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--- This message came from the IEEE 802.11 Working Group Reflector ---
Hi Rob, On 11/1/19, 12:57 PM, "Rob Sun" <Rob.Sun@xxxxxxxxxx> wrote: --- This message came from the IEEE 802.11 Working Group Reflector ---
Hi Rene: Glad to hear from you, for long time! The equation f(b/(n*a)):=b^3/(n*a)^3+ b/n + b, n is representing theta, and theta is nQR (non-quadratic residue), I think the f(u0) is also a nQR over GF(q).
Am I missing something here? The last of the criteria to find z says "is a quadratic residue." That has to be the case because it's assigned to x1 in the case that m = 0. It has to be square. regards, Dan. Thanks Rob From: Rene Struik [mailto:rstruik.ext@xxxxxxxxx]
--- This message came from the IEEE 802.11 Working Group Reflector ---
Hi Dan: One more glitch (which is also easy to correct): In my email of yesterday (see below), I suggested I did not know the rationale for requirement 1)iv) {resp. 2)iv)} of Section 14.4.3.2.3.
I checked the internet draft [1] you referenced and it seems that this specific condition is an escape clause in case one would otherwise divide by zero. The condition is then that for u0:=b/(theta*a), this would
indeed be a point of the Weierstrass curve with defining equation y^2=f(x):=x^3+a*x+b or, in other words, that f(u0) would be a square in GF(q). If so, the language in Section 14.4.3.2.3, 1)iv) should read - in your notation - f(b/(n*a)):=b^3/(n*a)^3+ b/n
+ b is a square in the field in question. {Note the b^3 instead of b here}. Similar changes elsewhere. BTW - a much simpler escape clause would be to map w:=z*u^2 to a fixed point P0 of the curve in case w is equal to 0 or -1 (or avoid this from happening by constraining the input values for u [if z=-1, this corresponds
to avoiding u=0 or u=1]). Ref: [1] draft-irtf-cfrg-hash-to-curve-04 Best regards, Rene On 10/31/2019 2:53 PM, Rene Struik wrote:
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