Recurrence relation for Stirling numbers of the second kind
$begingroup$
I was solving a recurrence relation for Stirling numbers of the second kind:
$$S(n,k)=kS(n-1,k)+S(n-1,k-1)$$
For $k=1$ or $k=n $ $ S(n,k)=1$
For $k=0$ or $k>n $ $ S(n,k)=0$
Substitution method is not working here because at every time $k$ the value changes.
Can anyone tell me what will be the correct method for this? I just want to calculate the time complexity.
combinatorics recurrence-relations stirling-numbers
edited Dec 20 '18 at 14:41
Zacky
6,6451958
asked Dec 20 '18 at 13:46
akashkingakashking
33
$endgroup$
|
show 3 more comments
$begingroup$
I was solving a recurrence relation for Stirling numbers of the second kind:
$$S(n,k)=kS(n-1,k)+S(n-1,k-1)$$
For $k=1$ or $k=n $ $ S(n,k)=1$
For $k=0$ or $k>n $ $ S(n,k)=0$
Substitution method is not working here because at every time $k$ the value changes.
Can anyone tell me what will be the correct method for this? I just want to calculate the time complexity.
combinatorics recurrence-relations stirling-numbers
edited Dec 20 '18 at 14:41
Zacky
6,6451958
asked Dec 20 '18 at 13:46
akashkingakashking
33
$endgroup$
$begingroup$
I have edited abit of your question, let me know if I did it right!
$endgroup$
– Zacky
Dec 20 '18 at 14:42
$begingroup$
thank you very much @Zacky
$endgroup$
– akashking
Dec 20 '18 at 15:06
$begingroup$
Also, you want a proof for that recurrence too, or ??
$endgroup$
– Zacky
Dec 20 '18 at 15:08
$begingroup$
yes if you can provide me dear
$endgroup$
– akashking
Dec 20 '18 at 15:14
2
$begingroup$
It's unclear what you are asking (esp the bit about "time complexity"). If you want the solution to the recurrence, well, see the explicit form of the Stirling numbers of the second kind , as an alternating sum. You can check that it verifies the recurrence (with the initial conditions). If you want a general recipe for solving that kind of recurrence... I doubt you'll find anything. The normal way of getting the formula is by combinatorial reasoning - and the combinatorial interpretation can be linked to the recursion also.
$endgroup$
– leonbloy
Dec 20 '18 at 15:57
|
show 3 more comments
$begingroup$
I was solving a recurrence relation for Stirling numbers of the second kind:
$$S(n,k)=kS(n-1,k)+S(n-1,k-1)$$
For $k=1$ or $k=n $ $ S(n,k)=1$
For $k=0$ or $k>n $ $ S(n,k)=0$
Substitution method is not working here because at every time $k$ the value changes.
Can anyone tell me what will be the correct method for this? I just want to calculate the time complexity.
combinatorics recurrence-relations stirling-numbers
edited Dec 20 '18 at 14:41
Zacky
6,6451958
asked Dec 20 '18 at 13:46
akashkingakashking
33
$endgroup$
I was solving a recurrence relation for Stirling numbers of the second kind:
$$S(n,k)=kS(n-1,k)+S(n-1,k-1)$$
For $k=1$ or $k=n $ $ S(n,k)=1$
For $k=0$ or $k>n $ $ S(n,k)=0$
Substitution method is not working here because at every time $k$ the value changes.
Can anyone tell me what will be the correct method for this? I just want to calculate the time complexity.
combinatorics recurrence-relations stirling-numbers
combinatorics recurrence-relations stirling-numbers
edited Dec 20 '18 at 14:41
Zacky
6,6451958
asked Dec 20 '18 at 13:46
akashkingakashking
33
edited Dec 20 '18 at 14:41
Zacky
6,6451958
asked Dec 20 '18 at 13:46
akashkingakashking
33
edited Dec 20 '18 at 14:41
Zacky
6,6451958
edited Dec 20 '18 at 14:41
Zacky
6,6451958
edited Dec 20 '18 at 14:41
Zacky
6,6451958
6,6451958
asked Dec 20 '18 at 13:46
akashkingakashking
33
asked Dec 20 '18 at 13:46
akashkingakashking
33
asked Dec 20 '18 at 13:46
akashkingakashking
33
33
$begingroup$
I have edited abit of your question, let me know if I did it right!
$endgroup$
– Zacky
Dec 20 '18 at 14:42
$begingroup$
thank you very much @Zacky
$endgroup$
– akashking
Dec 20 '18 at 15:06
$begingroup$
Also, you want a proof for that recurrence too, or ??
