Bijection between $mathsf{Sh}(p+1, q)$ and $mathsf{Sh}(p, q)sqcup mathsf{Sh}(p-1, q+1)$?
$begingroup$
Let $p$ and $q$ be two positive integers. Define $$mathsf{Sh}(p, q):={sigmainmathsf{S}_{p+q}: sigma(1)<ldots<sigma(p) textrm{and} sigma(p+1)<ldots<sigma(p+q)},$$ where $mathsf{S}_{p+q}$ is the set of permutations of the first $p+q$ integers, i.e, bijections of ${1, ldots, p+q}$. An element of $mathsf{Sh}(p, q)$ is called a $(p, q)$-shuffle.
It can be seen that $$#mathsf{Sh}(p, q)=binom{p+q}{p}.$$
In particular, we see that $$mathsf{Sh}(p+1, q)=#mathsf{Sh}(p, q)+#mathsf{Sh}(p+1, q-1),$$ and therefore $$#mathsf{Sh}(p+1, q)quad textrm{and}quad mathsf{Sh}(p, q)sqcup mathsf{Sh}(p+1, q-1)$$
are in bijection. How to construct an explicit bijection between those sets?
Thanks.
combinatorics
asked Jun 3 '17 at 23:15
PtFPtF
3,94921733
$endgroup$
add a comment |
$begingroup$
Let $p$ and $q$ be two positive integers. Define $$mathsf{Sh}(p, q):={sigmainmathsf{S}_{p+q}: sigma(1)<ldots<sigma(p) textrm{and} sigma(p+1)<ldots<sigma(p+q)},$$ where $mathsf{S}_{p+q}$ is the set of permutations of the first $p+q$ integers, i.e, bijections of ${1, ldots, p+q}$. An element of $mathsf{Sh}(p, q)$ is called a $(p, q)$-shuffle.
It can be seen that $$#mathsf{Sh}(p, q)=binom{p+q}{p}.$$
In particular, we see that $$mathsf{Sh}(p+1, q)=#mathsf{Sh}(p, q)+#mathsf{Sh}(p+1, q-1),$$ and therefore $$#mathsf{Sh}(p+1, q)quad textrm{and}quad mathsf{Sh}(p, q)sqcup mathsf{Sh}(p+1, q-1)$$
are in bijection. How to construct an explicit bijection between those sets?
Thanks.
combinatorics
asked Jun 3 '17 at 23:15
PtFPtF
3,94921733
$endgroup$
add a comment |
$begingroup$
Let $p$ and $q$ be two positive integers. Define $$mathsf{Sh}(p, q):={sigmainmathsf{S}_{p+q}: sigma(1)<ldots<sigma(p) textrm{and} sigma(p+1)<ldots<sigma(p+q)},$$ where $mathsf{S}_{p+q}$ is the set of permutations of the first $p+q$ integers, i.e, bijections of ${1, ldots, p+q}$. An element of $mathsf{Sh}(p, q)$ is called a $(p, q)$-shuffle.
It can be seen that $$#mathsf{Sh}(p, q)=binom{p+q}{p}.$$
In particular, we see that $$mathsf{Sh}(p+1, q)=#mathsf{Sh}(p, q)+#mathsf{Sh}(p+1, q-1),$$ and therefore $$#mathsf{Sh}(p+1, q)quad textrm{and}quad mathsf{Sh}(p, q)sqcup mathsf{Sh}(p+1, q-1)$$
are in bijection. How to construct an explicit bijection between those sets?
Thanks.
combinatorics
asked Jun 3 '17 at 23:15
PtFPtF
3,94921733
$endgroup$
Let $p$ and $q$ be two positive integers. Define $$mathsf{Sh}(p, q):={sigmainmathsf{S}_{p+q}: sigma(1)<ldots<sigma(p) textrm{and} sigma(p+1)<ldots<sigma(p+q)},$$ where $mathsf{S}_{p+q}$ is the set of permutations of the first $p+q$ integers, i.e, bijections of ${1, ldots, p+q}$. An element of $mathsf{Sh}(p, q)$ is called a $(p, q)$-shuffle.
