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<font style="" face="Courier New">Could someone provide an elegant solution to Bird problem 4.2.13?</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"><br></font><font style="" face="Courier New">Here are the problem and my inelegant solution:</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"><br></font><font style="" face="Courier New">Problem</font><font style="" face="Courier New"><br></font><font style="" face="Courier New">-------</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"><br></font><font style="" face="Courier New">Since concatenation seems such a basic operation on lists, we can try to construct a data type that captures</font><font style="" face="Courier New"><br></font><font style="" face="Courier New">concatenation as a primitive.</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"><br></font><font style="" face="Courier New">For example,</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"><br></font><font style="" face="Courier New">data (CatList a) = CatNil</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"> | Wrap a</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"> | Cat (CatList a) (CatList a)</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"><br></font><font style="" face="Courier New">The intention is that CatNil represents [], Wrap x represents [x], and Cat x y represents</font><font style="" face="Courier New"><br></font><font style="" face="Courier New">x ++ y.</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"><br></font><font style="" face="Courier New">However, since "++" is associative, the expressions "Cat xs (Cat ys zs)" and "Cat (Cat xs ys) zs"</font><font style="" face="Courier New"> should be regarded as equal.</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"><br></font><font style="" face="Courier New">Define appropriate instances of "Eq" and "Ord" for "CatList".</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"><br></font><font style="" face="Courier New">Inelegant Solution</font><font style="" face="Courier New"><br></font><font style="" face="Courier New">------------------</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"><br></font><font style="" face="Courier New">The following solution works:</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"><br></font><font style="" face="Courier New">instance (Eq a) => Eq (CatList a) where</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"> CatNil == CatNil = True</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"> CatNil == Wrap z = False</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"> CatNil == Cat z w = ( z == CatNil && w == CatNil )</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"><br></font><font style="" face="Courier New"> Wrap x == CatNil = False</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"> Wrap x == Wrap z = x == z</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"> Wrap x == Cat z w = ( Wrap x == z && w == CatNil ) ||<br> ( Wrap x == w && z == CatNil )</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"><br></font><font style="" face="Courier New"> Cat x y == CatNil = x == CatNil && y == CatNil</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"> Cat x y == Wrap z = ( x == Wrap z && y == CatNil ) ||<br> ( x == CatNil && y == Wrap z )</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"> Cat x y == Cat z w = unwrap (Cat x y) == unwrap (Cat z w)</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"><br></font><font style="" face="Courier New">unwrap :: CatList a -> [a]</font><font style="" face="Courier New"><br></font><font style="" face="Courier New">unwrap CatNil = []</font><font style="" face="Courier New"><br></font><font style="" face="Courier New">unwrap (Wrap x) = [x]</font><font style="" face="Courier New"><br></font><font style="" face="Courier New">unwrap (Cat x y) = unwrap x ++ unwrap y</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"><br></font><font style="" face="Courier New">instance (Eq a, Ord a) => Ord (CatList a) where</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"> x < y = unwrap x < unwrap y</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"><br></font><font style="" face="Courier New">This solution correctly recognizes the equality of the following, including nested lists</font><font style="" face="Courier New">(represented, for example, by Wrap (Wrap 1), which corresponds to [[1]]):</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"><br></font><font style="" face="Courier New">Wrap 1 == Cat (Wrap 1) CatNil</font><font style="" face="Courier New"><br></font><font style="" face="Courier New">Cat (Wrap 1) (Cat (Wrap 2) (Wrap 3)) == Cat (Wrap 1) (Cat (Wrap 2) (Wrap 3))</font><font style="" face="Courier New"><br></font><font style="" face="Courier New">Wrap (Wrap 1) == Wrap (Cat (Wrap 1) CatNil)</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"><br></font><font style="" face="Courier New">Although this solution works, it's a hack, because unwrap converts CatLists to lists</font><font style="" face="Courier New">. The question clearly seeks a pure solution that does not</font><font style="" face="Courier New"> rely on Haskell's built-in lists.<br></font><font style="" face="Courier New"><br></font><font style="" face="Courier New">What's the pure solution that uses cases and recursion on</font><font style="" face="Courier New"><br></font><font style="" face="Courier New">CatList, not Haskell's built-in lists, to capture the equality of nested CatLists?</font><font style="" face="Courier New"><br></font><font style="" face="Courier New"><br></font><br /><hr />Windows Live™: Life without walls. <a href='http://windowslive.com/explore?ocid=TXT_TAGLM_WL_allup_1a_explore_032009' target='_new'>Check it out.</a></body>
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