Re: [Haskell-beginners] Cartesian Product in Standard Haskell Libraries

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 wrote:

On Mon, Dec 24, 2012 at 8:01 AM, Jay Sulzberger 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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