Die Dirichlet-Faltung ist eine spezielle Art der Faltung , die in der Zahlentheorie als sehr nützliches Werkzeug erscheint. Es arbeitet mit dem Satz von arithmetischen Funktionen .

Herausforderung

Wenn zwei arithmetische Funktionen $f,g$ (dh Funktionen $f,g: \mathbb N \to \mathbb R$ ) gegeben sind, berechnen Sie die Dirichlet-Faltung $(f * g): \mathbb N \to \mathbb R$ wie unten definiert.

Einzelheiten

Wir verwenden die Konvention $0 \notin \mathbb N = \{1,2,3,\ldots \}$ .
Die Dirichlet-Faltung $f*g$ zweier arithmetischer Funktionen $f,g$ ist wiederum eine arithmetische Funktion und definiert als $(f * g) (n) = \sum_{d | n} f (\frac{n}{d}) \cdot g (d) = \sum_{i \cdot j = n} f (i) \cdot g (j) .$ $(f * g)(n) = \sum_\limits{d|n} f\left(\frac{n}{d}\right)\cdot g(d) = \sum_{i\cdot j = n} f(i)\cdot g(j).$ (Beide Summen sind äquivalent. Der Ausdruck $d|n$ bedeutet $d \in \mathbb N$ dividiert $n$ , daher ist die Summation über den natürlichenTeilernvon $n$ . Ebenso können wir $i = \frac{n}{d} \in \mathbb N, j =d \in \mathbb N$ und wir erhalten die zweite äquivalente Formulierung. Wenn Sie nicht an diese Notation gewöhnt sind, finden Sie im Folgenden ein schrittweises Beispiel.) Nur zur Erläuterung (dies ist für diese Herausforderung nicht direkt relevant): Die Definition stammt aus der Berechnung des Produkts derDirichlet-Reihe: $(\sum_{n \in N} \frac{f (n)}{n^{s}}) \cdot (\sum_{n \in N} \frac{g (n)}{n^{s}}) = \sum_{n \in N} \frac{(f * g) (n)}{n^{s}}$ $\left(\sum_{n\in\mathbb N}\frac{f(n)}{n^s}\right)\cdot \left(\sum_{n\in\mathbb N}\frac{g(n)}{n^s}\right) = \sum_{n\in\mathbb N}\frac{(f * g)(n)}{n^s}$
Die Eingabe erfolgt als zwei Blackbox-Funktionen . Alternativ können Sie auch eine unendliche Liste, einen Generator, einen Stream oder ähnliches verwenden, um eine unbegrenzte Anzahl von Werten zu erzeugen.
Es gibt zwei Ausgabemethoden: Entweder wird eine Funktion $f*g$ zurückgegeben, oder Sie können eine zusätzliche Eingabe $n \in \mathbb N$ und $(f*g)(n)$ direkt zurückgeben.
Der Einfachheit halber können Sie davon ausgehen, dass jedes Element von $\mathbb N$ zB mit einem positiven 32-Bit-Int dargestellt werden kann.
Der Einfachheit halber kann man auch davon ausgehen, dass jeder Eintrag $\mathbb R$ zB durch eine einzige reelle Gleitkommazahl dargestellt werden kann.

Beispiele

Definieren wir zunächst einige Funktionen. Beachten Sie, dass die Liste der Zahlen unter jeder Definition die ersten Werte dieser Funktion darstellt.

