Permutations and terminology
Say I have the following permutation
$$sigma ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&8&4&5&6&7end{pmatrix}}$$
which consists to let unchanged some first elements from $1$ to $k$ and to apply a circular shift on the elements ${k+1, ldots, n}$.
What is the name for such transformation?
Thank you
permutations permutation-cycles
edited Dec 2 '18 at 19:25
Bernard
118k639112
asked Dec 2 '18 at 19:09
Adam54
435
|
show 3 more comments
Say I have the following permutation
$$sigma ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&8&4&5&6&7end{pmatrix}}$$
which consists to let unchanged some first elements from $1$ to $k$ and to apply a circular shift on the elements ${k+1, ldots, n}$.
What is the name for such transformation?
Thank you
permutations permutation-cycles
edited Dec 2 '18 at 19:25
Bernard
118k639112
asked Dec 2 '18 at 19:09
Adam54
435
2
The name is "a $5$-cycle in $S_8$", namely $sigma=(48765)$.
– Dietrich Burde
Dec 2 '18 at 19:10
Thank you @DietrichBurde Does this (n-k)-cycle let unchanged the first $k$ elements?
– Adam54
Dec 2 '18 at 19:19
Yes, it keeps $1$,$2$, $3$ fixed (unchanged). Have a look at the cycle notation for permutations.
– Dietrich Burde
Dec 2 '18 at 19:20
1
Have a look at the cycle notation: $(45)$ means that $4$ goes to $5$ and $5$ goes to $4$. Then $(48765)$ means that $4mapsto 8mapsto 7mapsto 6mapsto 5mapsto 4mapstocdots $
– Dietrich Burde
Dec 2 '18 at 19:23
1
You should read en.wikipedia.org/wiki/Cyclic_permutation and en.wikipedia.org/wiki/Permutation#Cycle_notation
– Jean-Claude Arbaut
Dec 2 '18 at 19:31
|
show 3 more comments
Say I have the following permutation
$$sigma ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&8&4&5&6&7end{pmatrix}}$$
which consists to let unchanged some first elements from $1$ to $k$ and to apply a circular shift on the elements ${k+1, ldots, n}$.
What is the name for such transformation?
Thank you
permutations permutation-cycles
edited Dec 2 '18 at 19:25
Bernard
118k639112
asked Dec 2 '18 at 19:09
Adam54
435
Say I have the following permutation
$$sigma ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&8&4&5&6&7end{pmatrix}}$$
which consists to let unchanged some first elements from $1$ to $k$ and to apply a circular shift on the elements ${k+1, ldots, n}$.
What is the name for such transformation?
Thank you
permutations permutation-cycles
permutations permutation-cycles
edited Dec 2 '18 at 19:25
Bernard
118k639112
asked Dec 2 '18 at 19:09
Adam54
435
edited Dec 2 '18 at 19:25
Bernard
118k639112
asked Dec 2 '18 at 19:09
Adam54
435
edited Dec 2 '18 at 19:25
Bernard
118k639112
edited Dec 2 '18 at 19:25
Bernard
118k639112
edited Dec 2 '18 at 19:25
Bernard
118k639112
118k639112
asked Dec 2 '18 at 19:09
Adam54
435
asked Dec 2 '18 at 19:09
Adam54
435
asked Dec 2 '18 at 19:09
Adam54
435
435
2
The name is "a $5$-cycle in $S_8$", namely $sigma=(48765)$.
– Dietrich Burde
Dec 2 '18 at 19:10
Thank you @DietrichBurde Does this (n-k)-cycle let unchanged the first $k$ elements?
– Adam54
Dec 2 '18 at 19:19
Yes, it keeps $1$,$2$, $3$ fixed (unchanged). Have a look at the cycle notation for permutations.
