Randomly pick one number that is different than two other random numbers

Question

Two random numbers A and B have been generated to be either 1, 2, or 3

your job is to randomly pick a third number C that can also be 1,2 or 3. But, C cannot equal A or B.

• And Yes, A can equal B.
• If A = B, then C has only two numbers left it can be.
• If A does not equal B, C has only one number it can be.
• Assume A and B have already been chosen for you

This is how A and B would be created in Python

A = random.randrange(1,4)
B = random.randrange(1,4)

Assume this is already in your code.

This is the shortest I've come up with in Python

while True:
    C = random.randrange(1,4)
    if C != A and C != B:
        break

This is what A, B and C can equal.

• 1,2,3
• 1,1,2
• 2,3,1
• 3,3,2

This is what A, B and C can't equal

• 1,2,1
• 2,3,3
• 1,1,1
• 3,2,3

Answer 1

Ruby, 22 characters

([1,2,3]-[A,B]).sample

Still not sure if I understood the question correctly ...

Answer 2

C,26

a-b?6-a-b:(rand()%2+a)%3+1

If I've understood the question correctly:

If a and b are different, there is no random. the answer must be the only one of 1,2,3 that is unused: 6-a-b.

IF a and b are the same there are 2 choices:

a=b=           1 2 3
            return value    
rand()%2=0     2 3 1
rand()%2=1     3 1 2

Answer 3

Befunge (156 89 85 74)

Okay, this is terrible, I know. But it's my first Befunge attempt ever, so I'm still pretty happy that it even works. I'm sure there's a much, much better solution.

<v1p90&p80&
<<@.g70_v#-g70_v#-g70g90g80p70
  v     <      <
^1?v
^3<2
^  <

Answer 4

Python - 35

C=random.sample({1,2,3}-{A,B},1)[0]

Assumes random is imported, which seems to be specified in the question.

PYG - 25

C=RSm({1,2,3}-{A,B},1)[0]

Answer 5

GolfScript, 13 chars

~0]4,^.,rand=

This is a complete GolfScript program that reads two whitespace-separated numbers (each assumed to be either 1, 2, or 3) from standard input, and outputs a random number from the set {1, 2, 3} that does not equal any of the input numbers.

Try it online. (Note: the link is to the previous version; I'm on a mobile device and can't fix it.)

Here's a commented version of the program:

~         # eval the input, pushing the input numbers onto the stack
0         # push the number 0 onto the stack
]         # collect all the numbers on the stack into an array
4,        # create another array containing the numbers 0, 1, 2 and 3
^         # xor the arrays (i.e. take their symmetric set difference)
.,rand=   # choose a random element from the array

If you'd prefer a named function that takes the two numbers as arguments on the stack, that takes a few more chars:

{[\0]4,^.,rand=}:f;

The actual body of the function is only one char longer than the stand-alone code (because we need the [ to make sure we consume only two arguments), but the overhead of wrapping the code in a block and assigning it to a symbol takes five more chars, for a total of 19.

Alternatively, if you literally have the two numbers assigned into the variables A and B, and want the third number assigned to C, that can also be done in 19 chars:

4,[0A B]^.,rand=:C;

(If leaving the third number on the stack instead is acceptable, you can leave the :C; off the end.)

Ps. Thanks for the suggestion to use ^, Howard.

Answer 6

Python, 14 characters

I tried it for every 9 possible cases and it seems to work fine!

C=A^B or A^1|2

(edit): As edc65 pointed out, this isn't valid since it's not random... I missed that part of the question and I feel stupid right now.

Answer 7

Befunge - 99 bytes

&:01p&:11p-!!#v_v
   @,g2+g11g10< "
   321 vv*2g<"v ^
 2v v v 5v*2^10<"
 v?v?v?vp5     ^<
 2 3 1 2<        
 > > > >.@       

Not very impressive.

Answer 8

PowerShell, 21

1..3-ne$A-ne$B|random

Very straightforward. Abusing the fact that comparison operators act differently with an array as their left operand.

Answer 9

Mathematica, 37 bytes

RandomChoice@DeleteCases[{1,2,3},a|b]

Basically the same as the Ruby answer, but considerably longer thanks to Mathematica's function names. I'm using lower-case variables, because upper-case names might clash with built-ins (they don't, in this case, but you simply don't do it in Mathematica).

Answer 10

R, 42 chars

x=c(1,1,1);x[c(A,B)]=0;C=sample(1:3,1,p=x)

Vector x is the vector of probability weights for obtaining the elements of the vector being sampled. It is set to 1 for each at first, then elements corresponding to A and B are set to 0, hence they have no chance of being picked.

Answer 11

Rebol - 40 chars

random/only difference[1 2 3]reduce[A B]

Answer 12

CJam - 12

4,[AB0]-mr0=

This assumes the variables A and B have already been set, according to the question.

You can try it at http://cjam.aditsu.net/

To test it with random numbers, use:

"A="3mr):A", B="3mr):B", C="
4,[AB0]-mr0=

To test it with specific values, use (for example):

"A="1:A", B="1:B", C="
4,[AB0]-mr0=

Explanation:

4, creates the array [0 1 2 3]
[AB0]- removes the numbers A, B and 0 from the array
mr shuffles the remaining array
0= takes the first element

In a future CJam version this program will be 2 bytes shorter :)

Answer 13

C 67

int C(int a,int b){int c=0;while(c!=a&&c!=b)c=rand()%3+1;return c;}

Answer 14

JS, 35

inspired by Brandon Anzaldi's answer

A=1; // init
B=3; // init
do{C=1+new Date%3}while(C==A||C==B) // 35b

Answer 15

Julia, 32 or 56 depending on the rules

julia> r()=rand(1:3);f(x...)=(i=r();i in x?f(x...):i)
julia> f(r(),r())

32 if I don't need to generate a and b.

