Merge pull request #140 from status-im/simplify-merkle-tree
simplify merkle tree chunking
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98312f40b5
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@ -402,40 +402,32 @@ Return the hash of the serialization of the value.
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First, we define some helpers and then the Merkle tree function. The constant `CHUNK_SIZE` is set to 128.
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```python
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# Returns the smallest power of 2 equal to or higher than x
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def next_power_of_2(x):
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return x if x == 1 else next_power_of_2((x+1) // 2) * 2
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# Extends data length to a power of 2 by minimally right-zero-padding
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def extend_to_power_of_2(data):
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return data + b'\x00' * (next_power_of_2(len(data)) - len(data))
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# Concatenate a list of homogeneous objects into data and pad it
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def list_to_glob(lst):
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if len(lst) == 0:
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return b''
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if len(lst[0]) != next_power_of_2(len(lst[0])):
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lst = [extend_to_power_of_2(x) for x in lst]
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data = b''.join(lst)
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# Pad to chunksize
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data += b'\x00' * (CHUNKSIZE - (len(data) % CHUNKSIZE or CHUNKSIZE))
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return data
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# Merkle tree hash of a list of items
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# Merkle tree hash of a list of homogenous, non-empty items
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def merkle_hash(lst):
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# Turn list into padded data
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data = list_to_glob(lst)
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# Store length of list (to compensate for non-bijectiveness of padding)
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datalen = len(lst).to_bytes(32, 'big')
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# Convert to chunks
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chunkz = [data[i:i+CHUNKSIZE] for i in range(0, len(data), CHUNKSIZE)]
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if len(lst) == 0:
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# Handle empty list case
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chunkz = [b'\x00' * CHUNKSIZE]
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elif len(lst[0]) < CHUNKSIZE:
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# See how many items fit in a chunk
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items_per_chunk = CHUNKSIZE // len(lst[0])
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# Build a list of chunks based on the number of items in the chunk
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chunkz = [b''.join(lst[i:i+items_per_chunk]) for i in range(0, len(lst), items_per_chunk)]
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else:
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# Leave large items alone
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chunkz = lst
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# Tree-hash
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while len(chunkz) > 1:
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if len(chunkz) % 2 == 1:
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chunkz.append(b'\x00' * CHUNKSIZE)
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chunkz = [hash(chunkz[i] + chunkz[i+1]) for i in range(0, len(chunkz), 2)]
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# Return hash of root and length data
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return hash((chunkz[0] if len(chunks) > 0 else b'\x00' * 32) + datalen)
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return hash(chunkz[0] + datalen)
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```
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To `tree_hash` a list, we simply do:
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