> the result is of type [()] but for a cartesian n-product, you would like [[a]]

Right. So what we have here is a product over a count of 0 sets, which is isomorphic to the function space that has the null set as domain. The latter has exactly one element: the trivial function.

My apologies for misreading what OP wrote:

> This looks to me to be a violation of the rule that the Cartesian
> product of an empty list of lists is a list with one element in
> it.

I thought he meant something along the lines of sequence ["","x"].


-- Kim-Ee


On Mon, Dec 24, 2012 at 4:42 PM, Chaddaï Fouché <chaddai.fouche@gmail.com> wrote:
On Mon, Dec 24, 2012 at 8:01 AM, Jay Sulzberger <jays@panix.com> wrote:
>
>   > sequence []
>   []
>   it :: [()]
>
> This looks to me to be a violation of the rule that the Cartesian
> product of an empty list of lists is a list with one element in
> it.  It looks to be a violation because "[]" looks like a name
> for an empty list.  But we also have
>
>   > length (sequence [])
>   1
>   it :: Int
>
> which almost reassures me.
>

Well the type of the first response is a dead give-away : the result
is of type [()] but for a cartesian n-product, you would like [[a]]
(with a maybe instantiated to a concrete type) ...
What's happening here is that sequence is not "the cartesian
n-product" in general, it is only that in the list monad but in
"sequence []" there's nothing to indicate that we're in the list
monad, so GHC default to the IO monad and unit () so sequence has the
type "[IO ()] -> IO [()]" and there's no IO action in the list
parameter, so there's nothing in the result list.

Try :
> sequence [] :: [[Int]]
and you should be reassured.

--
Jedaï

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