$endgroup$
– Zacky
Dec 20 '18 at 15:08
$begingroup$
yes if you can provide me dear
$endgroup$
– akashking
Dec 20 '18 at 15:14
2
$begingroup$
It's unclear what you are asking (esp the bit about "time complexity"). If you want the solution to the recurrence, well, see the explicit form of the Stirling numbers of the second kind , as an alternating sum. You can check that it verifies the recurrence (with the initial conditions). If you want a general recipe for solving that kind of recurrence... I doubt you'll find anything. The normal way of getting the formula is by combinatorial reasoning - and the combinatorial interpretation can be linked to the recursion also.
$endgroup$
– leonbloy
Dec 20 '18 at 15:57
|
show 3 more comments
$begingroup$
I have edited abit of your question, let me know if I did it right!
$endgroup$
– Zacky
Dec 20 '18 at 14:42
$begingroup$
thank you very much @Zacky
$endgroup$
– akashking
Dec 20 '18 at 15:06
$begingroup$
Also, you want a proof for that recurrence too, or ??
$endgroup$
– Zacky
Dec 20 '18 at 15:08
$begingroup$
yes if you can provide me dear
$endgroup$
– akashking
Dec 20 '18 at 15:14
2
$begingroup$
It's unclear what you are asking (esp the bit about "time complexity"). If you want the solution to the recurrence, well, see the explicit form of the Stirling numbers of the second kind , as an alternating sum. You can check that it verifies the recurrence (with the initial conditions). If you want a general recipe for solving that kind of recurrence... I doubt you'll find anything. The normal way of getting the formula is by combinatorial reasoning - and the combinatorial interpretation can be linked to the recursion also.
$endgroup$
– leonbloy
Dec 20 '18 at 15:57
$begingroup$
I have edited abit of your question, let me know if I did it right!
$endgroup$
– Zacky
Dec 20 '18 at 14:42
$begingroup$
I have edited abit of your question, let me know if I did it right!
$endgroup$
– Zacky
Dec 20 '18 at 14:42
$begingroup$
thank you very much @Zacky
$endgroup$
– akashking
Dec 20 '18 at 15:06
$begingroup$
thank you very much @Zacky
$endgroup$
– akashking
Dec 20 '18 at 15:06
$begingroup$
Also, you want a proof for that recurrence too, or ??
$endgroup$
– Zacky
Dec 20 '18 at 15:08
$begingroup$
Also, you want a proof for that recurrence too, or ??
$endgroup$
– Zacky
Dec 20 '18 at 15:08
$begingroup$
yes if you can provide me dear
$endgroup$
– akashking
Dec 20 '18 at 15:14
$begingroup$
yes if you can provide me dear
$endgroup$
– akashking
Dec 20 '18 at 15:14
2
2
$begingroup$
It's unclear what you are asking (esp the bit about "time complexity"). If you want the solution to the recurrence, well, see the explicit form of the Stirling numbers of the second kind , as an alternating sum. You can check that it verifies the recurrence (with the initial conditions). If you want a general recipe for solving that kind of recurrence... I doubt you'll find anything. The normal way of getting the formula is by combinatorial reasoning - and the combinatorial interpretation can be linked to the recursion also.
$endgroup$
– leonbloy
Dec 20 '18 at 15:57
$begingroup$
It's unclear what you are asking (esp the bit about "time complexity"). If you want the solution to the recurrence, well, see the explicit form of the Stirling numbers of the second kind , as an alternating sum. You can check that it verifies the recurrence (with the initial conditions). If you want a general recipe for solving that kind of recurrence... I doubt you'll find anything. The normal way of getting the formula is by combinatorial reasoning - and the combinatorial interpretation can be linked to the recursion also.
$endgroup$
– leonbloy
Dec 20 '18 at 15:57
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Welcome to MSE! The proof I know uses a Pascal argument, i.e., the set of $k$-partitions of the set $[n]={1,ldots,n}$ is divided into two sets such that $A$ is the set of $k$-partitions of $[n]$ which contain ${n}$ as an element and $B$ is the remaining set of $k$-partitions of $[n]$. Then $S(n,k) = |A|+|B|$, since both sets are disjoint.
Each element of $A$, $P={P_1,ldots,P_{k-1},{n}}$, becomes a $k-1$-partition of $[n-1]$ by deleting the element ${n}$. This gives a bijection between $A$ and the set of $k-1$-partitions of $[n-1]$, i.e., $|A|= S(n-1,k-1)$.
In view of the set $B$, each $k$-partition of $[n-1]$, $P={P_1,ldots,P_k}$, can be ''extended'' to a $k$-partition $P^{(i)}$ of $[n]$ by adding the element $n$ to one element $P_i$. The assignment $(i,P)mapsto P^{(i)}$ gives a bijection $[k]times (mbox{set of $k$-partitions of $[n-1]$})rightarrow B$. Thus $|B|=kcdot S(n-1,k)$.
