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144 sats \ 5 replies \ @SimpleStacker 11 Feb \ parent \ on: The SN home page is an auction meta

I see. You're wanting to update the sum incrementally by adding the right-most component each time. Newer zaps have a larger zap_time and thus a higher weight. But the weights are enormous because of the exponent.

Wasn't quite sure what you meant by only having to store the time of the last zap and the moving sum though.

53 sats \ 4 replies \ @k00b 11 Feb

Yes that's right. I store the sum because it's like the mantissa and the time because it's like the exponent. When a new zap comes in, the time allows me to discount the old sum.

Rank_{i} \leftarrow Rank_{i} \cdot e^{- λ (zap_time_{i j_{new}} - zap_time_{ij})} + zap_amount_{i j_{new}}

I think that's right.

47 sats \ 2 replies \ @SimpleStacker 11 Feb

I think you'd have to update the rank of all the other items at the same time that item $i$ gets zapped, otherwise item $i$ 's rank relative to other items might actually go down, because its old rank gets down-weighted via the exponent.

(As an example, imagine item $i$ gets a zap of just 1 sat. Its old rank gets down-weighted for time, by the increase from the 1 sat is negligible.)

68 sats \ 1 reply \ @k00b 11 Feb

That’s not the full sort key. My bad.

I’ll try to write it in a way I understand tomorrow.

74 sats \ 0 replies \ @k00b 11 Feb

Last night what I shared as $Rank_{i}$ is hot_s.

Rank_{i} = j \sum zap_amount_{ij} e^{λ zap_time_{ij}} where λ = \frac{ln 2}{4 h} .

We store hot_t and hot_s:

Rank_{i} = e^{λ hot_t_{i}} hot_s_{i}

with

hot_t_{i} = j max zap_time_{ij}, hot_s_{i} = j \sum zap_amount_{ij} e^{λ (zap_time_{ij} - hot_t_{i})} .

Substituting those definitions directly:

Rank_{i} = e^{λ m a x_{j} zap_time_{ij}} j \sum zap_amount_{ij} e^{λ (zap_time_{ij} - m a x_{j} zap_time_{ij})} .

1 sat \ 0 replies \ @k00b 11 Feb

(This mostly MathGPT btw. I'm barely hanging on trying to read forms like this.)