def _set(self, node_hash, keypath, value, if_delete_subtrie=False):
"""
If if_delete_subtrie is set to True, what it will do is that it take in a keypath
and traverse til the end of keypath, then delete the whole subtrie of that node.
Note: keypath should be in binary array format, i.e., encoded by encode_to_bin()
"""
# Empty trie
if node_hash == BLANK_HASH:
if value:
return self._hash_and_save(
encode_kv_node(
keypath, self._hash_and_save(encode_leaf_node(value))))
else:
return BLANK_HASH
nodetype, left_child, right_child = parse_node(self.db[node_hash])
# Node is a leaf node
if nodetype == LEAF_TYPE:
# Keypath must match, there should be no remaining keypath
if keypath:
raise NodeOverrideError(
"Fail to set the value because the prefix of it's key"
" is the same as existing key")
if if_delete_subtrie:
return BLANK_HASH
return self._hash_and_save(
encode_leaf_node(value)) if value else BLANK_HASH
# node is a key-value node
elif nodetype == KV_TYPE:
# Keypath too short
if not keypath:
if if_delete_subtrie:
return BLANK_HASH
else:
raise NodeOverrideError(
"Fail to set the value because it's key"
" is the prefix of other existing key")
return self._set_kv_node(keypath, node_hash, nodetype, left_child,
right_child, value, if_delete_subtrie)
# node is a branch node
elif nodetype == BRANCH_TYPE:
# Keypath too short
if not keypath:
if if_delete_subtrie:
return BLANK_HASH
else:
raise NodeOverrideError(
"Fail to set the value because it's key"
" is the prefix of other existing key")
return self._set_branch_node(keypath, nodetype, left_child,
right_child, value, if_delete_subtrie)
raise Exception("Invariant: This shouldn't ever happen")
def _set_kv_node(self,
keypath,
node_hash,
node_type,
left_child,
right_child,
value,
if_delete_subtrie=False):
# Keypath prefixes match
if if_delete_subtrie:
if len(keypath) < len(
left_child) and keypath == left_child[:len(keypath)]:
return BLANK_HASH
if keypath[:len(left_child)] == left_child:
# Recurse into child
subnode_hash = self._set(
right_child,
keypath[len(left_child):],
value,
if_delete_subtrie,
)
# If child is empty
if subnode_hash == BLANK_HASH:
return BLANK_HASH
subnodetype, sub_left_child, sub_right_child = parse_node(
self.db[subnode_hash])
# If the child is a key-value node, compress together the keypaths
# into one node
if subnodetype == KV_TYPE:
return self._hash_and_save(
encode_kv_node(left_child + sub_left_child,
sub_right_child))
else:
return self._hash_and_save(
encode_kv_node(left_child, subnode_hash))
# Keypath prefixes don't match. Here we will be converting a key-value node
# of the form (k, CHILD) into a structure of one of the following forms:
# 1. (k[:-1], (NEWCHILD, CHILD))
# 2. (k[:-1], ((k2, NEWCHILD), CHILD))
# 3. (k1, ((k2, CHILD), NEWCHILD))
# 4. (k1, ((k2, CHILD), (k2', NEWCHILD))
# 5. (CHILD, NEWCHILD)
# 6. ((k[1:], CHILD), (k', NEWCHILD))
# 7. ((k[1:], CHILD), NEWCHILD)
# 8. (CHILD, (k[1:], NEWCHILD))
else:
common_prefix_len = get_common_prefix_length(
left_child, keypath[:len(left_child)])
# New key-value pair can not contain empty value
# Or one can not delete non-exist subtrie
if not value or if_delete_subtrie:
return node_hash
# valnode: the child node that has the new value we are adding
# Case 1: keypath prefixes almost match, so we are in case (1), (2), (5), (6)
if len(keypath) == common_prefix_len + 1:
valnode = self._hash_and_save(encode_leaf_node(value))
# Case 2: keypath prefixes mismatch in the middle, so we need to break
# the keypath in half. We are in case (3), (4), (7), (8)
else:
valnode = self._hash_and_save(
encode_kv_node(
keypath[common_prefix_len + 1:],
self._hash_and_save(encode_leaf_node(value)),
))
# oldnode: the child node the has the old child value
# Case 1: (1), (3), (5), (6)
if len(left_child) == common_prefix_len + 1:
oldnode = right_child
# (2), (4), (6), (8)
else:
oldnode = self._hash_and_save(
encode_kv_node(left_child[common_prefix_len + 1:],
right_child))
# Create the new branch node (because the key paths diverge, there has to
# be some "first bit" at which they diverge, so there must be a branch
# node somewhere)
if keypath[common_prefix_len:common_prefix_len + 1] == BYTE_1:
newsub = self._hash_and_save(
encode_branch_node(oldnode, valnode))
else:
newsub = self._hash_and_save(
encode_branch_node(valnode, oldnode))
# Case 1: keypath prefixes match in the first bit, so we still need
# a kv node at the top
# (1) (2) (3) (4)
if common_prefix_len:
return self._hash_and_save(
encode_kv_node(left_child[:common_prefix_len], newsub))
# Case 2: keypath prefixes diverge in the first bit, so we replace the
# kv node with a branch node
# (5) (6) (7) (8)
else:
return newsub
def test_encode_binary_trie_leaf_node(value, expected_output):
if expected_output:
assert expected_output == encode_leaf_node(value)
else:
with pytest.raises(ValidationError):
encode_leaf_node(value)