Recursive Data Structures and Grammars  Themes  Recursive Description of Data Structures  Recursive Definitions of Properties of Data Structures  Recursive Algorithms for Manipulating and Traversing Data Structures  Structural induction  Examples  Lists  Trees  Expressions and Expression Trees

List Processing  Recursive Definition  List := nil | (cons All List)  List := nil (cons Type List )  Process lists using these two cases and use recursion to recursively process lists in cons  Use first and rest to access components of cons  Size is the number of conses (generally number of constructors)  Prove properties by inducting on size – assume property holds for smaller size objects and show that this implies it holds for object at hand 2

Member (defun member (x l) (cond ((equal l nil) nil) ((equal x (first l)) t) ((member x (rest l))))) 3

Append (defun append (x y) (if (equal x nil) y (cons (first x) (append (rest x) y))))  Properties 1.(append x nil) = x 2.(length (append x y)) = (+ (length x) (length y)) 3.(append x (append y z)) = (append (append x y) z) 4

Structural Induction  When using induction on recursively defined data structures like lists you can induct on the size of the data structure = to the number of calls to the constructors.  When trying to show a property for a data structure of a given size, you can assume that the property holds when making a recursive call on a smaller data structure. You must make sure that the property holds for all constructors including base cases.  With lists (rest …) will return a smaller data structure (at least one fewer cons)  Structural induction allows you to induct on the recursive data structure without being explicit about the size provided the IH is applied to smaller objects.

Proof of Property 1  Show (append x nil) = x using structural induction  Base case. x = nil. In this case, (append nil nil) returns nil = x.  By induction assume recursive call satisfies the property [note (rest x) is smaller than x]  I.E. (append (rest x) nil) = (rest x)  Thus (append x nil) returns (cons (first x) (rest x)) = x

Proof of Property 2  Show (length (append x y) = (+ (length x) (length y)) using structural induction on x  Base case. x = nil. (append nil y) = y and (length y) = (+ (length nil) (length y))  By induction assume recursive call satisfies the property  (length (append (rest x) y) = (+ (length (rest x)) (length y))  Thus (length (append x y)) = (length (cons (first x) (append (rest x) y)) = (+ 1 (length (rest x)) + (length y)) = (+ (length x) (length y))

Proof of Property 3  Show (append x (append y z)) = (append (append x y) z)  Base case. x = nil. (append nil (append y z)) = (append y z) = (append (append nil y) z)  Assume property holds for (rest x)  (append (append x y) z)  (append (cons (first x) (append (rest x) y)) z) [by def]  (cons (first x) (append (append (rest x) y) z)) [by def]  (cons (first x) (append (rest x) (append y z))) [by IH]  (append (cons (first x) (rest x)) (append y z)) [by def]  (append x (append y z)) [by property of cons]

Reverse (defun reverse (l) (if (equal l nil) nil (append (reverse (rest l)) (cons (first l) nil))))  Properties  (length (reverse x)) = (length x)  (reverse (append x y)) = (append (reverse y) (reverse x))  (reverse (reverse x)) = x 9

Proof of Property 2  Show (rev (append x y)) = (append (rev y) (rev x))  Base case. x = nil. (rev (append nil y)) = (rev y) = (append (rev y) nil) = (append (rev y) (rev nil))  Assume property holds for (rest x)  (rev (append x y))  (rev (cons (first x) (append (rest x) y)) [def apppend]  (append (rev (append (rest x) y)) (cons (first x) nil)) [def rev]  (append (append (rev y) (rev (rest x))) (cons (first x) nil)) [IH]  (append (rev y) (append (rev (rest x)) (cons (first x) nil))) [prop app]  (append (rev y) (rev x)) [def of rev]

Proof of Property 3  Show (rev (rev x)) = x  Base case. x = nil. (rev (rev nil)) = (rev nil) = nil  Assume property holds for (rest x)  (rev (rev x))  (rev (append (rev (rest x)) (cons (first x) nil))) [def rev]  (append (rev (cons (first x) nil)) (rev (rev (rest x)))) [property 2 of rev]  (append (cons (first x) nil) (rev (rev (rest x)))) [def of rev]  (append (cons (first x) nil) (rest x)) [IH]  (cons (first x) (append nil (rest x))) [def of app]  (cons (first x) (rest x)) = x [def of app and prop of cons]

n-Trees  Finite collection of nodes  Each node has exactly one parent, but for the root node  Each node may have up to n children  A leaf node has no children  An interior node is a node that is not a leaf  Each subtree is a tree rooted at a given node  Tree := nil | (All List )  E.G. (1 ((2 nil) (3 nil) (4 nil)))

Binary Trees  A binary tree is a 2-tree  A binary tree is  Empty, or  Consists of a node with 3 attributes:  value  left, which is a binary tree  right, which is a binary tree BinTree := (Type BinTree BinTree )  E.G. (1 (2 nil nil) (3 (4 nil nil) (5 nil nil))) (BinTree 1 (BinTree 2 nil nil) (BinTree 3 (BinTree 4 nil nil) (BinTree 5 nil nil)))

