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12:00 AM
it's a math term
 
Clausura transitiva
 
In mathematics, the transitive closure of a binary relation R on a set X is the smallest relation on X that contains R and is transitive. For example, if X is a set of airports and xRy means "there is a direct flight from airport x to airport y" (for x and y in X), then the transitive closure of R on X is the relation R+ such that x R+ y means "it is possible to fly from x to y in one or more flights". Informally, the transitive closure gives you the set of all places you can get to from any starting place. More formally, the transitive closure of a binary relation R on a set X is the transitive...
check the graph theory part
 
@ypercubeᵀᴹ like a "path for all possible points that somehow are related" ?
 
it's late, a tomorrow-work. nite all
 
12:08 AM
Or the airplane connections example
> One example of a non-transitive relation is "city x can be reached via a direct flight from city y" on the set of all cities.
> The transitive closure of this relation is a different relation, namely "there is a sequence of direct flights that begins at city x and ends at city y".
 
 
2 hours later…
1:47 AM
As a contrarian, I say: good night.
 
0
Q: Why does the search function not work as advertised?

mustaccioWhen you click in the search bar it shows you these hints: I take to understand the highlighted one as this: the phrase in double quotes will search for the exact quoted literal. Not so, as the examples below show. Am I wrong thinking that this does not work as a reasonable person would expect...

 
 
7 hours later…
9:25 AM
@ypercubeᵀᴹ have a look, I think I understood the concept (more or less)
 
@McNets nice one. +1
Did you mean to add an 'Approach 2' heading?
To add a it more explanation, the original relation has only the immediate connections, which can be seen as paths of length 1 (2 nodes each).
Approach 1 finds all paths of length as it applies a method of connecting once.
To find paths of length N, you have to apply the methods N-1 times.
To find all paths of arbitrary lengths (alas the transitive closure), you need a recursive solution.
Or a while loop.
It can't be done with simple SQL, ie one query without CTEs.
Correction: Approach 1 finds all paths of length 2 (3 nodes)
 
10:26 AM
@ypercubeᵀᴹ thanks for the lesson.
 

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