LLM will NEVER be AGI: The Proof

The proof is trivial with a little help of a necessary condition of complexity theories. All LLM runs under the complexity of \(O(n)\) , where \(n\) is the length of the output. Suppose LLM is AGI, then it is able to solve any problem that a human can solve. Consider the following problem: Given a string \(s\) of length \(n\) , determine whether \(s\) is a palindrome. Answer "Y" if it is and "N"…