It can be seen that $$#mathsf{Sh}(p, q)=binom{p+q}{p}.$$
In particular, we see that $$mathsf{Sh}(p+1, q)=#mathsf{Sh}(p, q)+#mathsf{Sh}(p+1, q-1),$$ and therefore $$#mathsf{Sh}(p+1, q)quad textrm{and}quad mathsf{Sh}(p, q)sqcup mathsf{Sh}(p+1, q-1)$$
are in bijection. How to construct an explicit bijection between those sets?
Thanks.
combinatorics
combinatorics
asked Jun 3 '17 at 23:15
PtFPtF
3,94921733
asked Jun 3 '17 at 23:15
PtFPtF
3,94921733
edited Dec 4 '18 at 23:11
PtF
asked Jun 3 '17 at 23:15
PtFPtF
3,94921733
asked Jun 3 '17 at 23:15
PtFPtF
3,94921733
asked Jun 3 '17 at 23:15
PtFPtF
3,94921733
3,94921733
add a comment |
add a comment |
2 Answers
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$begingroup$
As stated, you have a wrong identity. Notice that $Sh(3+1,2)=binom{6}{4}=15,Sh(3,2)=binom{5}{3}=10$ and $Sh(2,3)=binom{5}{2}=10$ but $15neq 10+10.$
I am going to make the assumption that you wanted to use Pascal's recursion, in which case the correct identity is $$|Sh(p,q+1)|=|Sh(p,q)|+|Sh(p-1,q+1)|.$$
In order to prove the later, let $sigma in Sh(p,q+1),$ then there are $2$ options:
$bullet $ $sigma ^{-1}(p+q+1)<=p$: if you remove $p+q+1$ you end up with a permutation in $Sh(pcolor{red}{-1},q+1).$
$bullet$ $sigma ^{-1}(p+q+1)>p$:if you remove $p+q+1$ you end up with a permutation in $Sh(p,q+1color{red}{-1})=Sh(p,q).$
It should be clear the bijection from here.
answered Jun 4 '17 at 5:12
PhicarPhicar
2,6251915
$endgroup$
$begingroup$
Sorry, I mistyped. Thanks for your answer =)
$endgroup$
– PtF
Jun 4 '17 at 12:38
$begingroup$
Sure, glad to help.
$endgroup$
– Phicar
Jun 4 '17 at 16:20
add a comment |
$begingroup$
I wrote down the details so I'm going to share just in case someone else needs it someday.
The first thing to be noticed is that $sigma^{-1}({1})={1, p+2}$, hence:
$$mathsf{Sh}(p+1, q)={sigmain mathsf{Sh}(p+1, q): sigma(1)=1}sqcup {sigmainmathsf{Sh}(p+1, q): sigma(p+2)=1}.$$
This allows us to define the bijection easily setting:
$$mathsf{Sh}(p, q)longrightarrow mathsf{Sh}(p+1, q)\ sigmalongmapsto left(begin{array}{ccccccc}1 & 2& ldots & p+2 & ldots &p+q+1\ 1 &sigma(1)+1 & ldots &sigma(p+1)+1 & ldots & sigma(p+q)+1 end{array}right)$$
and
$$ mathsf{Sh}(p+1, q-1)longrightarrow mathsf{Sh}(p+1, q)\ sigmalongmapsto left(begin{array}{ccccccc}1 & ldots &p+1 & p+2 & p+3& ldots & p+q+1\ sigma(1)+1 & ldots & sigma(p+1)+1 & 1 & sigma(p+2)+1 & ldots &sigma(p+q)+1 end{array}right).$$ The inverse should be clear from the definitions.
