Binary relation finding the transitive closure
$begingroup$
Let $A={a,b,c}$ and the relation given by the following matrix: $mathcal M_r=begin{pmatrix}1&0&1\0&1&0\1&1&0end{pmatrix}$
The task is to find the smallest transitive relation $E$ such that $rsubseteq E$. Which is equivalent of saying find the transitive closure of $r$.
so what I know is that: $t(r)=cup_{ngeq0}r^n, r^n=rcirc rcirc rcirc...circ r, n$ times.
so $mathcal M_{t(r)}=sum_{ninmathbb{N}}mathcal (M_r)^n$ and this gives me $$mathcal M_{t(r)}=begin{pmatrix}1&1&1\0&1&0\1&1&1end{pmatrix}$$
Is there something wrong in my approach?
relations binary
edited Jan 18 at 19:21
Kevin Long
3,58121431
asked Jan 18 at 13:50
C. CristiC. Cristi
1,655318
$endgroup$
add a comment |
$begingroup$
Let $A={a,b,c}$ and the relation given by the following matrix: $mathcal M_r=begin{pmatrix}1&0&1\0&1&0\1&1&0end{pmatrix}$
The task is to find the smallest transitive relation $E$ such that $rsubseteq E$. Which is equivalent of saying find the transitive closure of $r$.
so what I know is that: $t(r)=cup_{ngeq0}r^n, r^n=rcirc rcirc rcirc...circ r, n$ times.
so $mathcal M_{t(r)}=sum_{ninmathbb{N}}mathcal (M_r)^n$ and this gives me $$mathcal M_{t(r)}=begin{pmatrix}1&1&1\0&1&0\1&1&1end{pmatrix}$$
Is there something wrong in my approach?
relations binary
edited Jan 18 at 19:21
Kevin Long
3,58121431
asked Jan 18 at 13:50
C. CristiC. Cristi
1,655318
$endgroup$
$begingroup$
It's not clear how do you calculate $mathcal M_r^n$ and their sum?
$endgroup$
– Berci
Jan 18 at 14:28
$begingroup$
@Berci Using logic operators, $+$ is denoted as OR and $.$ is denoted as AND. And $n$ is the cardinal of set $A$
$endgroup$
– C. Cristi
Jan 18 at 14:41
$begingroup$
The result is correct. For this specific small example there's a shorter way to solve, though.
$endgroup$
– Berci
Jan 18 at 14:45
$begingroup$
@Berci Which way?
$endgroup$
– C. Cristi
Jan 18 at 14:45
add a comment |
$begingroup$
Let $A={a,b,c}$ and the relation given by the following matrix: $mathcal M_r=begin{pmatrix}1&0&1\0&1&0\1&1&0end{pmatrix}$
The task is to find the smallest transitive relation $E$ such that $rsubseteq E$. Which is equivalent of saying find the transitive closure of $r$.
so what I know is that: $t(r)=cup_{ngeq0}r^n, r^n=rcirc rcirc rcirc...circ r, n$ times.
so $mathcal M_{t(r)}=sum_{ninmathbb{N}}mathcal (M_r)^n$ and this gives me $$mathcal M_{t(r)}=begin{pmatrix}1&1&1\0&1&0\1&1&1end{pmatrix}$$
Is there something wrong in my approach?
relations binary
edited Jan 18 at 19:21
Kevin Long
3,58121431
asked Jan 18 at 13:50
C. CristiC. Cristi
1,655318
$endgroup$
Let $A={a,b,c}$ and the relation given by the following matrix: $mathcal M_r=begin{pmatrix}1&0&1\0&1&0\1&1&0end{pmatrix}$
The task is to find the smallest transitive relation $E$ such that $rsubseteq E$. Which is equivalent of saying find the transitive closure of $r$.
so what I know is that: $t(r)=cup_{ngeq0}r^n, r^n=rcirc rcirc rcirc...circ r, n$ times.
so $mathcal M_{t(r)}=sum_{ninmathbb{N}}mathcal (M_r)^n$ and this gives me $$mathcal M_{t(r)}=begin{pmatrix}1&1&1\0&1&0\1&1&1end{pmatrix}$$
Is there something wrong in my approach?
relations binary
relations binary
edited Jan 18 at 19:21
Kevin Long
3,58121431
asked Jan 18 at 13:50
C. CristiC. Cristi
1,655318
edited Jan 18 at 19:21
Kevin Long
3,58121431
asked Jan 18 at 13:50
C. CristiC. Cristi
1,655318
edited Jan 18 at 19:21
Kevin Long
3,58121431
edited Jan 18 at 19:21
Kevin Long
3,58121431
edited Jan 18 at 19:21
Kevin Long
3,58121431
3,58121431
asked Jan 18 at 13:50
C. CristiC. Cristi
1,655318
asked Jan 18 at 13:50
C. CristiC. Cristi
1,655318
asked Jan 18 at 13:50
C. CristiC. Cristi
1,655318
1,655318
$begingroup$
It's not clear how do you calculate $mathcal M_r^n$ and their sum?
