'selfish' set to be a set which has its own cardinality (number of elements) as an element
Define a selfish set to be a set which has its own cardinality (number of elements) as an element. Find, with proof, the number of subsets of ${1, 2, ldots, n}$ which are minimal selfish sets, that is, selfish sets none of whose proper subsets is selfish.
My Attempt:
Assume $textbf{A}$ to be a selfish set. If the cardinality of $textbf{A}$ is $c$, then can $textbf{A}$ contain $1,2,3....c-1$. Definitely answer is no. because if it contains $k<c$ then deleting $c-k$ elements except $k$ from $textbf{A}$ gives a subset of k elements contradicting the fact that $textbf{A}$ is minimal selfish.
Thus $textbf{A}$ must contain elements greater than or equal to $c$. But how do I find the minimal selfish sets with order $c$?
combinatorics discrete-mathematics
edited Dec 8 at 10:20
Mutantoe
562411
add a comment |
Define a selfish set to be a set which has its own cardinality (number of elements) as an element. Find, with proof, the number of subsets of ${1, 2, ldots, n}$ which are minimal selfish sets, that is, selfish sets none of whose proper subsets is selfish.
My Attempt:
Assume $textbf{A}$ to be a selfish set. If the cardinality of $textbf{A}$ is $c$, then can $textbf{A}$ contain $1,2,3....c-1$. Definitely answer is no. because if it contains $k<c$ then deleting $c-k$ elements except $k$ from $textbf{A}$ gives a subset of k elements contradicting the fact that $textbf{A}$ is minimal selfish.
Thus $textbf{A}$ must contain elements greater than or equal to $c$. But how do I find the minimal selfish sets with order $c$?
combinatorics discrete-mathematics
edited Dec 8 at 10:20
Mutantoe
562411
Reference note: This question, including the terminology, was problem B1 on the 1996 Putnam.
– jmerry
Dec 8 at 7:51
add a comment |
Define a selfish set to be a set which has its own cardinality (number of elements) as an element. Find, with proof, the number of subsets of ${1, 2, ldots, n}$ which are minimal selfish sets, that is, selfish sets none of whose proper subsets is selfish.
My Attempt:
Assume $textbf{A}$ to be a selfish set. If the cardinality of $textbf{A}$ is $c$, then can $textbf{A}$ contain $1,2,3....c-1$. Definitely answer is no. because if it contains $k<c$ then deleting $c-k$ elements except $k$ from $textbf{A}$ gives a subset of k elements contradicting the fact that $textbf{A}$ is minimal selfish.
Thus $textbf{A}$ must contain elements greater than or equal to $c$. But how do I find the minimal selfish sets with order $c$?
combinatorics discrete-mathematics
edited Dec 8 at 10:20
Mutantoe
562411
Define a selfish set to be a set which has its own cardinality (number of elements) as an element. Find, with proof, the number of subsets of ${1, 2, ldots, n}$ which are minimal selfish sets, that is, selfish sets none of whose proper subsets is selfish.
My Attempt:
Assume $textbf{A}$ to be a selfish set. If the cardinality of $textbf{A}$ is $c$, then can $textbf{A}$ contain $1,2,3....c-1$. Definitely answer is no. because if it contains $k<c$ then deleting $c-k$ elements except $k$ from $textbf{A}$ gives a subset of k elements contradicting the fact that $textbf{A}$ is minimal selfish.
Thus $textbf{A}$ must contain elements greater than or equal to $c$. But how do I find the minimal selfish sets with order $c$?
combinatorics discrete-mathematics
combinatorics discrete-mathematics
edited Dec 8 at 10:20
Mutantoe
562411
edited Dec 8 at 10:20
Mutantoe
562411
edited Dec 8 at 10:20
Mutantoe
562411
edited Dec 8 at 10:20
Mutantoe
562411
edited Dec 8 at 10:20
Mutantoe
562411
562411
asked Dec 8 at 6:03
user408906
Reference note: This question, including the terminology, was problem B1 on the 1996 Putnam.
– jmerry
Dec 8 at 7:51
add a comment |
Reference note: This question, including the terminology, was problem B1 on the 1996 Putnam.
– jmerry
Dec 8 at 7:51
Reference note: This question, including the terminology, was problem B1 on the 1996 Putnam.
