updates solution overview

rln-research/merkle-tree-update.md

@@ -1,9 +1,13 @@
 # Solution Overview
+
+In the solution explained below, we enable a peer to to perform the operations described in [Formal Problem Definition](#Formal-Problem-Definition) by only holding O(logn) many tree nodes (but not the entire tree).
+
+The solution relies on the following statement:
 > In a Merkle Tree with the capacity of `n` leaves , one can compute the root of the tree by maintaining the root nodes of log(n) number of complete Merkle trees. 
 
-We use the preciding  observation and define `F = [(L0,H0,index0), ..., (Ld,Hd,indexd)]` to be an array of size log(n)+1 holding the root of the complete Merkle trees (all positioned on the left side of the tree) for levels `[0, d=log(n)+1]`. Each entry of `F`  holds a tuple `(L,index, H)` `H`is the root of the complete subtree at level `L`, and  `index` indicates the index of the leaf node whose insertion resulted in `H`.
+We use the preciding  observation and define `F = [(L:0, H0, leafIndex0), ..., (L:d, Hd, leafIndexd)]` to be an array of size log(n)+1 holding the root of the complete Merkle trees (all positioned on the left side of the tree) for levels `[0, ..., d=log(n)]`. Each entry of `F`  is a tuple `(L, leafIndex, H)` in which `H`is the root of the complete subtree at level `L`, and  `leafIndex` indicates the index of the leaf node whose insertion resulted in `H`. The storage of `leafIndex` in each tuple will later enables the efficient support of deletion operation.  Each peer shall store `F` locally.
 
-For the Merkle Tree shown in Figure below, `F = [(L0, N12, leaf5), (L1, N6, leaf6), (L2, N2, leaf4), (L3, N1, leaf6)]` is highlighted in green. Note that none of the gray nodes are stored as part of `F`.
+For the Merkle Tree shown in Figure below, `F = [(L:0, N12, leafIndex:5), (L:1, N6, leafIndex:6), (L:2, N2, leafIndex:4), (L:3, N1, leafIndex:6)]` is highlighted in green. Note that `F` only contains the green nodes but none of the gray nodes.
 
 ![Tree](F.png)
 
@@ -12,44 +16,56 @@ For the Merkle Tree shown in Figure below, `F = [(L0, N12, leaf5), (L1, N6, leaf
 
 ## Computing the root after deletion
 
-Consider the deletion of the `leafi` with the authentication path / (membership proof) of the follwoing form `authpath = [(L0,H0), ..., (Ld,Hd)]` where `H` values represent the value of the merkle tree nodes at the indicated level `L`. 
-The authentication path of `leaf2`  is illustrated in the following figure and consists of `authpath2 = [(L0, N8), (L1,N5), (L2,N3), (L3,N1)]`.
+Given that `F` is stored by each peer locally, we demonestrate how deletion can be supported relying on `F`.
+
+When a node gets deleted, its authentication path is available, and the following solution relies on that.
+
+Consider the deletion of the `leafIndex:i` with the authentication path / (membership proof) of the follwoing form `authpath = [(L:0, H0), ..., (L:d, Hd)]` where in each tuple `H`  represents the value of the merkle tree node along the authentication path of leaf `i` at the corresponding level `L`. Note that `d` varies from `0` to `logn`. The last entry of `authpath` i.e., `(L:d, Hd)` is indeed the tree root.
+
+The authentication path of `leafIndex:2`  is illustrated in the following figure (highlighted in yellow) and consists of `authpath2 = [(L:0, N8), (L:1, N5), (L:2, N3), (L:3, N1)]`.
 
 ![authPath](authPath.png)
 
 We need to update `F` based on `authpath2`. In specific, we need to determine whether any of the nodes whose values get altered as the result of deletion of a leaf node intersect with the nodes in `F`, and if this is the case the corresponding nodes in `F` shall get updated too.
 
-Lets clarify it by the help of an example. Consider `leaf2`, the deletion of `leaf2` impacts `N9, N4, N2` and `N1`(root) of the tree. 
+Lets clarify it by the help of an example. Consider `leafIndex:2`, the deletion of `leafIndex:2` impacts `N9 (level 0), N4 (level 1), N2 (level 2)` and  `N1 (level 3)`(root) of the tree, as illustrated below with the dark circles. 
+
 ![Deletion](del.png)
 
