Cse502 Functional Programming 1 6/26/2015 Lecture #11, Nov. 4, 2002 Todays Topics –Using Calculation to prove 2 functions are equal »An example from class drawPic –Laws about Monads –Laws about sequence –Using Calculation to improve functions »An example from the first Homework assignment Read Chapter 11 - Proofs by induction Read Chapter 12 – Qualified types

Cse502 Functional Programming 2 6/26/2015 Proof from Class drawPic :: Window -> Picture -> IO () drawPic w (Region c r) = drawReg w (c,r) drawPic w (p1 `Over` p2) = do { drawPic w p2 ; drawPic w p1 } drawPic w EmptyPic = return () flattenPict :: Picture -> [(Color,Region)] flattenPict EmptyPic = [] flattenPict (Region c r) = [(c,r)] flattenPict (p1 `Over` p2) = flattenPict p2 ++ flattenPict p1 draw2 w p = sequence (map (drawReg w) (flattenPict p))

Cse502 Functional Programming 3 6/26/2015 We would like to Prove drawPict w p = draw2 w p We can draw a picture OR We can flatten a picture, then draw each of the regions in the order they appear in the flattened list. This is important, because for some applications we want to maintain the list of regions for other reasons. Recall the structure of a picture: data Picture = Region Color Region | Picture `Over` Picture | EmptyPic

Cse502 Functional Programming 4 6/26/2015 Structure of the proof To Prove : drawPic w p = draw2 w p Prove 3 cases –case 1) p = EmptyPic prove: drawPic w EmptyPic = draw2 w EmptyPic –case 2) p = (Region c r) prove: drawPic w (Region c r) = draw2 w (Region c r) –case 3) p = (p1 `Over` p2) assume: a) drawPic w p1 = draw2 w p1 b) drawPic w p2 = draw2 w p2 prove: drawPic w (p1 `Over` p2) = draw2 w (p1 `Over` p2)

Cse502 Functional Programming 5 6/26/2015 First a few laws Laws about do and return (we saw these when we first studied monads) Monad1: do { a <- x; return a } = x Monad2: do { x <- return y; e } = e[y/x] Monad3: do { a <- do { b <- e; c}; d } = do { b <- e; a <- c; d } provided b not in FreeVars(d) Monad4: do { x<- a; b} = do {a; b} provided x not in FreeVars(b)

Cse502 Functional Programming 6 6/26/2015 Draw2 uses sequence Definition sequence :: [ IO () ] -> IO () S1: sequence [] = return () S2: sequence (x:xs) = do { x; sequence xs } Lemmas Lemma 1: sequence [ x ] = x Lemma 2: sequence(a ++ b) = do {sequence a; sequence b} Lemma 3: map f x ++ map f y = map f (x ++ y)

Cse502 Functional Programming 7 6/26/2015 Proof Lemma1 Lemma 1: sequence [ x ] = x sequence [ x ] = By def of [ x ] syntax sequence (x : []) = by def of sequence rule s2 do { x; sequence [] } = by def of sequence rule S1 do { x; return () } = by Monad4 backwards do { a <- x; return () } = Since a has type (), and () is unique do { a <- x; return a } = by Monad1 x

Cse502 Functional Programming 8 6/26/2015 Lemma 2: sequence(a ++ b) = do {sequence a; sequence b} P{a} = sequence(a ++ b) = do {sequence a; sequence b} Proof by induction over the structure of a Base Case - Prove: sequence([] ++ b) = do {sequence []; sequence b} sequence([] ++ b) = by def of ++ sequence(b) = by Monad2 (backwards) do { x <- return (); sequence b } = by def sequence do { x <- sequence []; sequence b } = Monad4 do {sequence []; sequence b }

Cse502 Functional Programming 9 6/26/2015 Induction Step Assume: sequence(xs ++ b) = do {sequence xs; sequence b} Prove: sequence((x:xs) ++ b) = do {sequence (x:xs); sequence b} sequence((x:xs) ++ b) = by def of ++ sequence( x : (xs ++ b)) = by def of sequence do { x; sequence (xs ++ b) } = by IH do { x; do {sequence xs; sequence b} } = by Monad3 do { do {x; sequence xs}; sequence b } = def of sequence do { sequence (x:xs); sequence b }

Cse502 Functional Programming 10 6/26/2015 Back to the main proof To Prove : drawPic w p = draw2 w p Prove 3 cases –case 1) p = EmptyPic prove: drawPic w EmptyPic = draw2 w EmptyPic –case 2) p = (Region c r) prove: drawPic w (Region c r) = draw2 w (Region c r) –case 3) p = (p1 `Over` p2) assume: a) drawPic w p1 = draw2 w p1 b) drawPic w p2 = draw2 w p2 prove: drawPic w (p1 `Over` p2) = draw2 w (p1 `Over` p2)

Cse502 Functional Programming 11 6/26/2015 case 1. p = EmptyPic drawPic w EmptyPic = draw2 w EmptyPic drawPic w EmptyPic = by def drawPict return () = by def sequence sequence [] = by def map sequence (map (drawReg w) []) = by def flattenPict sequence (map (drawReg w) (flattenPict EmptyPic)) = by def draw2 draw2 w EmptyPic

Cse502 Functional Programming 12 6/26/2015 case 2. p = (Region c r) drawPic w (Region c r) = draw2 w (Region c r) drawPic w (Region c r) = by def drawPic drawReg w (c,r) = by Lemma1 sequence [ drawReg w (c,r) ] by def of map sequence (map (drawRegion w) [(c,r)]) by def flattenPict sequence (map (drawRegion w) (flattenPict (Region c r))) by def draw2 draw2 w (Region c r)