As a hint, for the bijection concerning $B$ it suffices to consider $k$-partitions of $[n-1]$ in standard-form.
A partition $P={P_1,ldots,P_k}$ of $[n]$ is in standard-form if $1in P_1$ and for each $igeq 1$, $P_{i+1}$ contains the smallest number not contained in $P_1cupldotscup P_i$. E.g., if $P={{1,2},{3,6},{5,8},{4,7}}$, then $P_1={1,2}$, $P_2={3,6}$, $P_3={4,7}$, and $P_4={5,8}$.
answered Dec 20 '18 at 14:36
WuestenfuxWuestenfux
4,7001413
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3047552%2frecurrence-relation-for-stirling-numbers-of-the-second-kind%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Welcome to MSE! The proof I know uses a Pascal argument, i.e., the set of $k$-partitions of the set $[n]={1,ldots,n}$ is divided into two sets such that $A$ is the set of $k$-partitions of $[n]$ which contain ${n}$ as an element and $B$ is the remaining set of $k$-partitions of $[n]$. Then $S(n,k) = |A|+|B|$, since both sets are disjoint.
Each element of $A$, $P={P_1,ldots,P_{k-1},{n}}$, becomes a $k-1$-partition of $[n-1]$ by deleting the element ${n}$. This gives a bijection between $A$ and the set of $k-1$-partitions of $[n-1]$, i.e., $|A|= S(n-1,k-1)$.
In view of the set $B$, each $k$-partition of $[n-1]$, $P={P_1,ldots,P_k}$, can be ''extended'' to a $k$-partition $P^{(i)}$ of $[n]$ by adding the element $n$ to one element $P_i$. The assignment $(i,P)mapsto P^{(i)}$ gives a bijection $[k]times (mbox{set of $k$-partitions of $[n-1]$})rightarrow B$. Thus $|B|=kcdot S(n-1,k)$.
As a hint, for the bijection concerning $B$ it suffices to consider $k$-partitions of $[n-1]$ in standard-form.
A partition $P={P_1,ldots,P_k}$ of $[n]$ is in standard-form if $1in P_1$ and for each $igeq 1$, $P_{i+1}$ contains the smallest number not contained in $P_1cupldotscup P_i$. E.g., if $P={{1,2},{3,6},{5,8},{4,7}}$, then $P_1={1,2}$, $P_2={3,6}$, $P_3={4,7}$, and $P_4={5,8}$.
answered Dec 20 '18 at 14:36
WuestenfuxWuestenfux
4,7001413
$endgroup$
add a comment |
$begingroup$
Welcome to MSE! The proof I know uses a Pascal argument, i.e., the set of $k$-partitions of the set $[n]={1,ldots,n}$ is divided into two sets such that $A$ is the set of $k$-partitions of $[n]$ which contain ${n}$ as an element and $B$ is the remaining set of $k$-partitions of $[n]$. Then $S(n,k) = |A|+|B|$, since both sets are disjoint.
Each element of $A$, $P={P_1,ldots,P_{k-1},{n}}$, becomes a $k-1$-partition of $[n-1]$ by deleting the element ${n}$. This gives a bijection between $A$ and the set of $k-1$-partitions of $[n-1]$, i.e., $|A|= S(n-1,k-1)$.
In view of the set $B$, each $k$-partition of $[n-1]$, $P={P_1,ldots,P_k}$, can be ''extended'' to a $k$-partition $P^{(i)}$ of $[n]$ by adding the element $n$ to one element $P_i$. The assignment $(i,P)mapsto P^{(i)}$ gives a bijection $[k]times (mbox{set of $k$-partitions of $[n-1]$})rightarrow B$. Thus $|B|=kcdot S(n-1,k)$.
As a hint, for the bijection concerning $B$ it suffices to consider $k$-partitions of $[n-1]$ in standard-form.
A partition $P={P_1,ldots,P_k}$ of $[n]$ is in standard-form if $1in P_1$ and for each $igeq 1$, $P_{i+1}$ contains the smallest number not contained in $P_1cupldotscup P_i$. E.g., if $P={{1,2},{3,6},{5,8},{4,7}}$, then $P_1={1,2}$, $P_2={3,6}$, $P_3={4,7}$, and $P_4={5,8}$.
answered Dec 20 '18 at 14:36
WuestenfuxWuestenfux
4,7001413
$endgroup$
add a comment |
$begingroup$
Welcome to MSE! The proof I know uses a Pascal argument, i.e., the set of $k$-partitions of the set $[n]={1,ldots,n}$ is divided into two sets such that $A$ is the set of $k$-partitions of $[n]$ which contain ${n}$ as an element and $B$ is the remaining set of $k$-partitions of $[n]$. Then $S(n,k) = |A|+|B|$, since both sets are disjoint.
