25 KiB
Merklist¶
Using matrix multiplication's associativity and non-commutativity properties provides a natural definition of a cryptographic hash / digest / summary of an ordered list of elements. Due to the non-commutativity property, lists that only differ in element order result in a different summary. Due to the associativity property, arbitrarily divided adjacent sub-lists can be summarized independently and combined to quickly find the summary of their concatenation. This definition provides exactly the properties needed to define a list, and does not impose any unnecessary structure that could cause two equivalent lists to produce different summaries. The name Merklist is intended to be reminicent of other hash-based data structures like Merkle Tree and Merklix Tree.
Definition¶
This definition of a hash of a list of elements is pretty simple:
- A list element is an arbitrary buffer of bytes. Any length, any content. Just bytes.
- A list, then, is a sequence of such elements.
- The hash of a list element is the cryptographic hash of its bytes, formatted into a square matrix with byte elements. (More details later.)
- The hash of a list is reduction by matrix multiplication of the hashes of all the list elements in the same order as they appear in the list.
- The hash of a list with 0 elements is the identity matrix.
This construction has a couple notable concequences:
- The hash of a list with only one item is just the hash of the item itself.
- You can calculate the hash of any list concatenated with itself by matrix multiplication of the the hash with itself. This works for single elements as well as arbitrarily long lists.
- A list can have multiple copies of the same list item, and swapping them does not affect the list hash. Consider how swapping the first two elements in
[1, 1, 2]doesn't change it. - Concatenating two lists is accomplished by matrix multiplication of their hashes, in the correct order.
- Appending or prepending lists of 0 elements yields the same hash, as expected.
Lets explore this definition in more detail with a simple implementation in python+numpy.
# setup import hashlib import numpy as np from functools import reduce def assert_equal(a, b): return np.testing.assert_equal(a, b) def assert_not_equal(a, b): return np.testing.assert_raises(AssertionError, np.testing.assert_equal, a, b)
The hash of a list element - hash_m/1¶
The function hash_m/1 takes a buffer of bytes as its first argument, and returns the sha512 hash of the bytes formatted as an 8×8 2-d array of 8-bit unsigned integers with wrapping overflow. This is the hash of a list element consisting of those bytes. Based on a shallow wikipedia dive, someone familiar with linear algebra might say it's a matrix ring, $R_{256}^{8×8}$. Not coincidentally, sha512 outputs 512 bits = 64 bytes = 8 * 8 array of bytes, how convenient. (In fact, that might even be the primary reason why I chose sha512!)
def hash_m(e): hash_bytes = list(hashlib.sha512(e).digest())[:64] # hash the bytes e, convert the digest into a list of 64 bytes return np.array(hash_bytes, dtype=np.uint8).reshape((8,8)) # convert the digest bytes into a numpy array with the appropriate data type and shape
8×8 seems big compared to 3×3 or 4×4 matrixes. The values are as random as you might expect a cryptographic hash to be, and range from 0-255:
print(hash_m(b"Hello A")) print() print(hash_m(b"Hello B"))
[[ 14 184 108 217 131 164 222 93] [132 227 82 144 111 178 195 109] [ 25 250 155 17 131 183 151 217] [212 60 138 36 0 60 115 181] [ 51 0 87 43 93 252 56 61] [108 239 175 222 23 142 41 216] [203 98 234 13 65 169 255 240] [ 46 127 15 167 112 153 222 94]] [[ 63 144 188 5 57 146 32 56] [ 27 189 98 140 113 194 70 87] [115 21 136 27 116 167 85 48] [ 29 162 119 29 104 32 145 241] [166 197 57 165 132 213 50 202] [ 48 71 33 19 230 26 58 164] [242 172 65 202 193 50 193 141] [206 110 165 129 52 132 250 73]]
