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differences between OR/XOR in Heuristic Miner

lauraUnivpm
lauraUnivpm
edited February 2014 in Process Mining

Hi guys,
I have some doubts about the semantic of the relationships in the causal net generated by the Heuristic Miner. In particular I didn't understand the semantic of OR/XOR relationships. Reading some papers about C-net, e.g. 
http://www.processmining.org/blogs/pub2011/causal_nets_a_modeling_language_tailored_towards_process_discovery
or the
 "Flexible Heuristic Miner" (available in tue repository)
I have found several example of interpretation of input/output bindings, but seems that there is no difference between OR and XOR. For instance in the second paper  (in section 3) they speak about various input/output of some example tasks; let's take  K, for which Input={{ J,H }, { J,D }}. When they describe K inputs, they say that there is an "or" relation between the subsets ( J and H) , (J and D) : however a OR relation should mean that I can execute the firs set, the second, or both, so I actually could execute J,H,D together. But it's not the same thing of declare that I have to execute only one of the two paths, that should be the actual interpretations of the binding sets, namely a XOR. Moreover, when I translate in a Petri net, I obtain an OR, thus missing the original interpretation.
Can you help me in understanding such interpretation? Thank you all!

Best Answers

  • jribeiro
    jribeiro
    Accepted Answer

    Hi!

    {{ J,H }, { J,D }} means that K follows one of those input sets. Remark that neither OR nor XOR are correct. The correct XOR semantic only happen when we have two cases. For instance, XOR of {A,B,C} is
    (A and not B and not C) or
    (not A and B and not C) or
    (not A and not B and C) or
    (A and B and C).

    Cheers,
    Joel 

  • jribeiro
    jribeiro
    Accepted Answer

     Hi Laura,

     Yes, if you have the set {A,B} then you can only have either A or B, but not both (i.e., you can only have one element from the set). If you have the 3-element set {(A,B),(C,D),(D,E)} then you can only have (A,B) or (C,D) or (D,E) or the 3 elements together (so, no, it does not become an OR). The most intuitive way to understand this is to consider that you can only have an odd number of elements from the set. 

     Cheers,
    Joel 

Answers

  • lauraUnivpm
    lauraUnivpm

    Hi,

    Thank you for the answer. Can you better explain what do you mean with "two cases"? If I understand: if I have (as input) the set {A,B}, this is a real xor, i.e. I can only have (A) XOR (B). But if I have more thatn two alternatives, it becomes a OR ?  and what if I have {(A,B),(C,D),(D,E)} for example?

    Thank you very much,

    have a nice day.

    Laura

  • lauraUnivpm
    lauraUnivpm

    Hi Joel,

    ok, I see. But I still have some "theoric" doubts. It seems that a special construct is introduced, that is not a XOR neither a OR at all, right? But in such a way, how can I translate in another representation like, for instance BPMN, where (to the best of my knowledge) this kind of construct doesn't exist? I've found a plugin for this in ProM, but actually it seems to suffer of such problem (i.e. during translation it seems that some relationships are "lost", i.e. not represented, when there are input/output bindings with more than two elements). Am I wrong?

    Thank you very much,

    Cheers,

    Laura

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