What is the best way to organize the structure of this code?

Question

Algorithms

Advise how it is better to organize the given code. The task is essentially simple:
Given A, B, C - numeric variables. There can be the following relationships between them:
A > B > C
A < B < C
Also given a numeric variable X. This variable can be equal to A, B, C (3 options), can be in the intervals between them (2 options), and go beyond the boundaries (to be less than A, or greater than C (2 options)). A total of 7 options. Considering that there can be both A > B > C and A < B < C, we multiply the number of options by 2, for a total of 14 possible options for X.
I solved this problem only using if's:

if (A < B && B < C) {
  if (X < A) { 
    // что-то делаем
  }
  if (X == A) {
    // делаем другое
  }
  if (X > A && X < B) {
    // делаем третье
  }
  // и так далее, все 14 вариантов вхождения для X
}

This code works, but it is quite cumbersome (a lot of if's) and not very readable (in reality, the names of the variables are different and there is more code). How can such code be organized differently? Is it possible to choose a data structure here that will make the code more concise and simple? Or another way to describe conditional checks?

Answer 1

If the actions are symmetrical, i.e. for X > (C > B > A), the actions are the same as for X > (A > B > C),
you can bring the variables to a uniquely increasing option - swap A and C if A > C.
So there will be 7 options.

if (A > C) [A, C] = [C, A]; // поменять местами значения A и C
                            // теперь точно A < B < C
if (X < A) {          // вар. 0
} else if (X == A) {  // вар. 1 
} else if (X < B) {   // вар. 2
} else if (X == B) {  // вар. 3
} else if (X < C) {   // вар. 4
} else if (X == C) {  // вар. 5
} else { // X > C     // вар. 6
}

Another way is with a binary tree, which intuitively but incorrectly tried to suggest xmoonlight
There are 2 * 7 = 14 options in total. The hypothesis that the actions are mirrored for the two options seems A > B > Cto A < B < Chave been confirmed, so I will leave the first step, where we swap A and C so that the sequence is definitely increasing.
There are 7 options left. 3 bits are enough to encode any of them. Each bit is an answer to 1 question, a test of one condition. Those. just three if's are enough to determine the desired block of code:

0  1  2  3  4  5  6 
---o-----o-----o---   
   A     B     C      

0  1  0  1  0  1  0  1
0  0  1  1  0  0  1  1
0  0  0  0  1  1  1  1

More or less like this:

if (A > C) [A, C] = [C, A]; // поменять местами значения A и C
                            // теперь точно A < B < C
if (X > B) {
  if (X > C) {    // вар. 6
  } else {
    if (X == C) { // вар. 5
    } else {      // вар. 4
    }
  }
} else {
  if (X > A) {
    if (X == B) { // вар. 3
    } else {      // вар. 2
    }
  } else {
    if (X == A) { // вар. 1
    } else {      // вар. 0
    }
  }
}

So less visually and readable, but saved on matches - fewer checks.
No more than 3 condition checks to determine where you are.

Answer 2

I will say right away that the answer of Sergey Sokolov is an extremely slow solution in terms of performance and, in general, his answer is illogical and even erroneous !
It has a lot of redundant conditions and checks that were simply incorrectly designed logically from the very beginning, which led to chaos in its code.
He tried to fix it using my answer, but unfortunately it didn't help the situation.
Do it right like this :

/* проверяем все центры */
1. X==A
2. X==B
3. X==C

/* проверяем левую ветку: X<B */
4. X<A
5. else ... //X>A

/* проверяем правую ветку: else */
6. X<C
7. else ... //X>C

With this approach, the code time spent on matching X to the desired value (A,B,C) or range is MINIMUM and OPTIMUM.

Answer 3

If actions cannot be somehow grouped according to conditions, then if ... else is perhaps the best construction for js.