answered Jun 8 '17 at 12:24
PtFPtF
3,94921733
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
As stated, you have a wrong identity. Notice that $Sh(3+1,2)=binom{6}{4}=15,Sh(3,2)=binom{5}{3}=10$ and $Sh(2,3)=binom{5}{2}=10$ but $15neq 10+10.$
I am going to make the assumption that you wanted to use Pascal's recursion, in which case the correct identity is $$|Sh(p,q+1)|=|Sh(p,q)|+|Sh(p-1,q+1)|.$$
In order to prove the later, let $sigma in Sh(p,q+1),$ then there are $2$ options:
$bullet $ $sigma ^{-1}(p+q+1)<=p$: if you remove $p+q+1$ you end up with a permutation in $Sh(pcolor{red}{-1},q+1).$
$bullet$ $sigma ^{-1}(p+q+1)>p$:if you remove $p+q+1$ you end up with a permutation in $Sh(p,q+1color{red}{-1})=Sh(p,q).$
It should be clear the bijection from here.
answered Jun 4 '17 at 5:12
PhicarPhicar
2,6251915
$endgroup$
$begingroup$
Sorry, I mistyped. Thanks for your answer =)
$endgroup$
– PtF
Jun 4 '17 at 12:38
$begingroup$
Sure, glad to help.
$endgroup$
– Phicar
Jun 4 '17 at 16:20
add a comment |
$begingroup$
As stated, you have a wrong identity. Notice that $Sh(3+1,2)=binom{6}{4}=15,Sh(3,2)=binom{5}{3}=10$ and $Sh(2,3)=binom{5}{2}=10$ but $15neq 10+10.$
I am going to make the assumption that you wanted to use Pascal's recursion, in which case the correct identity is $$|Sh(p,q+1)|=|Sh(p,q)|+|Sh(p-1,q+1)|.$$
In order to prove the later, let $sigma in Sh(p,q+1),$ then there are $2$ options:
$bullet $ $sigma ^{-1}(p+q+1)<=p$: if you remove $p+q+1$ you end up with a permutation in $Sh(pcolor{red}{-1},q+1).$
$bullet$ $sigma ^{-1}(p+q+1)>p$:if you remove $p+q+1$ you end up with a permutation in $Sh(p,q+1color{red}{-1})=Sh(p,q).$
It should be clear the bijection from here.
answered Jun 4 '17 at 5:12
PhicarPhicar
2,6251915
$endgroup$
$begingroup$
Sorry, I mistyped. Thanks for your answer =)
$endgroup$
– PtF
Jun 4 '17 at 12:38
$begingroup$
Sure, glad to help.
$endgroup$
– Phicar
Jun 4 '17 at 16:20
add a comment |
$begingroup$
As stated, you have a wrong identity. Notice that $Sh(3+1,2)=binom{6}{4}=15,Sh(3,2)=binom{5}{3}=10$ and $Sh(2,3)=binom{5}{2}=10$ but $15neq 10+10.$
I am going to make the assumption that you wanted to use Pascal's recursion, in which case the correct identity is $$|Sh(p,q+1)|=|Sh(p,q)|+|Sh(p-1,q+1)|.$$
In order to prove the later, let $sigma in Sh(p,q+1),$ then there are $2$ options:
$bullet $ $sigma ^{-1}(p+q+1)<=p$: if you remove $p+q+1$ you end up with a permutation in $Sh(pcolor{red}{-1},q+1).$
$bullet$ $sigma ^{-1}(p+q+1)>p$:if you remove $p+q+1$ you end up with a permutation in $Sh(p,q+1color{red}{-1})=Sh(p,q).$
It should be clear the bijection from here.
