Mixing
What is the complexity of subset of the power set of a set
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The complexity is $2^n$, if I have a set $B$ and want to find the power set $mathcal{P}(B)$.
But what if I want to find only sets of max size $i$ of $mathcal{P}(B)$? So a subset of $mathcal{P}(B)$ where there only exist sets with sizes less or equal to $i$
Something points in the direction that the complexity is $n^i$ but I have trouble understanding why.
combinatorics discrete-mathematics
asked Jan 3 at 8:22
Adam SoelAdam Soel
32
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add a comment |
$begingroup$
The complexity is $2^n$, if I have a set $B$ and want to find the power set $mathcal{P}(B)$.
But what if I want to find only sets of max size $i$ of $mathcal{P}(B)$? So a subset of $mathcal{P}(B)$ where there only exist sets with sizes less or equal to $i$
Something points in the direction that the complexity is $n^i$ but I have trouble understanding why.
combinatorics discrete-mathematics
asked Jan 3 at 8:22
Adam SoelAdam Soel
32
$endgroup$
add a comment |
$begingroup$
The complexity is $2^n$, if I have a set $B$ and want to find the power set $mathcal{P}(B)$.
But what if I want to find only sets of max size $i$ of $mathcal{P}(B)$? So a subset of $mathcal{P}(B)$ where there only exist sets with sizes less or equal to $i$
Something points in the direction that the complexity is $n^i$ but I have trouble understanding why.
combinatorics discrete-mathematics
asked Jan 3 at 8:22
Adam SoelAdam Soel
32
$endgroup$
The complexity is $2^n$, if I have a set $B$ and want to find the power set $mathcal{P}(B)$.
But what if I want to find only sets of max size $i$ of $mathcal{P}(B)$? So a subset of $mathcal{P}(B)$ where there only exist sets with sizes less or equal to $i$
Something points in the direction that the complexity is $n^i$ but I have trouble understanding why.
combinatorics discrete-mathematics
combinatorics discrete-mathematics
asked Jan 3 at 8:22
Adam SoelAdam Soel
32
asked Jan 3 at 8:22
Adam SoelAdam Soel
32
edited Jan 3 at 8:37
Adam Soel
asked Jan 3 at 8:22
Adam SoelAdam Soel
32
asked Jan 3 at 8:22
Adam SoelAdam Soel
32
asked Jan 3 at 8:22
Adam SoelAdam Soel
32
32
add a comment |
add a comment |
1 Answer
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$begingroup$
Provided that the cardinality of a finite set $mathbf B$ is $n$ the number of subsets of $mathbf B$ with cardinality $j$ equals $binom{n}{j}$, so the number of subset with cardinality that does not exceed $i$ equals:$$sum_{j=0}^ibinom{n}j$$
There is no closed form of this expression.
Substituting $i=n$ observe that $sum_{j=0}^nbinom{n}j=2^n$ which is in accordance with your observation that the cardinality of $wp(mathbf B)$ equals $2^n$.
answered Jan 3 at 8:35
drhabdrhab
98.9k544130
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1 Answer
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$begingroup$
Provided that the cardinality of a finite set $mathbf B$ is $n$ the number of subsets of $mathbf B$ with cardinality $j$ equals $binom{n}{j}$, so the number of subset with cardinality that does not exceed $i$ equals:$$sum_{j=0}^ibinom{n}j$$
There is no closed form of this expression.
Substituting $i=n$ observe that $sum_{j=0}^nbinom{n}j=2^n$ which is in accordance with your observation that the cardinality of $wp(mathbf B)$ equals $2^n$.
answered Jan 3 at 8:35
drhabdrhab
98.9k544130
$endgroup$
add a comment |
$begingroup$
Provided that the cardinality of a finite set $mathbf B$ is $n$ the number of subsets of $mathbf B$ with cardinality $j$ equals $binom{n}{j}$, so the number of subset with cardinality that does not exceed $i$ equals:$$sum_{j=0}^ibinom{n}j$$
There is no closed form of this expression.
Substituting $i=n$ observe that $sum_{j=0}^nbinom{n}j=2^n$ which is in accordance with your observation that the cardinality of $wp(mathbf B)$ equals $2^n$.
answered Jan 3 at 8:35
drhabdrhab
98.9k544130
$endgroup$
add a comment |
$begingroup$
Provided that the cardinality of a finite set $mathbf B$ is $n$ the number of subsets of $mathbf B$ with cardinality $j$ equals $binom{n}{j}$, so the number of subset with cardinality that does not exceed $i$ equals:$$sum_{j=0}^ibinom{n}j$$
There is no closed form of this expression.
Substituting $i=n$ observe that $sum_{j=0}^nbinom{n}j=2^n$ which is in accordance with your observation that the cardinality of $wp(mathbf B)$ equals $2^n$.
answered Jan 3 at 8:35
drhabdrhab
98.9k544130
$endgroup$
Provided that the cardinality of a finite set $mathbf B$ is $n$ the number of subsets of $mathbf B$ with cardinality $j$ equals $binom{n}{j}$, so the number of subset with cardinality that does not exceed $i$ equals:$$sum_{j=0}^ibinom{n}j$$
There is no closed form of this expression.
Substituting $i=n$ observe that $sum_{j=0}^nbinom{n}j=2^n$ which is in accordance with your observation that the cardinality of $wp(mathbf B)$ equals $2^n$.
answered Jan 3 at 8:35
drhabdrhab
98.9k544130
answered Jan 3 at 8:35
drhabdrhab
98.9k544130
answered Jan 3 at 8:35
drhabdrhab
98.9k544130
answered Jan 3 at 8:35
drhabdrhab
98.9k544130
98.9k544130
add a comment |
add a comment |
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