– jmerry
Dec 8 at 7:51
Reference note: This question, including the terminology, was problem B1 on the 1996 Putnam.
– jmerry
Dec 8 at 7:51
add a comment |
2 Answers
2
active
oldest
votes
Your argument is correct.
Lets see if recursion helps.
Let $[n]$ denote the set ${1,2,ldots,n}$, and let $f_n$ denote the
number of minimal selfish subsets of $[n]$. Then the number of
minimal selfish subsets of $[n]$ not containing $n$ is equal to
$f_{n-1}$. On the other hand, for any minimal selfish subset of $[n]$
containing $n$, by subtracting 1 from each element, and then taking
away the element $n-1$ from the set, we obtain a minimal selfish
subset of $[n-2]$ (since $1$ and $n$ cannot both occur in a selfish
set). Conversely, any minimal selfish subset of $[n-2]$ gives rise to
a minimal selfish subset of $[n]$ containing $n$ by the inverse
procedure. Hence the number of minimal selfish subsets of $[n]$
containing $n$ is $f_{n-2}$. Thus we obtain $f_n=f_{n-1}+f_{n-2}$.
Since $f_1=f_2=1$, we have $f_n=F_n$, where $F_n$ denotes the $n$th
term of the Fibonacci sequence.
answered Dec 8 at 6:10
Rakesh Bhatt
851114
add a comment |
Your logic so far is fine. So what you know is that, since $c$ is in the set, then the other $c-1$ elements must all be at least $c+1$. There are $binom{n-c}{c-1}$ ways to choose them.
Summing over these gives you the total count. It turns out that this gives you the $n^{th}$ Fibonacci number, which you can prove by induction (hint: use Pascal’s identity).
answered Dec 8 at 6:08
platty
3,360320
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
Your argument is correct.
Lets see if recursion helps.
Let $[n]$ denote the set ${1,2,ldots,n}$, and let $f_n$ denote the
number of minimal selfish subsets of $[n]$. Then the number of
minimal selfish subsets of $[n]$ not containing $n$ is equal to
$f_{n-1}$. On the other hand, for any minimal selfish subset of $[n]$
containing $n$, by subtracting 1 from each element, and then taking
away the element $n-1$ from the set, we obtain a minimal selfish
subset of $[n-2]$ (since $1$ and $n$ cannot both occur in a selfish
set). Conversely, any minimal selfish subset of $[n-2]$ gives rise to
a minimal selfish subset of $[n]$ containing $n$ by the inverse
procedure. Hence the number of minimal selfish subsets of $[n]$
containing $n$ is $f_{n-2}$. Thus we obtain $f_n=f_{n-1}+f_{n-2}$.
Since $f_1=f_2=1$, we have $f_n=F_n$, where $F_n$ denotes the $n$th
term of the Fibonacci sequence.
answered Dec 8 at 6:10
Rakesh Bhatt
851114
add a comment |
Your argument is correct.
Lets see if recursion helps.
Let $[n]$ denote the set ${1,2,ldots,n}$, and let $f_n$ denote the
number of minimal selfish subsets of $[n]$. Then the number of
minimal selfish subsets of $[n]$ not containing $n$ is equal to
$f_{n-1}$. On the other hand, for any minimal selfish subset of $[n]$
containing $n$, by subtracting 1 from each element, and then taking
away the element $n-1$ from the set, we obtain a minimal selfish
subset of $[n-2]$ (since $1$ and $n$ cannot both occur in a selfish
set). Conversely, any minimal selfish subset of $[n-2]$ gives rise to
a minimal selfish subset of $[n]$ containing $n$ by the inverse
procedure. Hence the number of minimal selfish subsets of $[n]$
containing $n$ is $f_{n-2}$. Thus we obtain $f_n=f_{n-1}+f_{n-2}$.
Since $f_1=f_2=1$, we have $f_n=F_n$, where $F_n$ denotes the $n$th
term of the Fibonacci sequence.
answered Dec 8 at 6:10
Rakesh Bhatt
851114
add a comment |
Your argument is correct.
Lets see if recursion helps.
