Conver to affine representation?

[skardas]

Could you please help me on converting extended point to a affine point (x: bigInt, y:bigInt)

[bwesterb]

x = X/Z, y = Y/Z given we're not dealing with the point-at-infinity. Why does this come up working with Ristretto?

[skardas]

I want to implement elliptic.Curve interface with ristretto via followings:

func GenerateKey(curve Curve, rand io.Reader) (priv []byte, x, y *big.Int, err error)
func Marshal(curve Curve, x, y *big.Int) []byte
func MarshalCompressed(curve Curve, x, y *big.Int) []byte
func Unmarshal(curve Curve, data []byte) (x, y *big.Int)
func UnmarshalCompressed(curve Curve, data []byte) (x, y *big.Int)
type Curve
func Ristretto() Curve //over ed25519
type CurveParams
func (curve *CurveParams) Add(x1, y1, x2, y2 *big.Int) (*big.Int, *big.Int)
func (curve *CurveParams) Double(x1, y1 *big.Int) (*big.Int, *big.Int)
func (curve *CurveParams) IsOnCurve(x, y *big.Int) bool
func (curve *CurveParams) Params() *CurveParams
func (curve *CurveParams) ScalarBaseMult(k []byte) (*big.Int, *big.Int)
func (curve *CurveParams) ScalarMult(Bx, By *big.Int, k []byte) (*big.Int, *big.Int)

[skardas]

x = X/Z, y = Y/Z given we're not dealing with the point-at-infinity. Why does this come up working with Ristretto?

Yes, it is. what about from affine (x,y) to extended (X,Y,T,Z)?
X=x
Y=y
T=x*y
Z=1
are correct ?

[bwesterb]

But why would you want to implement this interface?