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By box, history, 4 months ago, In English

Consider the following problem (general form of original Min25 sieve):

We are given some multiplicative function $$$f(n)$$$, described at prime powers by $$$f(p^k)=g(p,k)$$$, where $$$g(p,1)$$$ is a polynomial of low degree, and we can calculate $$$g(p,k)$$$ for any singular $$$p$$$ and $$$k$$$ in $$$O(1)$$$.

How do we calculate the prefix sum of $$$f(n)$$$ at the distinct positive integer values of $$$\lfloor\frac Nk\rfloor$$$, where $$$N\le 10^{10}$$$? Clearly, brute force or regular sieves are not going to cut it here.


Prerequisite: For any integer $$$N$$$, for all integers $$$1\le k\le N$$$ there are at most $$$2\sqrt N$$$ distinct values of $$$\lfloor\frac Nk\rfloor$$$. Call these $$$2\sqrt N$$$ such values the blocks of $$$N$$$. (A note on notation: In the following text, $$$\frac Nk$$$ will be used to describe regular division, and $$$N/k$$$ will be used to denote floored division.)

Proof

Our workflow is going to be similar to this:

  1. We iterate over each nonzero exponent $$$T$$$ of $$$g(n,1)$$$. We first use the prefix sum of $$$x^T$$$ to estimate the contribution of $$$x^T$$$ to each prefix sum.
  2. Then, we sieve away composite values to ensure that our new prefix sum only counts prime $$$n$$$.
  3. After summing the result of the step 3 sieve over all $$$T$$$, we have obtained the prefix sum of $$$f(n)[n\in\mathbb P]$$$, where $$$\mathbb P$$$ is the set of primes.
  4. Then, we sieve back composite values to restore a correct answer.

The first step is done trivially by an interpolation of your choice.

The second step asks us to evaluate the following function at the blocks of $$$N$$$:

$$$S(n)=\sum\limits_{i=1}^n[i\in\mathbb P]i^T$$$

Let $$$p_1,p_2,\dots$$$ be the sequence of primes, $$$p_0=1$$$, and $$$\mathbb P(i)$$$ be the minimum prime factor of $$$i$$$.

Consider the following, modified, version of $$$S$$$, which "simulates" sieving away primes:

$$$S(n,j)=\sum\limits_{i=1}^n[i\in\mathbb P\lor\mathbb P(i)>p_j]i^T$$$

As $$$j$$$ tends to infinity, each $$$n$$$ satisfies $$$S(n)=S(n,j)$$$. In particular, when $$$p_j^2>n$$$, we have $$$S(n)=S(n,j)$$$, because if $$$x$$$ is composite its minimum prime factor must not be greater than $$$\sqrt x$$$! This means that we don't need to check all primes up to $$$N$$$, but instead only up to $$$\sqrt N$$$. Now, we have the edge case:

$$$S(n,0)=\sum\limits_{i=2}^ni^T$$$

The problem becomes: how do we convert $$$S(n,j-1)$$$ evaluated at the blocks of $$$N$$$ to $$$S(n,j)$$$ evaluated at the blocks of $$$N$$$? We effectively need to sieve away the composites with a minimum prime factor of exactly $$$p_j$$$.

In fact, the integers with a minimum prime factor of exactly $$$p_j$$$ are part of $$$p_j^TS(n/p_j,j-1)$$$, because this fixes at least one power of $$$p_j$$$, but unfortunately this includes primes less than $$$p_j$$$. So we simply subtract away them, and get what we need to sieve away: $$$p_j^T(S(n/p_j,j-1)-S(p_{j-1},j-1))$$$.

This gives us an recurrence:

$$$S(n,j)=S(n,j-1)-p_j^T(S(n/p_j,j-1)-S(p_{j-1},j-1))$$$

Note that $$$S(p_{j-1},j-1)$$$ is simply the sum of the $$$T$$$-th powers of the first $$$j-1$$$ primes, so this can be precalculated. We replace this with $$$W_{j-1}$$$:

$$$S(n,j)=S(n,j-1)-p_j^T(S(n/p_j,j-1)-W_{j-1})$$$

It is now apparent that the values of $$$S(n,j)$$$ at blocks of $$$N$$$ only depend on values of $$$S(n,j-1)$$$ at blocks of $$$N$$$; hence we maintain $$$S(n,j)$$$ at blocks of $$$N$$$ as we slowly increase $$$j$$$ until $$$p_j^2>N$$$.

What is the time complexity of this?

