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Why is the integration upper bound t=1 in the MNIST example? #260

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xliu99 opened this issue Jan 7, 2025 · 1 comment
Open

Why is the integration upper bound t=1 in the MNIST example? #260

xliu99 opened this issue Jan 7, 2025 · 1 comment

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@xliu99
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xliu99 commented Jan 7, 2025

Hi, Thank you so much for such an amazing work. In the MNIST example, the integration interval is from 0 to 1. I understand that the initial condition is h(0), which is also the input tensor. But why using 1 for the other evaluation point as the output hidden state of the ODE? Since t is a continuous value (which may not represent discrete layer index anymore), why do we have to use 1? What if we use other values, like 1.5?

Thank you so much for any guidance.

@mattsthilaire
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I'm not a contributor, and not an expert on this by any means, however, I'm probably in the same boat as you trying to learn all this neural ODE stuff (at least I think). From what I've understood so far, it is arbitrary, where 1 represents the hidden state that is the "goal". For something like flow matching, it makes more intuitive sense to me because [0, 1] represents how close you are to a target distribution, 1 being you've made it to the target distribution and 0 being your at the initial distribution.

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