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490 lines
19 KiB
Python
490 lines
19 KiB
Python
# -*- test-case-name: allmydata.test.test_hashtree -*-
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"""
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Read and write chunks from files.
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Version 1.0.0.
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A file is divided into blocks, each of which has size L{BLOCK_SIZE}
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(except for the last block, which may be smaller). Blocks are encoded
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into chunks. One publishes the hash of the entire file. Clients
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who want to download the file first obtain the hash, then the clients
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can receive chunks in any order. Cryptographic hashing is used to
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verify each received chunk before writing to disk. Thus it is
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impossible to download corrupt data if one has the correct file hash.
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One obtains the hash of a complete file via
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L{CompleteChunkFile.file_hash}. One can read chunks from a complete
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file by the sequence operations of C{len()} and subscripting on a
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L{CompleteChunkFile} object. One can open an empty or partially
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downloaded file with L{PartialChunkFile}, and read and write chunks
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to this file. A chunk will fail to write if its contents and index
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are not consistent with the overall file hash passed to
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L{PartialChunkFile} when the partial chunk file was first created.
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The chunks have an overhead of less than 4% for files of size
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less than C{10**20} bytes.
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Benchmarks:
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- On a 3 GHz Pentium 3, it took 3.4 minutes to first make a
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L{CompleteChunkFile} object for a 4 GB file. Up to 10 MB of
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memory was used as the constructor ran. A metafile filename
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was passed to the constructor, and so the hash information was
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written to the metafile. The object used a negligible amount
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of memory after the constructor was finished.
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- Creation of L{CompleteChunkFile} objects in future runs of the
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program took negligible time, since the hash information was
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already stored in the metafile.
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@var BLOCK_SIZE: Size of a block. See L{BlockFile}.
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@var MAX_CHUNK_SIZE: Upper bound on the size of a chunk.
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See L{CompleteChunkFile}.
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free (adj.): unencumbered; not under the control of others
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Written by Connelly Barnes in 2005 and released into the
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public domain with no warranty of any kind, either expressed
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or implied. It probably won't make your computer catch on fire,
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or eat your children, but it might. Use at your own risk.
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Ported to Python 3.
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"""
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from __future__ import absolute_import
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from __future__ import division
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from __future__ import print_function
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from __future__ import unicode_literals
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from future.utils import PY2
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if PY2:
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from builtins import filter, map, zip, ascii, chr, hex, input, next, oct, open, pow, round, super, bytes, dict, list, object, range, str, max, min # noqa: F401
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from allmydata.util import mathutil # from the pyutil library
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from allmydata.util import base32
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from allmydata.util.hashutil import tagged_hash, tagged_pair_hash
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__version__ = '1.0.0-allmydata'
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BLOCK_SIZE = 65536
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MAX_CHUNK_SIZE = BLOCK_SIZE + 4096
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def roundup_pow2(x):
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"""
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Round integer C{x} up to the nearest power of 2.
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"""
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ans = 1
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while ans < x:
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ans *= 2
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return ans
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class CompleteBinaryTreeMixin(object):
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"""
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Adds convenience methods to a complete binary tree.
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Assumes the total number of elements in the binary tree may be
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accessed via C{__len__}, and that each element can be retrieved
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using list subscripting.
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Tree is indexed like so::
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0
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/ \
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1 2
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/ \ / \
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3 4 5 6
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/ \ / \ / \ / \
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7 8 9 10 11 12 13 14
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"""
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def parent(self, i):
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"""
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Index of the parent of C{i}.
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"""
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if i < 1 or (hasattr(self, '__len__') and i >= len(self)):
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raise IndexError('index out of range: ' + repr(i))
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return (i - 1) // 2
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def lchild(self, i):
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"""
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Index of the left child of C{i}.
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"""
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ans = 2 * i + 1
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if i < 0 or (hasattr(self, '__len__') and ans >= len(self)):
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raise IndexError('index out of range: ' + repr(i))
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return ans
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def rchild(self, i):
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"""
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Index of right child of C{i}.
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"""
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ans = 2 * i + 2
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if i < 0 or (hasattr(self, '__len__') and ans >= len(self)):
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raise IndexError('index out of range: ' + repr(i))
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return ans
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def sibling(self, i):
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"""
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Index of sibling of C{i}.
