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| ▲ | 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: total = 0
for i in range(len(a)):
for j in range(len(b)):
total += a[i] * b[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. |
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| ▲ | 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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| ▲ | actionfromafar 2 hours ago | parent [-] | | Doesn't that have to do with how many bits you allow in the actual calculation in physical reality? | | |
| ▲ | hellohello2 35 minutes ago | parent [-] | | Well, for multiplication complexity is defined in terms of on the number of digits/bits digits directly. For attention, complexity is defined on terms of the number of input vectors which are all at fixed precision. I don't understand what happens to the method proposed in the paper at higher precision (since I don't understand the paper), but in reality in doesn't matter since there is no value in anything over float16 for machine learning. |
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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. |
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| ▲ | naasking an hour ago | parent | prev [-] |
| Your argument just assumes there is no latent structure that can be exploited. That's a big assumption. |
