How would you define a superproblem?












0












$begingroup$


Assuming that we have a problem, how would one classify its superproblem? There could be multiple ways to generalize a problem and for example two superproblems (although including the original as a subproblem) may not be compatible.



I am not sure whether there is a formal concept of a superproblem. But can you guide me to any formal source on this type of generalization?






problem-solving







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    $A$ is a subproblem of $B$ if and only if $B$ is a superproblem of $A$?
    $endgroup$
    – Cheerful Parsnip
    Jan 9 at 19:25










  • $begingroup$
    Perhaps, I used the word 'subproblem' in wrong context. What I meant was, has-a versus is-a distinction? A wider problem may contain the original problem but, there may not be an is-a relationship, only has-a? Is this possible?
    $endgroup$
    – yigoli
    Jan 9 at 19:29










  • $begingroup$
    This seems analogous to $A$ being a subset of various sets. The sets can't be disjoint (since all have $A$ as a subset) but might have disjoint subsets. Each superset of $A$ corresponds to a different way of generalising the problem represented by set $A$.
    $endgroup$
    – timtfj
    Jan 9 at 20:30










  • $begingroup$
    I don't think a problem is equivalent to a set though. At least there has to be an input-output relation, so more similar to a function maybe? So in the superproblem there maybe an additional hidden parameter that changes the output/solution, but at some value of this parameter this boils down to original problem.
    $endgroup$
    – yigoli
    Jan 9 at 21:32


















0












$begingroup$


Assuming that we have a problem, how would one classify its superproblem? There could be multiple ways to generalize a problem and for example two superproblems (although including the original as a subproblem) may not be compatible.



I am not sure whether there is a formal concept of a superproblem. But can you guide me to any formal source on this type of generalization?






problem-solving







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    $A$ is a subproblem of $B$ if and only if $B$ is a superproblem of $A$?
    $endgroup$
    – Cheerful Parsnip
    Jan 9 at 19:25










  • $begingroup$
    Perhaps, I used the word 'subproblem' in wrong context. What I meant was, has-a versus is-a distinction? A wider problem may contain the original problem but, there may not be an is-a relationship, only has-a? Is this possible?
    $endgroup$
    – yigoli
    Jan 9 at 19:29










  • $begingroup$
    This seems analogous to $A$ being a subset of various sets. The sets can't be disjoint (since all have $A$ as a subset) but might have disjoint subsets. Each superset of $A$ corresponds to a different way of generalising the problem represented by set $A$.
    $endgroup$
    – timtfj
    Jan 9 at 20:30










  • $begingroup$
    I don't think a problem is equivalent to a set though. At least there has to be an input-output relation, so more similar to a function maybe? So in the superproblem there maybe an additional hidden parameter that changes the output/solution, but at some value of this parameter this boils down to original problem.
    $endgroup$
    – yigoli
    Jan 9 at 21:32
















0












0








0


2



$begingroup$


Assuming that we have a problem, how would one classify its superproblem? There could be multiple ways to generalize a problem and for example two superproblems (although including the original as a subproblem) may not be compatible.



I am not sure whether there is a formal concept of a superproblem. But can you guide me to any formal source on this type of generalization?






problem-solving







share|cite|improve this question









$endgroup$




Assuming that we have a problem, how would one classify its superproblem? There could be multiple ways to generalize a problem and for example two superproblems (although including the original as a subproblem) may not be compatible.



I am not sure whether there is a formal concept of a superproblem. But can you guide me to any formal source on this type of generalization?






problem-solving




problem-solving






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 9 at 19:24









yigoliyigoli

177




177








  • 3




    $begingroup$
    $A$ is a subproblem of $B$ if and only if $B$ is a superproblem of $A$?
    $endgroup$
    – Cheerful Parsnip
    Jan 9 at 19:25










  • $begingroup$
    Perhaps, I used the word 'subproblem' in wrong context. What I meant was, has-a versus is-a distinction? A wider problem may contain the original problem but, there may not be an is-a relationship, only has-a? Is this possible?
    $endgroup$
    – yigoli
    Jan 9 at 19:29










  • $begingroup$
    This seems analogous to $A$ being a subset of various sets. The sets can't be disjoint (since all have $A$ as a subset) but might have disjoint subsets. Each superset of $A$ corresponds to a different way of generalising the problem represented by set $A$.
    $endgroup$
    – timtfj
    Jan 9 at 20:30










  • $begingroup$
    I don't think a problem is equivalent to a set though. At least there has to be an input-output relation, so more similar to a function maybe? So in the superproblem there maybe an additional hidden parameter that changes the output/solution, but at some value of this parameter this boils down to original problem.
    $endgroup$
    – yigoli
    Jan 9 at 21:32
















  • 3




    $begingroup$
    $A$ is a subproblem of $B$ if and only if $B$ is a superproblem of $A$?
    $endgroup$
    – Cheerful Parsnip
    Jan 9 at 19:25