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"""
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parent = self.parent(i)
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if self.lchild(parent) == i:
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return self.rchild(parent)
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else:
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return self.lchild(parent)
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def needed_for(self, i):
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"""
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Return a list of node indices that are necessary for the hash chain.
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"""
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if i < 0 or i >= len(self):
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raise IndexError('index out of range: 0 >= %s < %s' % (i, len(self)))
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needed = []
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here = i
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while here != 0:
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needed.append(self.sibling(here))
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here = self.parent(here)
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return needed
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def depth_first(self, i=0):
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yield i, 0
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try:
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for child,childdepth in self.depth_first(self.lchild(i)):
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yield child, childdepth+1
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except IndexError:
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pass
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try:
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for child,childdepth in self.depth_first(self.rchild(i)):
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yield child, childdepth+1
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except IndexError:
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pass
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def dump(self):
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lines = []
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for i,depth in self.depth_first():
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value = base32.b2a_or_none(self[i])
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if value is not None:
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value = str(value, "utf-8")
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lines.append("%s%3d: %s" % (" "*depth, i, value))
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return "\n".join(lines) + "\n"
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def get_leaf_index(self, leafnum):
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return self.first_leaf_num + leafnum
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def get_leaf(self, leafnum):
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return self[self.first_leaf_num + leafnum]
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def depth_of(i):
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"""Return the depth or level of the given node. Level 0 contains node 0
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Level 1 contains nodes 1 and 2. Level 2 contains nodes 3,4,5,6."""
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return mathutil.log_floor(i+1, 2)
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def empty_leaf_hash(i):
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return tagged_hash(b'Merkle tree empty leaf', b"%d" % i)
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def pair_hash(a, b):
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return tagged_pair_hash(b'Merkle tree internal node', a, b)
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class HashTree(CompleteBinaryTreeMixin, list):
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"""
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Compute Merkle hashes at any node in a complete binary tree.
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Tree is indexed like so::
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0
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/ \
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1 2
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/ \ / \
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3 4 5 6
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/ \ / \ / \ / \
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7 8 9 10 11 12 13 14 <- List passed to constructor.
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"""
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def __init__(self, L):
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"""
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Create complete binary tree from list of hash strings.
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The list is augmented by hashes so its length is a power of 2, and
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then this is used as the bottom row of the hash tree.
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The augmenting is done so that if the augmented element is at index
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C{i}, then its value is C{hash(tagged_hash('Merkle tree empty leaf',
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'%d'%i))}.
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"""
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# Augment the list.
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start = len(L)
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end = roundup_pow2(len(L))
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self.first_leaf_num = end - 1
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L = L + [None] * (end - start)
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for i in range(start, end):
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L[i] = empty_leaf_hash(i)
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# Form each row of the tree.
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rows = [L]
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while len(rows[-1]) != 1:
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last = rows[-1]
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rows += [[pair_hash(last[2*i], last[2*i+1])
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for i in range(len(last)//2)]]
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# Flatten the list of rows into a single list.
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rows.reverse()
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self[:] = sum(rows, [])
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def needed_hashes(self, leafnum, include_leaf=False):
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"""Which hashes will someone need to validate a given data block?
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I am used to answer a question: supposing you have the data block
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that is used to form leaf hash N, and you want to validate that it,
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which hashes would you need?
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I accept a leaf number and return a set of 'hash index' values, which
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are integers from 0 to len(self). In the 'hash index' number space,
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hash[0] is the root hash, while hash[len(self)-1] is the last leaf
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hash.
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This method can be used to find out which hashes you should request
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from some untrusted source (usually the same source that provides the
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data block), so you can minimize storage or transmission overhead. It
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can also be used to determine which hashes you should send to a
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remote data store so that it will be able to provide validatable data
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in the future.
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I will not include '0' (the root hash) in the result, since the root
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is generally stored somewhere that is more trusted than the source of
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the remaining hashes. I will include the leaf hash itself only if you
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ask me to, by passing include_leaf=True.
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"""
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needed = set(self.needed_for(self.first_leaf_num + leafnum))
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if include_leaf:
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needed.add(self.first_leaf_num + leafnum)
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return needed
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class NotEnoughHashesError(Exception):
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pass
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class BadHashError(Exception):
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pass
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class IncompleteHashTree(CompleteBinaryTreeMixin, list):
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"""I am a hash tree which may or may not be complete. I can be used to
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validate inbound data from some untrustworthy provider who has a subset
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of leaves and a sufficient subset of internal nodes.