answered Jun 4 '17 at 5:12
PhicarPhicar
2,6251915
$endgroup$
As stated, you have a wrong identity. Notice that $Sh(3+1,2)=binom{6}{4}=15,Sh(3,2)=binom{5}{3}=10$ and $Sh(2,3)=binom{5}{2}=10$ but $15neq 10+10.$
I am going to make the assumption that you wanted to use Pascal's recursion, in which case the correct identity is $$|Sh(p,q+1)|=|Sh(p,q)|+|Sh(p-1,q+1)|.$$
In order to prove the later, let $sigma in Sh(p,q+1),$ then there are $2$ options:
$bullet $ $sigma ^{-1}(p+q+1)<=p$: if you remove $p+q+1$ you end up with a permutation in $Sh(pcolor{red}{-1},q+1).$
$bullet$ $sigma ^{-1}(p+q+1)>p$:if you remove $p+q+1$ you end up with a permutation in $Sh(p,q+1color{red}{-1})=Sh(p,q).$
It should be clear the bijection from here.
answered Jun 4 '17 at 5:12
PhicarPhicar
2,6251915
answered Jun 4 '17 at 5:12
PhicarPhicar
2,6251915
answered Jun 4 '17 at 5:12
PhicarPhicar
2,6251915
answered Jun 4 '17 at 5:12
PhicarPhicar
2,6251915
2,6251915
$begingroup$
Sorry, I mistyped. Thanks for your answer =)
$endgroup$
– PtF
Jun 4 '17 at 12:38
$begingroup$
Sure, glad to help.
$endgroup$
– Phicar
Jun 4 '17 at 16:20
add a comment |
$begingroup$
Sorry, I mistyped. Thanks for your answer =)
$endgroup$
– PtF
Jun 4 '17 at 12:38
$begingroup$
Sure, glad to help.
$endgroup$
– Phicar
Jun 4 '17 at 16:20
$begingroup$
Sorry, I mistyped. Thanks for your answer =)
$endgroup$
– PtF
Jun 4 '17 at 12:38
$begingroup$
Sorry, I mistyped. Thanks for your answer =)
$endgroup$
– PtF
Jun 4 '17 at 12:38
$begingroup$
Sure, glad to help.
$endgroup$
– Phicar
Jun 4 '17 at 16:20
$begingroup$
Sure, glad to help.
$endgroup$
– Phicar
Jun 4 '17 at 16:20
add a comment |
$begingroup$
I wrote down the details so I'm going to share just in case someone else needs it someday.
The first thing to be noticed is that $sigma^{-1}({1})={1, p+2}$, hence:
$$mathsf{Sh}(p+1, q)={sigmain mathsf{Sh}(p+1, q): sigma(1)=1}sqcup {sigmainmathsf{Sh}(p+1, q): sigma(p+2)=1}.$$
This allows us to define the bijection easily setting:
$$mathsf{Sh}(p, q)longrightarrow mathsf{Sh}(p+1, q)\ sigmalongmapsto left(begin{array}{ccccccc}1 & 2& ldots & p+2 & ldots &p+q+1\ 1 &sigma(1)+1 & ldots &sigma(p+1)+1 & ldots & sigma(p+q)+1 end{array}right)$$
and
$$ mathsf{Sh}(p+1, q-1)longrightarrow mathsf{Sh}(p+1, q)\ sigmalongmapsto left(begin{array}{ccccccc}1 & ldots &p+1 & p+2 & p+3& ldots & p+q+1\ sigma(1)+1 & ldots & sigma(p+1)+1 & 1 & sigma(p+2)+1 & ldots &sigma(p+q)+1 end{array}right).$$ The inverse should be clear from the definitions.
answered Jun 8 '17 at 12:24
PtFPtF
3,94921733
$endgroup$
add a comment |
$begingroup$
I wrote down the details so I'm going to share just in case someone else needs it someday.
