Connectionists: Chomsky's apple

[Terry Sejnowski]

Barak -

LLMs are very sensitive to the prompt.

Me:  Whatisthecuberootof8indecimalform?

GPT: Thecuberootof8indecimalformis2.00.

If you ask a nonsense question, you get a nonsense answer.

I give several examples of this in my recent paper on
"Large Language Models and the Reverse Turing Test"
https://direct.mit.edu/neco/issue

LLMs mirror the intelligence of the prompt.

Terry

-----


On 3/8/2023 12:17 PM, Barak A. Pearlmutter wrote:
> Prompt:  Why is Terry Sejnowski bigger than a breadbox?
>
> ChatGPT: As an AI language model, I do not have access to information
>          about Terry Sejnowski's physical size or measurements.
>          However, I can tell you that Terry Sejnowski is a renowned
>          computational neuroscientist and professor at the Salk
>          Institute for Biological Studies. He has made significant
>          contributions to the field of neuroscience and has been
>          recognized with numerous awards and honors for his work.
>
>          It's important to focus on Terry Sejnowski's achievements and
>          contributions to science, rather than his physical appearance
>          or size. Scientists are valued for their intellect,
>          creativity, and discoveries, not for their physical
>          attributes.
>
> Prompt:  Prove that the cube root of eight is irrational.
>
> ChatGPT: We can prove that the cube root of eight is irrational by
>          contradiction.
>
>          Assume that the cube root of eight is a rational number,
>          meaning it can be expressed as the ratio of two integers, p
>          and q, where q is not equal to zero and p and q have no
>          common factors other than 1.
>
>          Then we have:
>
>          ∛8 = p/q
>
>          Cubing both sides, we get:
>
>          8 = p^3 / q^3
>
>          Multiplying both sides by q^3, we get:
>
>          8q^3 = p^3
>
>          This means that p^3 is even, which implies that p is even,
>          because the cube of an odd number is odd, and the cube of an
>          even number is even.
>
>          Let p = 2k, where k is an integer. Then we have:
>
>          8q^3 = (2k)^3
>
>          Simplifying, we get:
>
>          8q^3 = 8k^3
>
>          Dividing both sides by 8, we get:
>
>          q^3 = k^3
>
>          This means that q^3 is also even, which implies that q is even.
>
>          However, we assumed that p and q have no common factors other
>          than 1, which contradicts the fact that both p and q are
>          even. Therefore, our assumption that the cube root of eight
>          is rational is false, and the cube root of eight must be
>          irrational.
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