Amacınız verilen bir sayının nen az baytta asal olup olmadığını belirlemektir . Ancak, kodunuz yalnızca aşağıdakilerden oluşan sayılar üzerinde tek bir Python 2 ifadesi olmalıdır

• operatörler
• giriş değişkeni n
• tamsayı sabitleri
• parantez

Döngü yok, atama yok, yerleşik işlev yok, yalnızca yukarıda listelenenleri. Evet mümkün.

Operatörler

İşte Python 2'deki aritmetik, bitsel ve mantıksal işleçleri içeren tüm işleçlerin bir listesi :

+    adddition
-    minus or unary negation
*    multiplication
**   exponentiation, only with non-negative exponent
/    floor division
%    modulo
<<   bit shift left
>>   bit shift right
&    bitwise and
|    bitwise or
^    bitwise xor
~    bitwise not
<    less than
>    greater than
<=   less than or equals
>=   greater than or equals
==   equals
!=   does not equal

Tüm ara değerler tamsayılar (veya 0 ve 1'e tamamen eşit olan False / True). Üfleme negatif kayanlarla kullanılamaz, çünkü bunlar yüzebilir. Not /Python 3 farklı olarak zeminden bölünmesini yapar, böylece //gerekli değildir.

Python'u tanımasanız bile, operatörler oldukça sezgisel olmalıdır. Dilbilgisinin ayrıntılı bir belirtimi için operatör önceliği ve bu bölüm ve altındaki bu tabloya bakın . Şunları yapabilirsiniz TIO üzerinde Python 2 koşmak .

I / O

Giriş:n En az 2 olan pozitif bir tamsayı .

Çıktı: Eğer nasal ise 1 , aksi takdirde 0. Trueve Falseayrıca kullanılabilir. En az bayt kazanır.

Kodunuz bir ifade olduğundan, bir snippet olacaktır, giriş değerini olarak depolanan nve istenen çıktısını değerlendirir.

Kodunuz, nkeyfi olarak büyük, bir kenara sistem sınırları için çalışması gerekir . Python'un tam sayı türü sınırsız olduğundan, operatörlerde sınır yoktur. Kodunuzun çalışması uzun sürebilir.

43 bayt

(4**n+1)**n%4**n**2/n&2**(2*n*n+n)/-~2**n<1

Çevrimiçi deneyin!

Yöntem Dennis'in ikinci (silinmiş) cevabına benzer, ancak bu cevabın doğru olması daha kolaydır.

Kanıt

Kısa form

En yüksek değer basamağı (4**n+1)**n%4**n**2tabanında tarafından bölünebilir değildir n sonraki (daha az önemli) basamağını yapacak sonra (yani "bir sonraki basamak" fraksiyonel kısmında değilse) Sıfır dışındaki bir bit maskesi ile$2^n$$n$(4**n+1)**n%4**n**2/n&2**(2*n*n+n)/-~2**n check yürütüldüğünde Tek pozisyondaki herhangi bir rakam sıfır değilse.

Uzun formu

$[a_n,\dots,a_1,a_0]_b$ be the number having that base $b$ representation, i.e., $a_nb^n+\dots+a_1b^1+a_0b^0$, and $a_i$ be the digit at "position" $i$ in base $b$ representation.

• $\texttt{2**(2*n*n+n)/-~2**n} =\lfloor{2^{(2n+1)n}\over1+2^n}\rfloor =\lfloor{4^{n^2}\times 2^n\over1+2^n}\rfloor =\lfloor{{(4^{n^2}-1)\times 2^n\over1+2^n} +{2^n\over1+2^n}}\rfloor$.

Because $2^n\times{4^{n^2}-1\over1+2^n} =2^n(2^n-1)\times{(4^n)^n-1\over4^n-1} =[2^n-1,0,2^n-1,0,2^n-1,0]_{2^n}$ (with $n$ $2^n-1$s) is an integer, and $\lfloor{2^n\over1+2^n}\rfloor=0$, 2**(2*n*n+n)/-~2**n = $[2^n-1,0,2^n-1,0,2^n-1,0]_{2^n}$.

