x-only just makes it more confusing, but it's a separate thing right. The general pattern for DLEQ is:
proving P_1 and P_2 have same DL for A,B:
s = r + ex, where e = H(R_1,R_2,P_1, P_2) where R_1 = rA and R_2 = rB and P_1 = xA and P_2=xB.
Then return s, e, R_1, R_2
Then verify with
sA =?= R_1 + eP_1, sB =?= R_2 + eP_2, with e calculated as above.
waxwing /
npub1va…knuu7
2023-03-24 21:43:09
in reply to nevent1q…qf8v
Author Public Key
npub1vadcfln4ugt2h9ruwsuwu5vu5am4xaka7pw6m7axy79aqyhp6u5q9knuu7Published at
2023-03-24 21:43:09Event JSON
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"created_at": 1679694189,
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"content": "x-only just makes it more confusing, but it's a separate thing right. The general pattern for DLEQ is:\n\nproving P_1 and P_2 have same DL for A,B:\n\ns = r + ex, where e = H(R_1,R_2,P_1, P_2) where R_1 = rA and R_2 = rB and P_1 = xA and P_2=xB.\n\nThen return s, e, R_1, R_2\n\nThen verify with \n\nsA =?= R_1 + eP_1, sB =?= R_2 + eP_2, with e calculated as above.",
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