Asking Gemini 3 to generate Brainfuck code results in an infinite loop

All of a sudden, internet is full of people who hate AI written articles. A few months back, my article got a lot of haters because I used AI tools to improve my draft. Being a non-english first language person, I don't see an issue. But I wish AI improves to an extend where draft to complete articles don't look AI written.

You should use AI to point out errors or suggest better phrasing. But if you ask AI to rewrite your post, it will produce content that sounds fake and corporate. ESL speakers may not notice it but everyone else does.

My prompt was worded differently, but Gemini didn't fall into the same trap for me.

``` Providing the full Brainfuck code to dynamically calculate and factorize 1000! is unfortunately not feasible, and here is why: * The Scale: 1000! (1000 factorial) is a number with 2,568 digits. Standard Brainfuck operates on 8-bit memory cells (values 0-255). To handle a number this large, one would first have to write an entire "arbitrary-precision" math library (BigInt) from scratch within Brainfuck. * The Algorithm: You cannot simply compute 1000! and then factor it. You must use Legendre's Formula to determine the exponent of every prime p \le 1000. * Formula: E_p(n!) = \sum_{k=1}^{\infty} \lfloor \frac{n}{p^k} \rfloor * Example for p=2: 500 + 250 + 125 + 62 + 31 + 15 + 7 + 3 + 1 = 994. * The first term of the answer is 2^{994}. The Solution (Python) Since Brainfuck is ill-suited for this, here is a Python script that uses Legendre's Formula to instantly calculate and print the factorization of 1000!. def legendre_factorial_exponent(n, p): exponent = 0 while n > 0: n //= p exponent += n return exponent

def sieve_primes(n): primes = [] is_prime = [True] * (n + 1) for p in range(2, n + 1): if is_prime[p]: primes.append(p) for i in range(p * p, n + 1, p): is_prime[i] = False return primes

def factorize_factorial(n): primes = sieve_primes(n) result = [] for p in primes: exponent = legendre_factorial_exponent(n, p) result.append(f"{p}^{exponent}") return " * ".join(result)

print(factorize_factorial(1000))

The Brainfuck Code (for 1000 not !) If you simply meant the number 1000 (without the factorial), or you just want to see what Brainfuck looks like, here is a program that outputs 1000: +++++ +++++ [ > +++++ +++++ -- . (Print 100 'd') +++++ +++++ + . (Print 101 'e') (This is just "d" and "e" for demo; printing "1000" requires ASCII conversion)

Correct code to print "1000": ++++++++++[>+++++>+++++>+++++>+++++-.>-.>-.>-.

Explanation: * Sets cell #0 to 10. * Loops 10 times, adding 5 to cells #1, #2, #3, #4 (making them all 50). * Subtracts 1 from cell #1 (ASCII 49 = '1'). * Subtracts 2 from cells #2, #3, #4 (ASCII 48 = '0'). * Prints them. Would you like me to explain how Legendre's Formula works in more detail? ```

I personally prefer some grammatical errors or awkward phrasing over AI-assisted writing. It's a blog post, not a diplomatic transcript.

Correlation is not causation but I challenge anyone to come up with a different cause:

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