Tripling-oriented Doche–Icart–Kohel curve

Given ${\displaystyle P_{1}=(x_{1},y_{1})}$ and ${\displaystyle P_{2}=(x_{2},y_{2})}$ on ${\displaystyle T_{a}}$, the point ${\displaystyle P_{3}=(x_{3},y_{3})=P_{1}+P_{2}}$ has coordinates:

${\displaystyle {\begin{aligned}x_{3}&={\frac {1}{(x_{2}-x_{1})^{2}}}\left\{-x_{1}^{3}+(x_{2}-3a)x_{1}^{2}+(x_{2}^{2}+6ax_{2})x_{1}+(y_{1}^{2}-2y_{2}y_{1}+(-x_{2}^{3}-3ax_{2}^{2}+y_{2}^{2}))\right\}\\y_{3}&={\frac {1}{(x_{2}-x_{1})^{3}}}\left\{(-y_{1}+2y_{2})x_{1}^{3}+(-3ay_{1}-3y_{2}x_{2}+3ay_{2})x_{1}^{2}+(3x_{2}^{2}y_{1}+6ax_{2}y_{1}-6ay_{2}x_{2})x_{1}\right.\\&\qquad \qquad \qquad \qquad \left.+(y_{1}^{3}-3y_{2}y_{1}^{2}+(-2x_{2}^{3}-3ax_{2}^{2}+3y_{2}^{2})y_{1}+(y_{2}x_{2}^{3}+3ay_{2}x_{2}^{2}-y_{2}^{3}))\right\}\end{aligned}}}$

Given a point ${\displaystyle P_{1}=(x_{1},y_{1})}$ on ${\displaystyle T_{a}}$, the point ${\displaystyle P_{3}=(x_{3},y_{3})=2P_{1}}$ has coordinates:

${\displaystyle {\begin{aligned}x_{3}&={\frac {9}{4y_{1}^{2}x_{1}^{4}}}+{\frac {9}{y_{1}^{2}ax_{1}^{3}}}+\left({\frac {9}{y_{1}^{2}a^{2}}}+{\frac {9}{y_{1}^{2}a}}\right)x_{1}^{2}+\left({\frac {18}{y_{1}^{2}a^{2}}}-2\right)x_{1}+{\frac {9}{y_{1}^{2}a^{2}-3a}}\\y_{3}&=-{\frac {27}{8y_{1}^{3}x_{1}^{6}}}-{\frac {81}{4y_{1}^{3}ax_{1}^{5}}}+\left(-{\frac {81}{2y_{1}^{3}a^{2}}}-{\frac {81}{4y_{1}^{3}a}}\right)x_{1}^{4}+\left(-{\frac {27}{y_{1}^{3}a^{3}}}-{\frac {81}{y_{1}^{3}a^{2}}}+{\frac {9}{2y_{1}}}\right)x_{1}^{3}+\left(-{\frac {81}{y_{1}^{3}a^{3}}}-{\frac {81}{2y_{1}^{3}}}a^{2}+{\frac {27}{2y_{1}a}}\right)x_{1}^{2}\\&\qquad \qquad \qquad \qquad +\left(-{\frac {81}{y_{1}^{3}a^{3}}}+{\frac {9}{y_{1}a^{2}}}+{\frac {9}{y_{1}a}}\right)x_{1}+\left(-{\frac {27}{y_{1}^{3}a^{3}}}+{\frac {9}{y_{1}a^{2}}}-y_{1}\right)\end{aligned}}}$

Given a point ${\displaystyle P_{1}=(x_{1},y_{1})}$ on ${\displaystyle T_{a}}$, its negation with respect to the neutral element ${\displaystyle (0:1:0)}$ is ${\displaystyle -P_{1}=(x_{1},-y_{1})}$.

There are also other formulas given in [2] for Tripling-oriented Doche–Icart–Kohel curves for fast tripling operation and mixed-addition.

The following algorithm represents the sum of two points ${\displaystyle P_{1}}$ and ${\displaystyle P_{2}}$ on an elliptic curve in the Tripling-oriented Doche-Icart-Kohel form. The result is a point ${\displaystyle P_{3}=(X_{3},Y_{3},Z_{3},ZZ_{3})}$. It is assumed that ${\displaystyle Z_{2}=1}$ and that ${\displaystyle a_{3}=3a}$. The cost of this implementation is 7M + 4S + 1*a3 + 10add + 3*2 + 1*4, where M indicates the multiplications, S the squarings, a3 indicates the multiplication by the constant a₃, add represents the number of additions required.

