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Merge pull request #129 from jonasschnelli/bip32_fenn_parenthesis

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Bip32, fix missing parenthesis
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Wladimir J. van der Laan 2014-12-18 08:19:52 +00:00
commit fbc7e67783
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@ -104,7 +104,7 @@ This is not possible.
The next step is cascading several CKD constructions to build a tree. We start with one root, the master extended key m. By evaluating CKDpriv(m,i) for several values of i, we get a number of level-1 derived nodes. As each of these is again an extended key, CKDpriv can be applied to those as well. The next step is cascading several CKD constructions to build a tree. We start with one root, the master extended key m. By evaluating CKDpriv(m,i) for several values of i, we get a number of level-1 derived nodes. As each of these is again an extended key, CKDpriv can be applied to those as well.
To shorten notation, we will write CKDpriv(CKDpriv(CKDpriv(m,3<sub>H</sub>),2),5) as m/3<sub>H</sub>/2/5. Equivalently for public keys, we write CKDpub(CKDpub(CKDpub(M,3),2,5) as M/3/2/5. This results in the following identities: To shorten notation, we will write CKDpriv(CKDpriv(CKDpriv(m,3<sub>H</sub>),2),5) as m/3<sub>H</sub>/2/5. Equivalently for public keys, we write CKDpub(CKDpub(CKDpub(M,3),2),5) as M/3/2/5. This results in the following identities:
* N(m/a/b/c) = N(m/a/b)/c = N(m/a)/b/c = N(m)/a/b/c = M/a/b/c. * N(m/a/b/c) = N(m/a/b)/c = N(m/a)/b/c = N(m)/a/b/c = M/a/b/c.
* N(m/a<sub>H</sub>/b/c) = N(m/a<sub>H</sub>/b)/c = N(m/a<sub>H</sub>)/b/c. * N(m/a<sub>H</sub>/b/c) = N(m/a<sub>H</sub>/b)/c = N(m/a<sub>H</sub>)/b/c.
However, N(m/a<sub>H</sub>) cannot be rewritten as N(m)/a<sub>H</sub>, as the latter is not possible. However, N(m/a<sub>H</sub>) cannot be rewritten as N(m)/a<sub>H</sub>, as the latter is not possible.