How to determine if a set relation is Transitive?
$begingroup$
So, as far as binary relations go, I have a firm grasp on the symmetric, anti-symmetric, and reflexive relations. However, when it comes to transitive relations, I run into a block.
In this problem, for example, I had to determine the transitive closures for the given set that contained: (1,1) (1,2) (2,3) (3,2) (4,4)
The additional elements that would enable this set to be a transitive relation were: (1,3) (2,2) (3,3)
I think I understand that if there is a (1,2) and a (2,3) then there must also be a (1,3). I see that there is a (2,3), but why doesn't there have to be a (3, 4) and a (2,4)? I need some enlightenment as to how I can properly go about determining transitive closures, or determining if a completed relation set includes the right elements to make it transitive.
discrete-mathematics relations
edited Dec 8 '18 at 1:31
Andrés E. Caicedo
65.4k8158249
asked Dec 7 '18 at 23:37
ZepheriahZepheriah
83
$endgroup$
add a comment |
$begingroup$
So, as far as binary relations go, I have a firm grasp on the symmetric, anti-symmetric, and reflexive relations. However, when it comes to transitive relations, I run into a block.
In this problem, for example, I had to determine the transitive closures for the given set that contained: (1,1) (1,2) (2,3) (3,2) (4,4)
The additional elements that would enable this set to be a transitive relation were: (1,3) (2,2) (3,3)
I think I understand that if there is a (1,2) and a (2,3) then there must also be a (1,3). I see that there is a (2,3), but why doesn't there have to be a (3, 4) and a (2,4)? I need some enlightenment as to how I can properly go about determining transitive closures, or determining if a completed relation set includes the right elements to make it transitive.
discrete-mathematics relations
edited Dec 8 '18 at 1:31
Andrés E. Caicedo
65.4k8158249
asked Dec 7 '18 at 23:37
ZepheriahZepheriah
83
$endgroup$
add a comment |
$begingroup$
So, as far as binary relations go, I have a firm grasp on the symmetric, anti-symmetric, and reflexive relations. However, when it comes to transitive relations, I run into a block.
In this problem, for example, I had to determine the transitive closures for the given set that contained: (1,1) (1,2) (2,3) (3,2) (4,4)
The additional elements that would enable this set to be a transitive relation were: (1,3) (2,2) (3,3)
I think I understand that if there is a (1,2) and a (2,3) then there must also be a (1,3). I see that there is a (2,3), but why doesn't there have to be a (3, 4) and a (2,4)? I need some enlightenment as to how I can properly go about determining transitive closures, or determining if a completed relation set includes the right elements to make it transitive.
discrete-mathematics relations
edited Dec 8 '18 at 1:31
Andrés E. Caicedo
65.4k8158249
asked Dec 7 '18 at 23:37
ZepheriahZepheriah
83
$endgroup$
So, as far as binary relations go, I have a firm grasp on the symmetric, anti-symmetric, and reflexive relations. However, when it comes to transitive relations, I run into a block.
In this problem, for example, I had to determine the transitive closures for the given set that contained: (1,1) (1,2) (2,3) (3,2) (4,4)
The additional elements that would enable this set to be a transitive relation were: (1,3) (2,2) (3,3)
I think I understand that if there is a (1,2) and a (2,3) then there must also be a (1,3). I see that there is a (2,3), but why doesn't there have to be a (3, 4) and a (2,4)? I need some enlightenment as to how I can properly go about determining transitive closures, or determining if a completed relation set includes the right elements to make it transitive.