The hash of a list - mul_m/2¶
Ok so we've got our element hashes, how do we combine them to construct the hash of a list? We defined the hash of the list to be reduction by matrix multiplication of the hash of each element:
def mul_m(he1, he2): return np.matmul(he1, he2, dtype=np.uint8) # just, like, multiply them
Consider an example:
# list1 contains 3 elements elements = [b"A", b"Hello", b"World"] # first hash each element element_hashes = [hash_m(e) for e in elements] # get the hash of the list by reducing the hashes by matrix multiplication list_hash1 = mul_m(mul_m(element_hashes[0], element_hashes[1]), element_hashes[2]) # an alternative way to write the reduction list_hash2 = reduce(mul_m, element_hashes)
print("List of elements:") print(elements) print("\nHash of each element:") print(element_hashes) print("\nHash of full list:") print(list_hash1) assert_equal(list_hash1, list_hash2)
List of elements:
[b'A', b'Hello', b'World']
Hash of each element:
[array([[ 33, 180, 244, 189, 158, 100, 237, 53],
[ 92, 62, 182, 118, 162, 142, 190, 218],
[246, 216, 241, 123, 220, 54, 89, 149],
[179, 25, 9, 113, 83, 4, 64, 128],
[ 81, 107, 208, 131, 191, 204, 230, 97],
[ 33, 163, 7, 38, 70, 153, 76, 132],
[ 48, 204, 56, 43, 141, 197, 67, 232],
[ 72, 128, 24, 59, 248, 86, 207, 245]], dtype=uint8), array([[ 54, 21, 248, 12, 157, 41, 62, 215],
[ 64, 38, 135, 249, 75, 34, 213, 142],
[ 82, 155, 140, 199, 145, 111, 143, 172],
[127, 221, 247, 251, 213, 175, 76, 247],
[119, 211, 215, 149, 167, 160, 10, 22],
[191, 126, 127, 63, 185, 86, 30, 233],
[186, 174, 72, 13, 169, 254, 122, 24],
[118, 158, 113, 136, 107, 3, 243, 21]], dtype=uint8), array([[142, 167, 115, 147, 164, 42, 184, 250],
[146, 80, 15, 176, 119, 169, 80, 156],
[195, 43, 201, 94, 114, 113, 46, 250],
[ 17, 110, 218, 242, 237, 250, 227, 79],
[187, 104, 46, 253, 214, 197, 221, 19],
[193, 23, 224, 139, 212, 170, 239, 113],
[ 41, 29, 138, 172, 226, 248, 144, 39],
[ 48, 129, 208, 103, 124, 22, 223, 15]], dtype=uint8)]
Hash of full list:
[[178 188 57 157 60 136 190 127]
[ 40 234 254 224 38 46 250 52]
[156 72 193 136 219 98 28 4]
[197 2 43 132 132 232 254 198]
[ 93 64 113 215 2 246 130 192]
[ 91 107 85 13 149 60 19 173]
[ 84 77 244 98 0 239 123 17]
[ 58 112 98 250 163 20 27 6]]
What does this give us? Generally speaking, multiplying two square matrixes $M_1×M_2$ gives us at least these two properties:
- Associativity - Associativity enables you to reduce a computation using any partitioning because all partitionings yield the same result. Addition is associative $(1+2)+3 = 1+(2+3)$, subtraction is not $(5-3)-2\neq5-(3-2)$. (Associative property)
- Non-Commutativity - Commutativity allows you to swap elements without affecting the result. Addition is commutative $1+2 = 2+1$, but division is not $1\div2 \neq2\div1$. And neither is matrix multiplication. (Commutative property)
This is an unusual combination of properties for an operation, at least not a combination encountered under normal algebra operations:
| associative | commutative | |
|---|---|---|
| + | ✅ | ✅ |
| * | ✅ | ✅ |
| - | ❌ | ❌ |
| / | ❌ | ❌ |
| exp | ❌ | ❌ |
| M×M | ✅ | ❌ |
Upon consideration, these are the exact properties that one would want in order to define the hash of a list of items. Non-commutativity enables the order of elements in the list to be well-defined, since swapping different elements produces a different hash. Associativity enables caching the summary of an arbitrary sublist; I expect that doing this heirarchally on a huge list enables an algorithm to calculate the hash of any sublist at the cost of O(log(N)) time and space.
Lets sanity-check that these properties can hold for the construction described above.
