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Can an AI trained in English solve math problems in other languages without extra training?

35,770 次观看 • 1 年前 •via X (Twitter)

11 条评论

Sophia Yang, Ph.D. 的头像
Sophia Yang, Ph.D.1 年前

Full paper: Authored by: @yong_zhengxin @faridlazuarda @M_Jonibek @ruochenz_ @Muennighoff @CarstenEickhoff @gentaiscool @stevebach @AlhamFikri

Places Visited & Pictures Taken 的头像
Places Visited & Pictures Taken2 年前

Wondering what happens when AI is used? Here’s the answer 🙂

Aiden 的头像
Aiden1 年前

Cool paper! That compute insight for cross-lingual math is def interesting. On jenova ai, users can leverage our multi-step agent architecture or even build Custom AI Agents with specific models to tackle complex reasoning – could be powerful for these multilingual scenarios.

Mourad GHAFIRI 的头像
Mourad GHAFIRI1 年前

Nice question :) The short answer is yes in latent-space but couldn’t spell the answer in symbols(tokens) not in the vocab :) So practically it is a wrong question to ask, the correct question maybe, does the LLM create an token-independent representation to math problems :)

Alain GOUDEY 的头像
Alain GOUDEY1 年前

I guess yes as it seems you can extract reasoning processes from vast amount of data in a specific language... mathematics are universal. Will read. 😀

Sorbus 🌊 的头像
Sorbus 🌊1 年前

Or just go to the easy way, automatically translate the prompt before send it and translate the answer before show it to the final user. Extra time and computing cost but could give better answers. Who knows if it’s worth it.

Solomon | Multi-language Tech Support 的头像
Solomon | Multi-language Tech Support1 年前

I didn’t get the point. What do you mean by extra training, and what exactly are you referring to as AI here? If it’s about LLMs, they’re multilingual, so that shouldn’t be an issue.

Sophia Yang, Ph.D. 的头像
Sophia Yang, Ph.D.1 年前

English-centric reasoning language models

Abe 的头像
Abe1 年前

Love your videos!

Sophia Yang, Ph.D. 的头像
Sophia Yang, Ph.D.1 年前

thank you!

Arsen Ibragimov 的头像
Arsen Ibragimov1 年前

sounds like a cool hack but prob not without some extra tuning

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