Double the unpaired one. https://en.bitcoin.it/wiki/Protocol_specification#Merkle_Trees
Ahh right. I read that to mean that the "c" input was used twice rather than saying how the tree is constructed.
Thanks.
Topic: merkle tree order of hashes (Read 948 times)
Ahh right. I read that to mean that the "c" input was used twice rather than saying how the tree is constructed.
Thanks.
According to the paper, if you have 4 transactions, then the procedure is
12 = Hash(1,2)
34 = Hash(3,4)
root = hash(12,34)
However, it doesn't say how to handle non-power of 2 trees.
For 5 transactions, it could be something like:
Pass 1:
12 = Hash(1,2)
34 = Hash(3,4)
5_ = 5
Pass 2:
1234 = Hash(12,34)
5___ = 5_
Pass 3
root = Hash(1234, 5___)
However, that gives an unbalanced tree.
Another option would be
1,2,3,4,5
becomes
5,12,34
and then
512,34
and then
51234
Is there some flexibility? I couldn't see how the tree was defined in block explorer.
12 = Hash(1,2)
34 = Hash(3,4)
root = hash(12,34)
However, it doesn't say how to handle non-power of 2 trees.
For 5 transactions, it could be something like:
Pass 1:
12 = Hash(1,2)
34 = Hash(3,4)
5_ = 5
Pass 2:
1234 = Hash(12,34)
5___ = 5_
Pass 3
root = Hash(1234, 5___)
However, that gives an unbalanced tree.
Another option would be
1,2,3,4,5
becomes
5,12,34
and then
512,34
and then
51234
Is there some flexibility? I couldn't see how the tree was defined in block explorer.
Double the unpaired one. https://en.bitcoin.it/wiki/Protocol_specification#Merkle_Trees
According to the paper, if you have 4 transactions, then the procedure is
12 = Hash(1,2)
34 = Hash(3,4)
root = hash(12,34)
However, it doesn't say how to handle non-power of 2 trees.
For 5 transactions, it could be something like:
Pass 1:
12 = Hash(1,2)
34 = Hash(3,4)
5_ = 5
Pass 2:
1234 = Hash(12,34)
5___ = 5_
Pass 3
root = Hash(1234, 5___)
However, that gives an unbalanced tree.
Another option would be
1,2,3,4,5
becomes
5,12,34
and then
512,34
and then
51234
Is there some flexibility? I couldn't see how the tree was defined in block explorer.
12 = Hash(1,2)
34 = Hash(3,4)
root = hash(12,34)
However, it doesn't say how to handle non-power of 2 trees.
For 5 transactions, it could be something like:
Pass 1:
12 = Hash(1,2)
34 = Hash(3,4)
5_ = 5
Pass 2:
1234 = Hash(12,34)
5___ = 5_
Pass 3
root = Hash(1234, 5___)
However, that gives an unbalanced tree.
Another option would be
1,2,3,4,5
becomes
5,12,34
and then
512,34
and then
51234
Is there some flexibility? I couldn't see how the tree was defined in block explorer.
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