Associativity¶
f1 = hash_m(b"Hello A") f2 = hash_m(b"Hello B") f3 = hash_m(b"Hello C")
# x is calculated by association ((f1 × f2) × f3) x = np.matmul(np.matmul(f1, f2), f3) # y is calculated by association (f1 × (f2 × f3)) y = np.matmul(f1, np.matmul(f2, f3)) # observe that they produce the same result assert_equal(x, y)
print(x) print() print(y)
[[ 58 12 144 134 100 158 159 51] [ 73 206 202 190 87 79 223 2] [210 122 142 117 37 148 106 45] [175 146 187 223 235 171 64 226] [149 85 203 87 92 251 243 206] [ 18 252 160 103 125 251 181 133] [191 132 220 104 213 154 34 154] [127 197 95 87 166 3 22 3]] [[ 58 12 144 134 100 158 159 51] [ 73 206 202 190 87 79 223 2] [210 122 142 117 37 148 106 45] [175 146 187 223 235 171 64 226] [149 85 203 87 92 251 243 206] [ 18 252 160 103 125 251 181 133] [191 132 220 104 213 154 34 154] [127 197 95 87 166 3 22 3]]
Non-Commutativity¶
# x is f1 × f2 x = np.matmul(f1, f2) # y is f2 × f1 y = np.matmul(f2, f1) # observe that they produce different results assert_not_equal(x, y)
print(x) print() print(y)
[[ 87 79 149 131 148 247 195 90] [249 84 195 58 142 133 211 15] [177 93 69 254 240 234 97 37] [ 46 84 76 253 55 200 43 236] [ 21 84 99 157 55 148 170 2] [168 123 6 250 64 144 54 242] [230 78 164 76 30 29 214 68] [ 47 183 156 239 157 177 192 184]] [[149 18 239 238 84 188 191 109] [239 150 214 235 59 161 9 133] [ 89 174 59 14 70 113 124 243] [ 66 113 176 124 227 247 17 25] [247 138 152 181 177 143 184 97] [113 249 199 153 154 75 45 105] [121 201 225 42 249 213 180 244] [ 85 31 72 28 181 182 140 176]]
Other functions¶
# Create a list of 1024 elements and reduce them one by one list1 = [hash_m(b"A") for _ in range(0, 1024)] hash1 = reduce(mul_m, list1) # Take a starting element and square/double it 10 times. With 1 starting element over 10 doublings = 1024 elements hash2 = reduce((lambda m, _ : mul_m(m, m)), range(0, 10), hash_m(b"A")) # Observe that these two methods of calculating the hash have the same result assert_equal(hash1, hash2) # lets call it double def double_m(m, d=1): return reduce((lambda m, _ : mul_m(m, m)), range(0, d), m) assert_equal(hash1, double_m(hash_m(b"A"), 10)) def identity_m(): return np.identity(8, dtype=np.uint8) # generalize to any length, not just doublings, performed in ln(N) matmuls def repeat_m(m, n): res = identity_m() while n > 0: # concatenate the current doubling iff the bit representing this doubling is set if n & 1: res = mul_m(res, m) n >>= 1 m = mul_m(m, m) # double matrix m # print(s) return res # repeat_m can do the same as double_m assert_equal(hash1, repeat_m(hash_m(b"A"), 1024)) # but it can also repeat any number of times hash3 = reduce(mul_m, (hash_m(b"A") for _ in range(0, 3309))) assert_equal(hash3, repeat_m(hash_m(b"A"), 3309)) # Even returns a sensible result when requesting 0 elements assert_equal(identity_m(), repeat_m(hash_m(b"A"), 0)) # make helper for reducing an iterable of hashes def reduce_m(am): return reduce(mul_m, am)
print(hash1) print() print(hash2) print() print(np.identity(8,"B"))
[[ 68 252 159 3 14 52 199 199] [136 124 6 34 58 174 206 54] [ 3 234 2 13 120 240 7 163] [102 47 66 61 87 234 246 72] [ 19 135 80 115 75 242 242 5] [244 165 250 28 76 43 188 254] [233 46 187 39 151 241 175 130] [132 138 6 215 20 132 89 33]] [[ 68 252 159 3 14 52 199 199] [136 124 6 34 58 174 206 54] [ 3 234 2 13 120 240 7 163] [102 47 66 61 87 234 246 72] [ 19 135 80 115 75 242 242 5] [244 165 250 28 76 43 188 254] [233 46 187 39 151 241 175 130] [132 138 6 215 20 132 89 33]] [[1 0 0 0 0 0 0 0] [0 1 0 0 0 0 0 0] [0 0 1 0 0 0 0 0] [0 0 0 1 0 0 0 0] [0 0 0 0 1 0 0 0] [0 0 0 0 0 1 0 0] [0 0 0 0 0 0 1 0] [0 0 0 0 0 0 0 1]]
Fun with associativity¶
Does the hash of a list change even when swapping two elements in the middle of a very long list?
