Replacing old UTXOS with new one in place: an idea/suggested complexity reductio

[Shymaa-Arafat]

Getting back to the main topic, I think I should add this in reply (like it is there)
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Neglecting the effect on the proof size (which depends on the tree height)at least for now,  I think replacement reduces the size/amount of "change" in proofs. When less branches r modified, less internal nodes change their concatenated hashes. This could be helpful, especially when u r caching some of them.
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For example in the run results we were discussing, the numbers suggest an av of +9 (165251/180000) insertions, 6-7 deletions per block (surely this is a small test data), resulting in 2-3 increase in the no of UTXOs per block.
When inserting then deleting u change 15-17 branches along with the corresponding internal nodes hashes, while if u modified when possible u only mess with 9-10 branches.

[Shymaa-Arafat]

They finally replied
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he -> Hashes ever. All the hashing ever done.
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Yesterday when I was checking if they answered my issue, I found this
https://github.com/mit-dci/utreexo/issues/276
I think, my understanding that could be wrong I don't mind reading other should, could be the real reason for the output "he" having that unexplained/not logical value in the run...

Line 84 in the function that keeps running forever (probably meaning till the end of the run, manually as u did)

  84  curHeight++

This value is probably added to "height" in  line 99 here

 99 for ; height <= knownTipHeight && !stop; height++ {

 "he" when u stop the run