| ▲ | logicchains 3 hours ago | ||||||||||||||||
It can't be successful at that any more than 1+1 can equal 3. Fundamentally, if every token wants to be able to look at every previous token without loss of information, it must be O(n^2); N tokens looking at N tokens is quadratic. Any sub-quadratic attention must hence necessarily lose some information and be unable to support perfect recall on longer sequences. | |||||||||||||||||
| ▲ | orlp 38 minutes ago | parent | next [-] | ||||||||||||||||
> N tokens looking at N tokens is quadratic Convolving two arrays can be done perfectly accurately in O(n log n), despite every element being combined with every other element. Or consider the even more basic sum of products a[i] * b[j] for all possible i, j:
This can be computed in linear time as sum(a) * sum(b).Your logic that 'the result contains terms of all pairs, therefore the algorithm must be quadratic' simply doesn't hold. | |||||||||||||||||
| ▲ | hellohello2 2 hours ago | parent | prev | next [-] | ||||||||||||||||
I'm not saying if the paper is correct or not (since I can't tell), but I don't think your argument really holds. Consider applying it to multiplication: Fundamentally, multiplication need to look at every pair of integer from the two input numbers. It must be O(n^2); N digits looking at N other digits is quadratic. Any sub-quadratic multiplication must hence necessarily lose some information. | |||||||||||||||||
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| ▲ | oasisaimlessly 2 hours ago | parent | prev | next [-] | ||||||||||||||||
That argument could also be used to say that the FFT's time complexity of O(n log n) should be impossible. | |||||||||||||||||
| ▲ | naasking an hour ago | parent | prev [-] | ||||||||||||||||
Your argument just assumes there is no latent structure that can be exploited. That's a big assumption. | |||||||||||||||||