Let $[n]$ denote the set ${1,2,ldots,n}$, and let $f_n$ denote the
number of minimal selfish subsets of $[n]$. Then the number of
minimal selfish subsets of $[n]$ not containing $n$ is equal to
$f_{n-1}$. On the other hand, for any minimal selfish subset of $[n]$
containing $n$, by subtracting 1 from each element, and then taking
away the element $n-1$ from the set, we obtain a minimal selfish
subset of $[n-2]$ (since $1$ and $n$ cannot both occur in a selfish
set). Conversely, any minimal selfish subset of $[n-2]$ gives rise to
a minimal selfish subset of $[n]$ containing $n$ by the inverse
procedure. Hence the number of minimal selfish subsets of $[n]$
containing $n$ is $f_{n-2}$. Thus we obtain $f_n=f_{n-1}+f_{n-2}$.
Since $f_1=f_2=1$, we have $f_n=F_n$, where $F_n$ denotes the $n$th
term of the Fibonacci sequence.
answered Dec 8 at 6:10
Rakesh Bhatt
851114
Your argument is correct.
Lets see if recursion helps.
Let $[n]$ denote the set ${1,2,ldots,n}$, and let $f_n$ denote the
number of minimal selfish subsets of $[n]$. Then the number of
minimal selfish subsets of $[n]$ not containing $n$ is equal to
$f_{n-1}$. On the other hand, for any minimal selfish subset of $[n]$
containing $n$, by subtracting 1 from each element, and then taking
away the element $n-1$ from the set, we obtain a minimal selfish
subset of $[n-2]$ (since $1$ and $n$ cannot both occur in a selfish
set). Conversely, any minimal selfish subset of $[n-2]$ gives rise to
a minimal selfish subset of $[n]$ containing $n$ by the inverse
procedure. Hence the number of minimal selfish subsets of $[n]$
containing $n$ is $f_{n-2}$. Thus we obtain $f_n=f_{n-1}+f_{n-2}$.
Since $f_1=f_2=1$, we have $f_n=F_n$, where $F_n$ denotes the $n$th
term of the Fibonacci sequence.
answered Dec 8 at 6:10
Rakesh Bhatt
851114
answered Dec 8 at 6:10
Rakesh Bhatt
851114
answered Dec 8 at 6:10
Rakesh Bhatt
851114
answered Dec 8 at 6:10
Rakesh Bhatt
851114
851114
add a comment |
add a comment |
Your logic so far is fine. So what you know is that, since $c$ is in the set, then the other $c-1$ elements must all be at least $c+1$. There are $binom{n-c}{c-1}$ ways to choose them.
Summing over these gives you the total count. It turns out that this gives you the $n^{th}$ Fibonacci number, which you can prove by induction (hint: use Pascal’s identity).
answered Dec 8 at 6:08
platty
3,360320
add a comment |
Your logic so far is fine. So what you know is that, since $c$ is in the set, then the other $c-1$ elements must all be at least $c+1$. There are $binom{n-c}{c-1}$ ways to choose them.
Summing over these gives you the total count. It turns out that this gives you the $n^{th}$ Fibonacci number, which you can prove by induction (hint: use Pascal’s identity).
answered Dec 8 at 6:08
platty
3,360320
add a comment |
Your logic so far is fine. So what you know is that, since $c$ is in the set, then the other $c-1$ elements must all be at least $c+1$. There are $binom{n-c}{c-1}$ ways to choose them.
Summing over these gives you the total count. It turns out that this gives you the $n^{th}$ Fibonacci number, which you can prove by induction (hint: use Pascal’s identity).
answered Dec 8 at 6:08
platty
3,360320
Your logic so far is fine. So what you know is that, since $c$ is in the set, then the other $c-1$ elements must all be at least $c+1$. There are $binom{n-c}{c-1}$ ways to choose them.
Summing over these gives you the total count. It turns out that this gives you the $n^{th}$ Fibonacci number, which you can prove by induction (hint: use Pascal’s identity).
answered Dec 8 at 6:08
platty
3,360320
answered Dec 8 at 6:08
platty
3,360320
answered Dec 8 at 6:08
platty
3,360320
answered Dec 8 at 6:08
platty
3,360320
3,360320
add a comment |
add a comment |
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Reference note: This question, including the terminology, was problem B1 on the 1996 Putnam.
– jmerry
Dec 8 at 7:51