Is it possible to compute the first $n$ values of the Liouville function in linear time? Since we need to output $n$ values we clearly cannot do better than linear time, but the best I can figure out is something like $O(n \cdot \log{\log{n}})$: fill an array of size $n$ with ones, and for each prime power $p^a$, negate the value at the index of each of its multiples. I think it is possible to identify the prime powers as we count up using another table of $O(n)$ bits, essentially the sieve of Eratosthenes counting powers too, but there are still $\sum_{p^a \le n}{\frac{n}{p^a}} = n \cdot \log{\log{n}} + O(n)$ negation operations. Is it possible to do better than this?
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Dan Brumleve
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$\log \log n$ grows slowly enough that $n \log\log n$ is essentially linear. – marty cohen Dec 15 '15 at 05:04
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I know it doesn't matter for any practical purposes but what I'm curious about is whether it is really linear. An idea I had is that for any $x \gt 0$ almost all of the work takes place while flipping the bits of multiples of the first $n^x$ primes, but I'm not sure how to use that. – Dan Brumleve Dec 15 '15 at 05:07
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According to http://math.stackexchange.com/questions/84040/does-dtimeon-regular, on a single-tape Turing machine, $DTIME(o(n \cdot \log{n})) = REGULAR$. I believe the set of all finite prefixes of the sequence of values of the Liouville function cannot described by a regular expression, any use of * would admit an eventually periodic set. Does that prove it can't be done in $o(n \cdot \log{n})$ time on a single-tape TM? – Dan Brumleve Dec 22 '15 at 07:58
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I'm not sure it works. If you have a TM machine which generates Louville sequence prefix up to length $n$ in $o(n \log n)$ time, how could you generate a TM machine which checks if a given sequence of length $n$ is appropriate prefix in same time? Is it possible to compare two strings in $O(n)$ time on a single-tape machine? It is possible to do it on a multi-tape machine of course. – stgatilov Dec 22 '15 at 17:33
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If we expand the tape alphabet to a square of the size of the generator's we can do it: use the low state in each cell to supply the input, then run the generator modified to only operate on the high states, then pass through the tape verifying that the low state and high state in each cell are equal. This idea doesn't help us compare two strings that are concatenated on the tape, we have to be able to overlay them. – Dan Brumleve Dec 23 '15 at 04:50
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Ok, then you can surely create a "recognizer" machine with $o(n \log n)$ work time for the set of prefixes of your sequence -> contradiction, i.e. it is not possible to compute your sequence in linear time on single-tape TM machine. But anyway, this is only a "yet another evidence" that the model of Turing Machine seems rather useless when talking about time complexities like linear/quadratic. It is often used for polynomial time complexity of course, because it is not so awfully slow to turn polynomial algorithms into non-polynomial ones =) – stgatilov Dec 23 '15 at 18:08
4 Answers
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Your problem is related to the problem of finding all the primes up to $n$, and it can be solved by similar means.
Although the traditional sieve of Eratosthenes has complexity $O(n \log \log n)$, there are improved versions, which work in $O(n)$ time. It is achieved by crossing out each composite number exactly once. For the purpose of finding primes, these algorithms can even be optimized to $O(n /\log \log n)$ time by the so-called wheel optimization. You can find details in e.g. this paper.
In order to solve your problem, you have to calculate the least prime factor function (denoted by $lpf$) during sieving, not only the primes. When the $lpf$ function is ready, you can compute completely multiplicative functions in $O(n)$ easily by dynamic programming, e.g.: $$ \lambda(x) = \begin{cases} -1 & x = lpf(x) \\ -\lambda \left( \frac{x}{lpf(x)} \right) & \text{otherwise} \end{cases} $$
One of the algorithms which calculates least prime factors in $O(n)$ time is described here. Below you can see its implementation in C++, with primes being a sorted list of all primes found so far, and lpf being an array representing the same-named function.
vector<int> primes;
vector<int> lpf(n, -1);
for (int x = 2; x < n; x++) {
if (lpf[x] < 0) { //prime found
lpf[x] = x;
primes.push_back(x);
}
for (int i = 0; i < primes.size(); i++) {
int p1 = primes[i], y = p1 * x;
if (p1 > lpf[x] || y >= n)
break;
lpf[y] = p1;
}
}
Consider a number $x$ having least prime factor $p = lpf(x)$. For each prime $p_1 \le p$, it is easy to see that $lpf(p_1 \cdot x) = p_1$. The algorithm simply applies this crossing-out rule for each number $x$ in increasing order. Of course, it needs the sorted list of primes in order to iterate over all the necessary primes $p_1$.
