public ZephyrSL / DiscreteMathComputer05.hs Created 2012-01-07 — forked from kanak/DiscreteMathComputer05.hs

Discrete Math Using a Computer: Chapter 05 Trees notes and exercise solutions

DiscreteMathComputer05.hs
 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496 497 498 {- Discrete Mathematics Using a Computer Chapter 05: Trees-} -- =============================================================================-- 5.1 Components of a Tree {-Definition 1: A tree is either an empty tree or it is a node together with a sequence of trees. Definition 2: The node portion of a non-empty tree is called the root of a tree. The sequence of trees are called the children of the root Definition 3: A non-empty tree whose associated sequence of trees is empty is called a leaf. Definition 4: A tree s is said to be a subtree of t if: - s is the same as t (every tree is a subtree of itself) - if t is nonempty, s is a child of t or a subtree of a child of t - this can be simplified by saying "s is a subtree of a child of tree" because each tree is a subtree of itself Definition 5: A tree s is said to be a leaf of a tree t if s is a subtree of t and s is a leaf. Definition 6: A node n is said to occur in a tree t, written n \in t, if t consists of a node m together with a sequence of trees [s1,s2,...] and either n is the same as m or n occurs in one of the children of t Definition 7: A node n is said to be an interior node in the tree if n occurs in t and there is a sequence of trees [s1, s2, ...] such that n together with [s1, s2, ...] is a subtree of t.-} -- =============================================================================-- 5.2 Representing Trees in Haskell -- The book defines a binary tree, but I'm going to define tree based on the defn above. data Tree a = Leaf | Node a [Tree a] deriving (Eq, Show) ---------------------------------------------------------------------------------- Ex 1: Define a datatype Tree1 for a tree that contains a character and an integer in each node, along with exactly three subtrees data Tree1 = Leaf1 | Node1 (Char, Int) Tree1 Tree1 Tree1 deriving (Show) ---------------------------------------------------------------------------------- Ex 2: datatype Tree2 contains an integer at each node, and any number of subtrees type Tree2 = Tree Integer -- =============================================================================-- 5.3 Processing Trees with Recursion {- Preorder: visit root, then left subtree, then right subtree Inorder: visit left, then root, then right Postorder: visit left, then right, then root These aren't sensible for my Tree because my tree has no notion of "left"and "right". I guess I could have two lists one representing the nodes at the "left" of node, and another representing the nodes right of root.-} -- gen for "general" hahadata GenTree a = GenLeaf | GenNode a [GenTree a] [GenTree a] deriving (Eq, Show) -- draw out the following tree (easier to visualize)exampleGenTree :: GenTree IntegerexampleGenTree = GenNode 7 [genSubtree3, genSubtree5] [genSubtree9] where genSubtree3 = GenNode 3 [genSubtree1] [genSubtree4] genSubtree1 = GenNode 1 [GenLeaf] [GenLeaf] genSubtree4 = GenNode 4 [GenLeaf] [GenLeaf] genSubtree5 = GenNode 5 [GenLeaf] [GenNode 6 [GenLeaf] [GenLeaf]] genSubtree9 = GenNode 9 [GenNode 8 [GenLeaf] [GenLeaf]] [GenNode 10 [GenLeaf] [GenLeaf]] inorder:: GenTree a -> [a] inorder GenLeaf = []inorder (GenNode root left right) = inorder' left ++ [root] ++ inorder' right where inorder' :: [GenTree a] -> [a] inorder' = concat . map inorder -- > inorder exampleGenTree-- [1,3,4,5,6,7,8,9,10] preorder :: GenTree a -> [a]preorder GenLeaf = []preorder (GenNode root left right) = [root] ++ preorder' left ++ preorder' right where preorder' :: [GenTree a] -> [a] preorder' = concat . map preorder postorder :: GenTree a -> [a]postorder GenLeaf = []postorder (GenNode root left right) = postorder' left ++ postorder' right ++ [root] where postorder' = concat . map postorder ---------------------------------------------------------------------------------- Ex 3: Calculate the inorder traversal of exampleGenTree-- [1,3,4,5,6,7,8,9,10] ---------------------------------------------------------------------------------- Ex 4: Write a function inorderf :: (a -> b) -> GenTree a -> [b]-- Traverses in order and applies f to each value inorderf f = map f . inorder ---------------------------------------------------------------------------------- Processing Tree Structure-- Reflect returns a binary tree where everything is reflected left-to-right reflect :: GenTree a -> GenTree areflect GenLeaf = GenLeafreflect (GenNode root left right) = GenNode root (flipsub right) (flipsub left) where flipsub = reverse . map reflect -- *Main> inorder (reflect exampleGenTree)-- [10,9,8,7,6,5,4,3,1] height :: GenTree a -> Integerheight GenLeaf = 0height (GenNode _ left right) = 1 + max (maxsub left) (maxsub right) where maxsub = maximum . map height size :: GenTree a -> Integersize GenLeaf = 0size (GenNode _ left right) = 1 + sumsub left + sumsub right where sumsub = sum . map size -- *Main> height exampleGenTree-- 3-- *Main> size exampleGenTree-- 9 -- Note size exampleGenTree should be length . inorder/preorder/postorder-- because there can be many subtrees, i've defined balanced to mean that the maximum height of left subtree is equal to the maximum height of the right subtreebalanced :: GenTree a -> Boolbalanced GenLeaf = Truebalanced (GenNode _ left right) = balsub left && balsub right && (maximum (map height left) == maximum (map height right)) where balsub :: [GenTree a] -> Bool balsub = and . map balanced -- *Main> balanced exampleGenTree-- False-- *Main> balanced GenLeaf-- True-- *Main> balanced (GenNode 3 [GenNode 1 [GenLeaf] [GenLeaf]] [GenNode 4 [GenLeaf] [GenLeaf]])-- True ---------------------------------------------------------------------------------- Ex 5: highest and shortest binary trees containing 7 elements -- shortest is the fully balanced one-- highest is the one that looks like a chain ---------------------------------------------------------------------------------- Ex 6: we modify the balanced clause to be:-- balanced BinLeaf = True-- balanced (BinNode x t1 t2) = balanced t1 && balanced t2 -- Give a counterexample for this.{- 1 2 3 4 5 Notation: {1} means tree rooted at 1balanced {1}=> balanced {2} && balanced {3}=> (balanced {} && balanced {}) && balanced {3}=> (True && True) && balanced {3}=> balanced {3}=> balanced {4} && balanced {5}=> (balanced {} && balanced {}) && (balanced {} && balanced {})=> True WAIT! actually the following would work as well! balanced 1 2-} ---------------------------------------------------------------------------------- Ex 7: we modify balanced to be:{- balanced BinLeaf = True balanced (BinNode x t1 t2) = height t1 == height t2 Give a counterexample for this.  1 2 3 4 5 6 7balanced {1}= height {2} == height {3}= 3 == 3= True but the trees are not balanced. -} ---------------------------------------------------------------------------------- Binary Search Trees linearSearch :: Eq a => a -> [(a, b)] -> Maybe blinearSearch x [] = NothinglinearSearch key ((x,y):xs) | key == x = Just y | otherwise = linearSearch key xs -- using higher order functionslinearSearch' :: Eq a => a -> [(a, b)] -> Maybe blinearSearch' key = safeHead . filter ((== key) . fst) where safeHead :: [(a,b)] -> Maybe b safeHead [] = Nothing safeHead ((a,b):_) = Just b -- *Main> linearSearch' 10 (zip [1..20] ['a'..'z'])-- Just 'j'-- *Main> linearSearch 10 (zip [1..20] ['a'..'z'])-- Just 'j'  -- binary search on binary trees-- insertion on binary trees ---------------------------------------------------------------------------------- Ex 8: mapTree mapTree :: (a -> b) -> GenTree a -> GenTree bmapTree f GenLeaf = GenLeafmapTree f (GenNode val left right) = GenNode (f val) (map (mapTree f) left) (map (mapTree f) right) ---------------------------------------------------------------------------------- Ex 9: concatTree-- "takes a tree of lists and concatenates the lists in