a = hash_m(b"A") b = hash_m(b"B") a499 = repeat_m(a, 499) a500 = repeat_m(a, 500) # this should work because they're all a's assert_equal(reduce_m([a, a499]), a500) assert_equal(reduce_m([a499, a]), a500) # these are lists of 999 elements of a, with one b at position 500 (x) or 501 (y) x = reduce_m([a499, b, a500]) y = reduce_m([a500, b, a499]) # shifting the b by one element changed the hash assert_not_equal(x, y)
Flex that associativity (a × (a499 × b × a500) × (a500 × b × a499) × a) = (a500 × b × (a500 × a500) × b × a500)
assert_equal(reduce_m([a, x, y, a]), reduce_m([a500, b, repeat_m(a500, 2), b, a500]))
Unknowns¶
This appears to me to be a reasonable way to define the hash of a list. The mathematical definition of a list aligns very nicely with the properties offered by matrix multiplication. But is it appropriate to use for the same things that a Merkle Tree would be? The big questions are related to the valuable properties of hash functions:
- Given a merklist summary or sublist summaries of it, can you derive the hashes of elements or their order? (Elements themselves are protected by the preimage resistance of the underlying hash function.)
- If yes, when is that a problem?
- Given a merklist summary but not the elements, is it possible to produce a different list of elements that hash to the same summary? (~preimage resistance)
- Is it possible to predictably alter the merklist summary by concatenating it with some other sublist of real elements?
- Are there other desirable security properties that would be valuable for a list hash?
- Is there a better choice of hash function as a primitive than sha512?
- Is there a better choice of reduction function that still retains associativity+non-commutativity than simple matmul?
- Is there a more appropriate size than an 8x8 matrix / 64 bytes to represent merklist summaries?
Matrixes are well-studied objects, perhaps such information is already known. If you know something about deriving the preimage of the multiplication of a matrix ring, $R_{256}^{8×8}$, I would be very interested to know.
What's next?¶
***If** this construction has the appropriate security properties*, it seems to be a better merkle tree in all respects. Any use of a merkle tree could be replaced with this, and it could enable use-cases where merkle trees aren't useful. Some examples of what I think might be possible:
- Using a Merklist with a sublist summary tree structure enables creating a $O(1)$-sized 'Merklist Proof' that can verify the addition and subtraction of any number of elements at any single point in the list using only $O(log(N))$ time and $O(log(N))$ static space. As a bonus the proof generator and verifier can have totally different tree structures and can still communicate the proof successfully.
- Using a Merklist summary tree you can create a consistent hash of any ordered key-value store (like a btree) that can be maintained incrementally inline with regular node updates, e.g. as part of a LSM-tree. This could facilitate verification and sync between database replicas.
- The sublist summary tree structure can be as dense or sparse as desired. You could summarize down to pairs of elements akin to a merkle tree, but you could also summarize a compressed sublist of hundreds or even millions of elements with a single hash. Of course, calculating or verifying a proof of changes to the middle of that sublist would require rehashing the whole sublist, but this turns it from a fixed structure into a tuneable parameter.
- If all possible elements had an easily calculatable inverse, that would enable "subtracting" an element by inserting its inverse in front of it. That would basically extend the group from a ring into a field, and might have interesting implications.
- For example you could define a cryptographically-secure rolling hash where advancing either end can be calculated in
O(1)time.
- For example you could define a cryptographically-secure rolling hash where advancing either end can be calculated in
To be continued...