shymaa arafat [ARCHIVE] on Nostr: 📅 Original date posted:2021-09-11 📝 Original message:Allow me to introduce this ...

📅 Original date posted:2021-09-11
📝 Original message: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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