Every composite number $y$ is crossed out exactly once when considering number $x = \frac{y}{lpf(y)}$, so time complexity is $O(n)$. In practice however, it is slower than the traditional $O(n \log \log n)$ sieve, at least for the purpose of finding primes. I guess your approach would also be faster, especially with bitsets.
If you are interested in practical acceleration of your algorithm, you'd better think about memory/cache optimization, instead of improving asymptotic complexity by a double-log factor =)
stgatilov
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This is great, however it occurs to me that the analysis depends on the model of computation. On a multi-tape Turing machine, the y=p1*x calculation will take more than constant time, as will the x/lpf(x) calculation performed later. Of course we could be using a model of computation where arithmetic operations are constant time by definition (and I essentially implied this model in my question by talking about counting such operations) and in that case this algorithm is linear. Am I understanding this right? – Dan Brumleve Dec 20 '15 at 04:11
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Basically I'm still wondering if it is a plausible conjecture that this problem requires superlinear time on a TM. – Dan Brumleve Dec 20 '15 at 04:19
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@DanBrumleve: I always thought that random memory access is not present on Turing machine, so e.g. sieve of Erathosthenes would take about $O \left( \frac{n \sqrt{n}}{\log{n}} \right)$ time on it (moving through the whole tape for each prime). Please give a link to the definition of the machine you mean, so that I could probably optimize the problems out. P.S. By the way, random memory access would eat more time than division and multiplication on real computers. – stgatilov Dec 20 '15 at 06:55
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Well, random memory access is not present on Turing machines although I believe we can get closer to it on multi-tapes, I guess I am not sure what machine I really mean, maybe the RAM model? Your answer is quite an improvement over what I had in mind and now I am curious about in which models it is linear and which it may not be; I am not as familiar as I would like to be with the terminology and distinctions between computational models. I'm not interested in actually implementing this algorithm, I am just trying to explore the complexity theory aspects. Thanks for all your help! – Dan Brumleve Dec 20 '15 at 07:01
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1@DanBrumleve: My solution assumes that numbers up to $n$ fit into a machine word of $w$ bits, and all arithmetic operations on these words take $O(1)$ time. Perhaps you are asking about the case when the machine operates on bit level, i.e. $w = 1$ and only logical operations on bits take $O(1)$ time. Then traditional sieve of Erathosthenes takes $O(n \log n \log \log n)$ time, since $O(\log n)$ is spent on each addition. – stgatilov Dec 20 '15 at 07:06
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The number $N$ can not have more than one prime divisor $p$ that $p^2>N$. Such a divisor there, when $$N>\prod\limits_{\substack{p | N,\\ p^2\leq N}}p^a$$ This allows, similarly to the algorithm "Eratosthenes Sieve" (ES), to limit the enumeration prime divisors of N with ratio $p^2 \leq N$.
Differences algorithm compared with ES as follows:
1. The processing of each factor changes the sign of each corresponding $\lambda$, and each corresponding product multiplied by a factor $p$ (in ES - only the number zero).
2. In addition to the prime factors, involved the same processing passage of their degree $p^a\geq N$.
These changes complicate the algorithm proportionally, but do not increase the level of its complexity.
The proposed software implementation (PHP) pre-computed array primes dimension inter = 46340 (the SE algorithm), that allows to calculate the Liouville function for an arbitrary piece of data (val_from, val_to) in the range from 1 to maxval = 2147483647.
The test calculations for 1 000 000 points on the the most costly piece (maxval - 999999, maxval) on weak computer took 20 seconds.
In my opinion, such an algorithm is actually better..
