Computing Cardinality of Sumsets using Convolutions and FFT
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I read somewhere that you can compute the cardinality of sumsets by computing the convolutions of the characteristic vectors of the given sets, and that this can be done efficiently using Fast Fourier Transform. How would this work?
algorithms
asked Feb 25 at 4:28
Arnaud AvondetArnaud Avondet
83
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add a comment |
$begingroup$
I read somewhere that you can compute the cardinality of sumsets by computing the convolutions of the characteristic vectors of the given sets, and that this can be done efficiently using Fast Fourier Transform. How would this work?
algorithms
asked Feb 25 at 4:28
Arnaud AvondetArnaud Avondet
83
$endgroup$
add a comment |
$begingroup$
I read somewhere that you can compute the cardinality of sumsets by computing the convolutions of the characteristic vectors of the given sets, and that this can be done efficiently using Fast Fourier Transform. How would this work?
algorithms
asked Feb 25 at 4:28
Arnaud AvondetArnaud Avondet
83
$endgroup$
I read somewhere that you can compute the cardinality of sumsets by computing the convolutions of the characteristic vectors of the given sets, and that this can be done efficiently using Fast Fourier Transform. How would this work?
algorithms
algorithms
asked Feb 25 at 4:28
Arnaud AvondetArnaud Avondet
83
asked Feb 25 at 4:28
Arnaud AvondetArnaud Avondet
83
asked Feb 25 at 4:28
Arnaud AvondetArnaud Avondet
83
asked Feb 25 at 4:28
Arnaud AvondetArnaud Avondet
83
asked Feb 25 at 4:28
Arnaud AvondetArnaud Avondet
83
83
add a comment |
add a comment |
1 Answer
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Here is a sketch of the main ideas. Let $S,T subseteq mathbb{N}$ be two multisets of non-negative integers, and define $S+T$ to be the multiset
$$S+T = {s+t mid s in S, t in T}.$$
Let $chi_S$ represent the characteristic vector of a set $S$. Then
$$chi_{S+T} = chi_S * chi_T,$$
where $*$ is a convolution operator.
The Fourier transform $mathcal{F}$ has the property that
$$mathcal{F}(f*g) = mathcal{F}(f) times mathcal{F}(g),$$
where $times$ represents pointwise multiplication of functions. Therefore,
$$mathcal{F}(chi_{S+T}) = mathcal{F}(chi_S) times mathcal{F}(chi_T),$$
and in particular,
$$chi_{S+T} = mathcal{F}^{-1}(mathcal{F}(chi_S) times mathcal{F}(chi_T)).$$
This gives us a method to compute $S+T$. We can compute the characteristic vector $chi_{S+T}$ for the multiset $S+T$ by computing $mathcal{F}(chi_S)$, the Fourier transform of the characteristic vector for $S$, and $mathcal{F}(g)$; multiplying them; and then applying the inverse Fourier transform to the result. Each of these steps can be implemented efficiently using the Fast Fourier transform. The end result will be $chi_{S+T}$, from which we can reconstruct $S+T$ or its cardinality.
The overall running time will be $O(n log n)$, where $n$ is an upper bound on the largest element of $S,T$.
answered Feb 25 at 5:52
D.W.♦D.W.
102k12129293
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1 Answer
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1 Answer
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$begingroup$
Here is a sketch of the main ideas. Let $S,T subseteq mathbb{N}$ be two multisets of non-negative integers, and define $S+T$ to be the multiset
$$S+T = {s+t mid s in S, t in T}.$$
Let $chi_S$ represent the characteristic vector of a set $S$. Then
$$chi_{S+T} = chi_S * chi_T,$$
where $*$ is a convolution operator.
The Fourier transform $mathcal{F}$ has the property that
$$mathcal{F}(f*g) = mathcal{F}(f) times mathcal{F}(g),$$
where $times$ represents pointwise multiplication of functions. Therefore,
$$mathcal{F}(chi_{S+T}) = mathcal{F}(chi_S) times mathcal{F}(chi_T),$$
and in particular,
$$chi_{S+T} = mathcal{F}^{-1}(mathcal{F}(chi_S) times mathcal{F}(chi_T)).$$
This gives us a method to compute $S+T$. We can compute the characteristic vector $chi_{S+T}$ for the multiset $S+T$ by computing $mathcal{F}(chi_S)$, the Fourier transform of the characteristic vector for $S$, and $mathcal{F}(g)$; multiplying them; and then applying the inverse Fourier transform to the result. Each of these steps can be implemented efficiently using the Fast Fourier transform. The end result will be $chi_{S+T}$, from which we can reconstruct $S+T$ or its cardinality.
The overall running time will be $O(n log n)$, where $n$ is an upper bound on the largest element of $S,T$.
answered Feb 25 at 5:52
D.W.♦D.W.
