How would you define a superproblem?
$begingroup$
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
asked Jan 9 at 19:24
yigoliyigoli
177
$endgroup$
add a comment |
$begingroup$
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
asked Jan 9 at 19:24
yigoliyigoli
177
$endgroup$
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
add a comment |
$begingroup$
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
asked Jan 9 at 19:24
yigoliyigoli
177
$endgroup$
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
asked Jan 9 at 19:24
yigoliyigoli
177
asked Jan 9 at 19:24
yigoliyigoli
177
asked Jan 9 at 19:24
yigoliyigoli
177
asked Jan 9 at 19:24
yigoliyigoli
177
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Jan 10 at 0:52
marty cohenmarty cohen
74.9k549130
$endgroup$
add a comment |
$begingroup$
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 $A$.
answered Jan 9 at 19:36
David G. StorkDavid G. Stork
11.9k41735
$endgroup$
$begingroup$
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
add a comment |
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2 Answers
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active
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2 Answers
2
active
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active
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oldest
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$begingroup$
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.
answered Jan 10 at 0:52
marty cohenmarty cohen
74.9k549130
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 10 at 0:52
marty cohenmarty cohen
74.9k549130
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 10 at 0:52
marty cohenmarty cohen
74.9k549130
$endgroup$
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.
answered Jan 10 at 0:52
marty cohenmarty cohen
74.9k549130
answered Jan 10 at 0:52
marty cohenmarty cohen
74.9k549130
answered Jan 10 at 0:52
marty cohenmarty cohen
74.9k549130
answered Jan 10 at 0:52
marty cohenmarty cohen
74.9k549130
74.9k549130
add a comment |
add a comment |
$begingroup$
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 $A$.
answered Jan 9 at 19:36
David G. StorkDavid G. Stork
11.9k41735
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $A$.
answered Jan 9 at 19:36
David G. StorkDavid G. Stork
11.9k41735
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $A$.
answered Jan 9 at 19:36
David G. StorkDavid G. Stork
11.9k41735
$endgroup$
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 $A$.
answered Jan 9 at 19:36
David G. StorkDavid G. Stork
11.9k41735
answered Jan 9 at 19:36
David G. StorkDavid G. Stork
11.9k41735
answered Jan 9 at 19:36
David G. StorkDavid G. Stork
11.9k41735
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
add a comment |
$begingroup$
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
add a comment |
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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