Elliptic curve arithmetic: Difference between revisions
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→{{header|Julia}}
m (→{{header|Perl 6}}: given how clueless I am on the topic, please retract if I misunderstood anything; apparently the results from previous data set are not correct? for example, x=2, y=3.74165738677394 does not lie on y² = x³ + 7 ?) |
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Line 778:
a + b + d = Zero{Float64}
12345a = {Float64}(10.759, 35.387)</pre>
===Julia 1.x compatible version===
Adds assertion checks for points to be on the curve.
<lang julia>
using Printf
import Base.in
struct EllipticCurve{T <: AbstractFloat}
a::T
b::T
EllipticCurve(a::T, b::T) where T <: AbstractFloat = new{T}(a, b)
end
in(c::EllipticCurve, x, y) = (x == Inf || y == Inf || y^2 ≈ x^3 + c.a * x + c.b)
const Curve07 = EllipticCurve(0.0, 7.0)
struct EPoint{T <: AbstractFloat}
x::T
y::T
curve::EllipticCurve
end
EPoint{T}() where T <: AbstractFloat = EPoint{T}(Inf, Inf, Curve07)
EPoint() = EPoint{Float64}()
function EPoint(x, y, c::EllipticCurve=Curve07)
@assert in(c, x, y)
EPoint(a, b, c)
end
Base.show(io::IO, p::EPoint{T}) where T = iszero(p) ? print(io, "Zero{$T}") : @printf(io, "{%s}(%.3f, %.3f)", T, p.x, p.y)
Base.copy(p::EPoint) = EPoint(p.x, p.y, p.curve)
Base.iszero(p::EPoint{T}) where T = p.x in (-Inf, Inf, p.curve)
Base.:-(p::EPoint) = EPoint(p.x, -p.y, p.curve)
function dbl(p::EPoint{T}) where T
iszero(p) && return p
L = 3p.x ^ 2 / 2p.y
x = L ^ 2 - 2p.x
y = L * (p.x - x) - p.y
return EPoint{T}(x, y, p.curve)
end
function Base.:(==)(a::EPoint{T}, C::EPoint{T}) where T
@assert a.curve == C.curve
return (iszero(a) && iszero(C)) || (a.x == C.x && a.y == C.y)
end
function Base.:+(p::EPoint{T}, q::EPoint{T}) where T
@assert p.curve == q.curve
p == q && return dbl(p)
iszero(p) && return q
iszero(q) && return p
L = (q.y - p.y) / (q.x - p.x)
x = L ^ 2 - p.x - q.x
y = L * (p.x - x) - p.y
return EPoint{T}(x, y, p.curve)
end
function Base.:*(p::EPoint, n::Integer)
r = EPoint()
i = 1
while i ≤ n
if i & n != 0 r += p end
p = dbl(p)
i <<= 1
end
return r
end
Base.:*(n::Integer, p::EPoint) = p * n
const C = 7.0
function EPoint(y::AbstractFloat)
n = y ^ 2 - C
x = n ≥ 0 ? n ^ (1 / 3) : -((-n) ^ (1 / 3))
return EPoint{typeof(y)}(x, y, Curve07)
end
a = EPoint(1.0)
b = EPoint(2.0)
@show a b
@show c = a + b
@show d = -c
@show c + d
@show a + b + d
@show 12345a
</lang>
Output: Same as the original, Julia 0.6 version code.
=={{header|Kotlin}}==
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