Number of inversions in the permutation $(ijk)$.
$begingroup$
Find the number of inversions in the permutation $sigma= (ijk)$ where $1leq i<j<kleq n.$
My attempt: I counted as follows- first let's consider the tuples which are inverted with respect to $j$ then we have the following list $$(j,i+1),(j,i+2)...,(j,k)$$ of length $j-i.$ Next we look at the tuples which are inverted with respect with $k$, then we have the following list: $$(k,j+1),(k,j+2),...,(k,i)$$ which is of length $k-j.$ Finally we look at the tuples which are inverted with respect to $i$ then we have the following list $$(i,j),(i,j+1),...,(i,k-1)$$ which is of length $k-i.$ I subtract $2$ from this because I have double counted $(i,j)$ and $(i,k)$ so we get in total $$j-i+k-j+k-i-2=2(k-i-1)$$ inversions. Is this calculation correct? It gives the right answer in the sense that $(ijk)$ is a permutation of length $3$ and so the signature of $epsilon(sigma)=(-1)^{3-1}=(-1)^{N(sigma)}=(-1)^{2(k-j-1)}=1.$ But I am not sure.
combinatorics permutations
edited Dec 22 '18 at 12:31
Shaun
9,366113684
asked Jan 28 '18 at 22:58
Hello_WorldHello_World
4,14121931
$endgroup$
add a comment |
$begingroup$
Find the number of inversions in the permutation $sigma= (ijk)$ where $1leq i<j<kleq n.$
My attempt: I counted as follows- first let's consider the tuples which are inverted with respect to $j$ then we have the following list $$(j,i+1),(j,i+2)...,(j,k)$$ of length $j-i.$ Next we look at the tuples which are inverted with respect with $k$, then we have the following list: $$(k,j+1),(k,j+2),...,(k,i)$$ which is of length $k-j.$ Finally we look at the tuples which are inverted with respect to $i$ then we have the following list $$(i,j),(i,j+1),...,(i,k-1)$$ which is of length $k-i.$ I subtract $2$ from this because I have double counted $(i,j)$ and $(i,k)$ so we get in total $$j-i+k-j+k-i-2=2(k-i-1)$$ inversions. Is this calculation correct? It gives the right answer in the sense that $(ijk)$ is a permutation of length $3$ and so the signature of $epsilon(sigma)=(-1)^{3-1}=(-1)^{N(sigma)}=(-1)^{2(k-j-1)}=1.$ But I am not sure.
combinatorics permutations
edited Dec 22 '18 at 12:31
Shaun
9,366113684
asked Jan 28 '18 at 22:58
Hello_WorldHello_World
4,14121931
$endgroup$
$begingroup$
How do you define the number of inversions? Is it the smallest number of inversions needed to obtain the permutation?
$endgroup$
– Leaning
Dec 22 '18 at 12:38
add a comment |
$begingroup$
Find the number of inversions in the permutation $sigma= (ijk)$ where $1leq i<j<kleq n.$
My attempt: I counted as follows- first let's consider the tuples which are inverted with respect to $j$ then we have the following list $$(j,i+1),(j,i+2)...,(j,k)$$ of length $j-i.$ Next we look at the tuples which are inverted with respect with $k$, then we have the following list: $$(k,j+1),(k,j+2),...,(k,i)$$ which is of length $k-j.$ Finally we look at the tuples which are inverted with respect to $i$ then we have the following list $$(i,j),(i,j+1),...,(i,k-1)$$ which is of length $k-i.$ I subtract $2$ from this because I have double counted $(i,j)$ and $(i,k)$ so we get in total $$j-i+k-j+k-i-2=2(k-i-1)$$ inversions. Is this calculation correct? It gives the right answer in the sense that $(ijk)$ is a permutation of length $3$ and so the signature of $epsilon(sigma)=(-1)^{3-1}=(-1)^{N(sigma)}=(-1)^{2(k-j-1)}=1.$ But I am not sure.
combinatorics permutations
edited Dec 22 '18 at 12:31
Shaun
9,366113684
asked Jan 28 '18 at 22:58
Hello_WorldHello_World
4,14121931
$endgroup$
Find the number of inversions in the permutation $sigma= (ijk)$ where $1leq i<j<kleq n.$
My attempt: I counted as follows- first let's consider the tuples which are inverted with respect to $j$ then we have the following list $$(j,i+1),(j,i+2)...,(j,k)$$ of length $j-i.$ Next we look at the tuples which are inverted with respect with $k$, then we have the following list: $$(k,j+1),(k,j+2),...,(k,i)$$ which is of length $k-j.$ Finally we look at the tuples which are inverted with respect to $i$ then we have the following list $$(i,j),(i,j+1),...,(i,k-1)$$ which is of length $k-i.$ I subtract $2$ from this because I have double counted $(i,j)$ and $(i,k)$ so we get in total $$j-i+k-j+k-i-2=2(k-i-1)$$ inversions. Is this calculation correct? It gives the right answer in the sense that $(ijk)$ is a permutation of length $3$ and so the signature of $epsilon(sigma)=(-1)^{3-1}=(-1)^{N(sigma)}=(-1)^{2(k-j-1)}=1.$ But I am not sure.
combinatorics permutations
combinatorics permutations
edited Dec 22 '18 at 12:31
Shaun
9,366113684
asked Jan 28 '18 at 22:58
Hello_WorldHello_World
4,14121931
edited Dec 22 '18 at 12:31
Shaun
9,366113684
asked Jan 28 '18 at 22:58
Hello_WorldHello_World
4,14121931
edited Dec 22 '18 at 12:31
Shaun
9,366113684
edited Dec 22 '18 at 12:31
Shaun
9,366113684
edited Dec 22 '18 at 12:31
Shaun
9,366113684
9,366113684
asked Jan 28 '18 at 22:58
Hello_WorldHello_World
4,14121931
asked Jan 28 '18 at 22:58
Hello_WorldHello_World
4,14121931
asked Jan 28 '18 at 22:58
Hello_WorldHello_World
4,14121931
4,14121931
$begingroup$
How do you define the number of inversions? Is it the smallest number of inversions needed to obtain the permutation?
$endgroup$
– Leaning
Dec 22 '18 at 12:38
add a comment |
$begingroup$
How do you define the number of inversions? Is it the smallest number of inversions needed to obtain the permutation?
$endgroup$
– Leaning
Dec 22 '18 at 12:38
$begingroup$
How do you define the number of inversions? Is it the smallest number of inversions needed to obtain the permutation?
$endgroup$
– Leaning
Dec 22 '18 at 12:38
$begingroup$
How do you define the number of inversions? Is it the smallest number of inversions needed to obtain the permutation?
$endgroup$
– Leaning
Dec 22 '18 at 12:38
add a comment |
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$begingroup$
How do you define the number of inversions? Is it the smallest number of inversions needed to obtain the permutation?
$endgroup$
– Leaning
Dec 22 '18 at 12:38