– Dietrich Burde
Dec 2 '18 at 19:20
1
Have a look at the cycle notation: $(45)$ means that $4$ goes to $5$ and $5$ goes to $4$. Then $(48765)$ means that $4mapsto 8mapsto 7mapsto 6mapsto 5mapsto 4mapstocdots $
– Dietrich Burde
Dec 2 '18 at 19:23
1
You should read en.wikipedia.org/wiki/Cyclic_permutation and en.wikipedia.org/wiki/Permutation#Cycle_notation
– Jean-Claude Arbaut
Dec 2 '18 at 19:31
|
show 3 more comments
2
The name is "a $5$-cycle in $S_8$", namely $sigma=(48765)$.
– Dietrich Burde
Dec 2 '18 at 19:10
Thank you @DietrichBurde Does this (n-k)-cycle let unchanged the first $k$ elements?
– Adam54
Dec 2 '18 at 19:19
Yes, it keeps $1$,$2$, $3$ fixed (unchanged). Have a look at the cycle notation for permutations.
– Dietrich Burde
Dec 2 '18 at 19:20
1
Have a look at the cycle notation: $(45)$ means that $4$ goes to $5$ and $5$ goes to $4$. Then $(48765)$ means that $4mapsto 8mapsto 7mapsto 6mapsto 5mapsto 4mapstocdots $
– Dietrich Burde
Dec 2 '18 at 19:23
1
You should read en.wikipedia.org/wiki/Cyclic_permutation and en.wikipedia.org/wiki/Permutation#Cycle_notation
– Jean-Claude Arbaut
Dec 2 '18 at 19:31
2
2
The name is "a $5$-cycle in $S_8$", namely $sigma=(48765)$.
– Dietrich Burde
Dec 2 '18 at 19:10
The name is "a $5$-cycle in $S_8$", namely $sigma=(48765)$.
– Dietrich Burde
Dec 2 '18 at 19:10
Thank you @DietrichBurde Does this (n-k)-cycle let unchanged the first $k$ elements?
– Adam54
Dec 2 '18 at 19:19
Thank you @DietrichBurde Does this (n-k)-cycle let unchanged the first $k$ elements?
– Adam54
Dec 2 '18 at 19:19
Yes, it keeps $1$,$2$, $3$ fixed (unchanged). Have a look at the cycle notation for permutations.
– Dietrich Burde
Dec 2 '18 at 19:20
Yes, it keeps $1$,$2$, $3$ fixed (unchanged). Have a look at the cycle notation for permutations.
– Dietrich Burde
Dec 2 '18 at 19:20
1
1
Have a look at the cycle notation: $(45)$ means that $4$ goes to $5$ and $5$ goes to $4$. Then $(48765)$ means that $4mapsto 8mapsto 7mapsto 6mapsto 5mapsto 4mapstocdots $
– Dietrich Burde
Dec 2 '18 at 19:23
Have a look at the cycle notation: $(45)$ means that $4$ goes to $5$ and $5$ goes to $4$. Then $(48765)$ means that $4mapsto 8mapsto 7mapsto 6mapsto 5mapsto 4mapstocdots $
– Dietrich Burde
Dec 2 '18 at 19:23
1
1
You should read en.wikipedia.org/wiki/Cyclic_permutation and en.wikipedia.org/wiki/Permutation#Cycle_notation
– Jean-Claude Arbaut
Dec 2 '18 at 19:31
You should read en.wikipedia.org/wiki/Cyclic_permutation and en.wikipedia.org/wiki/Permutation#Cycle_notation
– Jean-Claude Arbaut
Dec 2 '18 at 19:31
|
show 3 more comments
1 Answer
1
active
oldest
votes
$1, 2, 3$ each map to themselves. $1to 1$. Period. $2to 2.$ Period. $3to 3.$ Period. We can represent this as $(1)(2)(3)$, but typically this is not customary.
