Definition of CCC and the logic that follows from it. about toponetx

[viyx]

Oh, I see, there is no contradicton now, I was wrong in the previous post. No more questions, thanks

[mhajij]

The definition has been related a little in the new version to meet Def 9 here

the above definition allows the creation of hypergraph when ranks are all equal.

from toponetx.classes import CombinatorialComplex as CCC

example = CCC()

example.add_cell([1,2], rank=1)
example.add_cell([1,2], rank=2)

when you add the second cell, the first one gets deleted and replaced by the second one.

You should access the cells with example.cells or example.skeleton(1) etc not with example.cells.hyperedge_dict

---

for the example

from toponetx.classes import CombinatorialComplex as CCC

example = CCC()

example.add_cell([1], rank=1)
example.add_cell([1], rank=2)

I just made merged a PR to address this issue to make it match the definition above. Let me know if this answers your question so that I close this issue.

[viyx]

Thanks for the quick response. I got the exception from your code, now it's clear that there cant be the same vertices and cells, vertices are prioritorized.

As I can see, there is still a contradiction with Def 9. From Def 9 we can get cells of rank>0 which are the vertices(rank=0). Even so could you clarify what is the definition of CCC in this repository now, to undestand how to work with CCC properly?

[mhajij]

I'm not sure what behavior you are seeking. Can you be specific?
Def 9 implies that cells with single elements can only have rank 0. This is why an exception is given.

[viyx]

Sorry, I'm confused again)

Is it legal CCC, where am I wrong?

Let S = {a,b,c}, X = {{a}, {bc}}, rk(a) = 3, rk(bc) = 1

It has vertice of rank=3 and at the same time satisfy Def 9.

[mhajij]

Hello,

No it's not legal. X is not valid here for ccc because it must contain all elements of S. Once that happens it cannot have a and in two different ranks, only the lower one, in this case rank zero.

[viyx]

Ok, I got that X is not valid, bad example from me

But this one looks legal
Let S = {a,b,c}, X = {{a}, {b}, {c}, {bc}}, rk(a) = rk(bc) = rk(b) = rk(c) = 3

[mhajij]

Ok, I got that X is not valid, bad example from me

But this one looks legal Let S = {a,b,c}, X = {{a}, {b}, {c}, {bc}}, rk(a) = rk(bc) = rk(b) = rk(c) = 3
the defintion forces

axiom 1 in def makes it impossible to create the above example/

rk(a) = rk(b) = rk(c) to be zero because they are singletons. All elements of S must have rank zero. Hence rk(b) and rk(c) cannot be 3.

[viyx]

Ok, I got that X is not valid, bad example from me

But this one looks legal Let S = {a,b,c}, X = {{a}, {b}, {c}, {bc}}, rk(a) = rk(bc) = rk(b) = rk(c) = 3

What definition do you use? Acoording to this one

Axiom 1 is satisfied, all singletons s in X
Axiom 2 is satisfied, 3 <= rk(bc) <= rk(b) <= rk(c) <= 3

[mhajij]

Ok I see the source of confusion. We actually consider remark

for our implemenation which forces the singularities to have zero ranks.
In other words, in TopoNetX, you can assume that the definition is 9 + the remark above. So in your example, a valid CCC :

rk(a) (must be) =0, rk(b) (must be) =0, rk(c) (must be) =0

[viyx]

Ok, and what is about axiom2, if I modify a bit my exmaple

Let S = {a,b,c}, X = {{a}, {b}, {c}, {bc}}, rk(a) = rk(bc) = rk(b) = rk(c) = 0

Again all axioms is satisfied, including remark 4.1

Axiom 1 is satisfied, all singletons s in X and their rank 0
Axiom 2 is satisfied, 0 <= rk(bc) <= rk(b) <= rk(c) <= 0