Observe that for some $$$j$$$, we only need to update values of $$$n$$$ where $$$p_j^2<n$$$, because at other positions, $$$S(n,j-1)$$$ is already a correct estimate of $$$S(n)$$$. (Without this optimization, the time complexity becomes $$$O(N/\log N)$$$.)

This means that the time complexity of this part is proportional to the sum of the number of primes less than $$$n$$$ for each $$$n$$$ in the blocks of $$$N$$$. Here it suffices to consider only the $$$n>\sqrt N$$$, as brute force works with the other half.

We have:

$$$\sum\limits_{k=1}^{\sqrt N}\pi(\sqrt{N/k})=\sum\limits_{k=1}^{\sqrt N}O(\frac{\sqrt{\frac Nk}}{\log\frac Nk})$$$
$$$=O(\frac{\int_{1}^{\sqrt N}\sqrt{\frac Nk}dk}{\log N})=O(\frac{N^{3/4}}{\log N})$$$

Now, we move on to the fourth step.

To sieve composite values back, we consider "reversing" the process we did in the second step. This suggests that we still use a function sieving based on the minimum prime factor. Define

$$$R(n,j)=\sum\limits_{i=2}^n[i\in\mathbb P\lor\mathbb P(i)\ge p_j]f(n)$$$

Our base case is now

$$$R(n,\pi(\sqrt n)+1)=S(n)$$$

Our problem becomes to transition from $$$j+1$$$ to $$$j$$$; that is, to sieve back in integers with a minimum prime factor of $$$p_j$$$.

The composite number in question, $$$x$$$, could fall into two categories.

  1. $$$x$$$ may have prime factors that are not $$$p_j$$$; in other words, $$$x=p_j^cy$$$ for some $$$1\le c$$$ and $$$p_j<\mathbb P(y)$$$.
  2. $$$x$$$ is a non-prime power of $$$p_j$$$; in other words, $$$x=p_j^c$$$ for some $$$1<c$$$. Note that, if $$$c=1$$$, then $$$x$$$ is not composite.

We count the contribution of all $$$c$$$ where $$$p_j^c\le n$$$. For the second type, we may count by evaluating $$$g(p_j,c)$$$ with brute force, so we care about deriving the value of the first type.

We must select $$$y$$$ such that $$$1\le y\le n/p_j^c$$$ and $$$\mathbb P(y)>p_j$$$. Like our calculation for the second step, this quantity is $$$R(n/p_j^c,j+1)-\text{included primes}=R(n/p_j^c,j+1)-S(p_j)$$$. If we aren't careful in our enumeration of $$$c$$$, it might be that $$$p_j>n/p_j^c$$$, so we take $$$\max(0,R(n/p_j^c)-S(p_j))$$$.

Remember that $$$f$$$ is multiplicative, so to account for the power $$$p_j^c$$$, we just multiply this by $$$g(p_j,c)$$$.

To summarize what we are sieving in:

$$$\sum\limits_{c=1,p_m^c\le n}g(p_j,c)\max(0,R(n/p_j^c)-S(p_j))+\sum\limits_{c=2,p_m^c\le n}g(p_j,c)$$$

Observe that for the first summation, $$$c$$$ contributes if and only if $$$p_m^{c+1}\le n$$$, so we can rewrite the summation and remove the maximum operand — to get a full recurrence:

$$$R(n,j)=R(n,j+1)+\sum\limits_{c=1,p_m^{c+1}\le n}g(p_j,c)(R(n/p_j^c)-S(p_j))+g(p_j,c+1)$$$

Iterating $$$j$$$ from $$$\pi(\sqrt n)+1$$$ to $$$0$$$, our answers are at $$$R(n,0)$$$.

The analysis of the time complexity of this part is practically the same as that of the first part.


Example: SPOJ ASSIEVE

(Yes, in fact this entire blog post was to advertise this problem of mine.)

First let's weaken the problem a bit: How do we calculate

$$$g(n,k,r)=\sum\limits_{i=1}^ni^k[L(i)\equiv r\pmod 4]$$$

where $$$L(i)$$$ is the integer log function, equal to the sum of the prime divisors of $$$i$$$, counting repetition?

At first it is not obvious at all how this is a multiplicative function. However, the Min25 sieve is not bound to functions in the integers. It works perfectly fine for polynomials too!

Consider the multiplicative function $$$\mathfrak L(i)$$$, defined by $$$\mathfrak L(p^c)=p^{ck}x^{(pc)\bmod 4}$$$, operating on $$$\mathbb Z/p\mathbb Z[x]/(x^4-1)$$$. This happens to be the cyclical convolution ring of polynomials of degree $$$4$$$ — in other words, convolution here is defined so that $$$x^{i+j}$$$ overflows to $$$x^{(i+j)\bmod 4}$$$. This is exactly what we need — addition modulo $$$4$$$!

