Recursive Data Structures and Grammars

ThemesRecursive Description of Data StructuresRecursive Definitions of Properties of Data StructuresRecursive Algorithms for Manipulating and Traversing

Data StructuresStructural induction

ExamplesListsTreesExpressions and Expression Trees

List Processing

Recursive DefinitionList := nil | (cons All List)List<Type> := nil (cons Type List<Type>)Process lists using these two cases and use

recursion to recursively process lists in consUse 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))))

Properties1. (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 xBase 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)) = xBase case. x = nil. (rev (rev nil)) = (rev nil) = nilAssume 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 nodesEach node has exactly one parent, but for the

root nodeEach node may have up to n childrenA leaf node has no childrenAn interior node is a node that is not a leaf

Each subtree is a tree rooted at a given nodeTree := nil | (All List<Tree>)

E.G. (1 ((2 nil) (3 nil) (4 nil)))

Binary Trees

A binary tree is a 2-treeA binary tree is

Empty, orConsists of a node with 3 attributes:

value left, which is a binary tree right, which is a binary tree

• BinTree<Type> := (Type BinTree<Type> BinTree<Type>)

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 emptyNnodes(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 leavesH(T) = -1 if T is emptymax( H(left(T)), H(right(T)) ) + 1

(defun Height (BT)

(if (equal BT nil)

-1

(+ (max (Height (left BT)) (Height (right BT))) 1)))

Correctness of Height

Show (Height BT) = length of the longest path in BT

Prove by structural inductionBase 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 emptyIPL( 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)

1

2 3

4 5

IPL = 6 = 0 + 2 + 4

Nnodes = 5IPL((left BT)) = 0IPL((right BT)) = 2

Correctness of IPLShow (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 Tn be the number of binary trees with n nodes.

T0 = 1, T1 = 1, T2 = 2, T3 = 5

TTT kn

n

kkn 1

1

0

Expression Trees

Basic arithmetic expressions can be represented by a binary treeInternal nodes are operatorsLeaf nodes are operands

Consider 2 * ( 3 + 5 ) :

*

2 +

3 5

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

variablesConsider x * ( z + 5 )

When x = 2 & z = 5 the value is 20When x = 0 & z =5 the value is 0When x = 2 & z = -1the value is 8…

*

x +

z 5

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 variableE.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)))))

Binary Search Trees

Binary Search Tree (BST)Binary TreeAll elements in left(T) are < value(T)All elements in right(T) are value(T)

Each subtree of a BST is a BST

Partially Ordered Trees

Heap (binary)Complete binary tree

So, height = Θ(lg n)

Nodes are assigned some measure, a priorityNo node has lower priority than its children

Inorder traversal

Recursively visit nodes in T.leftvisit rootRecursively visit nodes in T.right

An in order traversal of a BST lists the elements in sorted order. Proof by induction.

Expression Trees

Basic arithmetic expressions can be represented by a binary treeInternal nodes are operatorsLeaf nodes are operands

Consider 2 * ( 3 + 5 ) :

*

2 +

3 5Note that parentheses don’t appear. Expression trees are not ambiguous, as infix expressions are (can be).

Expression Tree – In-order Traversal

An in-order traversal yields 2 * 3 + 5

We put parentheses around every operation to make this correct: (2 * ( 3 + 5 ) ) (not all are needed)

*

2 +

3 5

Pre- and Post-Order Traversal

Always go left-rightPre (me first):

Visit node Traverse left subtree Traverse right subtree

Post (me last): Traverse left subtree Traverse right subtree Visit node

Expression Trees – Pre- and Post-Order

Pre: * 2 + 3 5Post: 2 3 5 + *

Note: Parentheses never needed

*

2 +

3 5