Hi.
Thanks for the reference, but I missed where you want save space with this
compression on the Merkle Tree.
Regards.
Vincent.
vincenzo.pala...@protonmail.com
https://github.com/vincenzopalazzo
‐‐‐‐‐‐‐ Original Message ‐‐‐‐‐‐‐
On Thursday, September 16th, 2021 at 5:15 AM, shymaa arafat
<shymaa.ara...@gmail.com> wrote:
> Allow me to introduce this simple idea that could be useful ...
>
> -The Intuition was some discussion on Utreexo project about storage saving
> and some traversing issues in handling the UTXOS Merkle Tree/ forest; that is
> N internal nodes need to be stored along with 2N pointers (left&right), +
> maybe 1 more pointer in the leaves special nodes to handle different
> traversing options (insert, delete, & differently proof fetch that traverse
> aunt or niece node according to your implementation
> https://github.com/mit-dci/utreexo/discussions/316)
> .
>
> Then, I thought of a simple idea that gets rid of all the pointers; specially
> appealing when we have all trees are full (complete) in the forest, but can
> work for any Merkle Tree:
>
> - 2D array with variable row size; R[j] is of length (N/2^j)
> -For example when N=8 nodes
> R[0]=0,1,2,...,7
> R[1]=8,9,10,11
> R[2]=12,13
> R[3]=14
> .
> -We can see that total storage is just 2N-1 nodes,
> no need for pointers, and traversing could be neat in any direction with the
> right formula:
>
> -Pseudo code to fetch proof[i] ...
>
> //direction to know + or -
> If ((i mod 2)==0) drct=1;
> else drct=-1;
> // first, the sibling node
> proof[i]=R[0,i+drct]
>
> //add the rest thru loop
> For(j=1; j≤logN; j++)
> { index= i/(2^j)+drct;
> proof[i]=Add(R[j,index]);
> }
>
> -In fact it's just the simple primitive approach of transforming a recursion
> to an iteration, and even if Utreexo team solved their problem differently I
> thought it is worth telling as it can work for any Merkle Tree
> .
> Thanks for your time,
> Shymaa M Arafat_______________________________________________
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