The prorgam is:
ini_set('memory_limit','256M');
$maxint = 2147483647;
$inter = 46340; // =[sqrt($maxint)]
$val_from = 1; //$maxint- 999;
$val_to = 1000; //$maxint;
function print_p($arr, $width, $from){
$curr = $from;
printf("<br>(%05d : %05d)", $curr, ($curr+=$width)-1);
$prn = $width;
foreach($arr as $item){
if ($prn-- == 0){
$prn = $width-1;
printf("<br>(%05d : %05d)",$curr, ($curr+=$width)-1);
}
printf(" %2d", $item);
};
print("<br>");
}
function base_sieve(){
global $inter, $p_base;
$p0 = array( 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71,
73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173,
179, 181, 191, 193, 197, 199, 211, 223);
$p_sieve = range(0,$inter-1);
array_walk($p0, function($item) use(&$p_sieve, $inter) {
for($j=$item*$item; $j<$inter; $j+=$item) $p_sieve[$j] = 0;
});
$p_sieve[1]=0;
$p_base = array();
array_walk($p_sieve, function ($item) use(&$p_base){
if($item) $p_base[]=$item;
});
}
function liouville($from, $to){
global $p_base;
if(is_int($from) && is_int($to)){
$lambda = array_fill($from, $to-$from+1, 1);
$prod = $lambda;
}else{
exit(1);
}
$count_base = count($p_base);
for($key = 0; $key < $count_base-1; $key++){
$factor = $p_base[$key];
$factor2 = pow($factor,2);
if($factor2 > $to) break;
$item_old = 1;
for($item = $factor; $item <= $to; $item *= $factor){
if($item <= $item_old) break; // overflow test
$item_old = $item;
for($j=$to-$to%$item; $j>= max($from, $factor2); $j-=$item) {
$prod[$j] *= $factor;
$lambda[$j] = -$lambda[$j];
}
}
};
printf("<br>");
for($key = $from; $key <= $to; $key++){
//printf("prod[%d]=%d ", $key, $prod[$key]);
if($prod[$key] != $key) $lambda[$key] *= -1;
}
return $lambda;
}
$p_base=0;
base_sieve();
print ("maxint=$maxint inter=$inter");
printf("<br> base_sieve: p_first = %d p_last = %d", reset($p_base), end($p_base));
printf("<br> liouville: val_from = %d val_to = %d<br>", $val_from, $val_to);
$lambda = liouville($val_from, $val_to);
print_p($lambda,25,$val_from);
Results:
maxint=2147483647 inter=46340 base_sieve: p_first = 2 p_last = 46337 liouville: val_from = 1 val_to = 1000 (00001 : 00025) 1 -1 -1 1 -1 1 -1 -1 1 1 -1 -1 -1 1 1 1 -1 -1 -1 -1 1 1 -1 1 1 (00026 : 00050) 1 -1 -1 -1 -1 -1 -1 1 1 1 1 -1 1 1 1 -1 -1 -1 -1 -1 1 -1 -1 1 -1 (00051 : 00075) 1 -1 -1 1 1 1 1 1 -1 1 -1 1 -1 1 1 -1 -1 -1 1 -1 -1 -1 -1 1 -1 (00076 : 00100) -1 1 -1 -1 -1 1 1 -1 1 1 1 1 1 -1 1 1 -1 1 1 1 1 -1 -1 -1 1 (00101 : 00125) -1 -1 -1 1 -1 1 -1 -1 -1 -1 1 -1 -1 -1 1 -1 -1 1 1 -1 1 1 1 -1 -1 (00126 : 00150) 1 -1 -1 1 -1 -1 1 1 1 1 1 -1 -1 -1 1 1 1 1 1 1 1 -1 -1 -1 1 (00151 : 00175) -1 1 -1 -1 1 1 -1 1 1 1 1 -1 -1 -1 -1 1 -1 -1 1 -1 -1 -1 -1 -1 -1 (00176 : 00200) -1 1 1 -1 -1 -1 -1 1 1 1 -1 1 -1 1 -1 -1 -1 -1 1 -1 1 -1 1 -1 -1 (00201 : 00225) 1 1 1 1 1 1 -1 -1 1 1 -1 -1 1 1 1 1 1 1 1 1 1 -1 -1 1 1 (00226 : 00250) 1 -1 1 -1 -1 -1 1 -1 1 1 -1 1 -1 -1 1 -1 -1 -1 -1 -1 -1 1 1 1 1 (00251 : 00275) -1 -1 1 1 -1 1 -1 -1 1 1 -1 1 -1 -1 1 -1 1 -1 -1 -1 -1 -1 -1 1 -1 (00276 : 00300) 1 -1 1 -1 -1 -1 -1 -1 -1 -1 -1 1 -1 1 -1 1 -1 -1 1 1 1 1 1 1 -1 (00301 : 00325) 1 1 1 -1 1 1 -1 1 1 -1 -1 -1 -1 1 1 -1 -1 -1 1 -1 1 -1 1 1 -1 (00326 : 00350) 1 1 1 1 1 -1 -1 -1 1 1 1 -1 -1 1 1 1 1 -1 1 -1 1 -1 1 -1 1 (00351 : 00375) 1 1 -1 -1 1 -1 -1 1 -1 1 1 1 -1 1 1 -1 -1 -1 -1 -1 1 1 -1 -1 1 (00376 : 00400) 1 1 -1 -1 1 1 1 -1 1 -1 1 -1 -1 -1 1 1 -1 1 1 1 -1 -1 1 -1 1 (00401 : 00425) -1 -1 1 -1 -1 -1 1 -1 -1 -1 1 -1 1 1 1 1 1 -1 -1 -1 -1 1 -1 1 -1 (00426 : 00450) -1 1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 -1 -1 -1 1 -1 -1 1 1 1 1 -1 -1 -1 (00451 : 00475) 1 -1 1 1 -1 -1 -1 1 1 1 -1 1 -1 -1 -1 1 -1 -1 1 -1 1 1 1 -1 -1 (00476 : 00500) 1 -1 1 -1 -1 1 1 -1 1 1 1 -1 1 1 1 -1 1 1 -1 1 -1 1 -1 -1 -1 (00501 : 00525) 1 1 -1 1 1 -1 -1 -1 -1 1 1 -1 1 1 1 1 1 -1 1 -1 -1 1 -1 -1 1 (00526 : 00550) 1 1 1 1 -1 -1 1 1 -1 1 1 1 1 -1 1 -1 1 1 1 1 1 -1 -1 -1 1 (00551 : 00575) 1 -1 1 1 -1 -1 -1 1 1 1 -1 1 -1 1 1 1 -1 1 -1 1 -1 1 1 -1 -1 (00576 : 00600) 1 -1 -1 1 1 1 -1 1 1 1 1 -1 -1 1 -1 1 -1 -1 -1 -1 -1 1 -1 -1 1 (00601 : 00625) -1 -1 -1 -1 -1 -1 -1 1 -1 -1 1 -1 -1 1 -1 -1 -1 -1 -1 1 1 1 1 1 1 (00626 : 00650) 1 -1 -1 1 -1 -1 1 1 1 1 1 -1 -1 -1 1 -1 -1 -1 1 -1 -1 -1 -1 1 1 (00651 : 00675) -1 -1 -1 -1 1 -1 -1 -1 -1 -1 -1 1 -1 1 -1 1 1 -1 1 -1 1 -1 -1 1 -1 (00676 : 00700) 1 -1 -1 1 -1 1 -1 -1 -1 1 1 1 -1 1 1 -1 -1 1 1 1 -1 1 1 1 -1 (00701 : 00725) -1 -1 1 -1 -1 1 1 1 -1 -1 -1 1 1 1 -1 -1 1 1 -1 -1 1 -1 1 -1 -1 (00726 : 00750) 1 -1 -1 1 -1 1 1 -1 1 1 1 1 1 -1 1 -1 -1 -1 -1 1 1 -1 1 1 -1 (00751 : 00775) -1 -1 1 -1 1 1 -1 1 -1 -1 -1 -1 1 -1 1 1 1 -1 -1 1 1 -1 -1 1 -1 (00776 : 00800) 1 -1 1 1 -1 1 -1 1 1 1 -1 -1 -1 1 -1 1 1 1 1 -1 -1 -1 1 1 -1 (00801 : 00825) -1 1 1 1 -1 -1 1 1 -1 1 -1 1 1 -1 1 1 1 1 1 1 -1 -1 -1 1 1 (00826 : 00850) -1 -1 -1 -1 -1 1 -1 -1 -1 1 1 1 1 -1 1 1 1 1 -1 -1 1 -1 -1 1 1 (00851 : 00875) 1 1 -1 -1 1 1 -1 1 -1 1 -1 1 -1 1 1 1 -1 1 1 1 1 1 -1 -1 1 (00876 : 00900) 1 -1 1 1 1 -1 -1 -1 1 -1 1 -1 -1 1 -1 -1 -1 1 -1 1 1 -1 1 1 1 (00901 : 00925) 1 -1 -1 1 1 -1 -1 -1 -1 1 -1 1 1 1 -1 -1 1 -1 -1 -1 1 1 1 -1 -1 (00926 : 00950) 1 -1 1 -1 1 -1 -1 1 1 -1 1 -1 -1 1 1 -1 -1 1 -1 -1 -1 -1 1 1 1 (00951 : 00975) 1 -1 -1 1 1 -1 -1 1 1 1 1 -1 -1 -1 1 1 -1 -1 -1 -1 -1 -1 1 1 1 (00976 : 01000) -1 -1 -1 1 -1 -1 1 -1 -1 1 -1 -1 1 1 -1 -1 1 1 -1 1 1 -1 1 1 1
Yuri Negometyanov
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This has the same issue as the algorithm I sketched out in the question, it has to do $n \cdot \log{\log{n}} + O(1)$ negations so it can't finish in linear time. – Dan Brumleve Dec 21 '15 at 19:10
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@Dan Brumleve It seems that $$\sum\limits_{p<n}\dfrac n{p-1} > > \sum\limits_{p<\sqrt n}\dfrac n{p-1} $$ First formula complexity is more than $n\log\log n$. Minus 1 in the denominator corresponds to decreasing exponentially for degrees – Yuri Negometyanov Dec 21 '15 at 19:34
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1Both of those formulas are $n \cdot \log{\log{n}} + O(n)$, same if you replace $\sqrt{n}$ with $n^x$ for any $x \gt 0$. – Dan Brumleve Dec 21 '15 at 19:43
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@Dan Brumleve For $n =10^8$ theoretical gain of 30%. Really - more because we get bargain cost of organizing $\dfrac n{ \log n}$ almost empty cycles. And this despite the fact that the issue price is equal to factor $\log\log n = 3$ – Yuri Negometyanov Dec 21 '15 at 20:18
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@DanBrumleve: I believe a problem comes from the fact that people see your question as a call for a practical solution. Obviously, this SE site is not very appropriate for this type of discussion. And we are anyway far behind the state-of-art practical solution: this solution finds all primes up to $n = 2^{32}$ in 200 ms on a commodity PC. Now you talking about 30% increase? to find $10^6$ of those primes in 20 seconds? Hmm... – stgatilov Dec 22 '15 at 06:57
