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Source: http://examples.gurobi.com/traveling-salesman-problem

I don't get this part: (look at the source)

$$\sum_{i,j\in\{1,2,3\},i\neq j} x_{ij}=3>2=|\{1,2,3\}|-1$$

I get that $x_{ij}$ is equal to 3, but why the "> 2" ?

And what is the deal with subtracting 1 from a set? How do you even do that?

How come $|\{1,2,3\}|-1 = 3 > 2$ ?!?

Okay so: $$|\{1,2,3\}|-1 = 2$$

So how is he allowed to write: $$|\{1,2,3\}|-1 = 3 > 2$$

?

That is basically the same as writing: (which is incorrect right?) $$2 = 3 > 2$$

I don't get this part at all, please elaborate on what happened in as simple language as possible. My level is high school final math level.

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  • $\begingroup$ $3 \gt 2$ is what they imply I think. The remaining part subtracts 1 from the cardinality of the set and not the set itself. The cardinality of the set is the number of elements in it. Here, the set has 3 elements so you get 2 if you subtract 1 from the cardinality $\endgroup$ – kauray Dec 9 '18 at 10:01
  • $\begingroup$ This is the very common practice of chaining (in)equalities. Think of it as $3 > 2$ and $2 = 3 - 1$ being chained together as $3 > 2 = 3 - 1$. $\endgroup$ – Alex Vong Dec 9 '18 at 18:43
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The point of these constraints is eliminating subtours, which the source explains quite clearly. So for every subset $S$ of the nodes, such as $\{1,2,3\}$, they add a constraint which says $\Sigma_{i,j \in S, i \neq j} x_{ij} \leq |S| - 1$. So when this constraint is satisfied, there is no way to form a cycle on the vertices in $S$.

Now, if this constraint was not satisfied (i.e., the number of edges was at least $|S|$), then a cycle could be formed like they show in their figures. For example, on $\{1,2,3\}$, you can form a triangle (which is a cycle) if you use 3 edges.

Particularly regarding your confusion, note that they have written $|S|-1$ (and not $S-1$). Here, $|S|$ refers to the size of the set $S$ (also known as the cardinality of $S$), so $|\{1,2,3\}| = 3$. Further, notice that they don't write $2 = 3 > 2$, but instead $3 > 2 = 3 - 1$. If it's clearer, you can also assume the constraint just says $3 > |\{1,2,3\}| - 1$.

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  • $\begingroup$ Question, What does "S≠∅" Mean? That the subset should not be none/empty? $\endgroup$ – Ryan Cameron Dec 9 '18 at 10:51
  • $\begingroup$ @Ryan $\emptyset$ stands for the set with no elements, i.e., the empty set. So you are exactly right. $\endgroup$ – Juho Dec 9 '18 at 10:56
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You seem to have misunderstood pretty much every part of the statement

$$\sum_{i,j\in\{1,2,3\},i\neq j} x_{ij}=3>2=|\{1,2,3\}|-1\,.$$

I get that $x_{ij}$ is equal to 3,

No, the sum of all values $x_{ij}$ where $i$ and $j$ are distinct values from $\{1,2,3\}$ is equal to $3$.

but why the "> 2" ?

Because three is bigger than two.

And what is the deal with subtracting 1 from a set? How do you even do that?

No, it's subtracting one from the cardinality of the set. Notice the $|\dots|$.

How come $|\{1,2,3\}|-1 = 3 > 2$ ?!?

It isn't. When we write something like $A=B>C=D$, it means that $A=B$, $B>C$ and $C=D$. You can't just re-order the terms and expect the statement to remain true, just as you can't reorder $3>2$ as $2<3$ and expect it to remain true.

So, the statement as a whole means:

  • The sum of the values $x_{ij}$ is equal to $3$.
  • Also, $3>2$.
  • Also, $2=|\{1,2,3\}|-1$.

So how is he allowed to write: $$|\{1,2,3\}|-1 = 3 > 2\,?$$

He isn't and he doesn't.

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  • $\begingroup$ Sorry, I am confused. Regarding the sum of all values xij where i and j are distinct values from {1,2,3} is equal to 3. So I get distinct values 1 and 3 from the set where sum is not 3. So I am not understanding, can you help please? What do you mean sum of all values xij where i and j are distinct? $\endgroup$ – Koray Tugay Dec 9 '18 at 16:01
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    $\begingroup$ @KorayTugay In this context, $x_{i,j}$ exists only when $i<j$, so the sum above is $x_{1,2}+x_{1,3}+x_{2,3}$ and the result of this is 3 (since it's $1+1+1$, in the considered example). You don't sum the indices, you sum the values of all the numbers $x_{i,j}$. $\endgroup$ – chi Dec 9 '18 at 16:05
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    $\begingroup$ @KorayTugay I think the underlying problem here is that you don't understand mathematical notation. I suggest you talk to your school maths teacher about that, because you need more interactive help than we can really give on this site. $\endgroup$ – David Richerby Dec 9 '18 at 16:06
  • $\begingroup$ @chi I see thanks I understand. You count the possible combinations. Thanks. $\endgroup$ – Koray Tugay Dec 9 '18 at 17:46
  • $\begingroup$ @RyanCameron My suggestion to understanding $\sum$ notation is to expand it out. By definition, we can write $$\sum_{i, j \in \{1, 2, 3\}, i \neq j} x_{ij} = x_{12} + x_{13} + x_{21} + x_{23} + x_{31} + x_{32}$$ I remember my number theory professor used to write $$\sum_{\substack{1 \le i, j \le 3 \\ i \neq j}} x_{ij}$$ which I think is less formal and more readable. $\endgroup$ – Alex Vong Dec 9 '18 at 19:04

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