discrete-mathematics relations
discrete-mathematics relations
edited Dec 8 '18 at 1:31
Andrés E. Caicedo
65.4k8158249
asked Dec 7 '18 at 23:37
ZepheriahZepheriah
83
edited Dec 8 '18 at 1:31
Andrés E. Caicedo
65.4k8158249
asked Dec 7 '18 at 23:37
ZepheriahZepheriah
83
edited Dec 8 '18 at 1:31
Andrés E. Caicedo
65.4k8158249
edited Dec 8 '18 at 1:31
Andrés E. Caicedo
65.4k8158249
edited Dec 8 '18 at 1:31
Andrés E. Caicedo
65.4k8158249
65.4k8158249
asked Dec 7 '18 at 23:37
ZepheriahZepheriah
83
asked Dec 7 '18 at 23:37
ZepheriahZepheriah
83
asked Dec 7 '18 at 23:37
ZepheriahZepheriah
83
83
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let's make a list of the pairs together with the ones that can be chained with them by transitivity
begin{array}{cccc}
1 & 2 & to \
hline
(1,1) & (1,1) & (1,1) \
(1,1) & (1,2) & (1,2) \
(1,2) & (2,3) & color{red}{(1,3)} \
(2,3) & (3,2) & color{red}{(2,2)} \
(3,2) & (2,3) & color{red}{(3,3)} \
(4,4) & (4,4) & (4,4)
end{array}
In the first column, a pair; in the second column a pair that can be chained to the first one; in the third column, what's necessary to satisfy transitivity. Red color means the needed pair is missing.
Now we have to repeat the check with the newly added pairs
begin{array}{cccc}
1 & 2 & to \
hline
(1,1) & (1,1) & (1,1) & *\
(1,1) & (1,2) & (1,2) & * \
(1,2) & (2,2) & (2,2) \
(1,2) & (2,3) & (1,3) & * \
(1,3) & (3,2) & (1,2) \
(1,3) & (3,3) & (1,3) \
(2,2) & (2,2) & (2,2) \
(2,2) & (2,3) & (2,3) \
(2,3) & (3,2) & (2,2) & * \
(2,3) & (3,3) & (2,3) \
(3,2) & (2,3) & (3,3) & * \
(3,3) & (3,2) & (3,2) \
(3,3) & (3,3) & (3,3) \
(4,4) & (4,4) & (4,4) & *
end{array}
None has been marked red, because all pairs in the third column belong to the “second stage” relation. The $*$ denotes a row that has already been considered in the first stage.
If at this stage still some pairs had been missing, a third stage would be necessary.
In this case, however, we're done: the transitive closure is
$$
{(1,1),(1,2),(1,3),(2,2),(2,3),(3,2),(3,3),(4,4)}
$$
answered Dec 8 '18 at 0:14
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
The trick here is to build chains: can you build a chain with these relationships that gets you from $2$ to $4$? If not, then you don't need $(2,4)$ to transitively close the relation. On the other hand, $(2,3)$ and $(3,2)$ can be chained together $(2,3,2)$ and so $(2,2)$ must be in the closure.
answered Dec 7 '18 at 23:45
Dan UznanskiDan Uznanski
6,62021427
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
Let's make a list of the pairs together with the ones that can be chained with them by transitivity
begin{array}{cccc}
1 & 2 & to \
hline
(1,1) & (1,1) & (1,1) \
(1,1) & (1,2) & (1,2) \
(1,2) & (2,3) & color{red}{(1,3)} \
(2,3) & (3,2) & color{red}{(2,2)} \
(3,2) & (2,3) & color{red}{(3,3)} \
(4,4) & (4,4) & (4,4)
end{array}
In the first column, a pair; in the second column a pair that can be chained to the first one; in the third column, what's necessary to satisfy transitivity. Red color means the needed pair is missing.
Now we have to repeat the check with the newly added pairs
begin{array}{cccc}
1 & 2 & to \
hline
(1,1) & (1,1) & (1,1) & *\
(1,1) & (1,2) & (1,2) & * \
(1,2) & (2,2) & (2,2) \
(1,2) & (2,3) & (1,3) & * \
(1,3) & (3,2) & (1,2) \
(1,3) & (3,3) & (1,3) \
(2,2) & (2,2) & (2,2) \
(2,2) & (2,3) & (2,3) \
(2,3) & (3,2) & (2,2) & * \
(2,3) & (3,3) & (2,3) \
(3,2) & (2,3) & (3,3) & * \
(3,3) & (3,2) & (3,2) \
(3,3) & (3,3) & (3,3) \
(4,4) & (4,4) & (4,4) & *
end{array}
None has been marked red, because all pairs in the third column belong to the “second stage” relation. The $*$ denotes a row that has already been considered in the first stage.