So we can start with $4$:
$4 to 8$, then
$8 to 7$, then
$7to 6$,
then $6to 5$, and finally,
$5 to 4$, completing the five cycle (bringing us back to the starting number for the cycle, $4$:
That gives us the permutation $sigma = (48765)$. This can also be written as $$sigma = (1)(2)(3)(48765) = (48765)$$
To get the cycle you are looking for, you'll need the following permutation:
$alpha ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&5&6&7&8&4end{pmatrix}} = (45678)$
The name for this transformation is merely "permutation" $alpha$ on the set $S$ of $8$ numbers, specifically, both $alpha, sigma in mathbb S_8$, the group of permutations on $S$.
If we were talking about vertices of an octagon, numbered 1-8, then $(12345678)$ could be described as a rotation of the octagon.
answered Dec 2 '18 at 19:26
amWhy
192k28224439
Thanks @amWhy As I have written to Dietrich, I think I am reffering to another kind of cycle.
– Adam54
Dec 2 '18 at 19:29
Top number goes to bottom number in each column.
– amWhy
Dec 2 '18 at 19:37
Thank you. So writing left shift is straightforward, but for writing a right shift we need to read from top to bottom?
– Adam54
Dec 2 '18 at 19:42
Yes, the numbers in the top row represent the 8 possible numbers to permute. Where each of those numbers goes is written immediately below the number, in the second row.
– amWhy
Dec 2 '18 at 19:46
Thanks again. So, if I have a sequence of 8 unknown numbers, but I know that when I will have these numbers, I will right shift the 5 last numbers. How can I write this function, terminologically?
– Adam54
Dec 2 '18 at 19:49
|
show 2 more comments
Your Answer
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$1, 2, 3$ each map to themselves. $1to 1$. Period. $2to 2.$ Period. $3to 3.$ Period. We can represent this as $(1)(2)(3)$, but typically this is not customary.
So we can start with $4$:
$4 to 8$, then
$8 to 7$, then
$7to 6$,
then $6to 5$, and finally,
$5 to 4$, completing the five cycle (bringing us back to the starting number for the cycle, $4$:
That gives us the permutation $sigma = (48765)$. This can also be written as $$sigma = (1)(2)(3)(48765) = (48765)$$
To get the cycle you are looking for, you'll need the following permutation:
$alpha ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&5&6&7&8&4end{pmatrix}} = (45678)$
The name for this transformation is merely "permutation" $alpha$ on the set $S$ of $8$ numbers, specifically, both $alpha, sigma in mathbb S_8$, the group of permutations on $S$.
If we were talking about vertices of an octagon, numbered 1-8, then $(12345678)$ could be described as a rotation of the octagon.
answered Dec 2 '18 at 19:26
amWhy
192k28224439
Thanks @amWhy As I have written to Dietrich, I think I am reffering to another kind of cycle.
– Adam54
Dec 2 '18 at 19:29
Top number goes to bottom number in each column.
– amWhy
Dec 2 '18 at 19:37
Thank you. So writing left shift is straightforward, but for writing a right shift we need to read from top to bottom?
– Adam54
Dec 2 '18 at 19:42
Yes, the numbers in the top row represent the 8 possible numbers to permute. Where each of those numbers goes is written immediately below the number, in the second row.
– amWhy
Dec 2 '18 at 19:46
Thanks again. So, if I have a sequence of 8 unknown numbers, but I know that when I will have these numbers, I will right shift the 5 last numbers. How can I write this function, terminologically?
– Adam54
Dec 2 '18 at 19:49
|
show 2 more comments
$1, 2, 3$ each map to themselves. $1to 1$. Period. $2to 2.$ Period. $3to 3.$ Period. We can represent this as $(1)(2)(3)$, but typically this is not customary.
So we can start with $4$:
$4 to 8$, then
$8 to 7$, then
$7to 6$,
then $6to 5$, and finally,
$5 to 4$, completing the five cycle (bringing us back to the starting number for the cycle, $4$:
That gives us the permutation $sigma = (48765)$. This can also be written as $$sigma = (1)(2)(3)(48765) = (48765)$$
To get the cycle you are looking for, you'll need the following permutation:
$alpha ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&5&6&7&8&4end{pmatrix}} = (45678)$
The name for this transformation is merely "permutation" $alpha$ on the set $S$ of $8$ numbers, specifically, both $alpha, sigma in mathbb S_8$, the group of permutations on $S$.
