correct array
print("begin of function sorted \n");
println(n);
println(" ");
//find duplicates
for (int i=0; i < n.length-1; i++)
{
//test against next number
if (n[i] == n[i+1])
{
print("numbet at index = " + i + " is repeated \n");
println(n);
println(" ");
// if equal draw a new and different one
while (newNumber == n[i])
{
newNumber = (int)random(range+1);
}
// change number in array
n[i+1] = newNumber;
print("new number is = " + n[i+1] +"\n");
println(n);
println(" ");
//here we go again...
testRepeated(n, range);
//return if duplicate ffound
return;
}
}
}

So i tried a function version, but still not working, the original array keeps the original values:

What am i missing?

code with int[] function:

int[] numbers = new int[6];
int range = 6;
public void setup()
{
for (int i = 0; i< numbers.length; i++)
{
numbers[i] = (int)random(1, range+1);
}
println(numbers);// before sort
numbers = testRepeated(numbers, range);
print("in setup: \n");
//array not modified here... ??
//i though it was a reference when passed to function
println(numbers);
}
public int[] testRepeated(int[] n, int range)
{
//sort
n = sort(n);
// to hold new number for testing and getting a different one
int newNumber = n[0];
//debug here we got the correct array
print("begin of function sorted \n");
println(n);
println(" ");
//find duplicates
for (int i=0; i < n.length-1; i++)
{
//test against next number
if (n[i] == n[i+1])
{
print("numbet at index = " + i + " is repeated \n");
println(n);
println(" ");
// if equal draw a new and different one
while (newNumber == n[i])
{
newNumber = (int)random(range+1);
}
// change number in array
n[i+1] = newNumber;
print("new number is = " + n[i+1] +"\n");
println(n);
println(" ");
//here we go again...
testRepeated(n, range);
//return if duplicate ffound
break;
}
}
return n;
}

not_55

Re: How to generate a sequence of random numbers

10 months ago

If you don't need a particular number (quantity) of random numbers, what's wrong with something like this that will produce chains of random numbers in the range you need:

int n=0;
for(int i=0; i<999999; i++){
n+=int(random(10))+1;
print(n % 60 + ", ");
}

v.k.

Re: How to generate a sequence of random numbers

10 months ago

he wants unique numbers.

In my code above I'm failing to exit the recursion...

Here a simplified version of the other approach, shuffling the numbers and picking first six. This is working fine:

int[] numbers = new int[8];// change the length of the array to set the range of numbers
int[] result = new int[6];
void setup() {
size(200,50);
for (int i = 0; i < numbers.length; i++) {
numbers[i]=i+1;
}
text ("press a key to draw numbers", 0, height/2);
}
void draw() {
}
void keyReleased() {
println(numbers);
shuffle(numbers);
println("\nthe same shuffled");
println(numbers);
for (int i = 0; i < result.length; i++) {
result[i]=numbers[i];
}
result=sort(result);
println(" result 6 unique numbers up to" + numbers.length);
println(result);
}
void shuffle(int[] array) {
// i is the number of items remaining to be shuffled.
for (int i = array.length; i > 1; i--) {
// Pick a random element to swap with the i-th element.
int j = (int)random(i); // 0 <= j <= i-1 (0-based array)
// Swap array elements.
int tmp = array[j];
array[j] = array[i-1];
array[i-1] = tmp;
}
}

not_55

Re: How to generate a sequence of random numbers

10 months ago

My code prints sequences of unique numbers although two sequences could be the same (but unlikely). It seems to fit the original criteria:

How can I generate an ordered set of random numbers between 1 and 60?
The numbers cannot repeat.

v.k.

Re: How to generate a sequence of random numbers

10 months ago

I got this from your code, i must be missing something... :) Also i think he needs it sorted.

Anyway I'd like to understand why my try to test recursively for repeated numbers is failing...

Can you explain me that?

thanks.