Next, consider

$$\begin{align} \texttt{(4**n+1)**n} &=(4^n+1)^n \\ &=\binom n04^{0n}+\binom n14^{1n}+\dots+\binom nn4^{n^2} \\ &=\left[\binom nn,0,\dots,0,\binom n1,0,\binom n0\right]_{2^n} \end{align}$$

$4^{n^2}=(2^n)^{2n}$, so %4**n**2 will truncate the number to $2n$ last digits - that excludes the $\binom nn$ (which is 1) but include all other binomial coefficients.

About /n:

• If $n$ is a prime, the result will be $\left[\binom n{n-1}/n,0,\dots,0,\binom n1/n,0,0\right]_{2^n}$. All digits at odd position are zero.

• If $n$ is not a prime:

  Let $a$ be the largest integer such that $n\nmid\binom na$ ($n>a>0$). Rewrite the dividend as

  $\left[\binom n{n-1},0,\binom n{n-2},0,\dots,\binom n{a+1}, 0,0,0,\dots,0,0,0\right]_{2^n} + \left[\binom na,0,\binom n{a-1},0,\dots,\binom n0\right]_{2^n}$

  The first summand has all digits divisible by $n$, and the digit at position $2a-1$ zero.

  The second summand has its most significant digit (at position $2a$) not divisible by $n$ and (the base) $2^n>n$, so the quotient when dividing that by $n$ would have the digit at position $2a-1$ nonzero.

  Therefore, the final result ((4**n+1)**n%4**n**2/n) should have the digit (base $2^n$, of course) at position $2a+1$ nonzero.

Finally, the bitwise AND (&) performs a vectorized bitwise AND on the digits in base $2^n$ (because the base is a power of 2), and because $a\texttt &0=0,a\texttt&(2^n-1)=a$ for all $0\le a<2^n$, (4**n+1)**n%4**n**2/n&2**(2*n*n+n)/-~2**n is zero iff (4**n+1)**n%4**n**2/n has all digits in first $n$ odd positions zero - which is equivalent to $n$ being prime.

Python 2, 56 bytes

n**(n*n-n)/(((2**n**n+1)**n**n>>n**n*~-n)%2**n**n)%n>n-2

Try it online!

This is a proof-of-concept that this challenge is doable with only arithmetic operators, in particular without bitwise |, &, or ^. The code uses bitwise and comparison operators only for golfing, and they can easily be replaced with arithmetic equivalents.

However, the solution is extremely slow, and I haven't been able to run $n=6$`, thanks to two-level exponents like $2^{n^n}$.

The main idea is to make an expression for the factorial $n!$, which lets us do a Wilson's Theorem primality test $(n-1)! \mathbin{\%} n > n-2$ where $\mathbin{\%}$ is the modulo operator.

We can make an expression for the binomial coefficient, which is made of factorials

$$\binom{m}{n} \ = \frac{m!}{n!(m-n)!}$$

But it's not clear how to extract just one of these factorials. The trick is to hammer apart $n!$ by making $m$ really huge.

$$\binom{m}{n} \ = \frac{m(m-1)\cdots(m-n+1)}{n!}= \frac{m^n}{n!}\cdot \left(1-\frac{1}{m}\right)\left(1-\frac{2}{m}\right)\cdots \left(1-\frac{n-1}{m}\right)$$

So, if we let $c$ be the product $\left(1-\frac{1}{m}\right)\left(1-\frac{2}{m}\right)\cdots \left(1-\frac{n-1}{m}\right)$, we have

$$n! = \frac{m^n}{\binom{m}{n}} \cdot c$$

If we could just ignore $c$, we'd be done. The rest of this post is looking how large we need to make $m$ to be able to do this.

