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You can reach the person managing the list at beginners-ow...@haskell.org When replying, please edit your Subject line so it is more specific than "Re: Contents of Beginners digest..." Today's Topics: 1. Re: Hutton 2016 ex8.3a (trent shipley) ---------------------------------------------------------------------- Message: 1 Date: Mon, 3 Sep 2018 23:00:53 -0700 From: trent shipley <trent.ship...@gmail.com> To: The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskell <beginners@haskell.org> Subject: Re: [Haskell-beginners] Hutton 2016 ex8.3a Message-ID: <caeflyb+jsgbnmtlyhsnzhxjxyvzmg6iw9qmbzeo6nvbl8he...@mail.gmail.com> Content-Type: text/plain; charset="utf-8" data Tree a = Leaf a | Node (Tree a) (Tree a) How could I modify this definition so that a Node could be: Node (Tree a) Null/Nothing/[] -- basically a zilch that would match anything or nothing, because it would make programming simple. Alternatively: Node (Tree a) -- Would I be looking at, data Tree a = Leaf a | Node (Tree a) Maybe (Tree a) -- Or possibly the best I can do is, data Tree a = Leaf a | Node (Tree a) (Tree a) | Node (Tree a) On Mon, Sep 3, 2018 at 3:06 AM Rishi Rajasekaran < rajasekaran.ri...@gmail.com> wrote: > Hi Trent > > For executing your approach, you can do the following: > 1. Compute the number of leaves in the left subtree first. > 2. Pass that computed value into leaves' call for the right subtree > > Regards > > Rishi > > > On Mon, 3 Sep, 2018, 2:36 PM trent shipley, <trent.ship...@gmail.com> > wrote: > >> Given >> data Tree a = Leaf a | Node (Tree a) (Tree a) >> >> Write a leaf counter. >> >> Hutton suggests: >> >> leaves :: Tree a -> Int >> leaves (Leaf _) = 1 >> leaves (Node l r) = leaves l + leaves r >> >> I tried: >> >> leavesTrent :: Tree a -> Int >> leavesTrent = leaves' 0 >> where >> leaves' n (Leaf a) = n + 1 >> leaves' n (Node l r) = (leaves' n l), (leaves' n r) >> >> The idea is: >> >> If it is a leaf, add one to the accumulator. (Following Hutton's >> explanation of how sum works if defined with foldl.) If it is a tree, >> proceed down the left subtree recursively, until you get to a leaf, then >> roll up to the right subtree. The problem (among the problems) is that I >> don't know how to tell the compiler to do all lefts, before backing up to >> go right. I only know how to do that using a real operator like "+" or foo >> (l, r). >> >> Is that kind of no-op recursive branching possible? >> >> Trent. >> > _______________________________________________ >> Beginners mailing list >> Beginners@haskell.org >> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners >> > _______________________________________________ > Beginners mailing list > Beginners@haskell.org > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners > -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://mail.haskell.org/pipermail/beginners/attachments/20180903/0e2064d7/attachment-0001.html> ------------------------------ Subject: Digest Footer _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners ------------------------------ End of Beginners Digest, Vol 123, Issue 2 *****************************************