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@stgatilov The proposed program, taking into account the programming language and the lack of optimization software should be considered as a layout illustrating the operation of the algorithm. But we are talking about the algorithms that can improve any program. And the practical side of the issue - in the most complete account of all the factors affecting the performance. Binding performance evaluation operations change the sign seems to me not quite accurate, but the question admits of discussion and clarification. And discussed algorithm isn't SE. – Yuri Negometyanov Dec 22 '15 at 13:01
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I know this question was posed 5 months ago but I thought I would add something. Not sure if this will increase computing time but,
$\lambda(n)=i^{\tau(n^{2})-1}$, where $\tau(n)$ is the divisor function and $i$ is the imaginary unit. From this of course the summatory Liouville function is $L(n)=\sum_{j=1}^{n}i^{\tau(j^{2})-1}$. Perhaps the divisor function can be calculated in less time.
e2theipi2026
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Actually this cannot be true because you can't get the prime (power)s in $O(n)$:
The sieve of eratosthenes takes at least $O(nloglogn)$ Operations.
Assume we have a list of all the primes up to $n$ given. Then checking each $k$ for membership in the list takes at least (if we only check the first $\pi(\sqrt{k})$ elements) $\pi(\sqrt{k})$ operations each, which amounts to more than $O(1)$ so that's out. Then we could check each number for primality but the fastest known primality test runs in $O(log(n)^6)$, also more than $O(1)$. Since for each number we either have to run it against a list or check it by hand that amounts to all possibilites and we conclude it's impossible to calculate the value for all $k \leq n$ in $O(n)$
obstkuchen
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1This isn't an adequate answer for two reasons. First, the run time of the sieve of Eratosthenes doesn't address the question of whether there's some other faster method of finding the primes. (I doubt there is, but still.) Second, it's not at all clear whether there's some way of calculating all these $\lambda$ values without singling out the primes among them. – Greg Martin Dec 17 '15 at 23:10
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I adress the question of whether a faster way to identify primes exists right below talking about the sieve in particular (i only adress the sieve because it's the alg he uses in the original post). Your second statement is valid if there were an algorithm for calculating $\lambda$ without calculating $\omega$ at least implicitly and that is even stronger than checking for primality. It is possible that it's possible but that'd require some identity about $(-1)^{\Omega(n)}$ I'm unaware of. If there actually is one then it may be possible to get the result in $O(n)$ – obstkuchen Dec 17 '15 at 23:14
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1I see that you attempt to, but your analysis is incorrect—it leads to a larger runtime for the sieve of Eratosthenes itself than it actually has, for example. The whole point of the sieve is that it's faster to holistically find all the primes in an interval than it is to check each number in the interval individually for primality. – Greg Martin Dec 17 '15 at 23:16
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Yesyes, however, that only generates you a list of all the primes up to n. If you then want to check each number against that liust you'll end up with higher runtime again I don't make any statement about the runtime of such an algorithm, i state it doesn't matter. – obstkuchen Dec 17 '15 at 23:18
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