zlacker

[parent] [thread] 4 comments
1. refulg+(OP)[view] [source] 2026-02-04 00:41:30
That math is for comparing all n-grams for all n <= N simultaneously, which isn't what was being discussed.

For any fixed n-gram size, the complexity is still O(N^2), same as standard attention.

replies(1): >>measur+O8
2. measur+O8[view] [source] 2026-02-04 01:39:15
>>refulg+(OP)
I was talking about all n-gram comparisons.
replies(1): >>refulg+Zb
◧◩
3. refulg+Zb[view] [source] [discussion] 2026-02-04 02:01:04
>>measur+O8
Thanks for clarifying. I was hoping to clarify the disconnect between you two, looked like on on "bigrams, trigrams, & so on." It reads idiomatically as enumerating fixed-n cases. Parsing "& so on" as "their simultaneous union" asks quite a bit of those two words. Either way, as ChatGPT showed you and you shared, all-ngram comparison brings us to O(N^3), still several exponents short of N^10 that started this thread.
replies(1): >>measur+Ac
◧◩◪
4. measur+Ac[view] [source] [discussion] 2026-02-04 02:06:38
>>refulg+Zb
This is getting tiresome. I can make the operations as complicated as necessary by comparing all possible permutations of the input string w/ every other permutation & that will not be reducible to standard attention comparisons. The n-gram was a simple example anyone should be able to understand. You can ask your favorite chatbot to compute the complexity for the permutation version.
replies(1): >>refulg+oj
◧◩◪◨
5. refulg+oj[view] [source] [discussion] 2026-02-04 03:01:22
>>measur+Ac
No worries! I enjoyed it fwiw, appreciate your time :) (The permutation version would be factorial, fwiw, not polynomial. Different beast entirely.)
[go to top]