[edited] deleted a lot of numbers:)

3, 12, 16, 22, 32, 35, 44, 54, 0, 8, 14, 15, 22, 31, 37, 39, 41, 51, 58, 59, 2, 6, 13, 23, 25, 33, 34, 43, 49, 57, 7, 13, 14, 21, 31, 41, 46, 48, 57, 4, 9, 16, 24, 26, 31, 34, 39, 45, 50, 57, 4, 8, 15, 18, 19, 22, 29, 31, 39, 48, 53, 55, 59, 8, 16, 22, 24, 28, 29, 30, 40, 43, 45, 49, 57, 58, 6, 16, 19, 27, 30, 38, 42, 44, 46, 54,

.

21, 31, 39, 40, 45, 46, 51, 0, 9, 16, 20, 24, 30, 32, 36, 42, 49, 58, 2, 5, 12, 19, 26, 33, 38, 43, 51, 0, 6, 8, 13, 14, 21, 30, 35, 44, 50, 54, 3, 5, 8, 16, 23, 27, 36, 40, 44, 52, 2, 5, 12, 22, 31, 40, 43, 52, 59, 7, 13, 23, 33, 43, 52, 53, 54, 0,

not_55

Re: How to generate a sequence of random numbers

10 months ago

Lol, well there are only 60 possibilities and the question didn't say how many numbers were required. As I said above, my code generates sequences of unique numbers between 0-60 and the sequences are ordered. I'll test you code...brb!

v.k.

Re: How to generate a sequence of random numbers

10 months ago

thanks for your patience, I still don't get the sequences you mean, if i try this with your code i still get repeated numbers.

int n=0;
int[] numbers = new int[60];
for(int i=0; i<60; i++){
n+=int(random(10))+1;
int number = n % 60;
print(n % 60 + ", ");
numbers [i] = number;
}
numbers = sort(numbers);
println("\nsorted");
println(numbers);

not_55

Re: How to generate a sequence of random numbers

10 months ago

From the big chunk you printed above the first few sequences are

3, 12, 16, 22, 32, 35, 44, 54
0, 8, 14, 15, 22, 31, 37, 39, 41, 51, 58, 59
2, 6, 13, 23, 25, 33, 34, 43, 49, 57
7, 13, 14, 21, 31, 41, 46, 48, 57

'22' appears in each of the first two sequences but all numbers in each sequence are unique and ordered. I thought that was what was required?

The code could easily be made to generate at least (n) numbers in each sequence (within the limits of the range 0-60 that is!).

v.k.

Re: How to generate a sequence of random numbers

10 months ago

I see, cool. I love simple solutions, but the sequences length are arbitrary? How would you make say 6 unique random numbers fom 0 to 8 ?

not_55

Re: How to generate a sequence of random numbers

10 months ago

0-8 doesn't give you much scope if you need 6 numbers! It would be a case of adding 1 or 2, lets think...

start at 0 adding 1...

0 1 2 3 4 5

start at 0 adding 1 five times and two once...

0 1 2 3 4 6

etc etc

A bit of thought would crack it for all the permutations I'm sure :)

As I mentioned, the original question would be better if it said whether a fixed quantity or numbers was required.

v.k.

Re: How to generate a sequence of random numbers

10 months ago

I don't even know if it is a requirement of the original question, I just got interested, and liked your approach. I picked 6 out of 8 just because is easy to test.

el_dippo

Re: How to generate a sequence of random numbers

10 months ago

You mean like this:

int n=0;
for (int i=0; i<6; i++) {
n+=(int)random(2);
print(n + ", ");
n++;
}

v.k.

Re: How to generate a sequence of random numbers

10 months ago

But this in not constrained up to 8, is it?

I'm making another topic with my recursion doubt. :)

not_55

Re: How to generate a sequence of random numbers

10 months ago

el_dippo: no, as vk says that won't quite work - you need additional logic to prevent 'n' from exceeding 8. The number in the for loop in my code determines the number of sequences not the quantity of numbers.