In ECDSA, why is (r,−s mod n) complementary to (r, s)?
I am trying to find resources in previous malleability posts, but couldn't find derivations/proofs of this fact or how the exact low-s value is derived. Any pointers would greatly appreciated.
ecdsa transaction-malleability
asked yesterday
James C.
888111
add a comment |
I am trying to find resources in previous malleability posts, but couldn't find derivations/proofs of this fact or how the exact low-s value is derived. Any pointers would greatly appreciated.
ecdsa transaction-malleability
asked yesterday
James C.
888111
add a comment |
I am trying to find resources in previous malleability posts, but couldn't find derivations/proofs of this fact or how the exact low-s value is derived. Any pointers would greatly appreciated.
ecdsa transaction-malleability
asked yesterday
James C.
888111
I am trying to find resources in previous malleability posts, but couldn't find derivations/proofs of this fact or how the exact low-s value is derived. Any pointers would greatly appreciated.
ecdsa transaction-malleability
ecdsa transaction-malleability
asked yesterday
James C.
888111
asked yesterday
James C.
888111
asked yesterday
James C.
888111
asked yesterday
James C.
888111
asked yesterday
James C.
888111
888111
add a comment |
add a comment |
1 Answer
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ECDSA signatures are pairs (r,s) such that r = x(m/sG + r/sP) mod n, where P is the public key and m is the message digest. x() in that equation means "the X coordinate of".
In that equation, if you substitute s = -s', you get *r = x(m/(-s')*G + r/(-s)P) mod n, or *r = x(-(m/s'*G + r/s'P)).
However, it is true that for any point Q, x(Q) = x(-Q), as negating a point only affects the Y coordinate. Thus, *r = x(m/s'*G + r/s'P) mod n, or (r,s') is valid signature whenever (r,s) is.
answered yesterday
Pieter Wuille
45.5k393154
Thank you very much. I understand negating a scalar over ff, but not why that negated scalar * point will result in a point with the same x-coord as scalar * point.
– James C.
yesterday
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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ECDSA signatures are pairs (r,s) such that r = x(m/sG + r/sP) mod n, where P is the public key and m is the message digest. x() in that equation means "the X coordinate of".
In that equation, if you substitute s = -s', you get *r = x(m/(-s')*G + r/(-s)P) mod n, or *r = x(-(m/s'*G + r/s'P)).
However, it is true that for any point Q, x(Q) = x(-Q), as negating a point only affects the Y coordinate. Thus, *r = x(m/s'*G + r/s'P) mod n, or (r,s') is valid signature whenever (r,s) is.
answered yesterday
Pieter Wuille
45.5k393154
Thank you very much. I understand negating a scalar over ff, but not why that negated scalar * point will result in a point with the same x-coord as scalar * point.
– James C.
yesterday
add a comment |
ECDSA signatures are pairs (r,s) such that r = x(m/sG + r/sP) mod n, where P is the public key and m is the message digest. x() in that equation means "the X coordinate of".
In that equation, if you substitute s = -s', you get *r = x(m/(-s')*G + r/(-s)P) mod n, or *r = x(-(m/s'*G + r/s'P)).
However, it is true that for any point Q, x(Q) = x(-Q), as negating a point only affects the Y coordinate. Thus, *r = x(m/s'*G + r/s'P) mod n, or (r,s') is valid signature whenever (r,s) is.
answered yesterday
Pieter Wuille
45.5k393154
Thank you very much. I understand negating a scalar over ff, but not why that negated scalar * point will result in a point with the same x-coord as scalar * point.
– James C.
yesterday
add a comment |
ECDSA signatures are pairs (r,s) such that r = x(m/sG + r/sP) mod n, where P is the public key and m is the message digest. x() in that equation means "the X coordinate of".
In that equation, if you substitute s = -s', you get *r = x(m/(-s')*G + r/(-s)P) mod n, or *r = x(-(m/s'*G + r/s'P)).
However, it is true that for any point Q, x(Q) = x(-Q), as negating a point only affects the Y coordinate. Thus, *r = x(m/s'*G + r/s'P) mod n, or (r,s') is valid signature whenever (r,s) is.
answered yesterday
Pieter Wuille
45.5k393154
ECDSA signatures are pairs (r,s) such that r = x(m/sG + r/sP) mod n, where P is the public key and m is the message digest. x() in that equation means "the X coordinate of".
In that equation, if you substitute s = -s', you get *r = x(m/(-s')*G + r/(-s)P) mod n, or *r = x(-(m/s'*G + r/s'P)).
However, it is true that for any point Q, x(Q) = x(-Q), as negating a point only affects the Y coordinate. Thus, *r = x(m/s'*G + r/s'P) mod n, or (r,s') is valid signature whenever (r,s) is.
answered yesterday
Pieter Wuille
45.5k393154
answered yesterday
Pieter Wuille
45.5k393154
answered yesterday
Pieter Wuille
45.5k393154
answered yesterday
Pieter Wuille
45.5k393154
45.5k393154
Thank you very much. I understand negating a scalar over ff, but not why that negated scalar * point will result in a point with the same x-coord as scalar * point.
– James C.
yesterday
add a comment |
Thank you very much. I understand negating a scalar over ff, but not why that negated scalar * point will result in a point with the same x-coord as scalar * point.
– James C.
yesterday
Thank you very much. I understand negating a scalar over ff, but not why that negated scalar * point will result in a point with the same x-coord as scalar * point.
– James C.
yesterday
Thank you very much. I understand negating a scalar over ff, but not why that negated scalar * point will result in a point with the same x-coord as scalar * point.
– James C.
yesterday
add a comment |
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