Question

Problem

I am just going through the You Could Have Invented Monads! tutorial, and in the section entitled A Container: Multivalued Functions, where the function types (of complex square and cube roots) are:

Complex Float -> [Complex Float]

Bind is defined as:

bind :: (Complex Double -> [Complex Double]) -> ([Complex Double] -> [Complex Double])
bind f x = concat (map f x)
--shortcut:
f * g = bind f . g 

and unit and lift are:

unit x = [x]    
lift f = unit . f

Now I am confused, is the f in lift function the same as the f in the bind function?
In other words what are the type of the unit and lift functions?

Paradox

Also, by my reasoning I come to a strange paradox: If lift can take functions such as:

realRoot :: Double -> Double
realRoot x =  sqrt(x)

and lift it into our monad so that it becomes of type:

lift realRoot :: [Complex Double]

wouldn’t I be able to do stuff like:

(lift realRoot * imaginaryRoot) -1

where

imaginaryRoot :: [Complex Double]

But how can I take real root of complex numbers?

Any help appreciated!

Answer 1

Now I am confused, is the f in lift function the same as the f in the bind function? In other words what are the type of the unit and lift functions?

No, it’s not the same. lift‘s argument is a function Complex Double -> Complex Double, so

unit :: Complex Double -> [Complex Double]
lift :: (Complex Double -> Complex Double) -> (Complex Double -> [Complex Double])

and thus your realRoot is not an acceptable argument for lift.