Non-repeating random numbers [closed]

Question

Find the shortest way to generate a random number between 0-9 without repeating the last generated number. Here is an example written in javascript.

var n=t=0;
while(n==t){
  t=Math.floor(Math.random()*9);
}
n=t;

I don't know if it is possible, but it would be interesting if there is a solution using only one variable.

To clarify the solution can be in any language. Based on the comments I am going to disallow using arrays since apparently that makes it very easy and this is code-golf so the shortest answer wins.

Answer 1

Pyth, 7 bytes

e=+ZhO9

Try it online: Demonstration

Explanation

          implicit: Z = 0
     O9   Generate a random int from [0, 1, ..., 8]
    h     add 1
 =+Z      add this to Z and update Z
e         print Z modulo 10

Answer 2

CJam, 10 8 bytes

Thanks @Dennis for a 2 bytes golf!

You can repeat this piece of code how many times you want.

A,W-mR:W

Explanation:

         e# At startup, W = -1
A,       e# Push [0 ... 9]
  W-     e# Remove W from the list
    mR   e# Random choice from list
      :W e# Assign to W
         e# The number is left on the stack

Try it online.

Answer 3

K5, 19 bytes

Not particularly clever:

t:-1;r:{t::*1?10^t}

Each time r is invoked, it picks 1 random number (*1?) from the set formed by taking the range 0-9 except (^) the previous random number and then stores the result in t (t::). We initialize t (t:-1;) such that the first result can be any digit 0-9.

The definition of r is purely for convenience in repeatedly invoking it, so you're really only using a single variable.

edit:

Just for fun, here's a k5 program which will generate a sequence of N numbers containing no immediate repetitions (as in the above problem) without using any variables:

1_(*1?10^)\[;-1]

For example,

  1_(*1?10^)\[;-1]10
1 8 0 6 8 5 1 0 5 1
  1_(*1?10^)\[;-1]3
4 0 7

Doesn't work quite right in my K implementation, so I suppose writing this and discovering a bug was a decent use of my time!

Answer 4

Java, 62 bytes

int a(int p){p+=9*(p+Math.random());return p%10==p/10?9:p%10;}

This is a function. You call it with the previously returned value. Uses only the variable supplied.

2 variables, 59 bytes

int a(int p){int a=(int)(Math.random()*9);return a==p?9:a;}

Answer 5

Julia - 25 18 bytes

t=(rand(1:9)+t)%10

Or, for a slightly longer, iterative solution:

while t==(t=rand(0:9))end

No temporary second variable necessary.

Answer 6

Python 2, 79 bytes

Here's a simple one you can repeatedly run in a REPL:

from random import*
try:p
except:p=n=0
while p==n:n=int(random()*9)
print n
p=n

Answer 7

Java 8, 89 bytes

As an IntStream...

// Int Stream (89 bytes)
IntStream.iterate(0,(o)->new Random().ints(0,10).filter(x->x!=o).findFirst().getAsInt());

Minimal usage...

// Usage (119 bytes)
IntStream.iterate(0,(o)->new Random().ints(0,10).filter(x->x!=o).findFirst().getAsInt()).forEach(System.out::println);

Answer 8

Octave, 24 bytes

Here's an anonymous function using one variable (plus the function handle):

r=@(n)mod(randi(9)+n,10)

Sample run:

>> n=5;
>> n=r(n)
n =  2

Answer 9

Javascript, 39 bytes

Not the most exciting answer, but it is in Javascript.

t=-1;while(t==(t=(Math.random()*9)|0));

To test it:

//set the value here, or it will produce repeated numbers:
t=-1;
for(var i=0; i < 10; i++){
    //generate a new random number, using the code in the answer
    while(t==(t=(Math.random()*9)|0));
  
    document.write(t + '<br>');
}

Answer 10

Python 3, 42 33 bytes

If your criteria for "random" is generating a number 0-9,

[...] without repeating the last generated number.

Something like this would work:

i=(i+1)%10if"i"in globals()else 0