"No instance for .. ", Haskell'98 code on GHC7.8.3 - haskell

Question

I'm trying out the attached code for linear regression with automatic differentiation.
It specifies a datatype [Dual][2] made of two Floats, and declares it to be instance of Num, Fractional and Floating.
As in all fitting/regression tasks, there's a scalar cost function, parametrized by the fitting parameters c and m, and an optimizer which improves on estimates of these two parameters by gradient descent.
Question
I'm using GHC 7.8.3, and the authors explicitly mention that this is H98 code (I mentioned it in the title because it's the only substantial difference I can think of between my setup and the Author's, however plz correct if wrong).
Why does it choke within the definition of the cost function?
My understanding is: the functions idD and constD map Floats to Duals, g is polymorphic (it can perform algebraic operations on Dual inputs because Dual inherits from Num, Fractional and Floating), and deriv maps Duals to Doubles.
The type signature for g (the eta-reduced cost function wrt the data) was inferred. I tried omitting it, and making it more general by substituting a Floating type constraint to the Fractional one.
Moreover, I tried converting the numeric types of c and m inline with (fromIntegral c :: Double), to no avail.
Specifically this code gives this error:
No instance for (Integral Dual) arising from a use of ‘g’
In the first argument of ‘flip’, namely ‘g’
In the expression: flip g (constD c)
In the second argument of ‘($)’, namely ‘flip g (constD c) $ idD m’
Any hints, please? I'm sure it's a very noob question, but I just don't get it.
The complete code is the following:
{-# LANGUAGE NoMonomorphismRestriction #-}
module ADfw (Dual(..), f, idD, cost) where
data Dual = Dual Double Double deriving (Eq, Show)
constD :: Double -> Dual
constD x = Dual x 0
idD :: Double -> Dual
idD x = Dual x 1.0
instance Num Dual where
fromInteger n = constD $ fromInteger n
(Dual x x') + (Dual y y') = Dual (x+y) (x' + y')
(Dual x x') * (Dual y y') = Dual (x*y) (x*y' + y*x')
negate (Dual x x') = Dual (negate x) (negate x')
signum _ = undefined
abs _ = undefined
instance Fractional Dual where
fromRational p = constD $ fromRational p
recip (Dual x x') = Dual (1.0 / x) (- x' / (x*x))
instance Floating Dual where
pi = constD pi
exp (Dual x x') = Dual (exp x) (x' * exp x)
log (Dual x x') = Dual (log x) (x' / x)
sqrt (Dual x x') = Dual (sqrt x) (x' / (2 * sqrt x))
sin (Dual x x') = Dual (sin x) (x' * cos x)
cos (Dual x x') = Dual (cos x) (x' * (- sin x))
sinh (Dual x x') = Dual (sinh x) (x' * cosh x)
cosh (Dual x x') = Dual (cosh x) (x' * sinh x)
asin (Dual x x') = Dual (asin x) (x' / sqrt (1 - x*x))
acos (Dual x x') = Dual (acos x) (x' / (-sqrt (1 - x*x)))
atan (Dual x x') = Dual (atan x) (x' / (1 + x*x))
asinh (Dual x x') = Dual (asinh x) (x' / sqrt (1 + x*x))
acosh (Dual x x') = Dual (acosh x) (x' / (sqrt (x*x - 1)))
atanh (Dual x x') = Dual (atanh x) (x' / (1 - x*x))
-- example
-- f = sqrt . (* 3) . sin
-- f' x = 3 * cos x / (2 * sqrt (3 * sin x))
-- linear fit sum-of-squares cost
-- cost :: Fractional s => s -> s -> [s] -> [s] -> s
cost m c x y = (/ (2 * (fromIntegral $ length x))) $
sum $ zipWith errSq x y
where
errSq xi yi = zi * zi
where
zi = yi - (m * xi + c)
-- test data
x_ = [1..10]
y_ = [a | a <- [1..20], a `mod` 2 /= 0]
-- learning rate
gamma = 0.04
g :: (Integral s, Fractional s) => s -> s -> s
g m c = cost m c x_ y_
deriv (Dual _ x') = x'
z_ = (0.1, 0.1) : map h z_
h (c, m) = (c - gamma * cd, m - gamma * md) where
cd = deriv $ g (constD m) $ idD c
md = deriv $ flip g (constD c) $ idD m
-- check for convergence
main = do
take 2 $ drop 1000 $ map (\(c, m) -> cost m c x_ y_) z_
take 2 $ drop 1000 $ z_
where the test data x_ and y_ are arrays and the learning rate gamma a scalar.
[2]: The two fields of a Dual object are in fact adjoint one with the other, if we see the derivative as an operator

Answer 1

First, (Integral s, Fractional s) makes no sense; Integral is for Euclidean domains (ones with div and mod), while Fractional is for fields (ones with /). If you have true division all your remainders are going to be zero... .
I think the problem is y_'s attempt to filter to odd numbers. Haskell 98 defines a 'stepped' range form for numbers, so you could write y_ as [1,3..19]. That should allow y_ to be used at the type [Dual], which should allow g to use it without needing the Integral constraint.
Edit: Ørjan Johansen points out that you need an Enum instance for Dual as well, which is actually fairly easy to implement (this is pretty standard for numeric types; I basically copied GHC's instance for Double (which is identical to its instance for Float, for example)):
instance Enum Dual where
succ x = x + 1
pred x = x - 1
toEnum = fromIntegral
fromEnum (Dual x _) = fromEnum x
enumFrom = numericEnumFrom
enumFromTo = numericEnumFromTo
enumFromThen = numericEnumFromThen
enumFromThenTo = numericEnumFromThenTo

Answer 2

In the original code, I don't see a type signature for g. In your code, you have specifically written
g :: (Integral s, Fractional s) => s -> s -> s
The error message says there's no Integral instance for Dual. The code manually defines instances for Num and Fractional, but not Integral.
I'm not actually sure why g needs to be Integral. If you remove that constraint, the code may even work...
EDIT: It seems the Integral instance is necessary because of your use of mod to generate test data. I'm not really sure what this huge block of code does, but I suspect if you apply fromIntegral to convert everything to (say) Double, then it may work.
(I suspect making Dual an instance of Integral is probably not what the original authors intended. Then again, I don't really understand the code, so...)