C9 Lectures: Dr. Erik Meijer - Functional Programming Fundamentals Chapter 6 of 13

The Discussion

• 

  Did Erik get sign-off on the poster-frame for chapter 6?

• 

  Any chances of Don Syme doing a similar training series on F#???

   

  I'd love to learn me some F#  

   

• 

  Yes, in fact. Stay tuned.

  C

• 

  I'd use the word 'Brilliant', Charles, but for that I think we need to break out 'Awesome'!

   

  I'm looking forward to that already!

   

• 

  The Don Syme lectures will be broken up into 3 parts. These won't air until after the PDC.

  C

• 

  I think it's truly great that you do these, I know Don has already spoken several places about F# but a concentrated lecture series is really something which will be even more beneficial to it.

   

  Allright, back to the lazy world

   

  I couldn't resist the Zip homework first, it looks extremely simple, first with tuple-zip

  public static IEnumerable<Tuple<A, B>> Zip<A, B>(this IEnumerable<A> ps, IEnumerable<B> qs) { var pe = ps.GetEnumerator(); var qe = qs.GetEnumerator(); while (pe.MoveNext() && qe.MoveNext()) yield return new Tuple<A, B>(pe.Current,
   qe.Current); }

  
  

  Then we can verify that this behavior is correct by matching up with the built-in behavior

  var ps = new[] { 1, 2, 3 }; var qs = new[] { 1, 2, 3, 4, 5, 6 }; var z1 = ps.Zip(qs, Tuple.Create); var z2 = ps.Zip(qs); foreach (var x in z1) { Console.Write(x); Console.Write(", "); } Console.WriteLine(); foreach (var x
   in z2) { Console.Write(x); Console.Write(", "); } Console.WriteLine(); Console.Read(); 

  
  

  It's trivial to rewrite the tuple definition to the generic function definition.

• 

  Very nice show on recursive, even the topic is a basic programming skill.

   

  Hugs is new for me. Sometimes I could not get examples in the show to work. Anyway, here are some interfactive examples I played during the time I watch the show.

   

  take 2 [1..] -- take first two items as a list from a unlimited list
  -- simlar as above with a condiction x > 3 in list -1 to 20
  take 2 x where x = filter (>3) [-1..20]
  

   

  I would like to see more codes related to the topics. Enjoy it!

• 

  It seems that the animation got lost in the full screen slides.

   

  Anyway, if anyone gets stuck on implementing a more efficient reverse function, here's a hint:

  • Define reverse in terms of an auxiliary function.
  • The auxiliary function uses an accumulator.

   

   About the presented Haskell zip function, its definition can be shorter if you reorder the clauses:

  zip (x:xs) (y:ys) = (x,y) : zip xs ys zip _ _ = []

   

• 

  Took a look at Bart's Zip blog post, it also mentions exceptions. How about an expression tree transformation that takes an exception-throwing iterator and rewrites it to force the exceptions first, defining a second helper iterator for the rest. Don't know if it's possible.

   

  A less ambitious (and alas less desirable) approach is to implement a first-is-strict combinator that forces and memoizes the first element of the iterator, if there is one, thus forcing the initial exception into the light.

   

  Nontheless, both of these approaches are inferior: they delay the inevitable: that some nested throws clause will pop up in the middle of an iterator. This sucks for exceptions as "control flow" rather than proper return types you can examine using regular if statements rather than try/catch.

   

  (Not appropos: see this video (from LtU post): - is the world of tomorrow a flashback to the past? Beautiful unicode in APL - when will this come to mainstream; waiting since the sixties seems cruel and inhuman to me! The video also contains a mention of funny puns on the APL name, like APL π, pronounced "apple pie"; and a quote: "changes of a fundamental kind take a long time" - I'll buy that for a dollar)

• 

  The factorial function defined with the (n+k) pattern will not go into an infinite loop when given a negative number. (n+k) patterns only match with numbers >=k. So you would rather get an exception about non-exhaustive patterns.

• 

  It's so cool to hear Erik saying that explicit recursion can be abstracted. I can't wait to see you mention the paper "Functional programming with bananas, lenses, envelopes and barbed wire"!

• 

  My book finally arrived, after just over a month of waiting...

• 

  Again, nice lecture.  I'm lloking forward to the rest.

