我必须编写一个程序来解决正方形,返回一个复数结果。
到目前为止,我已经定义了一个复数,声明它是num的一部分,所以+, - 和* - ing可以发生。
我还为二次方程定义了一种数据类型,但我现在仍然坚持二次方程式的实际求解。我的数学很差,所以任何帮助都会非常感激......
data Complex = C {
re :: Float,
im :: Float
} deriving Eq
-- Display complex numbers in the normal way
instance Show Complex where
show (C r i)
| i == 0 = show r
| r == 0 = show i++"i"
| r < 0 && i < 0 = show r ++ " - "++ show (C 0 (i*(-1)))
| r < 0 && i > 0 = show r ++ " + "++ show (C 0 i)
| r > 0 && i < 0 = show r ++ " - "++ show (C 0 (i*(-1)))
| r > 0 && i > 0 = show r ++ " + "++ show (C 0 i)
-- Define algebraic operations on complex numbers
instance Num Complex where
fromInteger n = C (fromInteger n) 0 -- tech reasons
(C a b) + (C x y) = C (a+x) (b+y)
(C a b) * (C x y) = C (a*x - b*y) (b*x + b*y)
negate (C a b) = C (-a) (-b)
instance Fractional Complex where
fromRational r = C (fromRational r) 0 -- tech reasons
recip (C a b) = C (a/((a^2)+(b^2))) (b/((a^2)+(b^2)))
root :: Complex -> Complex
root (C x y)
| y == 0 && x == 0 = C 0 0
| y == 0 && x > 0 = C (sqrt ( ( x + sqrt ( (x^2) + 0 ) ) / 2 ) ) 0
| otherwise = C (sqrt ( ( x + sqrt ( (x^2) + (y^2) ) ) / 2 ) ) ((y/(2*(sqrt ( ( x + sqrt ( (x^2) + (y^2) ) ) / 2 ) ) ) ) )
-- quadratic polynomial : a.x^2 + b.x + c
data Quad = Q {
aCoeff, bCoeff, cCoeff :: Complex
} deriving Eq
instance Show Quad where
show (Q a b c) = show a ++ "x^2 + " ++ show b ++ "x + " ++ show c
solve :: Quad -> (Complex, Complex)
solve (Q a b c) = STUCK!
编辑:我似乎错过了使用我自己的复数数据类型的全部意义,即了解自定义数据类型。我很清楚我可以使用complex.data。到目前为止,使用我的解决方案提供的任何帮助都将非常感激。\
编辑2:似乎我最初的问题措辞可怕。我知道二次公式会将两个(或只是一个)根返回给我。我遇到麻烦的地方是将这些根作为一个(复杂的,复杂的)元组返回上面的代码。
我很清楚我可以使用内置的二次函数,如下所示,但这不是练习。练习背后的想法,以及创建自己的复数数据类型,是要了解自定义数据类型。
答案 0 :(得分:6)
就像newacct所说,它只是二次方程式:
(-b +- sqrt(b^2 - 4ac)) / 2a
module QuadraticSolver where
import Data.Complex
data Quadratic a = Quadratic a a a deriving (Show, Eq)
roots :: (RealFloat a) => Quadratic a -> [ Complex a ]
roots (Quadratic a b c) =
if discriminant == 0
then [ numer / denom ]
else [ (numer + root_discriminant) / denom,
(numer - root_discriminant) / denom ]
where discriminant = (b*b - 4*a*c)
root_discriminant = if (discriminant < 0)
then 0 :+ (sqrt $ -discriminant)
else (sqrt discriminant) :+ 0
denom = 2*a :+ 0
numer = (negate b) :+ 0
实践中:
ghci> :l QuadraticSolver
Ok, modules loaded: QuadraticSolver.
ghci> roots (Quadratic 1 2 1)
[(-1.0) :+ 0.0]
ghci> roots (Quadratic 1 0 1)
[0.0 :+ 1.0,(-0.0) :+ (-1.0)]
并适应您的使用条款:
solve :: Quad -> (Complex, Complex)
solve (Q a b c) = ( sol (+), sol (-) )
where sol op = (op (negate b) $ root $ b*b - 4*a*c) / (2 * a)
虽然我没有测试过那段代码
答案 1 :(得分:5)
由于Haskell的sqrt
也可以处理复数,因此可以进一步简化rampion的解决方案:
import Data.Complex
-- roots for quadratic equations with complex coefficients
croots :: (RealFloat a) =>
(Complex a) -> (Complex a) -> (Complex a) -> [Complex a]
croots a b c
| disc == 0 = [solution (+)]
| otherwise = [solution (+), solution (-)]
where disc = b*b - 4*a*c
solution plmi = plmi (-b) (sqrt disc) / (2*a)
-- roots for quadratic equations with real coefficients
roots :: (RealFloat a) => a -> a -> a -> [Complex a]
roots a b c = croots (a :+ 0) (b :+ 0) (c :+ 0)
如果您更改类型以适合您的实施,也可以将此croots
函数与您自己的数据类型一起使用(并调用root
函数而不是sqrt
)。