  • $begingroup$
    Perhaps, I used the word 'subproblem' in wrong context. What I meant was, has-a versus is-a distinction? A wider problem may contain the original problem but, there may not be an is-a relationship, only has-a? Is this possible?
    $endgroup$
    – yigoli
    Jan 9 at 19:29










  • $begingroup$
    This seems analogous to $A$ being a subset of various sets. The sets can't be disjoint (since all have $A$ as a subset) but might have disjoint subsets. Each superset of $A$ corresponds to a different way of generalising the problem represented by set $A$.
    $endgroup$
    – timtfj
    Jan 9 at 20:30










  • $begingroup$
    I don't think a problem is equivalent to a set though. At least there has to be an input-output relation, so more similar to a function maybe? So in the superproblem there maybe an additional hidden parameter that changes the output/solution, but at some value of this parameter this boils down to original problem.
    $endgroup$
    – yigoli
    Jan 9 at 21:32










3




3




$begingroup$
$A$ is a subproblem of $B$ if and only if $B$ is a superproblem of $A$?
$endgroup$
– Cheerful Parsnip
Jan 9 at 19:25




$begingroup$
$A$ is a subproblem of $B$ if and only if $B$ is a superproblem of $A$?
$endgroup$
– Cheerful Parsnip
Jan 9 at 19:25












$begingroup$
Perhaps, I used the word 'subproblem' in wrong context. What I meant was, has-a versus is-a distinction? A wider problem may contain the original problem but, there may not be an is-a relationship, only has-a? Is this possible?
$endgroup$
– yigoli
Jan 9 at 19:29




$begingroup$
Perhaps, I used the word 'subproblem' in wrong context. What I meant was, has-a versus is-a distinction? A wider problem may contain the original problem but, there may not be an is-a relationship, only has-a? Is this possible?
$endgroup$
– yigoli
Jan 9 at 19:29












$begingroup$
This seems analogous to $A$ being a subset of various sets. The sets can't be disjoint (since all have $A$ as a subset) but might have disjoint subsets. Each superset of $A$ corresponds to a different way of generalising the problem represented by set $A$.
$endgroup$
– timtfj
Jan 9 at 20:30




$begingroup$
This seems analogous to $A$ being a subset of various sets. The sets can't be disjoint (since all have $A$ as a subset) but might have disjoint subsets. Each superset of $A$ corresponds to a different way of generalising the problem represented by set $A$.
$endgroup$
– timtfj
Jan 9 at 20:30












$begingroup$
I don't think a problem is equivalent to a set though. At least there has to be an input-output relation, so more similar to a function maybe? So in the superproblem there maybe an additional hidden parameter that changes the output/solution, but at some value of this parameter this boils down to original problem.
$endgroup$
– yigoli
Jan 9 at 21:32






$begingroup$
I don't think a problem is equivalent to a set though. At least there has to be an input-output relation, so more similar to a function maybe? So in the superproblem there maybe an additional hidden parameter that changes the output/solution, but at some value of this parameter this boils down to original problem.
$endgroup$
– yigoli
Jan 9 at 21:32












2 Answers
2






active

oldest

votes


















1












$begingroup$

I would say that
A is a superproblem of B
if B is a special case of A;
in other words,
A has a number of parameters such that,
when some constant values
are substituted for some of those parameters,
the resulting problem
is equivalent to B.



If you look at my questions and answers,
you will see that many of them
are generalizations of
someone else's problem.



The way I put this internally
is that I would
rather solve an infinite number of problems
than just one.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    This is clearly an informal term, but my view is that for a problem $A$, a superproblem $B$ of problem $A$ has the following properties:




    • Solving $B$ either solves $A$ directly or leads to a natural solution to $A$, but not vice versa.

    • Solving $B$ leads to a natural solution of other problems ($C, D, ldots$) but solving $A$ does not lead to a natural solution of those other problems.


    • $A$ is a special case of $B$ but $B$ is not a special case of $A$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      So this is analogous to, but the same as, $A$ being a subset of $B$? In which case we can maybe associate each problem with a set?
      $endgroup$
      – timtfj
      Jan 9 at 20:23










    • $begingroup$
      @timtfj: Informally... I guess so.
      $endgroup$
      – David G. Stork
      Jan 9 at 20:30












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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    I would say that
    A is a superproblem of B
    if B is a special case of A;
    in other words,
    A has a number of parameters such that,
    when some constant values
    are substituted for some of those parameters,
    the resulting problem
    is equivalent to B.



    If you look at my questions and answers,
    you will see that many of them
    are generalizations of
    someone else's problem.



    The way I put this internally
    is that I would
    rather solve an infinite number of problems
    than just one.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      I would say that
      A is a superproblem of B
      if B is a special case of A;
      in other words,
      A has a number of parameters such that,
      when some constant values
      are substituted for some of those parameters,
      the resulting problem
      is equivalent to B.



      If you look at my questions and answers,
      you will see that many of them
      are generalizations of
      someone else's problem.