${\displaystyle A=X_{2}ZZ_{1}}$

${\displaystyle B=Y_{2}ZZ_{1}Z_{1}}$

${\displaystyle C=X_{1}-A}$

${\displaystyle D=2(Y_{1}-B)}$

${\displaystyle F=C^{2}}$

${\displaystyle F_{4}=4F}$

${\displaystyle Z_{3}=(Z_{1}+C)^{2}-ZZ_{1}-F}$

${\displaystyle E={Z_{3}}^{2}}$

${\displaystyle G=CF_{4}}$

${\displaystyle H=AF_{4}}$

${\displaystyle X_{3}=D^{2}-G-2H-a_{3}E}$

${\displaystyle Y_{3}=D(H-X_{3})-2BG}$

${\displaystyle ZZ_{3}=E}$

Let ${\displaystyle P_{1}=(1,{\sqrt {13}})}$ and ${\displaystyle P_{2}=(0,{\sqrt {3}})}$ affine points on the elliptic curve over ${\displaystyle \mathbb {R} }$:

${\displaystyle y^{2}=x^{3}+3(x+1)^{2}}$.

Then:

${\displaystyle A=X_{2}ZZ_{1}=0}$

${\displaystyle B=Y_{2}ZZ_{1}Z_{1}={\sqrt {3}}}$

${\displaystyle C=X_{1}-A=1}$

${\displaystyle D=2(Y_{1}-B)=2({\sqrt {13}}-{\sqrt {3}})}$

${\displaystyle F=C^{2}=1}$

${\displaystyle F_{4}=4F=4}$

${\displaystyle Z_{3}=(Z_{1}+C)^{2}-ZZ_{1}-F=2}$

${\displaystyle E={Z_{3}}^{2}=4}$

${\displaystyle G=CF_{4}=4}$

${\displaystyle H=AF_{4}=0}$

${\displaystyle X_{3}=D^{2}-G-2H-a_{3}E=48-8{\sqrt {39}}}$

${\displaystyle Y_{3}=D(H-X_{3})-2BG=296{\sqrt {3}}-144{\sqrt {13}}}$

${\displaystyle ZZ_{3}=E=4}$

Notice that in this case ${\displaystyle Z_{1}=Z_{2}=1}$. The resulting point is ${\displaystyle P_{3}=(X_{3},Y_{3},Z_{3},ZZ_{3})=(48-8{\sqrt {39}},296{\sqrt {3}}-144{\sqrt {13}},2,4)}$, that in affine coordinates is ${\displaystyle P_{3}=(12-2{\sqrt {39}},37{\sqrt {3}}-18{\sqrt {13}})}$.

The following algorithm represents the doubling of a point ${\displaystyle P_{1}}$ on an elliptic curve in the Tripling-oriented Doche-Icart-Kohel form. It is assumed that ${\displaystyle a_{3}=3a}$, ${\displaystyle a_{2}=2a}$. The cost of this implementation is 2M + 7S + 1*a2 + 1*a3 + 12add + 2*2 + 1*3 + 1*8; here M represents the multiplications, S the squarings, a2 and a3 indicates the multiplications by the constants a₂ and a₃ respectively, and add indicates the additions.

${\displaystyle A={X_{1}}^{2}}$

${\displaystyle B=a_{2}ZZ_{1}(X_{1}+ZZ_{1})}$

${\displaystyle C=3(A+B)}$

${\displaystyle D={Y_{1}}^{2}}$

${\displaystyle E=D^{2}}$

${\displaystyle Z_{3}=(Y_{1}+Z_{1})^{2}-D-ZZ_{1}}$

${\displaystyle ZZ_{3}=Z_{3}^{2}}$

${\displaystyle F=2((X_{1}+D)^{2}-A-E)}$

${\displaystyle X_{3}=C^{2}-a_{3}ZZ_{3}-2F}$

${\displaystyle Y_{3}=C(F-X_{3})-8E}$

Let ${\displaystyle P_{1}=(0,{\sqrt {3}})}$ be a point on ${\displaystyle y^{2}=x^{3}+3(x+1)^{2}}$.

Then:

${\displaystyle A={X_{1}}^{2}=0}$

${\displaystyle B=a_{2}ZZ_{1}(X_{1}+ZZ_{1})=2}$

${\displaystyle C=3(A+B)=6}$

${\displaystyle D={Y_{1}}^{2}=3}$

${\displaystyle E=D^{2}=9}$

${\displaystyle Z_{3}=(Y_{1}+Z_{1})^{2}-D-ZZ_{1}=2{\sqrt {3}}}$

${\displaystyle ZZ_{3}=Z_{3}^{2}=12}$

${\displaystyle F=2((X_{1}+D)^{2}-A-E)=0}$

${\displaystyle X_{3}=C^{2}-a_{3}ZZ_{3}-2F=0}$

${\displaystyle Y_{3}=C(F-X_{3})-8E=-72}$

Notice that here the point is in affine coordinates, so ${\displaystyle Z_{1}=1}$. The resulting point is ${\displaystyle P_{3}=(0,-72,2{\sqrt {3}},12)}$, that in affine coordinates is ${\displaystyle P_{3}=(0,-{\sqrt {3}})}$.