order from left to right"concatTree :: GenTree [a] -> [a]concatTree = concat . inorder -- *Main> concatTree (mapTree (\ x -> [x]) exampleGenTree)-- [1,3,4,5,6,7,8,9,10]-- *Main> inorder exampleGenTree-- [1,3,4,5,6,7,8,9,10] ---------------------------------------------------------------------------------- Ex 10: zipTree -- the book talks about returning Nothing if the two trees don't have the same shape-- but i think a better thing to do is to return a GenLeaf for those trees. zipTree :: GenTree a -> GenTree b -> GenTree (a,b)zipTree _ GenLeaf = GenLeafzipTree GenLeaf _ = GenLeafzipTree (GenNode lval lleft lright) (GenNode rval rleft rright) = GenNode (lval, rval) (zipWith zipTree lleft rleft) (zipWith zipTree lright rright) -- *Main> zipTree exampleGenTree (mapTree (* 2) exampleGenTree)-- GenNode (7,14) [GenNode (3,6) [GenNode (1,2) [GenLeaf] [GenLeaf]] [GenNode (4,8) [GenLeaf] [GenLeaf]],GenNode (5,10) [GenLeaf] [GenNode (6,12) [GenLeaf] [GenLeaf]]] [GenNode (9,18) [GenNode (8,16) [GenLeaf] [GenLeaf]] [GenNode (10,20) [GenLeaf] [GenLeaf]]] ---------------------------------------------------------------------------------- Ex 11: zipWithTree zipWithTree :: (a -> b -> c) -> GenTree a -> GenTree b -> GenTree czipWithTree f t1 t2 = mapTree (uncurry f) \$ zipTree t1 t2 -- ============================================================================={- 5.4 Induction on Trees Principle of Induction on Trees Base Case: P(Leaf) Inductive Case: Assuming that P holds for the subtrees, does it hold for a node which has a particular value and subtrees where P holds? --------------------------------------------------------------------------------Theorem: Suppose t :: GenTree a, then reflect (reflect t) = t Proof is by Induction on t.  Base Case: t = GenLeaf Left Side: reflect (reflect GenLeaf) = reflect (GenLeaf) = GenLeaf Right Side: GenLeaf  Inductive Case: Suppose t = GenNode val [l1, ..., ln] [r1, ..., rn] Inductive Hypothesis: reflect (reflect foo) = foo where foo = l1 ... ln, r1 ... rn i.e. each individual tree itself is flipped correctly  Left Side: reflect (reflect (GenNode val [l1...ln] [r1...rn])) = reflect (GenNode val (flipsub right) (flipsub left)) = GenNode val (flipsub (flipsub left)) (flipsub (flipsub right)) = GenNode val left right (have to prove that flipsub (flipsub x) = x)  Proof is by induction on x, which is a list of trees Base Case: x = [] Left Side: flipsub (flipsub []) = fipsub (reverse . map reflect []) = flipsub (reverse []) = flipsub [] = reverse . map reflect [] = reverse [] = [] Right side: []  Inductive Case: xs = t:ts Inductive Hypothesis: flipsub (flipsub ts) = ts (also need inductive hypothesis from reflect) Left Side: flipsub (flipsub (t:ts)) = flipsub (reverse . map reflect (t:ts)) = flipsub (reverse . ((reflect t) : map reflect ts)) = flipsub ((map reflect ts) ++ [reflect t]) = reverse (map reflect ((map reflect ts) ++ [reflect t])) = reverse ((map reflect (map reflect ts)) ++ (map reflect [reflect t]) = (map reflect [reflect t]) ++ (map reflect (map reflect ts)) = reflect (reflect t) : (map reflect (map reflect ts)) (since [t] is a singleton list) = t : reverse (reverse (map reflect (map reflect ts))) (since reverse (reverse x) is x) = t : reverse (map reflect (reverse (map reflect ts))) (since reverse (map f xs) = map f (reverse xs)) = t : flipsub (flipsub ts) = t : ts (by inductive hypothesis) = xs --------------------------------------------------------------------------------Theorem : length (inorder t) = size t, where t :: GenTree a Proof is by induction on t Base Case: t = GenLeaf Left Side = length (inorder GenLeaf) = length [] = 0 Right Side = size GenLeaf = 0 Inductive Case: t = GenNode val x@[x1, ... , xn] y@[y1, ..., ym], where x_i, y_i :: GenTree a, val :: aInductive Hypothesis: length (inorder x_i) = size x_i length (inorder y_i) = size y_i forall i Left Side = length (inorder t) = length (inorder GenNode val x y) = length (inorder' x ++ [val] ++ inorder' y) = length ((concat . map inorder x) ++ [val] ++ (concat . map inorder y)) = length (concat . map inorder x) + 1 + length (concat . map inorder y) = sum (map (length . inorder) x) + 1 + sum (map (length . inorder) y) (since length (concat xs) = sum (map length) xs, and map f (map g xs) = map (f . g) xs) = sum ([size x_1, ..., size x_n]) + 1 + sum ([size y_1, ..., size y_n]) = SUM size x_i + 1 + SUM size y_i Right Side = size (GenNode val x@[x1, ..., xn] y@[y1, ... ym]) = 1 + sumsub x + sumsub y = 1 + (sum (map size x)) + (sum (map size y)) = 1 + (sum [size x_1, ... , size x_n]) + (sum [size y_1, ..., y_n]) = 1 + SUM size x_i + SUM size y_i -} -- =============================================================================-- 5.5 Improving Execution Time {- Timing Analysis of FunctionsUse equational reasoning to deduce how long a function would take e.g. inorder 1. time (inorder GenLeaf) = 0 (simplifying assumption, small amount of time taken to return the empty list) 2. time (inorder (GenNode root left right)) = time (inorder' left ++ [root] ++ inorder' right) How long does ++ take?Do equational reasoning on definition of ++get: time (xs ++ ys) = length xs So, time (inorder' left ++ [root] ++ inorder' right) = length (inorder' left) + 1 + time (inorder' left) + time (inorder' right) = 1 + length (concat (map inorder left)) + time (concat (map inorder left)) + time (concat (map inorder right)) = 1 + sum (map length (map inorder left))) + time (concat (map inorder left)) + time (concat (map inorder right)) [since length (concat xs) = sum (map length xs)] = 1 + sum (map (length . inorder) left) + time (concat (map inorder left)) + time (concat (map inorder right)) = 1 + SIZE_left + time (concat (map inorder left)) + time (concat (map inorder right)) = 1 + SIZE_left + (time to concatenate inorder results + time to traverse each left subtree) + (time to concatenate inorder results + time to traverse each right subtree) ~ 1 + SIZE_left + (SIZE_left + TIME_left) + (SIZE_right + TIME_right) [approximately equal because concatenation time is overestimated; we don't need to spend time proportional to the last child in the list, but we're still adding that for simplicity] = 1 + 2 * SIZE_left + TIME_left + TIME_right (Does the above make sense? Where do we get 2 * SIZE_left from? One is from when we're building the list corresponding to the tree (time taken is approximately SIZE_left). Then we need to add this list in front of the list from the right subtree, so another SIZE_left there.  ** This is a recurrence equation **Problem Case for our function: when the lists are left-heavy.-} -- =============================================================================-- 5.6 Flattening Trees in Linear Time -- Reason that inorder is slow is that it recopies the list repeatedly as it concatenates-- Trick: accumulate the result list in a collection of partial computations-- that permit pasting the results directly, avoiding costly concatenations-- The partial list is called a continuation -- second argument is the continuationinorderLinear :: GenTree a -> [a] -> [a]inorderLinear GenLeaf ks = ksinorderLinear (GenNode val left right) ks = inorderLinear' left (val : (inorderLinear' right ks)) where inorderLinear' :: [GenTree a] -> [a] -> [a] inorderLinear' [] ks = ks inorderLinear' (x:xs) ks = inorderLinear x (inorderLinear' xs ks) {- What is going on?My version is more complicated than the book's because my left/right is a list of subtrees and not just a single subtreeSo, I have two functions: one that performs traversal on a Tree, and another that does traversal on a list of trees.  To traverse a single tree: 1. If it is a leaf, then just perform the continuation 2. If it is a Node, then traverse the list of subtrees in the left. Once done, show the value of the Node, and then traverse the list of subtrees in the right. To traverse a list of trees (the reason I'm not simply mapping over, is that the continuation for the first subtree is traversing the list of rest of the subtrees; the continuation is to be applied only when the list is empty)  1. If the list is empty, apply the continuation. 