SDR question: What is the number $m$ of different sets $P_k$ in $mathcal P_n$ for a given $n$?
$begingroup$
Consider elements of $Z_{2n}$, that is, numbers $0,1,2,dots, 2n-1$. Let
$mathcal P_n={ P_1,P_2,dots, P_m}$ be a collection of unordered sets, where each $P_k$ consists of $n$ ordered pairs $(a_i,b_i)$ with the following properties:
$(1)$ $A={ a_i| 1leq ileq n}, B={ b_i| 1leq ileq n}$, and
$Acup B={0,1,2,dots,2n-1}$
$(2)$ For every $1leq ileq n$, we have $a_i < b_i$
$(3)$ For any $1leq i,jleq n, ineq j$ we have $a_i < a_j < b_i$ if and only if $a_i < b_j < b_i$
What is the number $m$ of different sets $P_k$ in $mathcal P_n$ for a given $n$?
combinatorics
asked Dec 3 '18 at 17:52
DummKorfDummKorf
335
$endgroup$
add a comment |
$begingroup$
Consider elements of $Z_{2n}$, that is, numbers $0,1,2,dots, 2n-1$. Let
$mathcal P_n={ P_1,P_2,dots, P_m}$ be a collection of unordered sets, where each $P_k$ consists of $n$ ordered pairs $(a_i,b_i)$ with the following properties:
$(1)$ $A={ a_i| 1leq ileq n}, B={ b_i| 1leq ileq n}$, and
$Acup B={0,1,2,dots,2n-1}$
$(2)$ For every $1leq ileq n$, we have $a_i < b_i$
$(3)$ For any $1leq i,jleq n, ineq j$ we have $a_i < a_j < b_i$ if and only if $a_i < b_j < b_i$
What is the number $m$ of different sets $P_k$ in $mathcal P_n$ for a given $n$?
combinatorics
asked Dec 3 '18 at 17:52
DummKorfDummKorf
335
$endgroup$
$begingroup$
What have you tried so far? How far have you gotten?
$endgroup$
– saulspatz
Dec 3 '18 at 18:27
add a comment |
$begingroup$
Consider elements of $Z_{2n}$, that is, numbers $0,1,2,dots, 2n-1$. Let
$mathcal P_n={ P_1,P_2,dots, P_m}$ be a collection of unordered sets, where each $P_k$ consists of $n$ ordered pairs $(a_i,b_i)$ with the following properties:
$(1)$ $A={ a_i| 1leq ileq n}, B={ b_i| 1leq ileq n}$, and
$Acup B={0,1,2,dots,2n-1}$
$(2)$ For every $1leq ileq n$, we have $a_i < b_i$
$(3)$ For any $1leq i,jleq n, ineq j$ we have $a_i < a_j < b_i$ if and only if $a_i < b_j < b_i$
What is the number $m$ of different sets $P_k$ in $mathcal P_n$ for a given $n$?
combinatorics
asked Dec 3 '18 at 17:52
DummKorfDummKorf
335
$endgroup$
Consider elements of $Z_{2n}$, that is, numbers $0,1,2,dots, 2n-1$. Let
$mathcal P_n={ P_1,P_2,dots, P_m}$ be a collection of unordered sets, where each $P_k$ consists of $n$ ordered pairs $(a_i,b_i)$ with the following properties:
$(1)$ $A={ a_i| 1leq ileq n}, B={ b_i| 1leq ileq n}$, and
$Acup B={0,1,2,dots,2n-1}$
$(2)$ For every $1leq ileq n$, we have $a_i < b_i$
$(3)$ For any $1leq i,jleq n, ineq j$ we have $a_i < a_j < b_i$ if and only if $a_i < b_j < b_i$
What is the number $m$ of different sets $P_k$ in $mathcal P_n$ for a given $n$?
combinatorics
combinatorics
asked Dec 3 '18 at 17:52
DummKorfDummKorf
335
asked Dec 3 '18 at 17:52
DummKorfDummKorf
335
asked Dec 3 '18 at 17:52
DummKorfDummKorf
335
asked Dec 3 '18 at 17:52
DummKorfDummKorf
335
asked Dec 3 '18 at 17:52
DummKorfDummKorf
335
335
$begingroup$
What have you tried so far? How far have you gotten?