Note that $c$ approaches $1$ from below as $m \to \infty$. We just need to make $m$ huge enough that omitting $c$ gives us a value with integer part $n!$ so that we may compute

$$n! = \left\lfloor \frac{m^n}{\binom{m}{n}} \right\rfloor$$

For this, it suffices to have $1 - c < 1/n!$ to avoid the ratio passing the next integer $n!+1$.

Observe that $c$ is a product of $n$ terms of which the smallest is $\left(1-\frac{n-1}{m}\right)$. So, we have

$$c > \left(1-\frac{n-1}{m}\right)^n > 1 - \frac{n-1}{m} n > 1-\frac{n^2}{m},$$

which means $1 - c < \frac{n^2}{m}$. Since we're looking to have $1 - c < 1/n!$, it suffices to take $m \geq n! \cdot n^2$.

In the code, we use $m=n^n$. Since Wilson's Theorem uses $(n-1)!$, we actually only need $m \geq (n-1)! \cdot (n-1)^2$. It's easy to see that $m=n^n$ satisfies the bound for the small values and quickly outgrows the right hand side asymptotically, say with Stirling's approximation.

This answer doesn't use any number-theoretic cleverness. It spams Python's bitwise operators to create a manual "for loop", checking all pairs $1 \leq i,j < n$ to see whether $i \times j = n$.

Python 2, way too many bytes (278 thanks to Jo King in the comments!)

((((((2**(n*n)/(2**n-1)**2)*(2**((n**2)*n)/(2**(n**2)-1)**2))^((n*((2**(n*n-n)/(2**n-1))*(2**((n**2)*(n-1))/(2**n**2-1))))))-((2**(n*n-n)/(2**n-1))*(2**((n**2)*(n-1))/(2**(n**2)-1))))&(((2**(n*(n-1))/(2**n-1))*(2**((n**2)*(n-1))/(2**(n**2)-1)))*(2**(n-1)))==0))|((1<n<6)&(n!=4))

Try it online!

This is a lot more bytes than the other answers, so I'm leaving it ungolfed for now. The code snippet below contains functions and variable assignment for clarity, but substitution turns isPrime(n) into a single Python expression.

def count(k, spacing):
    return 2**(spacing*(k+1))/(2**spacing - 1)**2
def ones(k, spacing):
    return 2**(spacing*k)/(2**spacing - 1)

def isPrime(n):
    x = count(n-1, n)
    y = count(n-1, n**2)
    onebits = ones(n-1, n) * ones(n-1, n**2)
    comparison = n*onebits
    difference = (x*y) ^ (comparison)
    differenceMinusOne = difference - onebits
    checkbits = onebits*(2**(n-1))
    return (differenceMinusOne & checkbits == 0 and n>1)or 1<n<6 and n!=4

Why does it work?

I'll do the same algorithm here in base 10 instead of binary. Look at this neat fraction:

$$\frac{1.0}{999^2} = 1.002003004005\dots$$

If we put a large power of 10 in the numerator and use Python's floor division, this gives an enumeration of numbers. For example, $10^{15}/(999^2) = 1002003004$ with floor division, enumerating the numbers $1,2,3,4$.

Let's say we multiply two numbers like this, with different spacings of zeroes. I'll place commas suggestively in the product.

$$1002003004 \times 1000000000002000000000003000000000004 =$$ $$1002003004,002004006008,003006009012,004008012016$$

The product enumerates, in three-digit sequences, the multiplication table up to 4 times 4. If we want to check whether the number 5 is prime, we just have to check whether $005$ appears anywhere in that product.

To do that, we XOR the above product by the number $005005005\dots005$, and then subtract the number $001001001\dots001$. Call the result $d$. If $005$ appeared in the multiplication table enumeration, it will cause the subtraction to carry over and put $999$ in the corresponding place in $d$.

To test for this overflow, we compute an AND of $d$ and the number $900900900\dots900$. The result is zero if and only if 5 is prime.