   

  Here is my take on Append:

  public static IEnumerable<T> Append<T>(this IEnumerable<T> a, IEnumerable<T> b) { foreach (var t in a) { yield return t; } foreach (var t in b) { yield return t; } } 

• 

  More homework:

  1) Prove that fac n = recfac n

   

  Definition of recfact n:

  (1)        recfac 0 = 1

  (2)        recfac (n+1) = (n+1) * recfact n

   

  Definition of fac n:

  (3)        fac n = product [1...n]

                          where

  (4)                                product [] = 1

  (5)                                product [1..n] = 1 * 2 * … (n-1) * n

   

  We want to prove that:

                      fact n = recfac n

   

  Using induction:

   

  Case n = 0

              fac 0 = recfac 0

   

              fac 0 = 1 (Eq. 1)

   

              recfac 0 = product [1 .. 0] = product [] = 1 (Eq. 4)

   

              1 = 1

   

  Case (m + 1):

  (7)        fac (m+1) = recfac (m+1)

   

  fac (m+1) = product [  1..(m+1)] = 1 * 2 *….* m * (m+1)

   

  1. Using Eq.5:       1 * 2 *….* m  = product [1..m]

   

              fac (m+1) = 1 * 2 *….* m * (m+1)  =(product [1..m]) * (m+1)

   

  (8)        fac (m+1) = (product [1..m]) * (m+1)

   

  Using Eq 2 :  recfac (m+1) = (m+1) * recfac m

   

   

  Replacing Eq 8 into Eq 7:

              fac (m+1) = (product [1..m]) * (m+1) = recfac (m+1)

   

  Using Eq 2 on the right side:

              (product [1..m]) * (m+1) = (m+1) * recfac m

   

  Dividing by (m+1) on both sides:

              product [1..m] = recfac m

   

  Using Eq 3 on the left side:

              fac m = recfac m

   

  Q.E.D.!

• 

  Concerning the Zip homework, a couple of things to keep in mind:

  • Do proper argument validation: both input IEnumerable objects and the zipper function should not be null. This is a bit tricky. (Tip: when does the exception get thrown by the iterator?)
  • IEnumerator objects implement IDisposable. Make sure the enumerator objects get disposed correctly under all circumstances. (Tip: don't overcomplicate it; the language can do lots of work for you)

  Additional brain gymnastics can be triggered by trying to define Zip in terms of other LINQ Standard Query Operators. Enjoy!

• 

  (I'm not at home right now so no thorough answer, but -) I know you need to force exception handling by using another method to do the actual implementation, otherwise the exceptions will be thrown when the stream is enumerated and for disposal, well we have using statements, need to check that later. Nice having you here, Bart.

   

  

• 

  Towards the end of the video Erik referred to the slides. Where are those slides???

  It's a hassle working on the homework off the video stream, would be much easier if looking thru the PPT.

  Can we get the download link?

   

  Thanks

• 

  http://www.cs.nott.ac.uk/~gmh/book.html#slides

• 

  OK, this is a quick write-up that would seem to fit the bill

   

  public static class Extensions { public static IEnumerable<Tuple<A, B>> Zip<A, B>(this IEnumerable<A> ps, IEnumerable<B> qs) { if (ps == null) throw new ArgumentNullException("ps"); if (qs == null) throw new ArgumentNullException("qs");
   return ps.DoZip(qs); } private static IEnumerable<Tuple<A, B>> DoZip<A, B>(this IEnumerable<A> ps, IEnumerable<B> qs) { using (var pe = ps.GetEnumerator()) { using (var qe = qs.GetEnumerator()) { while (pe.MoveNext() && qe.MoveNext()) yield return new Tuple<A,
   B>(pe.Current, qe.Current); } } } }

  • Using nested using statements for resource disposal, as prescibed by C# for easy exception safe resource disposal.
  • Using a separate method to ensure exceptions are thrown upon calling the method.

  This doesn't use the more generic version using a function but the pattern is the same (the lower-order Zip can of course be written in terms of the higher-order Zip using Tuple.Create).

  As for writing Zip in terms of other LINQ operators: haven't looked into it but you wrote about that, I remember, although not what your precise solution was.

• 

  I think it's amusing that you refer to Hughes paper on FP as such a good paper, because I'm taking a class of his right now, and he strongly recommends this video series And, yes, it's a great paper. Good read. Thanks for the awesome video lectures, they complement his course very well!