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Initially I am completely unpopulated. Over time, I will become filled
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with hashes, just enough to validate particular leaf nodes.
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If you desire to validate leaf number N, first find out which hashes I
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need by calling needed_hashes(N). This will return a list of node numbers
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(which will nominally be the sibling chain between the given leaf and the
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root, but if I already have some of those nodes, needed_hashes(N) will
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only return a subset). Obtain these hashes from the data provider, then
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tell me about them with set_hash(i, HASH). Once I have enough hashes, you
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can tell me the hash of the leaf with set_leaf_hash(N, HASH), and I will
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either return None or raise BadHashError.
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The first hash to be set will probably be 0 (the root hash), since this
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is the one that will come from someone more trustworthy than the data
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provider.
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"""
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def __init__(self, num_leaves):
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L = [None] * num_leaves
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start = len(L)
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end = roundup_pow2(len(L))
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self.first_leaf_num = end - 1
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L = L + [None] * (end - start)
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rows = [L]
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while len(rows[-1]) != 1:
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last = rows[-1]
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rows += [[None for i in range(len(last)//2)]]
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# Flatten the list of rows into a single list.
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rows.reverse()
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self[:] = sum(rows, [])
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def needed_hashes(self, leafnum, include_leaf=False):
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"""Which new hashes do I need to validate a given data block?
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I am much like HashTree.needed_hashes(), except that I don't include
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hashes that I already know about. When needed_hashes() is called on
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an empty IncompleteHashTree, it will return the same set as a
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HashTree of the same size. But later, once hashes have been added
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with set_hashes(), I will ask for fewer hashes, since some of the
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necessary ones have already been set.
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"""
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maybe_needed = set(self.needed_for(self.first_leaf_num + leafnum))
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if include_leaf:
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maybe_needed.add(self.first_leaf_num + leafnum)
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return set([i for i in maybe_needed if self[i] is None])
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def _name_hash(self, i):
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name = "[%d of %d]" % (i, len(self))
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if i >= self.first_leaf_num:
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leafnum = i - self.first_leaf_num
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numleaves = len(self) - self.first_leaf_num
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name += " (leaf [%d] of %d)" % (leafnum, numleaves)
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return name
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def set_hashes(self, hashes=None, leaves=None):
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"""Add a bunch of hashes to the tree.
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I will validate these to the best of my ability. If I already have a
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copy of any of the new hashes, the new values must equal the existing
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ones, or I will raise BadHashError. If adding a hash allows me to
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compute a parent hash, those parent hashes must match or I will raise
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BadHashError. If I raise BadHashError, I will forget about all the
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hashes that you tried to add, leaving my state exactly the same as
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before I was called. If I return successfully, I will remember all
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those hashes.
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I insist upon being able to validate all of the hashes that were
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given to me. If I cannot do this because I'm missing some hashes, I
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will raise NotEnoughHashesError (and forget about all the hashes that
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you tried to add). Note that this means that the root hash must
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either be included in 'hashes', or it must have been provided at some
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point in the past.
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'leaves' is a dictionary uses 'leaf index' values, which range from 0
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(the left-most leaf) to num_leaves-1 (the right-most leaf), and form
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the base of the tree. 'hashes' uses 'hash_index' values, which range
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from 0 (the root of the tree) to 2*num_leaves-2 (the right-most
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leaf). leaf[i] is the same as hash[num_leaves-1+i].
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The best way to use me is to start by obtaining the root hash from
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some 'good' channel and populate me with it:
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iht = IncompleteHashTree(numleaves)
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roothash = trusted_channel.get_roothash()
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iht.set_hashes(hashes={0: roothash})
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Then use the 'bad' channel to obtain data block 0 and the
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corresponding hash chain (a dict with the same hashes that
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needed_hashes(0) tells you, e.g. {0:h0, 2:h2, 4:h4, 8:h8} when
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len(L)=8). Hash the data block to create leaf0, then feed everything
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into set_hashes() and see if it raises an exception or not::
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otherhashes = untrusted_channel.get_hashes()
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# otherhashes.keys() should == iht.needed_hashes(leaves=[0])
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datablock0 = untrusted_channel.get_data(0)
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leaf0 = HASH(datablock0)
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# HASH() is probably hashutil.tagged_hash(tag, datablock0)
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iht.set_hashes(otherhashes, leaves={0: leaf0})
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If the set_hashes() call doesn't raise an exception, the data block
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was valid. If it raises BadHashError, then either the data block was
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corrupted or one of the received hashes was corrupted. If it raises
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NotEnoughHashesError, then the otherhashes dictionary was incomplete.
