Difference between revisions of "Hask"

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== '''Hask''' ==
 
== '''Hask''' ==
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=== Is '''Hask''' even a category? ===
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Consider:
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<haskell>
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undef1 = undefined :: a -> b
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undef2 = \_ -> undefined
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</haskell>
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Note that these are not the same value:
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<haskell>
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seq undef1 () = undefined
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seq undef2 () = ()
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</haskell>
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This might be a problem, because <hask>undef1 . id = undef2</hask>. In order to make '''Hask''' a category, we define two functions <hask>f</hask> and <hask>g</hask> as the same morphism if <hask>f a = g a</hask> for all <hask>a</hask>. Thus <hask>undef1</hask> and <hask>undef2</hask> are different ''values'', but the same ''morphism'' in '''Hask'''.
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=== '''Hask''' is not Cartesian closed ===
   
 
Actual '''Hask''' does not have sums, products, or an initial object, and <hask>()</hask> is not a terminal object. The Monad identities fail for almost all instances of the Monad class.
 
Actual '''Hask''' does not have sums, products, or an initial object, and <hask>()</hask> is not a terminal object. The Monad identities fail for almost all instances of the Monad class.

Revision as of 07:51, 22 August 2012

Hask refers to a category with types as objects and functions between them as morphisms. However, its use is ambiguous. Sometimes it refers to Haskell (actual Hask), and sometimes it refers to some subset of Haskell where no values are bottom and all functions terminate (platonic Hask). The reason for this is that platonic Hask has lots of nice properties that actual Hask does not, and is thus easier to reason in. There is a faithful functor from platonic Hask to actual Hask allowing programmers to think in the former to write code in the latter.

Hask

Is Hask even a category?

Consider:

undef1 = undefined :: a -> b
undef2 = \_ -> undefined

Note that these are not the same value:

seq undef1 () = undefined
seq undef2 () = ()

This might be a problem, because undef1 . id = undef2. In order to make Hask a category, we define two functions f and g as the same morphism if f a = g a for all a. Thus undef1 and undef2 are different values, but the same morphism in Hask.

Hask is not Cartesian closed

Actual Hask does not have sums, products, or an initial object, and () is not a terminal object. The Monad identities fail for almost all instances of the Monad class.

Why Hask isn't as nice as you'd thought.
Initial Object Terminal Object Sum Product
Definition There is a unique function


u :: Empty -> r

There is a unique function


u :: r -> ()

For any functions


f :: a -> r
g :: b -> r

there is a unique function u :: Either a b -> r

such that: u . Left = f
u . Right = g

For any functions


f :: r -> a
g :: r -> b

there is a unique function u :: r -> (a,b)

such that: fst . u = f
snd . u = g

Platonic candidate u1 r = case r of {} u1 _ = () u1 (Left a) = f a


u1 (Right b) = g b

u1 r = (f r,g r)
Example failure condition r ~ () r ~ () r ~ ()


f _ = ()
g _ = ()

r ~ ()


f _ = undefined
g _ = undefined

Alternative u u2 _ = () u2 _ = undefined u2 _ = () u2 _ = undefined
Difference u1 undefined = undefined


u2 undefined = ()

u1 _ = ()


u2 _ = undefined

u1 undefined = undefined


u2 undefined = ()

u1 _ = (undefined,undefined)


u2 _ = undefined

Result FAIL FAIL FAIL FAIL

Platonic Hask

Because of these difficulties, Haskell developers tend to think in some subset of Haskell where types do not have bottoms. This means that it only includes functions that terminate, and typically only finite values. The corresponding category has the expected initial and terminal objects, sums and products. Instances of Functor and Monad really are endofunctors and monads.

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