      The way I put this internally
      is that I would
      rather solve an infinite number of problems
      than just one.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        I would say that
        A is a superproblem of B
        if B is a special case of A;
        in other words,
        A has a number of parameters such that,
        when some constant values
        are substituted for some of those parameters,
        the resulting problem
        is equivalent to B.



        If you look at my questions and answers,
        you will see that many of them
        are generalizations of
        someone else's problem.



        The way I put this internally
        is that I would
        rather solve an infinite number of problems
        than just one.






        share|cite|improve this answer









        $endgroup$



        I would say that
        A is a superproblem of B
        if B is a special case of A;
        in other words,
        A has a number of parameters such that,
        when some constant values
        are substituted for some of those parameters,
        the resulting problem
        is equivalent to B.



        If you look at my questions and answers,
        you will see that many of them
        are generalizations of
        someone else's problem.



        The way I put this internally
        is that I would
        rather solve an infinite number of problems
        than just one.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 10 at 0:52









        marty cohenmarty cohen

        74.9k549130




        74.9k549130























            1












            $begingroup$

            This is clearly an informal term, but my view is that for a problem $A$, a superproblem $B$ of problem $A$ has the following properties:




            • Solving $B$ either solves $A$ directly or leads to a natural solution to $A$, but not vice versa.

            • Solving $B$ leads to a natural solution of other problems ($C, D, ldots$) but solving $A$ does not lead to a natural solution of those other problems.


            • $A$ is a special case of $B$ but $B$ is not a special case of $A$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              So this is analogous to, but the same as, $A$ being a subset of $B$? In which case we can maybe associate each problem with a set?
              $endgroup$
              – timtfj
              Jan 9 at 20:23










            • $begingroup$
              @timtfj: Informally... I guess so.
              $endgroup$
              – David G. Stork
              Jan 9 at 20:30
















            1












            $begingroup$

            This is clearly an informal term, but my view is that for a problem $A$, a superproblem $B$ of problem $A$ has the following properties:




            • Solving $B$ either solves $A$ directly or leads to a natural solution to $A$, but not vice versa.

            • Solving $B$ leads to a natural solution of other problems ($C, D, ldots$) but solving $A$ does not lead to a natural solution of those other problems.


            • $A$ is a special case of $B$ but $B$ is not a special case of $A$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              So this is analogous to, but the same as, $A$ being a subset of $B$? In which case we can maybe associate each problem with a set?
              $endgroup$
              – timtfj
              Jan 9 at 20:23










            • $begingroup$
              @timtfj: Informally... I guess so.
              $endgroup$
              – David G. Stork
              Jan 9 at 20:30














            1












            1








            1





            $begingroup$

            This is clearly an informal term, but my view is that for a problem $A$, a superproblem $B$ of problem $A$ has the following properties:




            • Solving $B$ either solves $A$ directly or leads to a natural solution to $A$, but not vice versa.

            • Solving $B$ leads to a natural solution of other problems ($C, D, ldots$) but solving $A$ does not lead to a natural solution of those other problems.


            • $A$ is a special case of $B$ but $B$ is not a special case of $A$.






            share|cite|improve this answer









            $endgroup$



            This is clearly an informal term, but my view is that for a problem $A$, a superproblem $B$ of problem $A$ has the following properties:




            • Solving $B$ either solves $A$ directly or leads to a natural solution to $A$, but not vice versa.

            • Solving $B$ leads to a natural solution of other problems ($C, D, ldots$) but solving $A$ does not lead to a natural solution of those other problems.


            • $A$ is a special case of $B$ but $B$ is not a special case of $A$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 9 at 19:36









            David G. StorkDavid G. Stork

            11.9k41735




            11.9k41735












            • $begingroup$
              So this is analogous to, but the same as, $A$ being a subset of $B$? In which case we can maybe associate each problem with a set?
              $endgroup$
              – timtfj
              Jan 9 at 20:23










            • $begingroup$
              @timtfj: Informally... I guess so.
              $endgroup$
              – David G. Stork
              Jan 9 at 20:30


















            • $begingroup$
              So this is analogous to, but the same as, $A$ being a subset of $B$? In which case we can maybe associate each problem with a set?
              $endgroup$
              – timtfj
              Jan 9 at 20:23










            • $begingroup$
              @timtfj: Informally... I guess so.
              $endgroup$
              – David G. Stork
              Jan 9 at 20:30
















            $begingroup$
            So this is analogous to, but the same as, $A$ being a subset of $B$? In which case we can maybe associate each problem with a set?
            $endgroup$
            – timtfj
            Jan 9 at 20:23




            $begingroup$
            So this is analogous to, but the same as, $A$ being a subset of $B$? In which case we can maybe associate each problem with a set?
            $endgroup$
            – timtfj
            Jan 9 at 20:23












            $begingroup$
            @timtfj: Informally... I guess so.
            $endgroup$
            – David G. Stork
            Jan 9 at 20:30




            $begingroup$
            @timtfj: Informally... I guess so.
            $endgroup$
            – David G. Stork
            Jan 9 at 20:30


















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