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"""
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if hashes is None:
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hashes = {}
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if leaves is None:
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leaves = {}
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assert isinstance(hashes, dict)
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for h in hashes.values():
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assert isinstance(h, bytes)
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assert isinstance(leaves, dict)
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for h in leaves.values():
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assert isinstance(h, bytes)
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new_hashes = hashes.copy()
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for leafnum,leafhash in leaves.items():
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hashnum = self.first_leaf_num + leafnum
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if hashnum in new_hashes:
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if new_hashes[hashnum] != leafhash:
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raise BadHashError("got conflicting hashes in my "
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"arguments: leaves[%d] != hashes[%d]"
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% (leafnum, hashnum))
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new_hashes[hashnum] = leafhash
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remove_upon_failure = set() # we'll remove these if the check fails
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# visualize this method in the following way:
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# A: start with the empty or partially-populated tree as shown in
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# the HashTree docstring
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# B: add all of our input hashes to the tree, filling in some of the
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# holes. Don't overwrite anything, but new values must equal the
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# existing ones. Mark everything that was added with a red dot
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# (meaning "not yet validated")
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# C: start with the lowest/deepest level. Pick any red-dotted node,
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# hash it with its sibling to compute the parent hash. Add the
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# parent to the tree just like in step B (if the parent already
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# exists, the values must be equal; if not, add our computed
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# value with a red dot). If we have no sibling, throw
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# NotEnoughHashesError, since we won't be able to validate this
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# node. Remove the red dot. If there was a red dot on our
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# sibling, remove it too.
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# D: finish all red-dotted nodes in one level before moving up to
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# the next.
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# E: if we hit NotEnoughHashesError or BadHashError before getting
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# to the root, discard every hash we've added.
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try:
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num_levels = depth_of(len(self)-1)
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# hashes_to_check[level] is set(index). This holds the "red dots"
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# described above
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hashes_to_check = [set() for level in range(num_levels+1)]
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# first we provisionally add all hashes to the tree, comparing
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# any duplicates
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for i,h in new_hashes.items():
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if self[i]:
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if self[i] != h:
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raise BadHashError("new hash %r does not match "
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"existing hash %r at %r"
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% (base32.b2a(h),
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base32.b2a(self[i]),
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self._name_hash(i)))
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else:
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level = depth_of(i)
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hashes_to_check[level].add(i)
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self[i] = h
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remove_upon_failure.add(i)
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for level in reversed(range(len(hashes_to_check))):
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this_level = hashes_to_check[level]
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while this_level:
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i = this_level.pop()
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if i == 0:
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# The root has no sibling. How lonely. You can't
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# really *check* the root; you either accept it
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# because the caller told you what it is by including
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# it in hashes, or you accept it because you
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# calculated it from its two children. You probably
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# want to set the root (from a trusted source) before
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# adding any children from an untrusted source.
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continue
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siblingnum = self.sibling(i)
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if self[siblingnum] is None:
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# without a sibling, we can't compute a parent, and
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# we can't verify this node
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raise NotEnoughHashesError("unable to validate [%d]"%i)
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parentnum = self.parent(i)
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# make sure we know right from left
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leftnum, rightnum = sorted([i, siblingnum])
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new_parent_hash = pair_hash(self[leftnum], self[rightnum])
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if self[parentnum]:
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if self[parentnum] != new_parent_hash:
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raise BadHashError("h([%d]+[%d]) != h[%d]" %
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(leftnum, rightnum, parentnum))
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else:
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self[parentnum] = new_parent_hash
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remove_upon_failure.add(parentnum)
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parent_level = depth_of(parentnum)
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assert parent_level == level-1
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hashes_to_check[parent_level].add(parentnum)
|
|
|
|
# our sibling is now as valid as this node
|
|
this_level.discard(siblingnum)
|
|
# we're done!
|
|
|
|
except (BadHashError, NotEnoughHashesError):
|
|
for i in remove_upon_failure:
|
|
self[i] = None
|
|
raise
|