die multiplikative Identität ( A000007 ) $ϵ (n) = {\begin{cases} 1 & n = 1 \\ 0 & n > 1 \end{cases}$ $\epsilon(n) = \begin{cases}1 & n=1 \\ 0 & n>1 \end{cases}$ 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...
die konstante Einheitsfunktion ( A000012 ) $1 (n) = 1 \forall n$ $\mathbb 1(n) = 1 \: \forall n$ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
die Identitätsfunktion ( A000027 ) $i d (n) = n \forall n$ $id(n) = n \: \forall n$ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, ...
die Möbius-Funktion ( A008683 ) $μ (n) = {\begin{cases} (- 1)^{k} & if n is squarefree and k is the number of Primefactors of n \\ 0 & otherwise \end{cases}$ $\mu(n) = \begin{cases} (-1)^k & \text{ if } n \text{ is squarefree and } k \text{ is the number of Primefactors of } n \\ 0 & \text{ otherwise } \end{cases}$ 1, -1, -1, 0, -1, 1, -1, 0, 0, 1, -1, 0, -1, 1, 1, 0, -1, 0, -1, ...
die Euler-Totientenfunktion ( A000010 ) $φ (n) = n \prod_{p | n} (1 - \frac{1}{p})$ $\varphi(n) = n\prod_{p|n} \left( 1 - \frac{1}{p}\right)$ 1, 1, 2, 2, 4, 2, 6, 4, 6, 4, 10, 4, 12, 6, 8, 8, 16, 6, 18, 8, ...
die Liouville-Funktion ( A008836 ) $λ (n) = (- 1)^{k}$ $\lambda (n) = (-1)^k$ wobei $k$ die Anzahl der Primfaktoren von $n$ die mit der Multiplizität gezählt werden 1, -1, -1, 1, -1, 1, -1, -1, 1, 1, -1, -1, -1, 1, 1, 1, -1, -1, -1, ...
die Divisorsummenfunktion ( A000203 ) $σ (n) = \sum_{d | n} d$ $\sigma(n) = \sum_{d | n} d$ 1, 3, 4, 7, 6, 12, 8, 15, 13, 18, 12, 28, 14, 24, 24, 31, 18, 39, 20, ...
die Divisor-Zählfunktion ( A000005 ) $τ (n) = \sum_{d | n} 1$ $\tau(n) = \sum_{d | n} 1$ 1, 2, 2, 3, 2, 4, 2, 4, 3, 4, 2, 6, 2, 4, 4, 5, 2, 6, 2, 6, 4, 4, 2, 8, ...
die charakteristische Funktion von Quadratzahlen ( A010052 ) $s q (n) = {\begin{cases} 1 & if n is a square number \\ 0 & otherwise \end{cases}$ $sq(n) = \begin{cases} 1 & \text{ if } n \text{ is a square number} \\ 0 & \text{otherwise}\end{cases}$ 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...

Dann haben wir folgende Beispiele:

$\epsilon = \mathbb 1 * \mu$
$f = \epsilon * f \: \forall f$
$\epsilon = \lambda * \vert \mu \vert$
$\sigma = \varphi * \tau$
$id = \sigma * \mu$ und $\sigma = id * \mathbb 1$
$sq = \lambda * \mathbb 1$ and $\lambda = \mu * sq$
$\tau = \mathbb 1 * \mathbb 1$ and $\mathbb 1 = \tau * \mu$
$id = \varphi * \mathbb 1$ and $\varphi = id * \mu$

The last for are a consequence of the Möbius inversion: For any $f,g$ the equation $g = f * 1$ is equivalent to $f = g * \mu$ .

Step by Step Example

This is an example that is computed step by step for those not familiar with the notation used in the definition. Consider the functions $f = \mu$ and $g = \sigma$ . We will now evaluate their convolution $\mu * \sigma$ at $n=12$ . Their first few terms are listed in the table below.

\begin{array}{cccccccccccccc} f & f (1) & f (2) & f (3) & f (4) & f (5) & f (6) & f (7) & f (8) & f (9) & f (10) & f (11) & f (12) \\ μ & 1 & - 1 & - 1 & 0 & - 1 & 1 & - 1 & 0 & 0 & 1 & - 1 & 0 \\ σ & 1 & 3 & 4 & 7 & 6 & 12 & 8 & 15 & 13 & 18 & 12 & 28 \end{array}

$\begin{array}{c|ccccccccccccc} f & f(1) & f(2) & f(3) & f(4) & f(5) & f(6) & f(7) & f(8) & f(9) & f(10) & f(11) & f(12) \\ \hline \mu & 1 & -1 & -1 & 0 & -1 & 1 & -1 & 0 & 0 & 1 & -1 & 0 \\ \sigma & 1 & 3 & 4 & 7 & 6 & 12 & 8 & 15 & 13 & 18 & 12 & 28 \\ \end{array}$

The sum iterates over all natural numbers $d \in \mathbb N$ that divide $n=12$ , thus $d$ assumes all the natural divisors of $n=12 = 2^2\cdot 3$ . These are $d =1,2,3,4,6,12$ . In each summand, we evaluate $g= \sigma$ at $d$ and multiply it with $f = \mu$ evaluated at $\frac{n}{d}$ . Now we can conclude

\begin{aligned} (μ * σ) (12) & = μ (12) σ (1) & + μ (6) σ (2) & + μ (4) σ (3) & + μ (3) σ (4) & + μ (2) σ (6) & + μ (1) σ (12) \\ = 0 \cdot 1 & + 1 \cdot 3 & + 0 \cdot 4 & + (- 1) \cdot 7 & + (- 1) \cdot 12 & + 1 \cdot 28 \\ = 0 & + 3 & 10 & - 7 & - 12 & + 28 \\ = 12 \\ = i d (12) \end{aligned}