If we were talking about vertices of an octagon, numbered 1-8, then $(12345678)$ could be described as a rotation of the octagon.
answered Dec 2 '18 at 19:26
amWhy
192k28224439
Thanks @amWhy As I have written to Dietrich, I think I am reffering to another kind of cycle.
– Adam54
Dec 2 '18 at 19:29
Top number goes to bottom number in each column.
– amWhy
Dec 2 '18 at 19:37
Thank you. So writing left shift is straightforward, but for writing a right shift we need to read from top to bottom?
– Adam54
Dec 2 '18 at 19:42
Yes, the numbers in the top row represent the 8 possible numbers to permute. Where each of those numbers goes is written immediately below the number, in the second row.
– amWhy
Dec 2 '18 at 19:46
Thanks again. So, if I have a sequence of 8 unknown numbers, but I know that when I will have these numbers, I will right shift the 5 last numbers. How can I write this function, terminologically?
– Adam54
Dec 2 '18 at 19:49
|
show 2 more comments
$1, 2, 3$ each map to themselves. $1to 1$. Period. $2to 2.$ Period. $3to 3.$ Period. We can represent this as $(1)(2)(3)$, but typically this is not customary.
So we can start with $4$:
$4 to 8$, then
$8 to 7$, then
$7to 6$,
then $6to 5$, and finally,
$5 to 4$, completing the five cycle (bringing us back to the starting number for the cycle, $4$:
That gives us the permutation $sigma = (48765)$. This can also be written as $$sigma = (1)(2)(3)(48765) = (48765)$$
To get the cycle you are looking for, you'll need the following permutation:
$alpha ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&5&6&7&8&4end{pmatrix}} = (45678)$
The name for this transformation is merely "permutation" $alpha$ on the set $S$ of $8$ numbers, specifically, both $alpha, sigma in mathbb S_8$, the group of permutations on $S$.
If we were talking about vertices of an octagon, numbered 1-8, then $(12345678)$ could be described as a rotation of the octagon.
answered Dec 2 '18 at 19:26
amWhy
192k28224439
$1, 2, 3$ each map to themselves. $1to 1$. Period. $2to 2.$ Period. $3to 3.$ Period. We can represent this as $(1)(2)(3)$, but typically this is not customary.
So we can start with $4$:
$4 to 8$, then
$8 to 7$, then
$7to 6$,
then $6to 5$, and finally,
$5 to 4$, completing the five cycle (bringing us back to the starting number for the cycle, $4$:
That gives us the permutation $sigma = (48765)$. This can also be written as $$sigma = (1)(2)(3)(48765) = (48765)$$
To get the cycle you are looking for, you'll need the following permutation:
$alpha ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&5&6&7&8&4end{pmatrix}} = (45678)$
The name for this transformation is merely "permutation" $alpha$ on the set $S$ of $8$ numbers, specifically, both $alpha, sigma in mathbb S_8$, the group of permutations on $S$.
If we were talking about vertices of an octagon, numbered 1-8, then $(12345678)$ could be described as a rotation of the octagon.
answered Dec 2 '18 at 19:26
amWhy
192k28224439
edited Dec 31 '18 at 21:32
answered Dec 2 '18 at 19:26
amWhy
192k28224439
answered Dec 2 '18 at 19:26
amWhy
192k28224439
answered Dec 2 '18 at 19:26
amWhy
192k28224439
192k28224439
Thanks @amWhy As I have written to Dietrich, I think I am reffering to another kind of cycle.
– Adam54
Dec 2 '18 at 19:29
Top number goes to bottom number in each column.
– amWhy
Dec 2 '18 at 19:37
Thank you. So writing left shift is straightforward, but for writing a right shift we need to read from top to bottom?
