All combinations/cases of equality of 3 variables

Question

I have three integer variables like:

c0, c1, c2

I don't have the focus and have a difficult time determining all the possible cases of their equality:

if c1 == c0 && c2 == c0 {
    // ...
} else if c1 != c0 && c2 != c0 && c1 != c2 {
    // ...
} else if c1 != c0 && c2 != c0 && c1 == c2 {
    // ...
} else if c1 == c0 && c2 != c0 {
    // ...
} else if ... ?

Can anybody help?

Answer 1

They are either all equal (one possibility), or two are equal and one isn't (three possibilities), or none are equal (one possibility). This gives us five unique possibilities in total:

if c0 == c1 && c1 == c2 {
    // ...
} else if c0 == c1 && c1 != c2 {
    // ...
} else if c1 == c2 && c0 != c1 {
    // ...
} else if c0 == c2 && c1 != c2 {
    // ...
} 

And now use else to account for the last possibility of none of them being equal:

else {
    // ...
}

Answer 2

1. All three are equal, that is, $c_0 = c_1 = c_2$

2. Only two of them are equal, that is, $c_i = c_j \neq c_k$ where $\{i,j,k\} = \{0,1,2\}$. This case has three possibilities without repetition, since when you choose $k$, the pair $(i,j)$ and $(j,i)$ corresponds to the same thing.

3. None of them are equal to one another.

You can program it as follows :

if c0 == c1 && c1 == c2 : 
  //
elif c0 == c1 && c1 != c2 : 
  // ..   
elif c1 == c2 && c2 != c3 : 
  // ..
elif c2 == c3 && c3 != c1 : 
  // ..    
else : 
  // ..

(the else part corresponds to the last possibility, when $c0 \ != c1 \land c1 \ != c2 \land c2 \ != c3$)