$endgroup$
– Berci
Jan 18 at 14:28
$begingroup$
@Berci Using logic operators, $+$ is denoted as OR and $.$ is denoted as AND. And $n$ is the cardinal of set $A$
$endgroup$
– C. Cristi
Jan 18 at 14:41
$begingroup$
The result is correct. For this specific small example there's a shorter way to solve, though.
$endgroup$
– Berci
Jan 18 at 14:45
$begingroup$
@Berci Which way?
$endgroup$
– C. Cristi
Jan 18 at 14:45
add a comment |
$begingroup$
It's not clear how do you calculate $mathcal M_r^n$ and their sum?
$endgroup$
– Berci
Jan 18 at 14:28
$begingroup$
@Berci Using logic operators, $+$ is denoted as OR and $.$ is denoted as AND. And $n$ is the cardinal of set $A$
$endgroup$
– C. Cristi
Jan 18 at 14:41
$begingroup$
The result is correct. For this specific small example there's a shorter way to solve, though.
$endgroup$
– Berci
Jan 18 at 14:45
$begingroup$
@Berci Which way?
$endgroup$
– C. Cristi
Jan 18 at 14:45
$begingroup$
It's not clear how do you calculate $mathcal M_r^n$ and their sum?
$endgroup$
– Berci
Jan 18 at 14:28
$begingroup$
It's not clear how do you calculate $mathcal M_r^n$ and their sum?
$endgroup$
– Berci
Jan 18 at 14:28
$begingroup$
@Berci Using logic operators, $+$ is denoted as OR and $.$ is denoted as AND. And $n$ is the cardinal of set $A$
$endgroup$
– C. Cristi
Jan 18 at 14:41
$begingroup$
@Berci Using logic operators, $+$ is denoted as OR and $.$ is denoted as AND. And $n$ is the cardinal of set $A$
$endgroup$
– C. Cristi
Jan 18 at 14:41
$begingroup$
The result is correct. For this specific small example there's a shorter way to solve, though.
$endgroup$
– Berci
Jan 18 at 14:45
$begingroup$
The result is correct. For this specific small example there's a shorter way to solve, though.
$endgroup$
– Berci
Jan 18 at 14:45
$begingroup$
@Berci Which way?
$endgroup$
– C. Cristi
Jan 18 at 14:45
$begingroup$
@Berci Which way?
$endgroup$
– C. Cristi
Jan 18 at 14:45
add a comment |
1 Answer
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Your solution is correct
If our graph is small like this, we can simply draw it (loops at $a$ and $b$ and arrows $ato c, bto c, cto a$), and then use that we have an arrow $xto y$ in the transitive closure iff there is a path $xleadsto y$ in the original graph.
answered Jan 18 at 18:37
BerciBerci
62k23776
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add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
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oldest
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votes
$begingroup$
Your solution is correct
If our graph is small like this, we can simply draw it (loops at $a$ and $b$ and arrows $ato c, bto c, cto a$), and then use that we have an arrow $xto y$ in the transitive closure iff there is a path $xleadsto y$ in the original graph.
answered Jan 18 at 18:37
BerciBerci
62k23776
$endgroup$
add a comment |
$begingroup$
Your solution is correct
If our graph is small like this, we can simply draw it (loops at $a$ and $b$ and arrows $ato c, bto c, cto a$), and then use that we have an arrow $xto y$ in the transitive closure iff there is a path $xleadsto y$ in the original graph.
answered Jan 18 at 18:37
BerciBerci
62k23776
$endgroup$
add a comment |
$begingroup$
Your solution is correct
If our graph is small like this, we can simply draw it (loops at $a$ and $b$ and arrows $ato c, bto c, cto a$), and then use that we have an arrow $xto y$ in the transitive closure iff there is a path $xleadsto y$ in the original graph.
answered Jan 18 at 18:37
BerciBerci
62k23776
$endgroup$
Your solution is correct
If our graph is small like this, we can simply draw it (loops at $a$ and $b$ and arrows $ato c, bto c, cto a$), and then use that we have an arrow $xto y$ in the transitive closure iff there is a path $xleadsto y$ in the original graph.
answered Jan 18 at 18:37
BerciBerci
62k23776
answered Jan 18 at 18:37
BerciBerci
62k23776
answered Jan 18 at 18:37
BerciBerci
62k23776
answered Jan 18 at 18:37
BerciBerci
62k23776
62k23776
add a comment |
add a comment |
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$begingroup$
It's not clear how do you calculate $mathcal M_r^n$ and their sum?
$endgroup$
– Berci
Jan 18 at 14:28
$begingroup$
@Berci Using logic operators, $+$ is denoted as OR and $.$ is denoted as AND. And $n$ is the cardinal of set $A$
$endgroup$
– C. Cristi
Jan 18 at 14:41
$begingroup$
The result is correct. For this specific small example there's a shorter way to solve, though.
$endgroup$
– Berci
Jan 18 at 14:45
$begingroup$
@Berci Which way?
$endgroup$
– C. Cristi
Jan 18 at 14:45