您的目标是确定给定数字是否以n最少的字节为质数。但是，您的代码必须是单个Python 2表达式，其数字仅包含

• 经营者
• 输入变量 n
• 整数常数
• 括号

没有循环，没有分配，没有内置函数，只有上面列出的内容。是的，有可能。

经营者

以下是Python 2中所有运算符的列表，其中包括算术运算符，按位运算符和逻辑运算符：

+    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

所有中间值都是整数（或False / True，隐式等于0和1）。负指数不能与指数相乘，因为这可能会产生浮点数。请注意/，与Python 3不同，它可以进行楼层划分，因此//不需要。

即使您不熟悉Python，操作员也应该非常直观。有关运算符优先级的信息，请参见下表；有关语法的详细说明，请参见本节和下文。您可以在TIO上运行Python 2。

输入输出

输入：n至少为2的正整数。

输出：如果n为质数则为1，否则为0。True并且False也可以使用。最少的字节数获胜。

由于您的代码是一个表达式，因此它将是一个代码段，期望输入值存储为n，并评估所需的输出。

您的代码必须适用于n任意大的系统限制。由于Python的整数类型是无界的，因此对运算符没有限制。您的代码可能需要花费很长时间才能运行。

43字节

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

在线尝试！

该方法类似于丹尼斯的第二个（已删除）答案，但是更容易证明该答案是正确的。

证明

简写

不能被n整除的(4**n+1)**n%4**n**2基数中的最高有效数字将使下一个（较低有效）数字为非零（如果“下一位”不在小数部分），则执行带位掩码的a 进行检查如果奇数位置的任何数字不为零。$2^n$$n$(4**n+1)**n%4**n**2/n&2**(2*n*n+n)/-~2**n

长表

令为具有该基数b表示的数字，即a n b n + ⋯ + a 1 b 1 + a 0 b 0，a i为“位置“ i”（以b为基数表示）。$[a_n,\dots,a_1,a_0]_b$$b$$a_nb^n+\dots+a_1b^1+a_0b^0$$a_i$$i$$b$

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

因为（与Ñ2Ñ-1s）为一个整数，并⌊2$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}$$n$ $2^n-1$， =[2Ñ-1，0，2Ñ-1，0，2Ñ-1，0]2Ñ。$\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}$

接下来，考虑

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

，因此会将数字截断为最后 2 个数字-不包括（$4^{n^2}=(2^n)^{2n}$%4**n**2$2n$（为1），但包括所有其他二项式系数。$\binom nn$

关于/n：

• 如果是素数，则结果为[ （$n$。奇数位置的所有数字均为零。$\left[\binom n{n-1}/n,0,\dots,0,\binom n1/n,0,0\right]_{2^n}$

• 如果不是素数：$n$

  让是最大的整数，ñ ∤ （$a$（n>a>0）。将股息改写为$n\nmid\binom na$$n>a>0$

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

  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.

&$2^n$$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.