Number of Nodes of a Binary Trees  Nnodes(T) = 0 if T is empty  Nnodes(T.left) + Nnodes(T.right) + 1 (defun Nnodes (BT) (if (equal BT nil) 0 (+ (Nnodes (left BT)) (Nnodes (right BT)) 1))) (defun left (BT) (second BT)) (defun right (BT) (third BT))

Height of Binary Trees  Height of tree T, H(T), is the max length of the paths from the root all of the leaves  H(T) = -1 if T is empty  max( H(left(T)), H(right(T)) ) + 1 (defun Height (BT) (if (equal BT nil) (+ (max (Height (left BT)) (Height (right BT))) 1)))

Correctness of Height  Show (Height BT) = length of the longest path in BT  Prove by structural induction  Base case. -1 for nil is needed to obtain the correct height of a tree with a single node.  Assume (Height (left BT)) and (Height (right BT)) compute the lengths of the longest paths in the left and right subtrees  The longest path in BT is one more than the longest path in either the left or right subtrees which by induction is one more than the maximum of (Height (left BT)) and (Height (right BT))

Internal Path Length (IPL)  The sum of the length of the paths from the root to every node in the tree  Recursively:  IPL( T ) = 0 if T is empty  IPL( T ) = IPL( left( T )) + IPL( right( T )) + num_nodes( T ) – 1 (defun IPL (BT) (if (equal BT nil) 0 (+ (IPL (left BT)) (IPL (right BT)) (- (Nnodes BT) 1))))

IPL Verification (check= (IPL nil) 0) (check= (let ((BT (consBinTree 1 (consBinTree 2 nil nil) (consBinTree 3 (consBinTree 4 nil nil) (consBinTree 5 nil nil))))) (IPL BT)) 6) IPL = 6 = Nnodes = 5 IPL((left BT)) = 0 IPL((right BT)) = 2

Correctness of IPL  Show (ipl BT) = sum of the lengths of the paths from the root to all nodes in the tree.  Prove by structural induction  Base case. 0 is needed to get the proper value for a tree with a single node.  Assume (ipl (left BT)) and (ipl (right BT)) compute the sum of the lengths of paths in the left and right subtrees  The length of each path in BT is one more than the lengths of the paths in the left and right subtrees. Thus one has to be added for each node in the left and right subtrees and since there are (- (Nnodes BT) 1) such nodes this must be added to the sum of (ipl (left BT)) and (ipl (right BT))

Recurrence for the Number of Binary Trees  Let T n be the number of binary trees with n nodes.  T 0 = 1, T 1 = 1, T 2 = 2, T 3 = 5

Expression Trees  Basic arithmetic expressions can be represented by a binary tree  Internal nodes are operators  Leaf nodes are operands  Consider 2 * ( ) : * 2+ 35

Expression Tree Definition  ExpTree := (Number Integer) | (Add ExprTree ExprTree) | (Mult ExprTree ExprTree) (defun NumNode (val) (list 'Number val)) (defun AddNode (E1 E2) (list 'Add E1 E2)) (defun MultNode (E1 E2) (list 'Mult E1 E2)) (MultNode (NumNode 2) (AddNode (NumNode 3) (NumNode 5)))

Expression Tree Evaluation  Check each case in the recursive definition, recursively evaluate subexpressions and apply appropriate operator (defun EvalExpr (E) (cond ((isNumber E) (Val E)) ((isAdd E) (+ (EvalExpr (Op1 E)) (EvalExpr (Op2 E)))) ((isMult E) (* (EvalExpr (Op1 E)) (EvalExpr (Op2 E)))))) (check= (let ((E (MultNode (NumNode 2) (AddNode (NumNode 3) (NumNode 5))))) (EvalExpr E)) 16)

Expression Trees with Variables  What if we allow variables in expression trees?  The value depends on the values of the variables  Consider x * ( z + 5 )  When x = 2 & z = 5 the value is 20  When x = 0 & z =5 the value is 0  When x = 2 & z = -1the value is 8  … * x+ z5

Environments  Need to look up the values of all variables occurring in the expression  Environment contains a list of bindings where a binding is a pair (name value) that binds a value to the name of the variable  E.G. env = ((x 2) (z 5))  lookup returns the value associated with a given variable in an environment  (lookup ‘x env)  2

Expression Tree Evaluation  Modify EvalExpr to include variables – need second argument containing an environment and must handle variable case (defun EvalExpr (E env) (cond ((isNumber E) (Val E)) ((isVariable E) (lookup E env)) ((isAdd E) (+ (EvalExpr (Op1 E) env) (EvalExpr (Op2 E) env))) ((isMult E) (* (EvalExpr (Op1 E) env) (EvalExpr (Op2 E) env)))))