– Adam54
Dec 2 '18 at 19:42
Yes, the numbers in the top row represent the 8 possible numbers to permute. Where each of those numbers goes is written immediately below the number, in the second row.
– amWhy
Dec 2 '18 at 19:46
Thanks again. So, if I have a sequence of 8 unknown numbers, but I know that when I will have these numbers, I will right shift the 5 last numbers. How can I write this function, terminologically?
– Adam54
Dec 2 '18 at 19:49
|
show 2 more comments
Thanks @amWhy As I have written to Dietrich, I think I am reffering to another kind of cycle.
– Adam54
Dec 2 '18 at 19:29
Top number goes to bottom number in each column.
– amWhy
Dec 2 '18 at 19:37
Thank you. So writing left shift is straightforward, but for writing a right shift we need to read from top to bottom?
– Adam54
Dec 2 '18 at 19:42
Yes, the numbers in the top row represent the 8 possible numbers to permute. Where each of those numbers goes is written immediately below the number, in the second row.
– amWhy
Dec 2 '18 at 19:46
Thanks again. So, if I have a sequence of 8 unknown numbers, but I know that when I will have these numbers, I will right shift the 5 last numbers. How can I write this function, terminologically?
– Adam54
Dec 2 '18 at 19:49
Thanks @amWhy As I have written to Dietrich, I think I am reffering to another kind of cycle.
– Adam54
Dec 2 '18 at 19:29
Thanks @amWhy As I have written to Dietrich, I think I am reffering to another kind of cycle.
– Adam54
Dec 2 '18 at 19:29
Top number goes to bottom number in each column.
– amWhy
Dec 2 '18 at 19:37
Top number goes to bottom number in each column.
– amWhy
Dec 2 '18 at 19:37
Thank you. So writing left shift is straightforward, but for writing a right shift we need to read from top to bottom?
– Adam54
Dec 2 '18 at 19:42
Thank you. So writing left shift is straightforward, but for writing a right shift we need to read from top to bottom?
– Adam54
Dec 2 '18 at 19:42
Yes, the numbers in the top row represent the 8 possible numbers to permute. Where each of those numbers goes is written immediately below the number, in the second row.
– amWhy
Dec 2 '18 at 19:46
Yes, the numbers in the top row represent the 8 possible numbers to permute. Where each of those numbers goes is written immediately below the number, in the second row.
– amWhy
Dec 2 '18 at 19:46
Thanks again. So, if I have a sequence of 8 unknown numbers, but I know that when I will have these numbers, I will right shift the 5 last numbers. How can I write this function, terminologically?
– Adam54
Dec 2 '18 at 19:49
Thanks again. So, if I have a sequence of 8 unknown numbers, but I know that when I will have these numbers, I will right shift the 5 last numbers. How can I write this function, terminologically?
– Adam54
Dec 2 '18 at 19:49
|
show 2 more comments
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2
The name is "a $5$-cycle in $S_8$", namely $sigma=(48765)$.
– Dietrich Burde
Dec 2 '18 at 19:10
Thank you @DietrichBurde Does this (n-k)-cycle let unchanged the first $k$ elements?
– Adam54
Dec 2 '18 at 19:19
Yes, it keeps $1$,$2$, $3$ fixed (unchanged). Have a look at the cycle notation for permutations.
– Dietrich Burde
Dec 2 '18 at 19:20
1
Have a look at the cycle notation: $(45)$ means that $4$ goes to $5$ and $5$ goes to $4$. Then $(48765)$ means that $4mapsto 8mapsto 7mapsto 6mapsto 5mapsto 4mapstocdots $
– Dietrich Burde
Dec 2 '18 at 19:23
1
You should read en.wikipedia.org/wiki/Cyclic_permutation and en.wikipedia.org/wiki/Permutation#Cycle_notation
– Jean-Claude Arbaut
Dec 2 '18 at 19:31