If at this stage still some pairs had been missing, a third stage would be necessary.
In this case, however, we're done: the transitive closure is
$$
{(1,1),(1,2),(1,3),(2,2),(2,3),(3,2),(3,3),(4,4)}
$$
answered Dec 8 '18 at 0:14
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
Let's make a list of the pairs together with the ones that can be chained with them by transitivity
begin{array}{cccc}
1 & 2 & to \
hline
(1,1) & (1,1) & (1,1) \
(1,1) & (1,2) & (1,2) \
(1,2) & (2,3) & color{red}{(1,3)} \
(2,3) & (3,2) & color{red}{(2,2)} \
(3,2) & (2,3) & color{red}{(3,3)} \
(4,4) & (4,4) & (4,4)
end{array}
In the first column, a pair; in the second column a pair that can be chained to the first one; in the third column, what's necessary to satisfy transitivity. Red color means the needed pair is missing.
Now we have to repeat the check with the newly added pairs
begin{array}{cccc}
1 & 2 & to \
hline
(1,1) & (1,1) & (1,1) & *\
(1,1) & (1,2) & (1,2) & * \
(1,2) & (2,2) & (2,2) \
(1,2) & (2,3) & (1,3) & * \
(1,3) & (3,2) & (1,2) \
(1,3) & (3,3) & (1,3) \
(2,2) & (2,2) & (2,2) \
(2,2) & (2,3) & (2,3) \
(2,3) & (3,2) & (2,2) & * \
(2,3) & (3,3) & (2,3) \
(3,2) & (2,3) & (3,3) & * \
(3,3) & (3,2) & (3,2) \
(3,3) & (3,3) & (3,3) \
(4,4) & (4,4) & (4,4) & *
end{array}
None has been marked red, because all pairs in the third column belong to the “second stage” relation. The $*$ denotes a row that has already been considered in the first stage.
If at this stage still some pairs had been missing, a third stage would be necessary.
In this case, however, we're done: the transitive closure is
$$
{(1,1),(1,2),(1,3),(2,2),(2,3),(3,2),(3,3),(4,4)}
$$
answered Dec 8 '18 at 0:14
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
Let's make a list of the pairs together with the ones that can be chained with them by transitivity
begin{array}{cccc}
1 & 2 & to \
hline
(1,1) & (1,1) & (1,1) \
(1,1) & (1,2) & (1,2) \
(1,2) & (2,3) & color{red}{(1,3)} \
(2,3) & (3,2) & color{red}{(2,2)} \
(3,2) & (2,3) & color{red}{(3,3)} \
(4,4) & (4,4) & (4,4)
end{array}
In the first column, a pair; in the second column a pair that can be chained to the first one; in the third column, what's necessary to satisfy transitivity. Red color means the needed pair is missing.
Now we have to repeat the check with the newly added pairs
begin{array}{cccc}
1 & 2 & to \
hline
(1,1) & (1,1) & (1,1) & *\
(1,1) & (1,2) & (1,2) & * \
(1,2) & (2,2) & (2,2) \
(1,2) & (2,3) & (1,3) & * \
(1,3) & (3,2) & (1,2) \
(1,3) & (3,3) & (1,3) \
(2,2) & (2,2) & (2,2) \
(2,2) & (2,3) & (2,3) \
(2,3) & (3,2) & (2,2) & * \
(2,3) & (3,3) & (2,3) \
(3,2) & (2,3) & (3,3) & * \
(3,3) & (3,2) & (3,2) \
(3,3) & (3,3) & (3,3) \
(4,4) & (4,4) & (4,4) & *
end{array}
None has been marked red, because all pairs in the third column belong to the “second stage” relation. The $*$ denotes a row that has already been considered in the first stage.
If at this stage still some pairs had been missing, a third stage would be necessary.
