Storing the Merkle Tree in a compact way
2 messages, 11 to 16 Sep 2021. Started by shymaa arafat.
shymaa arafat (2)
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
It could be viewed as the simple complete tree to 1D array with no pointers described in lecture 8 here https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-006-introduction-to-algorithms-spring-2020/lecture-notes/index.htm starting from min 15 in this video https://youtu.be/Xnpo1atN-Iw
Since all trees in Utreexo forest are full binary trees, this is perfect to use, and we can save *76*10⁶*2*size of pointer(probably4bytes)* *~600MB *with almost no effort.
However, I suggest to put it in a 2D array to make it more easy to handle (the indexing math) as we, different than the lecture, traverse in many ways ( normally to delete or insert, and the parent siblings for the proofs)
I wrote more details here https://bitcointalk.org/index.php?topic=5360009.0
On Thu, Sep 16, 2021, 14:37 Vincent <vincent.palazzo@protonmail•com> wrote:
Quoted text (62 lines)
> Hi. > > Thanks for the reference, but I missed where you want save space with this > compression on the Merkle Tree. > > Regards. > > Vincent. > vincenzo.palazzo@protonmail•com > https://github.com/vincenzopalazzo > ‐‐‐‐‐‐‐ Original Message ‐‐‐‐‐‐‐ > On Thursday, September 16th, 2021 at 5:15 AM, shymaa arafat < > shymaa.arafat@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 > > >