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
6,0601857
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
6,0601857
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
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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
6,0601857
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
6,0601857
asked Dec 20 '18 at 13:46
akashkingakashking
33
edited Dec 20 '18 at 14:41
Zacky
6,0601857
asked Dec 20 '18 at 13:46
akashkingakashking
33
edited Dec 20 '18 at 14:41
Zacky
6,0601857
edited Dec 20 '18 at 14:41
Zacky
6,0601857
edited Dec 20 '18 at 14:41
Zacky
6,0601857
6,0601857
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
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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
4,2771413
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add a comment |
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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
4,2771413
$endgroup$
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
4,2771413
$endgroup$
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
4,2771413
$endgroup$
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
4,2771413
answered Dec 20 '18 at 14:36
WuestenfuxWuestenfux
4,2771413
answered Dec 20 '18 at 14:36
WuestenfuxWuestenfux
4,2771413
answered Dec 20 '18 at 14:36
WuestenfuxWuestenfux
4,2771413
4,2771413
add a comment |
add a comment |
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$begingroup$
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