$endgroup$
– saulspatz
Dec 3 '18 at 18:27
add a comment |
$begingroup$
What have you tried so far? How far have you gotten?
$endgroup$
– saulspatz
Dec 3 '18 at 18:27
$begingroup$
What have you tried so far? How far have you gotten?
$endgroup$
– saulspatz
Dec 3 '18 at 18:27
$begingroup$
What have you tried so far? How far have you gotten?
$endgroup$
– saulspatz
Dec 3 '18 at 18:27
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $C_n$ be the number of such sets for a given $n$. Clearly, $C_0=C_1=1.$ To establish a recurrence, we ask which number can be paired with $0.$ Suppose $(0,b)$ is one of the pairs. Then in every other pair either both numbers are greater than $b$, or both are less than $b$. That is, $b$ must be odd. Further, the pairings of the numbers from $1$ to $b-1$ and the pairings of the numbers from $b+1$ to $2n-1$ must both obey rule $(3).$
Therefore, $$E_n=sum_{k=0}^{n-1}E_kE_{n-1-k}$$ and we recognize the recurrence for the Catalan numbers. (look at the "First Proof" on the wiki page.
answered Dec 3 '18 at 19:07
saulspatzsaulspatz
14.3k21329
$endgroup$
add a comment |
$begingroup$
With any $P_kin mathcal P_n$, we associate a legal sequence of n open and n closed parentheses by putting an open parentheses at position $a_i$ and a closed one at position $b_i$. Conversely, given any legal sequence of length $2n$, we can pair the positions of corresponding parentheses and the set of such pairs gives us an element of $mathcal P_n$.
Therefore, $$mid mathcal P_nmid= C_n=frac{1}{n+1}binom{2n}{n}$$
Footnotes:
1. https://en.wikipedia.org/wiki/Catalan_number
answered Dec 3 '18 at 19:35
Anubhab GhosalAnubhab Ghosal
83518
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $C_n$ be the number of such sets for a given $n$. Clearly, $C_0=C_1=1.$ To establish a recurrence, we ask which number can be paired with $0.$ Suppose $(0,b)$ is one of the pairs. Then in every other pair either both numbers are greater than $b$, or both are less than $b$. That is, $b$ must be odd. Further, the pairings of the numbers from $1$ to $b-1$ and the pairings of the numbers from $b+1$ to $2n-1$ must both obey rule $(3).$
Therefore, $$E_n=sum_{k=0}^{n-1}E_kE_{n-1-k}$$ and we recognize the recurrence for the Catalan numbers. (look at the "First Proof" on the wiki page.
answered Dec 3 '18 at 19:07
saulspatzsaulspatz
14.3k21329
$endgroup$
add a comment |
$begingroup$
Let $C_n$ be the number of such sets for a given $n$. Clearly, $C_0=C_1=1.$ To establish a recurrence, we ask which number can be paired with $0.$ Suppose $(0,b)$ is one of the pairs. Then in every other pair either both numbers are greater than $b$, or both are less than $b$. That is, $b$ must be odd. Further, the pairings of the numbers from $1$ to $b-1$ and the pairings of the numbers from $b+1$ to $2n-1$ must both obey rule $(3).$
Therefore, $$E_n=sum_{k=0}^{n-1}E_kE_{n-1-k}$$ and we recognize the recurrence for the Catalan numbers. (look at the "First Proof" on the wiki page.
answered Dec 3 '18 at 19:07
saulspatzsaulspatz
14.3k21329
$endgroup$
add a comment |
$begingroup$
Let $C_n$ be the number of such sets for a given $n$. Clearly, $C_0=C_1=1.$ To establish a recurrence, we ask which number can be paired with $0.$ Suppose $(0,b)$ is one of the pairs. Then in every other pair either both numbers are greater than $b$, or both are less than $b$. That is, $b$ must be odd. Further, the pairings of the numbers from $1$ to $b-1$ and the pairings of the numbers from $b+1$ to $2n-1$ must both obey rule $(3).$
Therefore, $$E_n=sum_{k=0}^{n-1}E_kE_{n-1-k}$$ and we recognize the recurrence for the Catalan numbers. (look at the "First Proof" on the wiki page.