$\begin{array}{rlccccc} (\mu * \sigma)(12) &= \mu(12)\sigma(1) &+\mu(6)\sigma(2) &+\mu(4)\sigma(3) &+\mu(3)\sigma(4) &+\mu(2)\sigma(6) &+\mu(1)\sigma(12) \\ &= 0\cdot 1 &+ 1\cdot 3 &+ 0 \cdot 4 &+ (-1)\cdot 7 &+ (-1) \cdot 12 &+ 1 \cdot 28 \\ &= 0 & + 3 & 1 0 & -7 & - 12 & + 28 \\ &= 12 \\ & = id(12) \end{array}$

code-golf math arithmetic number-theory functional-programming flawr
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4

Lean, 108 100 95 78 75 bytes

def d(f g:_->int)(n):=(list.iota n).foldr(λd s,ite(n%d=0)(s+f d*g(n/d))s)0

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More testcases with all of the functions.

Leaky Nun
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is lambda really more expensive than four bytes for fun ?

Mario Carneiro

lambda is three bytes, I suppose

Leaky Nun

I think it's two in UTF8 (greek is pretty low unicode)

Mario Carneiro

You're right. I also golfed the import

Leaky Nun

I also used cond to save 5 bytes

Leaky Nun

4

Haskell, 46 bytes

(f!g)n=sum[f i*g(div n i)|i<-[1..n],mod n i<1]

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Thanks to flawr for -6 bytes and a great challenge! And thanks to H.PWiz for another -6!

Mego
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Simpler is shorter here

H.PWiz

@H.PWiz That's pretty clever - I didn't even think of doing it that way!

Mego

3

Python 3, 59 bytes

lambda f,g,n:sum(f(d)*g(n//d)for d in range(1,n+1)if 1>n%d)

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Leaky Nun
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Is // really needed instead of /?

Mr. Xcoder

/ would produce floats right?

Leaky Nun

Because d is a divisor of n by definition, the fractional part of n/d is zero, so there shouldn't be any issues with floating point arithmetic. Floats with fractional part zero are close enough to ints for Pythonic purposes, and the output of the function is a real number, so doing n/d instead of n//d should be fine.

Mego

3

Wolfram Language (Mathematica), 17 bytes

Of course Mathematica has a built-in. It also happens to know many of the example functions. I've included some working examples.

DirichletConvolve

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Kelly Lowder
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2

Add++, 51 bytes

D,g,@~,$z€¦~¦*
D,f,@@@,@b[VdF#B]dbRzGb]$dbL$@*z€g¦+

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Takes two pre-defined functions as arguments, plus $n$ , and outputs $(f * g)(n)$

How it works

D,g,		; Define a helper function, $g
	@~,	; $g takes a single argument, an array, and splats that array to the stack
		; $g takes the argument e.g. [[τ(x) φ(x)] [3 4]]
		; STACK : 			[[τ(x) φ(x)] [3 4]]
	$z	; Swap and zip:			[[3 τ(x)] [4 φ(x)]]
	€¦~	; Reduce each by execution:	[[τ(3) φ(4)]]
	¦*	; Take the product and return:	τ(3)⋅φ(4) = 4

D,f,		; Define the main function, $f
	@@@,	; $f takes three arguments: φ(x), τ(x) and n (Let n = 12)
		; STACK:			[φ(x) τ(x) 12]
	@	; Reverse the stack:		[12 τ(x) φ(x)]
	b[V	; Pair and save:		[12]			Saved: [τ(x) φ(x)]
	dF#B]	; List of factors:		[[1 2 3 4 6 12]]
	dbR	; Copy and reverse:		[[1 2 3 4 6 12] [12 6 4 3 2 1]]
	z	; Zip together:			[[[1 12] [2 6] [3 4] [4 3] [6 2] [12 1]]]
	Gb]	; Push Saved:			[[[1 12] [2 6] [3 4] [4 3] [6 2] [12 1]] [[τ(x) φ(x)]]]
	$dbL	; Number of dividors:		[[[τ(x) φ(x)]] [[1 12] [2 6] [3 4] [4 3] [6 2] [12 1]] 6]
	$@*	; Repeat:			[[[1 12] [2 6] [3 4] [4 3] [6 2] [12 1]] [[τ(x) φ(x)] [τ(x) φ(x)] [τ(x) φ(x)] [τ(x) φ(x)] [τ(x) φ(x)] [τ(x) φ(x)]]]
	z	; Zip:				[[[τ(x) φ(x)] [1 12]] [[τ(x) φ(x)] [2 6]] [[τ(x) φ(x)] [3 4]] [[τ(x) φ(x)] [4 3]] [[τ(x) φ(x)] [6 2]] [[τ(x) φ(x)] [12 1]]]
	€g	; Run $g over each subarray:	[[4 4 4 6 4 6]]
	¦+	; Take the sum and return:	28

caird coinheringaahing
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2

R, 58 bytes

function(n,f,g){for(i in (1:n)[!n%%1:n])F=F+f(i)*g(n/i)
F}

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Takes n, f, and g. Luckily the numbers package has quite a few of the functions implemented already.

