feat: update rollup circuits and contracts in yp (#4536)

Updates the yellow paper to align with the changes made by #4250 + updates the rollup circuits to align with one kernel per base.

yellow-paper/docs/l1-smart-contracts/frontier.md

@@ -0,0 +1,159 @@
+---
+title: Frontier Merkle Tree
+---
+
+The Frontier Merkle Tree is an append only Merkle tree that is optimized for minimal storage on chain.
+By storing only the right-most non-empty node at each level of the tree we can always extend the tree with a new leaf or compute the root without needing to store the entire tree.
+We call these values the frontier of the tree.
+If we have the next index to insert at and the current frontier, we have everything needed to extend the tree or compute the root, with much less storage than a full merkle tree. 
+Note that we're not actually keeping track of the data in the tree: we only store what's minimally required in order to be able to compute the root after inserting a new element.
+
+We will go through a few diagrams and explanations to understand how this works.
+And then a pseudo implementation is provided.
+
+
+## Insertion
+Whenever we are inserting, we need to update the "root" of the largest subtree possible.
+This is done by updating the node at the level of the tree, where we have just inserted its right-most descendant.
+This can sound a bit confusing, so we will go through a few examples. 
+
+At first, say that we have the following tree, and that it is currently entirely empty.
+
+![alt text](images/frontier/image-1.png)
+
+### The first leaf
+
+When we are inserting the first leaf (lets call it A), the largest subtree is that leaf value itself (level 0).
+In this case, we simply need to store the leaf value in `frontier[0]` and then we are done.
+For the sake of visualization, we will be drawing the elements in the `frontier` in blue.
+
+![alt text](images/frontier/image-2.png)
+
+Notice that this will be the case whenever we are inserting a leaf at an even index.
+
+### The second leaf
+
+When we are inserting the second leaf (lets call it B), the largest subtree will not longer be at level 0.
+Instead it will be level 1, since the entire tree below it is now filled!
+Therefore, we will compute the root of this subtree, `H(frontier[0],B)` and store it in `frontier[1]`.
+
+Notice, that we don't need to store the leaf B itself, since we won't be needing it for any future computations.
+This is what makes the frontier tree efficient - we get away with storing very little data.
+
+![alt text](images/frontier/image-3.png)
+
+### Third leaf
+When inserting the third leaf, we are again back to the largest subtree being filled by the insertion being itself at level 0.
+The update will look similar to the first, where we only update `frontier[0]` with the new leaf.
+
+![alt text](images/frontier/image-4.png)
+
+### Fourth leaf
+
+When inserting the fourth leaf, things get a bit more interesting.
+Now the largest subtree getting filled by the insertion is at level 2.
+
+To compute the new subtree root, we have to compute `F = H(frontier[0], E)` and then `G = H(frontier[1], F)`.  
+G is then stored in `frontier[2]`.
+
+
+As before, notice that we are only updating one value in the frontier. 
+![alt text](images/frontier/image-5.png)
+
+## Figuring out what to update
+
+To figure out which level to update in the frontier, we simply need to figure out what the height is of the largest subtree that is filled by the insertion.
+While this might sound complex, it is actually quite simple. 
+Consider the following extension of the diagram.
+We have added the level to update, along with the index of the leaf in binary.
+Seeing any pattern?
+
+![alt text](images/frontier/image-6.png)
+
+The level to update is simply the number of trailing ones in the binary representation of the index.
+For a binary tree, we have that every `1` in the binary index represents a "right turn" down the tree. 
+Walking up the tree from the leaf, we can simply count the number of right turns until we hit a left-turn.
+
+## How to compute the root
+
+Computing the root based on the frontier is also quite simple. 
+We can use the last index inserted a leaf at to figure out how high up the frontier we should start.
+Then we know that anything that is at the right of the frontier has not yet been inserted, so all of these values are simply "zeros" values. 
+Zeros here are understood as the root for a subtree only containing zeros.
+
+For example, if we take the tree from above and compute the root for it, we would see that level 2 was updated last.
+Meaning that we can simply compute the root as `H(frontier[2], zeros[2])`.
+
+![alt text](images/frontier/image-7.png)
+
+For cases where we have built further, we simply "walk" up the tree and use either the frontier value or the zero value for the level.
+
+## Pseudo implementation
+```python
+class FrontierTree:
+  HEIGHT: immutable(uint256)
+  SIZE: immutable(uint256)
+
+  frontier: HashMap[uint256, bytes32] # level => node
+  zeros: HashMap[uint256, uint256] # level => root of empty subtree of height level
+
+  next_index: uint256 = 0
+
+  # Can entirely be removed with optimizations
+  def __init__(self, _height_: uint256):
+    self.HEIGHT = _height
+    self.SIZE = 2**_height
+    # Populate zeros
+
+  def compute_level(_index: uint256) -> uint256:
+    '''
+    We can get the right of the most filled subtree by 
+    counting the number of trailing ones in the index
+    '''
+    count = 0
+    x = _index
+    while (x & 1 == 1):
+      count += 1
+      x >>= 1
+    return count
+
+  def root() -> bytes32:
+    '''
+    Compute the root of the tree
+    '''
+    if self.next_index == 0:
+      return self.zeros[self.HEIGHT]
+    elif self.next_index == SIZE:
+      return self.frontier[self.HEIGHT]
+    else:
+      index = self.next_index - 1
+      level = self.compute_level(index)
+
+      temp: bytes32 = self.frontier[level]
+
+      bits = index >> level
+      for i in range(level, self.HEIGHT):
+        is_right = bits & 1 == 1
+        if is_right:
+          temp = sha256(frontier[i], temp)
+        else:
+          temp = sha256(temp, self.zeros[i])
+        bits >>= 1
+      return temp
+
+  def insert(self, _leaf: bytes32):
+    '''
+    Insert a leaf into the tree
+    '''
+    level = self.compute_level(next_index)
+    right = _leaf
+    for i in range(0, level):
+      right = sha256(frontier[i], right)
+    self.frontier[level] = right
+    self.next_index += 1
+```
+
+## Optimizations
+- The `zeros` can be pre-computed and stored in the `Inbox` directly, this way they can be shared across all of the trees.
+
+