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I try to describe the logic of some algorithms I perform on meshes (in finite element analysis).

suppose i have some elements called e within some domain $E_\Omega$, each being a set of nodes n. Those n can be mutually shared between several e. Example:

$E_\Omega= \left\{ n_1 ,n_2,n_3,n_4,n_5,n_6 \right\}$

$e_1=\left\{n_1,n_2\right\}$

$e_2=\left\{n_2,n_3\right\}$

$e_3=\left\{n_1,n_4\right\}$

$e_4=\left\{n_4,n_6\right\}$

$e_5=\left\{n_5,n_6\right\}$

For some $e_i$, I now try to find all other e that share a node (i.e. be neighbors) to create a set $E_{neb}$. In software this is rather simple, but how do I describe this mathematically correctly?

following above example, i may wish to find a set of all "neighbors" of $e_4$ which would be $E_{neb,e_4}=\left\{e_3,e_5\right\}$ in that case.

I was thinking something like this:

$E_{neb, e_i}= E_{\Omega} \cap e_i \neq \emptyset$

update: something like this:

$E_{neb,e_i}= \left\{ e_j \subset E_{\Omega} \vert e_j \cap e_i \neq \emptyset\right\}$

so, more specifically, I want the set of elements (which itself are sets of nodes), for which it is true that they share a node with $e_i$.

thanks for all input, really apprechiate your help.

\Update: example added in

Daniel
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  • Not sure this is clear. I gather you have a big set, $E_{\Omega}$ and that you have a collection of subsets ${e_i}$ of that, right? Now, you fix one of the subsets, $e_i$ and you want all the indices $j$ such that $e_i\cap e_j\neq \emptyset$. Is this right? Assuming I have it write, I'd just let $S_i$ be the set of all those indices $j$. It's important to make the dependence on $i$ clear. – lulu Jun 01 '24 at 13:36
  • Example (of my interpretation). Let $E_{\Omega}$ be the set of natural numbers. Let $e_1={1,2,3}$, $e_2={4,5,6}$, $e_3={1, 10, 11, 12}$, $e_4={3, 4,5}$. Then, using my notation, $S_1={1, 3,4}$. In particular, $i$ is always in $S_i$ unless $e_i$ is empty. – lulu Jun 01 '24 at 13:39
  • thank you. $e_i$ is an arbitrary element within $E_\Omega$. Think of it as acutal geometric elements that actually overlap at their nodes. I do not need indices of all overlapping elements but those that overlap just that one. However, I like your Idea of having a set of the indices, I'll think about it – Daniel Jun 01 '24 at 13:47
  • Your post is hard to follow. In particular, it's not at all clear what things are subsets and which are nodes. I suggest: Focus on clarity. Efficiency of notation is a worthy goal ,but clarity is essential. – lulu Jun 01 '24 at 13:48
  • $E_{neb}= "all" e_j \subset E_{\Omega} "where" e_j \cap e_i \neq \emptyset$ but this written properly – Daniel Jun 01 '24 at 13:53
  • Again, since your definition critically depends on the choice of $i$, whether $e_i$ is a subset or just a single node, your notation should involve $i$. – lulu Jun 01 '24 at 13:58
  • sorry, I don't get what you mean, which is why I cannot clarify. $i$ is just some index describing which element has been chosen, it is arbitrary. – Daniel Jun 01 '24 at 14:03
  • Again, your post is very vague so I'm trying to guess at what you mean. I thought you fixed $e_i$, whether that's one node or a subset of nodes. Then I thought you wanted to know all the subsets which overlapped with that specific choice $e_i$. I suggest: Edit your post to contain an explicit example of what you actually want. Say $E_{\Omega}={n_1, n_2, n_3,n_4, n_5, n_6}$, so you have six nodes all in all. Now pick your subsets however you like and write out what you mean. – lulu Jun 01 '24 at 14:11
  • thanks, updated accordingly – Daniel Jun 01 '24 at 15:04
  • Well, that seems in line with my guess, and therefore the thing you want depends on the choice of $i$ – lulu Jun 01 '24 at 15:23
  • One distinction: you appear to reject the notion that $e_i$ intersects itself. That's fine, of course, but if that's your rule you should state it. – lulu Jun 01 '24 at 15:37
  • I maybe understand now. Of course it is based upon i, I want to formally describe this process of getting $E_{nebs}$ for i. Also, the set is allowed to contain $e_i$ itself. – Daniel Jun 01 '24 at 18:06
  • Well, your example didn't include $e_4$. Like I say, clarity is critical. You can have whatever rules you want, but they need to be clear. – lulu Jun 01 '24 at 18:43
  • corrected the question as suggested. – Daniel Jun 01 '24 at 20:32

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