• 

  Here is my take on the homework:

   

  1.- Define:

   

  and :

  and' :: [Bool] -> Bool and' [] = True and' (False:_) = False and' (_:xs) = and xs 

   

  concat:

  concat' :: [[a]] -> [a] concat' [] = [] concat' (xs:xss) = xs ++ concat' xss

   

  replicate:

  replicate' :: Int -> a -> [a] replicate' 0 x = [] replicate' (n+1) x = x:replicate' n x

   

  Select the Nth statement of a list:

  (!!@) :: [a] -> Int -> a (!!@) (x:_) 0 = x (!!@) (_:xs) (n+1) = xs !!@ n

   

  elem:

  elem' :: (Eq a) => a -> [a] -> Bool elem' _ [] = False elem' x (y:ys) | x == y = True | otherwise = elem' x ys

   

  2. Define merge :: [Int] -> [Int] -> [Int]

  merge:: (Ord a) => [a] -> [a] -> [a] merge [] [] = [] merge xs [] = xs merge [] ys = ys merge (x:xs) (y:ys) | x < y = x:merge xs (y:ys) | x == y = x:y:merge xs ys | otherwise = y:merge (x:xs) ys

   

  3. Define merge sort:

  msort :: (Ord a) => [a] -> [a] msort [] = [] msort [x] = [x] msort xs = merge (msort ys) (msort zs) where [(ys,zs)] = halve xs halve :: [a] -> [([a],[a])] halve [] = [] halve xs = [(take n xs, drop n xs)] where n = (length xs `div` 2)

   

  I'm having a blast with this series.

• 

  It's probably not a complete coincidence that Charles speaks of the high priests of the Lambda calculus.

   

  - folded hands, closed eyes, tilted head, pointing to the sky

   

  May the monad be with you!

   

  

• 

  Indeed! You are right, brothers. That still shot captures a moment of extreme reverence, a moment in time where Dr. Meijer, a reverend of the Lambda, evangelist of the functional composition of the monadic universe, exudes his deep thanks and praises for the higher-order function in the sky.

   

  C

• 

  Somewhere closer to the end of the talk, Mr. Meijer says that segregation of common parts

  of a function from the distinct parts is only possible with lazy evaluation (with the two clouds

  drawn on the board). Didn't he really mean functional composition instead, which is more

  general than lazy evaluation? If the distinct parts have no side-effects and the common part has

  no influence on the distinct parts besides parametrization (as is the case with pure functions),

  then you can also call the common part as a function with distinct parts as pure functional or

  value parameters to that common function.

   

  Of course you might save some computations with lazy evaluation, but I am quite sure I have

  applied this technique with eager evaluation as well!

   

  Am I missing something?

• 

  Finally got some time and went through this lecture too...

   

  (!!) :: [a] -> Int -> a
  (a:as) !! 0 = a
  (a:as) !! (n + 1) = as Main.!! n
  
  (++) :: [a] -> [a] -> [a]
  [] ++ [] = []
  as ++ [] = as
  [] ++ bs = bs
  (a:as) ++ (bs) = a : (Main.++) as bs
  
  (&&) :: Bool -> Bool -> Bool
  True && True = True
  _ && _ = False
  
  length :: Num a => [a] -> Int
  length [] = 0
  length [x] = 1
  length (a:as) = 1 + Main.length as
  
  take :: Num a => Int -> [a] -> [a]
  take 0 _ = []
  take _ [] = []
  take (n + 1) (a:as) = a : Main.take n as
  
  drop :: Num a => Int -> [a] -> [a]
  drop 0 as = as
  drop _ [] = []
  drop (n + 1) (a:as) = Main.drop n as
  
  qsort :: Ord a => [a] -> [a]
  qsort [] = []
  qsort [a] = [a]
  qsort (a:as) = qsort smaller Main.++ [a] Main.++ qsort greater
       where
  smaller = [x | x <- as, x <= a]
  greater = [x | x <- as, a < x]
  
  and :: [Bool] -> Bool
  and [] = True
  and (a:as) = a Main.&& Main.and as
  
  concat :: [[a]] -> [a]
  concat [] = []
  concat (a:as) = a Main.++ Main.concat as
  
  replicate :: Int -> a -> [a]
  replicate 0 a = []
  replicate (n + 1) a = a : Main.replicate n a
  
  elem :: Eq a => [a] -> a -> Bool
  elem [] _ = False
  elem (a:as) s = if a == s then True else Main.elem as s
  
  merge :: [Int] -> [Int] -> [Int]
  merge [] [] = []
  merge [] bs = bs
  merge as [] = as
  merge as bs = qsort $ as Main.++ bs
  
  msort :: [Int] -> [Int]
  msort [] = []
  msort [a] = [a]
  msort as = merge (msort $ Main.take n as) (msort $ Main.drop n as)
        where
            n = Main.length as `div` 2

   

  Sohail Qayum Malik.