FFT리스

가정 우리는 주어진 다른 정수 1 , 2 , ... , N ,되도록 0 ≤ I ≤ K N 일부에 대한 상수 (K) > 0 , 및 모든 I .$n$$a_1, a_2, \dots, a_n$$0 \le a_i \le kn$$k \gt 0$$i$

우리는 가능한 모든 페어 와이즈 합계 의 개수를 찾는 데 관심이 있습니다 . ( 나는 = $S_{ij} = a_i + a_j$$i = j$ 가 허용됨).

한 알고리즘은 다항식 구성하는 정도 ≤ K , N , 및 푸리에 변환 방법 및 생성 다항식들은 계수들을 갖는 힘을 판독하여 그 사각형을 계산한다. 이것은 O ( n log n ) 시간 알고리즘입니다.$P(x) = \sum_{j=1}^{n} x^{a_j}$$\le kn$$O(n \log n)$

두 가지 질문이 있습니다.

• O ( n log n ) 가 있습니까$O(n \log n)$ FFT를 사용하지 않는 알고리즘은?

• 더 나은 알고리즘 (예 알려져있다 )? (FFT 허용).$o(n \log n)$

이 문제는 정수 / 다항식 제곱과 같습니다.

1. 다항식 곱셈 정수 곱셈에 해당하는 것으로 알려져있다 .

2. 당신은 이미 다항식 / 정수 제곱 법으로 문제를 줄였습니다. 그러므로이 문제는 거의 제곱만큼 어렵습니다.

이제이 문제에 대한 정수 제곱을 줄입니다.

알고리즘이 있다고 가정하십시오.

$$F(\mathbf{\vec a})\rightarrow P^2(x),\\ \text{where } P(x)=\sum_{a_i \in \mathbf{\vec a}} x^{a_i}$$

이 알고리즘은 본질적으로 질문에 요청한 알고리즘입니다. 따라서이 작업을 수행 할 수있는 마법 알고리즘이 있다면 정수 y 를 제곱 하는 함수 를 만들 수 있습니다 ( 오, 예, mathjax : P ).${\rm S{\small QUARE}}\left(y\right)$$y$

$$\begin{array}{rlr}\hline &\mathbf{\text{Algorithm 1}} \text{ Squaring}&\hspace{2em}&\\ \hline {\tiny 1.:}&\mathbf {\text{procedure }} {\rm S{\small QUARE}}\left(y\right):\\ {\tiny 2.:}&\hspace{2em}\mathbf {\vec a} \leftarrow () &&\triangleright~\mathbf {\vec a}\text{ starts as empty polynomial sequence}\\ {\tiny 3.:}&\hspace{2em}i \leftarrow 0\\ {\tiny 4.:}&\hspace{2em}\mathbf{while}~y\ne0~\mathbf{do} &&\triangleright~\text{break }y\text{ down into a polynomial of base }2\\ {\tiny 5.:}&\hspace{4em}\mathbf{if}~y~\&~1~\mathbf{then} &&\triangleright~\text{if lsb of }y\text{ is set}\\ {\tiny 6.:}&\hspace{6em}\mathbf{\vec a} \leftarrow \mathbf{\vec a}i &&\triangleright~\text{append }i\text{ to }\mathbf{\vec a}~\text{(appending }x^i\text{)}\\ {\tiny 7.:}&\hspace{4em}\mathbf{end~if}\\ {\tiny 8.:}&\hspace{4em}i \leftarrow i + 1\\ {\tiny 9.:}&\hspace{4em}y \leftarrow y \gg 1 &&\triangleright~\text{shift }y\text{ right by one}\\ {\tiny 10.:}&\hspace{2em}\mathbf{end~while}\\ {\tiny 11.:}&\hspace{2em}P^2(x) \leftarrow F\left(\mathbf{\vec a}\right) &&\triangleright~\text{obtain the squared polynomial via } F\left(\mathbf{\vec a}\right)\\ {\tiny 12.:}&\hspace{2em}\mathbf{return}~P^2(2) &&\triangleright~\text{simply sum up the polynomial}\\ {\tiny 13.:}&\mathbf {\text{end procedure}}\\ \hline &\end{array}$$

Python (test with codepad):

#/cs//q/11418/2755

def F(a):
    n = len(a)
    for i in range(n):
        assert a[i] >= 0

    # (r) => coefficient
    # coefficient \cdot x^{r}
    S = {}
    for ai in a:
        for aj in a:
            r = ai + aj

            if r not in S:
                S[r] = 0

            S[r] += 1

    return list(S.items())

def SQUARE(x):
    x = int(x)

    a = []
    i = 0
    while x != 0:
        if x & 1 == 1:
            a += [i]
        x >>= 1
        i += 1

    print 'a:',a
    P2 = F(a)

    print 'P^2:',P2

    s = 0
    for e,c in P2:
        s += (1 << e)*c
    return s

3. Thus, squaring is at most as hard as this problem.

4. Therefore, integer squaring is equivalent to this problem. (they are each at most as hard as each-other, due to (2,3,1))

Now it is unknown if integer/polynomial multiplication admits bounds better than $\mathcal O(n\log n)$; in fact the best multiplication algorithms currently all use FFT and have run-times like $\mathcal O(n \log n \log \log n)$ (Schönhage-Strassen algorithm) and $\mathcal O\left(n \log n\,2^{\mathcal O(\log^* n)}\right)$ (Fürer's algorithm). Arnold Schönhage and Volker Strassen conjectured a lower bound of $\Omega\left(n \log n\right)$, and so far this seems to be holding.

This doesn't mean your use of FFT is quicker; $\mathcal O\left(n\log n\right)$ for FFT is the number of operations (I think), not the bit complexity; hence it ignores some factors of smaller multiplications; when used recursively, it would become closer to the FFT multiplication algorithms listed above (see Where is the mistake in this apparently-O(n lg n) multiplication algorithm?).

5. Now, your problem is not exactly multiplication, it is squaring. So is squaring easier? Well, it is an open problem (no for now): squaring is not known to have a faster algorithm than multiplication. If you could find a better algorithm for your problem than using multiplication; then this would likely be a breakthrough.

So as of now, the answer to both your questions is: no, as of now, all the ~$\mathcal O(n\log n)$ multiplication algorithms use FFT; and as of now squaring is as hard as multiplication. And no, unless a faster algorithm for squaring is found, or multiplication breaks the $\mathcal O(n\log n)$ barrier, your problem cannot be solved faster than $\mathcal O(n \log n)$; in fact, it cannot currently be solved in $\mathcal O(n\log n)$ either, as the best multiplication algorithm only approaches that complexity.