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I am reading CPDAG, can anyone please explain why G1 and G3 are not equivalence classes as in the picture below? Thank you very much!

according to https://www.cs.cmu.edu/afs/cs/project/jair/pub/volume18/acid03a-html/node2.html .

"Theorem 1 [59] Two DAGs are equivalent if and only if they have the same skeleton and the same v-structures.

The skeleton of a DAG is the undirected graph that results from ignoring the directionality of every edge.

A v-structure in a DAG $H$ is an ordered triplet of nodes, $(x,y,z)$, such that (1) $H$ contains the arcs $x\rightarrow y$ and $y \leftarrow z$, and (2) the nodes $x$ and $z$ are not adjacent in $H$. A head-to-head pattern (shortened h-h) in a DAG $H$ is an ordered triplet of nodes, $(x,y,z)$, such that $H$ contains the arcs $x\rightarrow y$ and $y \leftarrow z$. Note that in an h-h pattern $(x,y,z)$ the nodes $x$ and $z$ can be adjacent."

The concept of equivalence of DAGs partitions the space of DAGs into a set of equivalence classes.

enter image description here

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    $\begingroup$ Could you please explain/define the term CPDAG in your post? Also what do you think yourself, why aren't G1 and G3 in equivalence class, any idea? $\endgroup$ – fade2black Aug 13 '17 at 11:17
  • $\begingroup$ It's also weird why you ask "why $G_1$ is not in class $G$" when I can read on the fine prints under the picture saying $G_1$ is in class $G$. $\endgroup$ – Apiwat Chantawibul Aug 13 '17 at 12:04
  • $\begingroup$ Thank you. I added the description in the question. Please have a look. Thank you! $\endgroup$ – user2842390 Aug 13 '17 at 14:55
  • $\begingroup$ What is a CPDAG? (You've expanded the acronym but not given a definition.) What does "equivalent" mean? $\endgroup$ – David Richerby Jan 11 '18 at 1:25
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They have different v-structures: $G_1$ has the v-structures $(x_3,y,x_4),(x_4,y,x_3)$, whereas $G_3$ has the v-structures $(x_1,x_4,y),(y,x_4,x_1)$.

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  • $\begingroup$ Thank you, however, I can say so with G1 and G4 (but the example said that G1 G4 are in its equivalence class): "$G_1$ has the v-structures $(x_3,y,x_4),(x_4,y,x_3)$ whereas $G_4$ has the v-structures $(x_1,x_4,y),(y,x_4,x_1)$". So why are they sill equivalent? $\endgroup$ – user2842390 Aug 14 '17 at 6:25
  • $\begingroup$ It could be a typo. Ask the authors. $\endgroup$ – Yuval Filmus Aug 14 '17 at 7:46

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