102k12129293
$endgroup$
add a comment |
$begingroup$
Here is a sketch of the main ideas. Let $S,T subseteq mathbb{N}$ be two multisets of non-negative integers, and define $S+T$ to be the multiset
$$S+T = {s+t mid s in S, t in T}.$$
Let $chi_S$ represent the characteristic vector of a set $S$. Then
$$chi_{S+T} = chi_S * chi_T,$$
where $*$ is a convolution operator.
The Fourier transform $mathcal{F}$ has the property that
$$mathcal{F}(f*g) = mathcal{F}(f) times mathcal{F}(g),$$
where $times$ represents pointwise multiplication of functions. Therefore,
$$mathcal{F}(chi_{S+T}) = mathcal{F}(chi_S) times mathcal{F}(chi_T),$$
and in particular,
$$chi_{S+T} = mathcal{F}^{-1}(mathcal{F}(chi_S) times mathcal{F}(chi_T)).$$
This gives us a method to compute $S+T$. We can compute the characteristic vector $chi_{S+T}$ for the multiset $S+T$ by computing $mathcal{F}(chi_S)$, the Fourier transform of the characteristic vector for $S$, and $mathcal{F}(g)$; multiplying them; and then applying the inverse Fourier transform to the result. Each of these steps can be implemented efficiently using the Fast Fourier transform. The end result will be $chi_{S+T}$, from which we can reconstruct $S+T$ or its cardinality.
The overall running time will be $O(n log n)$, where $n$ is an upper bound on the largest element of $S,T$.
answered Feb 25 at 5:52
D.W.♦D.W.
102k12129293
$endgroup$
add a comment |
$begingroup$
Here is a sketch of the main ideas. Let $S,T subseteq mathbb{N}$ be two multisets of non-negative integers, and define $S+T$ to be the multiset
$$S+T = {s+t mid s in S, t in T}.$$
Let $chi_S$ represent the characteristic vector of a set $S$. Then
$$chi_{S+T} = chi_S * chi_T,$$
where $*$ is a convolution operator.
The Fourier transform $mathcal{F}$ has the property that
$$mathcal{F}(f*g) = mathcal{F}(f) times mathcal{F}(g),$$
where $times$ represents pointwise multiplication of functions. Therefore,
$$mathcal{F}(chi_{S+T}) = mathcal{F}(chi_S) times mathcal{F}(chi_T),$$
and in particular,
$$chi_{S+T} = mathcal{F}^{-1}(mathcal{F}(chi_S) times mathcal{F}(chi_T)).$$
This gives us a method to compute $S+T$. We can compute the characteristic vector $chi_{S+T}$ for the multiset $S+T$ by computing $mathcal{F}(chi_S)$, the Fourier transform of the characteristic vector for $S$, and $mathcal{F}(g)$; multiplying them; and then applying the inverse Fourier transform to the result. Each of these steps can be implemented efficiently using the Fast Fourier transform. The end result will be $chi_{S+T}$, from which we can reconstruct $S+T$ or its cardinality.
The overall running time will be $O(n log n)$, where $n$ is an upper bound on the largest element of $S,T$.
answered Feb 25 at 5:52
D.W.♦D.W.
102k12129293
$endgroup$
Here is a sketch of the main ideas. Let $S,T subseteq mathbb{N}$ be two multisets of non-negative integers, and define $S+T$ to be the multiset
$$S+T = {s+t mid s in S, t in T}.$$
Let $chi_S$ represent the characteristic vector of a set $S$. Then
$$chi_{S+T} = chi_S * chi_T,$$
where $*$ is a convolution operator.
The Fourier transform $mathcal{F}$ has the property that
$$mathcal{F}(f*g) = mathcal{F}(f) times mathcal{F}(g),$$
where $times$ represents pointwise multiplication of functions. Therefore,
$$mathcal{F}(chi_{S+T}) = mathcal{F}(chi_S) times mathcal{F}(chi_T),$$
and in particular,
$$chi_{S+T} = mathcal{F}^{-1}(mathcal{F}(chi_S) times mathcal{F}(chi_T)).$$
This gives us a method to compute $S+T$. We can compute the characteristic vector $chi_{S+T}$ for the multiset $S+T$ by computing $mathcal{F}(chi_S)$, the Fourier transform of the characteristic vector for $S$, and $mathcal{F}(g)$; multiplying them; and then applying the inverse Fourier transform to the result. Each of these steps can be implemented efficiently using the Fast Fourier transform. The end result will be $chi_{S+T}$, from which we can reconstruct $S+T$ or its cardinality.
The overall running time will be $O(n log n)$, where $n$ is an upper bound on the largest element of $S,T$.
answered Feb 25 at 5:52
D.W.♦D.W.
102k12129293
answered Feb 25 at 5:52
D.W.♦D.W.
102k12129293
answered Feb 25 at 5:52
D.W.♦D.W.
102k12129293
answered Feb 25 at 5:52
D.W.♦D.W.
102k12129293
102k12129293
add a comment |
add a comment |
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