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Restore indirect relations within a transitive relation

  1. Feb 25, 2014 #1
    Hi,

    I have a transitive relation and wana build a complete set of pairs that reflect all (direct/indirect) relations among the pairs.

    Ex.: suppose I have this relation R = { (1,2), (2,3), (3,5), (5,7), (3,4) }

    I wana to produce this relation R oper R = { (1,2), (1,3), (1,4), (1,5), (1,7), (2,3), (2,4), (2,5), (2,7), (3,4), (3,5), (3,7), (5,7) }

    I tried to use the composite operator (°), but I got this R U (R ° R) = { (1,2), (2,3), (3,5), (5,7), (3,4), (1,3), (2,4), (2,5), (3,7) } which is not complete. In this case I need a loop operator until all pairs are restored.

    Is there an operator that I can used to reflect that?

    Thanks in advance.
     
  2. jcsd
  3. Feb 25, 2014 #2

    Stephen Tashi

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    Science Advisor

    That isn't a transitive relation. How does this example illustrate your question?
     
  4. Feb 25, 2014 #3
    Sorry, Math is not my suppject, but this is how I understand transitive relation:
    if A > B and B > C then A > C.

    In my example I have , e.g.,
    - (1,2) = 1 > 2
    - (2,3) = 2 > 3
    which conclude that 1 > 3 ==or==> (1,3)

    from this new pair I can create another two pairs
    - (1,3)
    - (3,5) & (3,4) from the relation
    I can produce two new pairs (1,5) & (1,4)
    an so on

    In my research, I have two relations , based on partial order, build independently and I have to compare them.
    suppose I have these two
    R1 = { (1,2), (2,4) }
    R2 = { (1,2), (2,4), (1,4) }
    For me, they are equivalent. but How I can use math to prove my case. I am trying to generate/expand the total/all relations among the existing pairs.
     
  5. Feb 25, 2014 #4
    Hi,

    I am seeking help to derive my formula in set theory.

    I will explain my request through the following example:

    suppose I have this transitive relation R = { (1,2), (2,3), (3,5), (5,7), (3,4) }

    I mean by transitive that since
    (1,2) ==> 1 > 2, and
    (2,3) ==> 2 > 3
    then I can conclude that
    1 > 3 or (1,3)
    Hence, I can add this pair explicitly to R.

    I wana to add all these implicit relations to reach the final relation:
    R` = { (1,2), (1,3), (1,4), (1,5), (1,7), (2,3), (2,4), (2,5), (2,7), (3,4), (3,5), (3,7), (5,7) }

    Currently, here are the steps I used to build my case

    s1 = R ° R = { (1,3), (2,4), (2,5), (3,7) } // compsite operator

    s2 = R U s1 = { (1,2), (2,3), (3,5), (5,7), (3,4), (1,3), (2,4), (2,5), (3,7) } // union operator

    s1 = s2 ° R = { (1,3), (2,4), (2,5), (3,7), (1,5), (1,4), (2,7) }

    s2 = s2 U s1 = { (1,2), (2,3), (3,5), (5,7), (3,4), (1,3), (2,4), (2,5), (3,7), (1,5), (1,4), (2,7) }

    s1 = s2 ° R = { (1,3), (2,4), (2,5), (3,7), (1,5), (1,4), (2,7), (1,7) }

    s2 = s2 U s1 = { (1,2), (2,3), (3,5), (5,7), (3,4), (1,3), (2,4), (2,5), (3,7), (1,5), (1,4), (2,7), (1,7) }

    s1 = s2 ° R = { (1,3), (2,4), (2,5), (3,7), (1,5), (1,4), (2,7), (1,7) }

    I stop when s1 produce the same previous relation, and my result is in s2.

    I built a computer algorithm for that, but I want a math formula to report my work in formal way.

    I donot know if there is an operator reflect the generation of all implicit indirect relations among the elements of set (I read three discrete math books and browsed several math pages), or there is a loop operator that reflect a recursiveness based on a condition (not number of occurrences).

    Thanks for any hint,
    Reference to papers/books/tutorials are appreciated
     
  6. Feb 25, 2014 #5

    Stephen Tashi

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    Science Advisor

    I think the correct statement statement of your question is:

    If I am given a relation S as a finite set, what algorithm can be used to generate the smallest set R that is a transitive relation and that contains S as a subset?

    You have proposed an algorithm. Are you asking for a proof that your algorithm terminates in a finite number of steps and that it produces the desired R?
     
  7. Feb 26, 2014 #6
    Thanks Stephen,

    I had implemented the algorithm and it works fine.

    I am trying to put a nice formula in my report to reflect this algorithm.
     
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