The first thing to be noticed is that $sigma^{-1}({1})={1, p+2}$, hence:
$$mathsf{Sh}(p+1, q)={sigmain mathsf{Sh}(p+1, q): sigma(1)=1}sqcup {sigmainmathsf{Sh}(p+1, q): sigma(p+2)=1}.$$
This allows us to define the bijection easily setting:
$$mathsf{Sh}(p, q)longrightarrow mathsf{Sh}(p+1, q)\ sigmalongmapsto left(begin{array}{ccccccc}1 & 2& ldots & p+2 & ldots &p+q+1\ 1 &sigma(1)+1 & ldots &sigma(p+1)+1 & ldots & sigma(p+q)+1 end{array}right)$$
and
$$ mathsf{Sh}(p+1, q-1)longrightarrow mathsf{Sh}(p+1, q)\ sigmalongmapsto left(begin{array}{ccccccc}1 & ldots &p+1 & p+2 & p+3& ldots & p+q+1\ sigma(1)+1 & ldots & sigma(p+1)+1 & 1 & sigma(p+2)+1 & ldots &sigma(p+q)+1 end{array}right).$$ The inverse should be clear from the definitions.
answered Jun 8 '17 at 12:24
PtFPtF
3,94921733
$endgroup$
add a comment |
$begingroup$
I wrote down the details so I'm going to share just in case someone else needs it someday.
The first thing to be noticed is that $sigma^{-1}({1})={1, p+2}$, hence:
$$mathsf{Sh}(p+1, q)={sigmain mathsf{Sh}(p+1, q): sigma(1)=1}sqcup {sigmainmathsf{Sh}(p+1, q): sigma(p+2)=1}.$$
This allows us to define the bijection easily setting:
$$mathsf{Sh}(p, q)longrightarrow mathsf{Sh}(p+1, q)\ sigmalongmapsto left(begin{array}{ccccccc}1 & 2& ldots & p+2 & ldots &p+q+1\ 1 &sigma(1)+1 & ldots &sigma(p+1)+1 & ldots & sigma(p+q)+1 end{array}right)$$
and
$$ mathsf{Sh}(p+1, q-1)longrightarrow mathsf{Sh}(p+1, q)\ sigmalongmapsto left(begin{array}{ccccccc}1 & ldots &p+1 & p+2 & p+3& ldots & p+q+1\ sigma(1)+1 & ldots & sigma(p+1)+1 & 1 & sigma(p+2)+1 & ldots &sigma(p+q)+1 end{array}right).$$ The inverse should be clear from the definitions.
answered Jun 8 '17 at 12:24
PtFPtF
3,94921733
$endgroup$
I wrote down the details so I'm going to share just in case someone else needs it someday.
The first thing to be noticed is that $sigma^{-1}({1})={1, p+2}$, hence:
$$mathsf{Sh}(p+1, q)={sigmain mathsf{Sh}(p+1, q): sigma(1)=1}sqcup {sigmainmathsf{Sh}(p+1, q): sigma(p+2)=1}.$$
This allows us to define the bijection easily setting:
$$mathsf{Sh}(p, q)longrightarrow mathsf{Sh}(p+1, q)\ sigmalongmapsto left(begin{array}{ccccccc}1 & 2& ldots & p+2 & ldots &p+q+1\ 1 &sigma(1)+1 & ldots &sigma(p+1)+1 & ldots & sigma(p+q)+1 end{array}right)$$
and
$$ mathsf{Sh}(p+1, q-1)longrightarrow mathsf{Sh}(p+1, q)\ sigmalongmapsto left(begin{array}{ccccccc}1 & ldots &p+1 & p+2 & p+3& ldots & p+q+1\ sigma(1)+1 & ldots & sigma(p+1)+1 & 1 & sigma(p+2)+1 & ldots &sigma(p+q)+1 end{array}right).$$ The inverse should be clear from the definitions.
answered Jun 8 '17 at 12:24
PtFPtF
3,94921733
answered Jun 8 '17 at 12:24
PtFPtF
3,94921733
answered Jun 8 '17 at 12:24
PtFPtF
3,94921733
answered Jun 8 '17 at 12:24
PtFPtF
3,94921733
3,94921733
add a comment |
add a comment |
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