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??. Programming with Streams Infinite lists v.s. Streams Normal order evaluation Recursive streams Stream Diagrams Lazy patterns memoization Inductive properties of infinite lists Reading assignment Chapter 14. Programming with Streams Chapter 15. A module of reactive animations.
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?? • Programming with Streams • Infinite lists v.s. Streams • Normal order evaluation • Recursive streams • Stream Diagrams • Lazy patterns • memoization • Inductive properties of infinite lists • Reading assignment • Chapter 14. Programming with Streams • Chapter 15. A module of reactive animations
Infinite lists v.s. Streams data Stream a = a :^ Stream a • A stream is an infinite list. It is never empty • We could define a stream in Haskell as written above. But we prefer to use lists. • This way we get to reuse all the polymorphic functions on lists.
Infinite lists and bottom twos = 2 : twos twos = 2 : (2 : twos) twos = 2 : (2 : (2 : twos)) twos = 2 : (2 : (2 : (2 : twos))) bot :: a bot = bot • What is the difference between twos and bot ? Sometimes we write z for bot
Normal Order evaluation • Why does head(twos) work? • Head (2 : twos) • Head(2 : (2 : twos)) • Head (2: (2 : (2 : twos))) • The outermost – leftmost rule. • Outermost • Use the body of the function before its arguments • Leftmost • Use leftmost terms: (K 4) (5 + 2) • Be careful with Infix: (m + 2) `get` (x:xs)
Normal order continued • Let let x = y + 2 z = x / 0 in if x=0 then z else w • Where f w = if x=0 then z else w where x = y + 2 z = x / 0 • Case exp’s • case f x of [] -> a ; (y:ys) -> b
Recursive streams fibA 0 = 1 fibA 1 = 1 fibA n = fibA(n-1) + fibA(n-2) • Unfold this a few times fibA 8 = fibA 7 + fibA 6 = (fibA 6 + fibA 5) + (fibA 5 + fibA 4) = ((fibA 5 + fibA 4) + (fibA 4 + fibA 3)) +((fibA 4 + fibA 3) + (fibA 3 + fibA 2))
Fibonacci Stream fibs :: [ Integer ] fibs = 1 : 1 : (zipWith (+) fibs (tail fibs)) This is much faster! And uses less resources. Why? 1 1 2 3 5 8 13 21 … fibonacci sequence + 1 2 3 5 8 13 21 34 … tail of fibonacci sequence 2 3 5 8 13 21 34 55 …tail of tail of fibonacci sequence
Add x y = zipWith (+) x y Abstract on tail of fibs fibs = 1 : 1 : (add fibs (tail fibs)) = 1 : tf where tf = 1 : add fibs (tail fibs) = 1 : tf where tf = 1 : add fibs tf Abstract on tail of tf = 1 : tf where tf = 1 : tf2 tf2 = add fibs tf Unfold add = 1 : tf where tf = 1 : tf2 tf2 = 2 : add tf tf2
Abstract and unfold again fibs = 1 : tf where tf = 1 : tf2 tf2 = 2 : add tf tf2 = 1 : tf where tf = 1 : tf2 tf2 = 2 : tf3 tf3 = add tf tf2 = 1 : tf where tf = 1 : tf2 tf2 = 2 : tf3 tf3 = 3 : add tf2 tf3 tf is used only once, so eliminate = 1 : 1 : tf2 where tf2 = 2 : tf3 tf3 = 3 : add tf2 tf3
Again • This can go on forever. But note how the sharing makes the inefficiencies of fibA go away. fibs = 1 : 1 : 2 : tf3 where tf3 = 3 : tf4 tf4 = 5 : add tf3 tf4 fibs = 1 : 1 : 2 : 3 : tf4 where tf4 = 5 : tf5 tf5 = 8 : add tf4 tf5
Stream Diagrams fibs = [1,1,2,3,5,8,…] • Streams are “wires” along • which values flow. • Boxes and circles take • wires as input and produce • values for new wires as • output. • Forks in a wire send their • values to both destinations • A stream diagram corresponds • to a Haskell function (usually • recursive) (:) [1,2,3,5,8, …] 1 (:) [2,3,5,8,…] 1 add
[0] +1 if* next out 0 inp Example Stream Diagram counter :: [ Bool ] -> [ Int ] counter inp = out where out = if* inp then* 0 else* next next = [0] followedBy map (+ 1) out
1... [0] 0... +1 if* 0... next out 0 F... Example • counter :: [ Bool ] -> [ Int ] • counter inp = out • where • out = if* inp then* 0 else* next • next = [0] followedBy map (+ 1) out
1:2.. [0] 0:1.. +1 if* 0:1.. next out 0 F:F.. Example • counter :: [ Bool ] -> [ Int ] • counter inp = out • where • out = if* inp then* 0 else* next • next = [0] followedBy map (+ 1) out
