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Fuentes et al. algorithm for computing hard part of final exponentiation #40

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Fuentes et al. algorithm for computing hard part of final exponentiation #40

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65 changes: 35 additions & 30 deletions optate.go
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Original file line number Diff line number Diff line change
Expand Up @@ -208,6 +208,7 @@ func miller(q *twistPoint, p *curvePoint) *gfP12 {
func finalExponentiation(in *gfP12) *gfP12 {
t1 := &gfP12{}

// Easy part
// This is the p^6-Frobenius
t1.x.Neg(&in.x)
t1.y.Set(&in.y)
Expand All @@ -219,41 +220,45 @@ func finalExponentiation(in *gfP12) *gfP12 {
t2 := (&gfP12{}).FrobeniusP2(t1)
t1.Mul(t1, t2)

fp := (&gfP12{}).Frobenius(t1)
fp2 := (&gfP12{}).FrobeniusP2(t1)
fp3 := (&gfP12{}).Frobenius(fp2)

fu := (&gfP12{}).Exp(t1, u)
fu2 := (&gfP12{}).Exp(fu, u)
fu3 := (&gfP12{}).Exp(fu2, u)

y3 := (&gfP12{}).Frobenius(fu)
fu2p := (&gfP12{}).Frobenius(fu2)
fu3p := (&gfP12{}).Frobenius(fu3)
y2 := (&gfP12{}).FrobeniusP2(fu2)
// Hard Part
// Follows Fuentes et al. algorithm (https://link.springer.com/content/pdf/10.1007/978-3-642-28496-0_25.pdf)
// for computing the hard part based on the fact that f ^ d' is also a pairing and can be computed atleast as efficiently as f ^ d
// where d' is a multiple of d and d = Φk(p)/r = (p4 − p2 + 1)/r

// see algorithm 6 from https://eprint.iacr.org/2015/192.pdf
y0 := (&gfP12{}).Exp(t1, u)
y0.Conjugate(y0)
y0.Square(y0)
y1 := (&gfP12{}).Square(y0)
y1.Mul(y0, y1)
y2 := (&gfP12{}).Exp(y1, u)
y2.Conjugate(y2) // f ^ 6u²

y3 := (&gfP12{}).Conjugate(y1)
y1.Mul(y2, y3)
y3.Square(y2)
y4 := (&gfP12{}).Exp(y3, u)
y4.Conjugate(y4)
y4.Conjugate(y4)
y4.Mul(y4, y1) // f ^ (6u + 6u² + 12u²) = f ^ a₂

y0 := &gfP12{}
y0.Mul(fp, fp2).Mul(y0, fp3)
y3.Mul(y4, y0) // f ^ 4u * f ^ 6u² * f ^ 12u³ = f ^ a₁

y1 := (&gfP12{}).Conjugate(t1)
y5 := (&gfP12{}).Conjugate(fu2)
y3.Conjugate(y3)
y4 := (&gfP12{}).Mul(fu, fu2p)
y4.Conjugate(y4)
y0.Mul(y2, y4)
y0.Mul(y0, t1) // a * f ^ 6u² * f = f ^ a₀

y6 := (&gfP12{}).Mul(fu3, fu3p)
y6.Conjugate(y6)
y2.Frobenius(y3)
y0.Mul(y2, y0)
y2.FrobeniusP2(y4)
y0.Mul(y2, y0)
y2.Conjugate(t1)
y2.Mul(y2, y3) // f ^ -1 * f ^ 4u * f ^ 6u² * f ^ 12u³ = f^a₃

t0 := (&gfP12{}).Square(y6)
t0.Mul(t0, y4).Mul(t0, y5)
t1.Mul(y3, y5).Mul(t1, t0)
t0.Mul(t0, y2)
t1.Square(t1).Mul(t1, t0).Square(t1)
t0.Mul(t1, y1)
t1.Mul(t1, y0)
t0.Square(t0).Mul(t0, t1)
y2.FrobeniusP2(y2)
y2.Frobenius(y2)
y0.Mul(y2, y0) // f ^ s * (p⁴ - p² + 1 ) / r

return t0
return y0
}

func optimalAte(a *twistPoint, b *curvePoint) *gfP12 {
Expand Down