Each element of $A$, $P={P_1,ldots,P_{k-1},{n}}$, becomes a $k-1$-partition of $[n-1]$ by deleting the element ${n}$. This gives a bijection between $A$ and the set of $k-1$-partitions of $[n-1]$, i.e., $|A|= S(n-1,k-1)$.
In view of the set $B$, each $k$-partition of $[n-1]$, $P={P_1,ldots,P_k}$, can be ''extended'' to a $k$-partition $P^{(i)}$ of $[n]$ by adding the element $n$ to one element $P_i$. The assignment $(i,P)mapsto P^{(i)}$ gives a bijection $[k]times (mbox{set of $k$-partitions of $[n-1]$})rightarrow B$. Thus $|B|=kcdot S(n-1,k)$.
As a hint, for the bijection concerning $B$ it suffices to consider $k$-partitions of $[n-1]$ in standard-form.
A partition $P={P_1,ldots,P_k}$ of $[n]$ is in standard-form if $1in P_1$ and for each $igeq 1$, $P_{i+1}$ contains the smallest number not contained in $P_1cupldotscup P_i$. E.g., if $P={{1,2},{3,6},{5,8},{4,7}}$, then $P_1={1,2}$, $P_2={3,6}$, $P_3={4,7}$, and $P_4={5,8}$.
answered Dec 20 '18 at 14:36
WuestenfuxWuestenfux
4,7001413
$endgroup$
Welcome to MSE! The proof I know uses a Pascal argument, i.e., the set of $k$-partitions of the set $[n]={1,ldots,n}$ is divided into two sets such that $A$ is the set of $k$-partitions of $[n]$ which contain ${n}$ as an element and $B$ is the remaining set of $k$-partitions of $[n]$. Then $S(n,k) = |A|+|B|$, since both sets are disjoint.
Each element of $A$, $P={P_1,ldots,P_{k-1},{n}}$, becomes a $k-1$-partition of $[n-1]$ by deleting the element ${n}$. This gives a bijection between $A$ and the set of $k-1$-partitions of $[n-1]$, i.e., $|A|= S(n-1,k-1)$.
In view of the set $B$, each $k$-partition of $[n-1]$, $P={P_1,ldots,P_k}$, can be ''extended'' to a $k$-partition $P^{(i)}$ of $[n]$ by adding the element $n$ to one element $P_i$. The assignment $(i,P)mapsto P^{(i)}$ gives a bijection $[k]times (mbox{set of $k$-partitions of $[n-1]$})rightarrow B$. Thus $|B|=kcdot S(n-1,k)$.
As a hint, for the bijection concerning $B$ it suffices to consider $k$-partitions of $[n-1]$ in standard-form.
A partition $P={P_1,ldots,P_k}$ of $[n]$ is in standard-form if $1in P_1$ and for each $igeq 1$, $P_{i+1}$ contains the smallest number not contained in $P_1cupldotscup P_i$. E.g., if $P={{1,2},{3,6},{5,8},{4,7}}$, then $P_1={1,2}$, $P_2={3,6}$, $P_3={4,7}$, and $P_4={5,8}$.
answered Dec 20 '18 at 14:36
WuestenfuxWuestenfux
4,7001413
answered Dec 20 '18 at 14:36
WuestenfuxWuestenfux
4,7001413
answered Dec 20 '18 at 14:36
WuestenfuxWuestenfux
4,7001413
answered Dec 20 '18 at 14:36
WuestenfuxWuestenfux
4,7001413
4,7001413
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3047552%2frecurrence-relation-for-stirling-numbers-of-the-second-kind%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
I have edited abit of your question, let me know if I did it right!
$endgroup$
– Zacky
Dec 20 '18 at 14:42
$begingroup$
thank you very much @Zacky
$endgroup$
– akashking
Dec 20 '18 at 15:06
$begingroup$
Also, you want a proof for that recurrence too, or ??
$endgroup$
– Zacky
Dec 20 '18 at 15:08
$begingroup$
yes if you can provide me dear
$endgroup$
– akashking
Dec 20 '18 at 15:14
2
$begingroup$
It's unclear what you are asking (esp the bit about "time complexity"). If you want the solution to the recurrence, well, see the explicit form of the Stirling numbers of the second kind , as an alternating sum. You can check that it verifies the recurrence (with the initial conditions). If you want a general recipe for solving that kind of recurrence... I doubt you'll find anything. The normal way of getting the formula is by combinatorial reasoning - and the combinatorial interpretation can be linked to the recursion also.
$endgroup$
– leonbloy
Dec 20 '18 at 15:57