Unfortunately, this function does not seem to satisfy an important requirement of the sieve: $$$p^kx^{p\bmod 4}$$$ is not a polynomial, nor a fully multiplicative function. A first instinct would be to do FFT/NTT — but this does not help either.

The solution would be to completely circumvent the first step. Instead, we take the true meaning of the function $$$p^kx^{p\mod 4}$$$. To find the prefix sum of this at $$$n$$$, what we really want is for each $$$z=0,1,2,3$$$, the sum of the $$$k$$$-th powers of all primes less than $$$n$$$, where said prime modulo $$$4$$$ is equivalent to $$$z$$$.

Surprisingly, this is doable.

To formalize, we need, at the blocks of $$$N$$$:

$$$z=0,1,2,3:\sum\limits_{i=1}^{n}[i\in\mathbb P\land i\equiv z\pmod 4]i^k$$$

Observe that for $$$z=0$$$ the answer is always $$$0$$$, and for $$$z=2$$$ the answer is $$$[2\le n]2^k$$$. Consider how to sieve away the answers for $$$z=1$$$ and $$$z=3$$$ in parallel.

Let us recall the recurrence for the second step of the sieve:

$$$S(n,j)=S(n,j-1)-p_j^T(S(n/p_j,j-1)-W_{j-1})$$$

In order to incorporate $$$z$$$, we need to augment this into a polynomial, with nonzero coefficients of $$$x^1$$$ and $$$x^3$$$. Let the current prime in question be $$$p=p_j$$$, and let the previous prime be $$$p'=p_{j-1}$$$. We want

$$$[x^{z|z\in\{1,3\}}]S(n,j)=\sum\limits_{i=1}^n[i\in\mathbb P\lor\mathbb P(i)>p_j][i\equiv z\pmod 4]i^k$$$

Remember that from $$$p'$$$ to $$$p$$$, we sieve away the composites with a minimum prime factor of $$$p$$$, so these composites can all be expressed in the form $$$pd$$$. Now, we can do casework on $$$p$$$ modulo $$$4$$$. In the following, $$$p,pd,d$$$ are all expressed modulo $$$4$$$.

  1. If $$$p=1$$$, then:
    1. For $$$pd=1$$$, we must have $$$d=1$$$. Hence we seive away $$$p^k[x^1](S(n/p_j,j-1)-W_{j-1})$$$ from $$$[x^1]S(n,j)$$$.
    2. For $$$pd=3$$$, we must have $$$d=3$$$. Hence we seive away $$$p^k[x^3](S(n/p_j,j-1)-W_{j-1})$$$ from $$$[x^3]S(n,j)$$$.
  2. If $$$p=3$$$, then:
    1. For $$$pd=1$$$, we must have $$$d=3$$$. Hence we seive away $$$p^k[x^3](S(n/p_j,j-1)-W_{j-1})$$$ from $$$[x^1]S(n,j)$$$.
    2. For $$$pd=3$$$, we must have $$$d=1$$$. Hence we seive away $$$p^k[x^1](S(n/p_j,j-1)-W_{j-1})$$$ from $$$[x^3]S(n,j)$$$.

To calculate $$$S(n,0)$$$, use any polynomial interpolation algorithm.

This recurrence now allows us to successfully complete the second step of sieving. The fourth step is virtually the same, except we operate in $$$\mathbb Z/p\mathbb Z[x]/(x^4-1)$$$.

Notice that the given function at primes has not changed in the original problem. We only need to modify the multiplication operation in the fourth step to complete our solution.


I hope you enjoyed this blog! If you have any interesting problems that can be solved with this sieve, let me know.

Have a nice day :)

 
 
 
 
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5 weeks ago, # |
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Is this a variation on the derivation of prime-counting (Meissel-Lehmer) algorithm?

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    5 weeks ago, # ^ |
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    I guess it counts as an extension. Meissel-Lehmer is a closely related algorithm that doesn't need step 3 and step 4, making it more amenable to optimizations.

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      4 weeks ago, # ^ |
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      I never worked through the math myself, I just used someone else's implementation. Eventually I will go through rhe math.

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5 weeks ago, # |
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It's a standard trick in this type of number theory harmonic-lemma algorithm to use the brute-force technique all the way up to $$$\tilde{O}(n^{2/3})$$$ instead of only up to $$$\sqrt{n}$$$. This results in a slightly better time complexity, $$$\tilde{O}(n^{2/3})$$$ instead of $$$\tilde{O}(n^{3/4})$$$.