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 Jan 5 at 20:45
James C.James C.
1,8751214
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 Jan 5 at 20:45
James C.James C.
1,8751214
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 Jan 5 at 20:45
James C.James C.
1,8751214
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 Jan 5 at 20:45
James C.James C.
1,8751214
asked Jan 5 at 20:45
James C.James C.
1,8751214
asked Jan 5 at 20:45
James C.James C.
1,8751214
asked Jan 5 at 20:45
James C.James C.
1,8751214
asked Jan 5 at 20:45
James C.James C.
1,8751214
1,8751214
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 Jan 5 at 22:20
Pieter WuillePieter Wuille
47.5k399160
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.
Jan 5 at 22:31
add a comment |
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1 Answer
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1 Answer
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oldest
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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 Jan 5 at 22:20
Pieter WuillePieter Wuille
47.5k399160
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.
Jan 5 at 22:31
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 Jan 5 at 22:20
Pieter WuillePieter Wuille
47.5k399160
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.
Jan 5 at 22:31
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 Jan 5 at 22:20
Pieter WuillePieter Wuille
47.5k399160
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 Jan 5 at 22:20
Pieter WuillePieter Wuille
47.5k399160
answered Jan 5 at 22:20
Pieter WuillePieter Wuille
47.5k399160
answered Jan 5 at 22:20
Pieter WuillePieter Wuille
47.5k399160
answered Jan 5 at 22:20
Pieter WuillePieter Wuille
47.5k399160
47.5k399160
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.
Jan 5 at 22:31
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.
Jan 5 at 22:31
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.
Jan 5 at 22:31
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.
Jan 5 at 22:31
add a comment |
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