-Thus, in order to update `F`, we need to update those entries that contain `N9, N4, N2` or `N1`. 
+Thus, in order to update `F`, we need to update those entries of `F` that contain `N9, N4, N2` or `N1`. 
+
+To do so, we determine whether for each tuple `(L, leafIndex, H)` in `F`,  the `leafIndex` and the deleted leaf node have the same ancestor at level `L`. We use `HasCommAnc` method to perform this check, it is later defined in [Common ancestor](###Common-Ancestor ) subsection. For example, `leaf2` and `leaf4` have the same ancestor at level 2, which is `N2`, thus `HasCommAnc(2,4,2)` return true.
+
+Following our previous example, in order to find out whether `N9, N4, N2` or `N1` belong to `F` we proceed as follows.
 - inputs: 
-  - `authpath2 = [(L0, N8), (L1,N5), (L2,N3), (L3,N1)]`
-  - `F = [(L0, N12, leaf5), (L1, N6, leaf6), (L2, N2, leaf4), (L3, N1, leaf6)]` 
-- Update procedure: 
-  - `N9` at level `L0`, HasCommAnc(leaf2,leaf5) = no, thus `F` does not change
-  - `N4'` at level `L1`, HasCommAnc(leaf2,leaf5) = no, thus `F` does not change
-  - `N2'` at level `L2`, HasCommAnc(leaf2,leaf4) = yes, thus `F = [(L0, N12, leaf5), (L1, N6, leaf6), (L2, N2', leaf4), (L3, N1, leaf6)]` 
-  - `N1` at level `L3`, HasCommAnc(leaf2,leaf6) = yes, thus `F = [(L0, N12, leaf5), (L1, N6, leaf6), (L2, N2', leaf4), (L3, N1', leaf6)]` 
+  - Index of the deleted node `leafIndex:2`
+  - `F = [(L:0, N12, leafIndex:5), (L:1, N6, leafIndex:6), (L:2, N2, leafIndex:4), (L:3, N1, leafIndex:6)]` 
+  <!-- - This is not part of the inputs, but just for the ease of explanation, Altered nodes `=[(L:0,N9'), (L:1,N4'), (L:2,N2'), (L:3,N1')`-->
+
+- Update procedure of `F`: 
+  - level `L:0`, HasCommAnc(leafIndex:2, leafIndex:5, L:0) = false, thus `F` does not change
+  - level `L:1`, HasCommAnc(leafIndex:2, leafIndex:6, L:1) = false, thus `F` does not change
+  - level `L:2`, HasCommAnc(leafIndex:2, leafIndex:4, L:2) = true, thus `F = [(L:0, N12, leafIndex:5), (L:1, N6, leafIndex:6), (L:2, N2', leafIndex:4), (L:3, N1, leafIndex:6)]` 
+  - level `L:3`, HasCommAnc(leafIndex:2, leafIndex:6, L:3) = true, thus `F = [(L:0, N12, leafIndex:5), (L:1, N6, leafIndex:6), (L:2, N2', leafIndex:4), (L:3, N1', leafIndex:6)]` 
 - Output
-  - `F' = [(L0, N12, leaf5), (L1, N6, leaf6), (L2, N2', leaf4), (L3, N1', leaf6)]` 
+  - `F' = [(L:0, N12, leafIndex:5), (L:1, N6, leafIndex:6), (L:2, N2', leafIndex:4), (L:3, N1', leafIndex:6)]` 
 
 ![Output](out.png)
 
-### Common Ancestor
+### Common Ancestor 
 In order to determine whether two nodes with indices `i` and `j` have common anscestor at a particular level `lev`, the followong formula can be applied
 
 check whether `floor( (i-1)/2^lev )` is equal to `floor( (j-1)/2^lev )`
 
 
-# Update algorithm
-- Inputs: 
+# Update tree root after deletion
+- Inputs (the levels i.e., `L` values are removed since the position of each entry can represent its level): 
   - `F = [(H0,index0), ..., (Hd,indexd)]`
   - `leafIndex` ( The index of the deleted leaf)
   -  `authpath = [H0, ..., Hd]` (The authentication path of the deleted leaf)
   -  `Z = [H(0), H(Z[0]||Z[0]), H(Z[1]||Z[1]), ..., H(Z[d-1]||Z[d-1])]`
-- Output: `F'`
-  
+- Output: `F` which is the updated `F`
+    
 ```
 path = binary representation of leafIndex - 1
 acc = Z[0]
@@ -66,3 +82,11 @@ for lev in 0..d # d inclusive
 HasCommAnc(i, j, lev) =
   return floor( (i-1)/2^lev ) == floor( (j-1)/2^lev )
 ```
+
+# Updating authentication paths
+
+Lets index the tree nodes following the formula below, for each tree node with index `i`, its right and left child have index `2*i` and `2*i+1`, respectively. The root has index `1`. The sample tree is illustrated below.
+
+When a node gets deleted, 
+
+While writing up the preceding solution, I came up with a simiplified version in which we do not need to store the the leaf indices, instead we can compute them on the fly.