answered Dec 3 '18 at 19:07
saulspatzsaulspatz
14.3k21329
$endgroup$
Let $C_n$ be the number of such sets for a given $n$. Clearly, $C_0=C_1=1.$ To establish a recurrence, we ask which number can be paired with $0.$ Suppose $(0,b)$ is one of the pairs. Then in every other pair either both numbers are greater than $b$, or both are less than $b$. That is, $b$ must be odd. Further, the pairings of the numbers from $1$ to $b-1$ and the pairings of the numbers from $b+1$ to $2n-1$ must both obey rule $(3).$
Therefore, $$E_n=sum_{k=0}^{n-1}E_kE_{n-1-k}$$ and we recognize the recurrence for the Catalan numbers. (look at the "First Proof" on the wiki page.
answered Dec 3 '18 at 19:07
saulspatzsaulspatz
14.3k21329
answered Dec 3 '18 at 19:07
saulspatzsaulspatz
14.3k21329
answered Dec 3 '18 at 19:07
saulspatzsaulspatz
14.3k21329
answered Dec 3 '18 at 19:07
saulspatzsaulspatz
14.3k21329
14.3k21329
add a comment |
add a comment |
$begingroup$
With any $P_kin mathcal P_n$, we associate a legal sequence of n open and n closed parentheses by putting an open parentheses at position $a_i$ and a closed one at position $b_i$. Conversely, given any legal sequence of length $2n$, we can pair the positions of corresponding parentheses and the set of such pairs gives us an element of $mathcal P_n$.
Therefore, $$mid mathcal P_nmid= C_n=frac{1}{n+1}binom{2n}{n}$$
Footnotes:
1. https://en.wikipedia.org/wiki/Catalan_number
answered Dec 3 '18 at 19:35
Anubhab GhosalAnubhab Ghosal
83518
$endgroup$
add a comment |
$begingroup$
With any $P_kin mathcal P_n$, we associate a legal sequence of n open and n closed parentheses by putting an open parentheses at position $a_i$ and a closed one at position $b_i$. Conversely, given any legal sequence of length $2n$, we can pair the positions of corresponding parentheses and the set of such pairs gives us an element of $mathcal P_n$.
Therefore, $$mid mathcal P_nmid= C_n=frac{1}{n+1}binom{2n}{n}$$
Footnotes:
1. https://en.wikipedia.org/wiki/Catalan_number
answered Dec 3 '18 at 19:35
Anubhab GhosalAnubhab Ghosal
83518
$endgroup$
add a comment |
$begingroup$
With any $P_kin mathcal P_n$, we associate a legal sequence of n open and n closed parentheses by putting an open parentheses at position $a_i$ and a closed one at position $b_i$. Conversely, given any legal sequence of length $2n$, we can pair the positions of corresponding parentheses and the set of such pairs gives us an element of $mathcal P_n$.
Therefore, $$mid mathcal P_nmid= C_n=frac{1}{n+1}binom{2n}{n}$$
Footnotes:
1. https://en.wikipedia.org/wiki/Catalan_number
answered Dec 3 '18 at 19:35
Anubhab GhosalAnubhab Ghosal
83518
$endgroup$
With any $P_kin mathcal P_n$, we associate a legal sequence of n open and n closed parentheses by putting an open parentheses at position $a_i$ and a closed one at position $b_i$. Conversely, given any legal sequence of length $2n$, we can pair the positions of corresponding parentheses and the set of such pairs gives us an element of $mathcal P_n$.
Therefore, $$mid mathcal P_nmid= C_n=frac{1}{n+1}binom{2n}{n}$$
Footnotes:
1. https://en.wikipedia.org/wiki/Catalan_number
answered Dec 3 '18 at 19:35
Anubhab GhosalAnubhab Ghosal
83518
answered Dec 3 '18 at 19:35
Anubhab GhosalAnubhab Ghosal
83518
answered Dec 3 '18 at 19:35
Anubhab GhosalAnubhab Ghosal
83518
answered Dec 3 '18 at 19:35
Anubhab GhosalAnubhab Ghosal
83518
83518
add a comment |
add a comment |
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$begingroup$
What have you tried so far? How far have you gotten?
$endgroup$
– saulspatz
Dec 3 '18 at 18:27