1:2:1.. [0] 0:1:2 +1 if* 0:1:0.. next out 0 F:F:T.. Example • counter :: [ Bool ] -> [ Int ] • counter inp = out • where • out = if* inp then* 0 else* next • next = [0] followedBy map (+ 1) out
Client, Server Example type Response = Integer type Request = Integer client :: [Response] -> [Request] client ys = 1 : ys server :: [Request] -> [Response] server xs = map (+1) xs reqs = client resps resps = server reqs Typical. A set of mutually recursive equations
client ys = 1 : ys server xs = map (+1) xs reqs = client resps resps = server reqs reqs = client resps = 1 : resps = 1 : server reqs Abstract on (tail reqs) = 1 : tr where tr = server reqs Use definition of server = 1 : tr where tr = 2 : server reqs abstract = 1 : tr where tr = 2 : tr2 tr2 = server reqs Since tr is used only once = 1 : 2 : tr2 where tr2 = server reqs Repeat as required
Lazy Patterns • Suppose client wants to test servers responses. clientB (y : ys) = if ok y then 1 :(y:ys) else error "faulty server" where ok x = True server xs = map (+1) xs • Now what happens . . . Reqs = client resps = client(server reqs) = client(server(client resps)) = client(server(client(server reqs)) • We can’t unfold
Solution 1 • Rewrite client client ys = 1 : (if ok (head ys) then ys else error "faulty server") where ok x = True • Pulling the (:) out of the if makes client immediately exhibit a cons cell • Using (head ys) rather than the pattern (y:ys) makes the evaluation of ys lazy
Solution 2 – lazy patterns In Haskell the ~ before a pattern makes it lazy client ~(y:ys) = 1 : (if ok y then y:ys else err) where ok x = True err = error "faulty server” • Calculate using where clauses Reqs = client resps = 1 : (if ok y then y:ys else err) where (y:ys) = resps = 1 : y : ys where (y:ys) = resps = 1 : resps
Memoization fibsFn :: () -> [ Integer ] fibsFn () = 1 : 1 : (zipWith (+) (fibsFn ()) (tail (fibsFn ()))) • Unfolding we get: fibsFn () = 1:1: add (fibsFn()) (tail (fibsFn ())) = 1 : tf where tf = 1:add(fibsFn())(tail(fibsFn())) • But we can’t proceed since we can’t identify tf with tail(fibsFn()). Further unfolding becomes exponential.
0 1 1 1 2 2 3 3 4 5 memo1 • We need a function that builds a table of previous calls. • Memo : (a -> b) -> (a -> b) Memo fib x = if x in_Table_at i then Table[i] else set_table(x,fib x); return fib x Memo1 builds a table with exactly 1 entry.
Using memo1 mfibsFn x = let mfibs = memo1 mfibsFn in 1:1:zipWith(+)(mfibs())(tail(mfibs())) Main> take 20 (mfibsFn()) [1,1,2,3,5,8,13,21,34,55,89,144,233,377, 610,987,1597,2584,4181,6765]
Inductive properties of infinite lists • Which properties are true of infinite lists • take n xs ++ drop n xs = xs • reverse(reverse xs) = xs • Recall that z is the error or non-terminating computation. Think ofz as an approximation to an answer. We can get more precise approximations by: ones = 1 : ones z 1 : z 1 : 1 : z 1 : 1 : 1 : z
Proof by induction • To do a proof about infinite lists, do a proof by induction where the base case is z, rather than [] since an infinite list does not have a [] case (because its infinite). • 1) Prove P{z} • 2) Assume P{xs} is true then prove P{x:xs} • Auxiliary rule: • Pattern match against z returns z. • I.e. case z of { [] -> e; y:ys -> f }
Example • Prove: P{x} == (x ++ y) ++ w = x ++ (y++w) • Prove P{z} (z ++ y) ++ w = z ++ (y++w) • Assume P{xs} (xs ++ y) ++ w = xs ++ (y++w) Prove P{x:xs} (x:xs ++ y) ++ w = x:xs ++ (y++w)
Base Case (z ++ y) ++ w = z ++ (y++w) (z ++ y) ++ w • pattern match in def of ++ z ++ w • pattern match in def of ++ z • pattern match in def of ++ z ++ (y++w)
Induction step • Assume P{xs} (xs ++ y) ++ w = xs ++ (y++w) Prove P{x:xs} (x:xs ++ y) ++ w = x:xs ++ (y++w) (x:xs ++ y) ++ w • Def of (++) (x:(xs ++ y)) ++ w • Def of (++) x :((xs ++ y) ++ w) • Induction hypothesis x : (xs ++ (y ++ w)) • Def of (++) (x:xs) ++ (y ++ w)