Cse502 Functional Programming 13 6/26/2015 case 3. p = (a `Over` b) assume: a) drawPic w a = draw2 w a b) drawPic w b = draw2 w b prove: drawPic w (a `Over` b) = draw2 w (a `Over` b) drawPic w (a `Over` b) = by def drawPic do { drawPic w b; drawPic w a } = by IH do { draw2 w b; draw2 w a } = by def draw2 do { sequence (map (drawReg w) (flattenPict b)); sequence (map (drawReg w) (flattenPict a)) } =

Cse502 Functional Programming 14 6/26/2015 Case 3. continued = by Lemma2 sequence ( (map (drawReg w) (flattenPict b)) ++ (map (drawReg w) (flattenPict a)) ) = by Lemma3 sequence (map (drawReg w) (flattenPict b ++ flattenPict a)) = def of flattenPict sequence (map (drawReg w) (flattenPict (a `Over` b))) = by def of draw2 draw2 (a `Over` b)

Cse502 Functional Programming 15 6/26/2015 Ex2. Recall String to Integer A function which converts a string of digits into an Int. Uses the function which ord which takes a Char as input and returns its ascii code ord ‘1’ --> 49 Follows the "pipeline" analysis "167" --> ['1','6','7'] --> [49,54,55] --> [1,6,7] --> [(1,100),(6,10),(7,1)] --> [100, 60, 7] --> 167

Cse502 Functional Programming 16 6/26/2015 String to Int (cont) Then str2int is an easy composition str2int :: String -> Int str2int = sum. map (uncurry (*)). explist. map (\ z -> z -(ord '0')). map ord The Key is the function explist explist [5,3,4] --> [(5,100),(3,10),(4,1)]

Cse502 Functional Programming 17 6/26/2015 Key Function Explist Useful intermediates reverse [1,10,100] [1,10,100] --> [100,10,1] zip [3,4] [100, 10, 1] --> [(3,100), (4,10)] Definition explist zs = zip zs [ power 10 x | x <- reverse [0.. n-1] ] where n = length zs

Cse502 Functional Programming 18 6/26/2015 Another explist (we use this one in the improvement) explist = fst. foldr g ([],1) where z `g` (zs,n) = ((z,n) : zs, n * 10) Suppose we start with [5,3,2] Folding g leaves –5 `g` (3 `g` (2 `g` ([],1))) –5 `g` (3 `g` ((2,1):[],10)) –5 `g` (3 `g` ([(2,1)],10)) –5 `g` ((3,10):[(2,1)],100) –5 `g` ([(3,10),(2,1)],100) –([(5,100),(3,10),(2,1)],1000)

Cse502 Functional Programming 19 6/26/2015 Using theory to improve functions Given the definitions: sum: sum = foldr (+) 0 uncurry: uncurry f (x,y) = f x y and the laws: map1: map f. map g = map (f. g) lambda1: (\ x -> e). f = (\ x -> e[(f x)/x]) lambda2: f. (\ x -> e) = (\ x -> f e) mapfoldr: foldr c e. map f = (let d x y = c (f x) y in foldr d e) Improve the definition of str2int

Cse502 Functional Programming 20 6/26/2015 Improvement 1 str2int = sum. map (uncurry (*)). explist. map (\ z -> z -(ord '0')). map ord »by map1 str2int = sum. map (uncurry (*)). explist. map ((\z -> z -(ord '0')). ord)

Cse502 Functional Programming 21 6/26/2015 Improvement 2 str2int = sum. map (uncurry (*)). explist. map ((\z -> z -(ord '0')). ord) »by lambda1 str2int = sum. map (uncurry (*)). explist. map (\ z -> ord z -(ord '0'))

Cse502 Functional Programming 22 6/26/2015 Improvement 3 » by definition of sum str2int = foldr (+) 0. map (uncurry (*)). explist. map (\ z -> ord z -(ord '0')) »by mapfoldr str2int = (let d x y = (+)(uncurry (*) x) y in foldr d 0). explist. map (\ z -> ord z -(ord '0'))

Cse502 Functional Programming 23 6/26/2015 Improvement 4 str2int = (let d x y = (+)(uncurry (*) x) y in foldr d 0). explist. map (\ z -> ord z -(ord '0')) »by operator use : (+) x y = x + y str2int = (let d x y = (uncurry (*) x) + y in foldr d 0). explist. map (\ z -> ord z -(ord '0'))

Cse502 Functional Programming 24 6/26/2015 Improvement 5 » patterns and definition of uncurry str2int = (let d (a,b) y = a*b + y in foldr d 0). explist. map (\ z -> ord z -(ord '0')) »rearranging str2int = (foldr d 0). explist. map f where d (a,b) y = a*b + y f z = ord z -(ord '0')

Cse502 Functional Programming 25 6/26/2015 Improvement 6 »Using def of explist explist = fst. foldr g ([],1) where z `g` (zs,n) = ((z,n) : zs, n * 10)

Cse502 Functional Programming 26 6/26/2015 Improvement 7 str2int = (foldr d 0). fst. (foldr g ([],1)). map f where d (a,b) y = a*b + y f z = ord z -(ord '0') z `g` (zs,n) = ((z,n) : zs, n * 10) »Using mapfoldr again str2int = (foldr d 0).fst.(foldr d2 ([],1)) where d (a,b) y = a*b + y d2 z (zs,n) =((ord z - ord ‘0’,n)::zs, n*10)