Answer 16

TI-BASIC, 23

Lbl 1:If C=A+B=A:Goto 1

Answer 17

JS, 43

for(C=0;~[0,A,B].indexOf(C);)C=1+new Date%3

Answer 18

J (21 19: too long for my liking)

({~?@#)(>:i.3)-.A,B

Are there any J wizards around to help remove that variable assignment? It's only 2 chars shorter...

Or, if it doesn't have to be random, you could do with this:

{:(i.4)-.A,B

12 chars.

Answer 19

Java - 126 123 83 85 (using the clever c=6-a-b)

int c;if(a==b){int r=(int)(Math.random()*2);c=a==1?r+2:a==2?2*r+1:r+1;}else{c=6-a-b;}

Full version:

public void test(int a, int b) {
    int c;
    if (a == b) {
        // Random 0 or 1.
        int r = (int)Math.random()*2;
        c = // 1 -> 2 or 3
                a == 1 ? r + 2
                // 2 -> 1 or 3
                : a == 2 ? 2 * r + 1
                // 3 -> 1 or 2
                : r + 1;
    } else {
        // Whichever is not taken.
        //int[][] r = {{0, 3, 2}, {3, 0, 1}, {2, 1, 0}};
        //c = r[a - 1][b - 1];
        // Using @steveverrill's clever
        c = 6 - a - b;
    }
    System.out.println("a=" + a + " b=" + b + " c=" + c);
}

Answer 20

R, 24 characters

Initialize with

a = sample(1:3,1)
b = sample(1:3,1)

Then

n=1:3;n[!n%in%c(a,b)][1]

Or just n=1:3;n[!n%in%c(a,b)] but then you return both numbers.

Answer 21

R, 31 characters

sample(rep((1:3)[-c(A,B)],2),1)

If you do sample(x) in R, then it is interpreted as a random sample from 1:x. Repeating the vector (1:3)[-c(A,B)] twice is one way of stopping this from happening.

Answer 22

Javascript - 76

r=(y,z)=>Math.floor(Math.random()*(z-y+1)+y);a=b=r(1,3);while(c==a)c=r(1,3);

Answer 23

JavaScript - 41 (up to 46) 37 35 34 30

Updated:

Managed to get it down to 30 chars by modifying it, inspired by stevevarrill's answer in C.

C=A-B?6-A-B:1+(A+new Date%2)%3

---

Thank you nyuszika7h for getting me down to 34~:

C=A;while(C==A|C==B)C=1+new Date%3

Borrowing from xem's answer to at least fall on par with him:

C=A;while(C==A||C==B)C=1+new Date%3

Thanks for reminding me that 1+new Date%3 === (new Date%3)+1!

Previous Solution:

C=A;while(C==A||C==B)C=(new Date%3)+1

Ensure conditions of while() are satisfied, and loop through until they're not.

---

Other solution:

C=A!=B?6-A-B:A%2!=0?4-B:new Date%2!=1?3:1;

This assumes that C has already been declared OR that the JavaScript interpreter can handle undeclared variables.

However, if the JS interpreter can handle EOL without a semicolon, it could be bumped down to 41.

C=A!=B?6-A-B:A%2!=0?4-B:new Date%2!=1?3:1

If C hasn't been declared, and there is no error correction for it, that will bring the tally up to 46 characters.

var C=A!=B?6-A-B:A%2!=0?4-B:new Date%2!=1?3:1;

Test Program:

var iterations = 100;

for(var i = 0;i<iterations;i++) {
    var A = Math.floor(Math.random() * 3) + 1;
    var B = Math.floor(Math.random() * 3) + 1;
    C=A!=B?6-A-B:A%2!=0?4-B:new Date%2!=1?3:1
    if (C === A || C === B || C > 3 || C < 1) {
        console.log('FAILURE!');
        console.log(A + ',' + B + ',' + C)
        return;
    }
    console.log(A+','+B+','+C);
}

Answer 24

C - 38

main(c){for(;c==a|c==b;c=rand()%2+1);}

Answer 25

Java, 264 bytes

Random r = new Random();ArrayList<Integer> s = new ArrayList<>();ArrayList<Integer> q = new ArrayList<>();for(int i=0; i<n; i++) s.add(r.nextInt(k));q.add(s.get(r.nextInt(n)));q.add(s.get(r.nextInt(n)));int x;do{x = s.get(r.nextInt()); }while(!q.contains(x));

This code generates n different random numbers from 0 to k.

Answer 26

Golfscript, 13 Characters

~]4,^.,rand)=

Answer 27

Befunge-98 (57 bytes)

This code assumes the numbers will be input on stdin. It will pick a random number if both of the first numbers are the same until it is different, otherwise it will choose the last available number.

6&::11p&:12pw>   ?1  >#<:11g-!_.@
     @.-g21-<>3;#[2#;^

Answer 28

Python, 54 41 characters

Quite long answer in Python but I like list comprehension, so I decided to post this here

// [0] means it is the first element of

C=[i for i in[1,2,3]if not(i in(A,B))][0]