2. If the list is not empty, traverse the first subtree, then traverse the list of rest of the subtrees.-} --------------------------------------------------------------------------------{- Theorem: inorderLinear t ks = inorder t ++ ks, where t is a finite GenTree a, ks :: [a]Proof is by structural induction on t Base Case: t = GenLeaf Left Side: inorderLinear GenLeaf ks = ks Right Side; inorder GenLeaf ++ ks = [] ++ ks = ks Inductive Case: t = GenNode val [left_1, ..., left_n] [right_1, ..., right_n] Inductive Hypothesis: inorderLinear left_i ks = inorder left_i ++ ks Left Side: inorderLinear GenNode val left right = inorderLinear' left (val : inorderLinear' right ks)  -- Lemma: inorderLinear' left ks = concat (map inorder left) ++ ks Proof is by induction on "left" Base Case: Left Side: inorderLinear' [] ks = ks Right Side: concat (map inorder []) ++ ks = concat [] ++ ks = [] ++ ks = ks  Inductive Case: left = l:ls Hypothesis: inorderLinear' ls ks = concat (map inorder ls) ++ ks Left Side: inorderLinear' (l:ls) ks = inorderLinear l (inorderLinear' ls ks) = inorderLinear l (concat (map inorder ls) ++ ks) = inorder l ++ (concat (map inorder ls) ++ ks) [ Assuming inorderLinear works] = (concat (inorder l):(map inorder ls) ++ ks) = concat (map inorder l:ls) ++ ks QED  -- back to the original theorem: inorderLinear' left (val : inorderLinear' right ks) = concat (map inorderLin left) ++ (val : inorderLinear' right ks) (by Lemma) = concat (map inorderLin left) ++ (val : (concat (map inorderLin right) ++ ks) = concat (map inorderLin left) ++ [val] ++ (concat (map inorderLin right) ++ ks) = inorder' left ++ [val] ++ inorder' right ++ ks = inorder t ++ ksQED ASIDE: Question about the soundness of the proofThe problem is that inorder needs to assume that inorder' works, which in turn needs to assume that inorder works. The inductive hypothesis for inorder' states that it works for ls in the (l:ls) partThe inductive hypothesis for inorder states that it works for the subtrees of the node i'm currently looking at. So, when i'm in inorder' and I assume that inorder works, I'm assuming that it works for an element, A, in a particular tree. Now I'm in inorder and I need to know if inorder works for A. But inorder works for A if inorder' works for lA and lB, which are left and right children of A. but these are "smaller" in a sense than the original list that inorder' was looking at. So by the inductive hypothesis for inorder', inorder' DOES work for lA and lB, so inorder DOES work for A.  This is why the assumption is justified.-} --------------------------------------------------------------------------------{- Theorem: time (inorderLinear t []) = size t , where t is a finite tree.Proof is by structural induction on the tree time (inorderLinear GenLeaf []) = 0 (since we just need to return a value) time (inorderLinear (GenNode val left right) [])= time (inorderLinear' left (val : (inorderLinear' right [])))  How much time to do inorderLinear' t [] ? Claim: it is sum (map size t) aka grand size of t Induction on t: When t = [], we just return a value so 0 When t = x:xs, time (inorderLinear x (inorderLinear' xs [])) = time (inorderLinear x) + time (inorderLinear' xs []) = time (inorderLinear x) + sum (map size xs) = size x + sum (map size xs) [ ASSUMING inorderLinear's timing done correctly] = sum (map size x:xs) which proves our hypothesis back to the original problem:= time (inorderLinear' left (val : (inorderLinear' right [])))= sum (map size left) + 1 + sum (map size right) [ + 1 is for consing val onto a list]= size t-} ---------------------------------------------------------------------------------- Ex 12: Write appendTree, a function that takes a binary tree and a list-- and appends the contents of the tree, traversed left to right to the front of the list-- Try to find an efficient solution that minimizes recopying. appendTree :: GenTree a -> [a] -> [a]appendTree t ks = inorderLinear t ks