In this case, however, we're done: the transitive closure is
$$
{(1,1),(1,2),(1,3),(2,2),(2,3),(3,2),(3,3),(4,4)}
$$
answered Dec 8 '18 at 0:14
egregegreg
181k1485203
$endgroup$
Let's make a list of the pairs together with the ones that can be chained with them by transitivity
begin{array}{cccc}
1 & 2 & to \
hline
(1,1) & (1,1) & (1,1) \
(1,1) & (1,2) & (1,2) \
(1,2) & (2,3) & color{red}{(1,3)} \
(2,3) & (3,2) & color{red}{(2,2)} \
(3,2) & (2,3) & color{red}{(3,3)} \
(4,4) & (4,4) & (4,4)
end{array}
In the first column, a pair; in the second column a pair that can be chained to the first one; in the third column, what's necessary to satisfy transitivity. Red color means the needed pair is missing.
Now we have to repeat the check with the newly added pairs
begin{array}{cccc}
1 & 2 & to \
hline
(1,1) & (1,1) & (1,1) & *\
(1,1) & (1,2) & (1,2) & * \
(1,2) & (2,2) & (2,2) \
(1,2) & (2,3) & (1,3) & * \
(1,3) & (3,2) & (1,2) \
(1,3) & (3,3) & (1,3) \
(2,2) & (2,2) & (2,2) \
(2,2) & (2,3) & (2,3) \
(2,3) & (3,2) & (2,2) & * \
(2,3) & (3,3) & (2,3) \
(3,2) & (2,3) & (3,3) & * \
(3,3) & (3,2) & (3,2) \
(3,3) & (3,3) & (3,3) \
(4,4) & (4,4) & (4,4) & *
end{array}
None has been marked red, because all pairs in the third column belong to the “second stage” relation. The $*$ denotes a row that has already been considered in the first stage.
If at this stage still some pairs had been missing, a third stage would be necessary.
In this case, however, we're done: the transitive closure is
$$
{(1,1),(1,2),(1,3),(2,2),(2,3),(3,2),(3,3),(4,4)}
$$
answered Dec 8 '18 at 0:14
egregegreg
181k1485203
answered Dec 8 '18 at 0:14
egregegreg
181k1485203
answered Dec 8 '18 at 0:14
egregegreg
181k1485203
answered Dec 8 '18 at 0:14
egregegreg
181k1485203
181k1485203
add a comment |
add a comment |
$begingroup$
The trick here is to build chains: can you build a chain with these relationships that gets you from $2$ to $4$? If not, then you don't need $(2,4)$ to transitively close the relation. On the other hand, $(2,3)$ and $(3,2)$ can be chained together $(2,3,2)$ and so $(2,2)$ must be in the closure.
answered Dec 7 '18 at 23:45
Dan UznanskiDan Uznanski
6,62021427
$endgroup$
add a comment |
$begingroup$
The trick here is to build chains: can you build a chain with these relationships that gets you from $2$ to $4$? If not, then you don't need $(2,4)$ to transitively close the relation. On the other hand, $(2,3)$ and $(3,2)$ can be chained together $(2,3,2)$ and so $(2,2)$ must be in the closure.
answered Dec 7 '18 at 23:45
Dan UznanskiDan Uznanski
6,62021427
$endgroup$
add a comment |
$begingroup$
The trick here is to build chains: can you build a chain with these relationships that gets you from $2$ to $4$? If not, then you don't need $(2,4)$ to transitively close the relation. On the other hand, $(2,3)$ and $(3,2)$ can be chained together $(2,3,2)$ and so $(2,2)$ must be in the closure.
answered Dec 7 '18 at 23:45
Dan UznanskiDan Uznanski
6,62021427
$endgroup$
The trick here is to build chains: can you build a chain with these relationships that gets you from $2$ to $4$? If not, then you don't need $(2,4)$ to transitively close the relation. On the other hand, $(2,3)$ and $(3,2)$ can be chained together $(2,3,2)$ and so $(2,2)$ must be in the closure.
answered Dec 7 '18 at 23:45
Dan UznanskiDan Uznanski
6,62021427
answered Dec 7 '18 at 23:45
Dan UznanskiDan Uznanski
6,62021427
answered Dec 7 '18 at 23:45
Dan UznanskiDan Uznanski
6,62021427
answered Dec 7 '18 at 23:45
Dan UznanskiDan Uznanski
6,62021427
6,62021427
add a comment |
add a comment |
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