If vectorized versions were available, which is possible by wrapping each with Vectorize, then the following 45 byte version is possible:

R, 45 bytes

function(n,f,g,x=1:n,i=x[!n%%x])f(i)%*%g(n/i)

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Giuseppe
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1

Jelly, 9 bytes

ÆDṚÇ€ḋÑ€Ʋ

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Line at the top is the main line of $f$ , line at the bottom is the main line of $g$ . $n$ is passed as an argument to this function.

Erik the Outgolfer
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1

Swift 4, 74 70 54 bytes

{n in(1...n).map{n%$0<1 ?f(n/$0)*g($0):0}.reduce(0,+)}

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Mr. Xcoder
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1

JavaScript (ES6), 47 bytes

Takes input as (f)(g)(n).

f=>g=>h=(n,d=n)=>d&&!(n%d)*f(n/d)*g(d)+h(n,d-1)

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Examples

liouville =
n => (-1) ** (D = (n, k = 2) => k > n ? 0 : (n % k ? D(n, k + 1) : 1 + D(n / k, k)))(n)

mobius =
n => (M = (n, k = 1) => n % ++k ? k > n || M(n, k) : n / k % k && -M(n / k, k))(n)

sq =
n => +!((n ** 0.5) % 1)

identity =
n => 1

// sq = liouville * identity
console.log([...Array(25)].map((_, n) => F(liouville)(identity)(n + 1)))

// liouville = mobius * sq
console.log([...Array(20)].map((_, n) => F(mobius)(sq)(n + 1)))

Arnauld
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1

APL (Dyalog Classic), 20 bytes

{(⍺⍺¨∘⌽+.×⍵⍵¨)∪⍵∨⍳⍵}

with ⎕IO←1

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Easy to solve, hard to test - generally not my type of challenge. Yet, I enjoyed this one very much!

{ } defines a dyadic operator whose operands ⍺⍺ and ⍵⍵ are the two functions being convolved; ⍵ is the numeric argument

∪⍵∨⍳⍵ are the divisors of ⍵ in ascending order, i.e. unique (∪) of the LCMs (∨) of ⍵ with all natural numbers up to it (⍳)

⍵⍵¨ apply the right operand to each

⍺⍺¨∘⌽ apply the left operand to each in reverse

+.× inner product - multiply corresponding elements and sum

The same in ngn/apl looks better because of Unicode identifiers, but takes 2 additional bytes because of 1-indexing.

ngn
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Pretty sure it takes 27 additional bytes in ngn/apl...

Erik the Outgolfer

1

C (gcc), 108 bytes

#define F float
F c(F(*f)(int),F(*g)(int),int n){F s=0;for(int d=0;d++<n;)if(n%d<1)s+=f(n/d)*g(d);return s;}

Straightforward implementation, shamelessly stolen from Leaky Nun's Python answer.

Ungolfed:

float c(float (*f)(int), float (*g)(int), int n) {
    float s = 0;
    for(int d = 1; d <= n;++d) {
        if(n % d == 0) {
            s += f(n / d) * g(d);
        }
    }
    return s;
}

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joH1
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1

F#, 72 bytes

let x f g n=Seq.filter(fun d->n%d=0){1..n}|>Seq.sumBy(fun d->f(n/d)*g d)

Takes the two functions f and g and a natural number n. Filters out the values of d that do not naturally divide into n. Then evaluates f(n/d) and g(d), multiples them together, and sums the results.

Ciaran_McCarthy
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0

Pari/GP, 32 bytes

(f,g,n)->sumdiv(n,d,f(n/d)*g(d))

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There is a built-in dirmul function, but it only supports finite sequences.

alephalpha
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Dirichlet-Faltung

Herausforderung

Einzelheiten

Beispiele

Step by Step Example

Antworten:

Lean, 108 100 95 78 75 bytes

Haskell, 46 bytes

Python 3, 59 bytes

Wolfram Language (Mathematica), 17 bytes

Add++, 51 bytes

How it works

R, 58 bytes

R, 45 bytes

Jelly, 9 bytes

Swift 4, 74 70 54 bytes

JavaScript (ES6), 47 bytes

Examples

APL (Dyalog Classic), 20 bytes

C (gcc), 108 bytes

F#, 72 bytes

Pari/GP, 32 bytes