confusion when calculating the least used weight vector in MANN about ntm

[snowkylin]

Notice that "where n is set to equal the number of reads to memory." below equation (6) in the MANN paper. So not only one position will be set to 1.

[shamanez]

yeah yeah. If there are 4 read heads we need to have four ones in the used vectors and others in to zero. Sorry I missed that part.

But what is the logic? Cannot find why they take this number?

[shamanez]

@snowkylin can you please explain what does it mean to get least n indices which n is equal to the number of heads?

[snowkylin]

Considered that, if this is a conventional Turing Machine, we have 4 heads and want to store 4 different vectors into the memory by these heads, it means we need to write to 4 different positions in the memory, so we would like to find 4 proper slots (better if these slots are least used), make it empty and write data in them. NTM is more differentiable than TM, but the main point is roughly the same.

[shamanez]

Yeah I got it. Anyway it's more like giving a chance to use 4 empty slots for the 4 read heads. Anyway, it's like 4 read heads finds 4 different things they can write that down to 4 different slots. Otherwise, if they find something similar they have the option to write everything in to the same also. Am I correct?

[snowkylin]

Yes, I think the model just give the "chance" for the 4 heads to write to different slots without conflict by preparing 4 empty slots for them. However, it is still possible that some of them write to the same slot.