1. Announcements Exam 2 on Friday, November 2nd Topics
You are allowed 2 “cheat” pages only
Practice test under Course Materials on Submitty
New rule: DO NOT LEAVE YOUR SEAT during last half hour 3:30 – 4pm
Topics
Binding and scoping
Attribute grammars
Scheme: lists, recursion, higher-order functions (especially map and fold), tail recursion, let expressions, scoping
Lambda calculus

2. Announcements HW8 is posted, due Tuesday November 6th
Some lambda calculus questions
Use as additional practice problems for test!
Fall 18 CSCI 4430, A Milanova

3. Last Class Lambda calculus Introduction Syntax and semantics
Free and bound variables
Substitution
Fall 18 CSCI 4430, A Milanova/BG Ryder

4. Lambda Calculus
Reading: Scott, Ch. 11 on CD

5. Today’s Lecture Outline
Lambda calculus, continued
Substitution, review
Rules of the lambda calculus
Normal forms
Reduction strategies
Review for test
Fall 18 CSCI 4430, A Milanova

6. Syntax of Pure Lambda Calculus
Convention:
notation f, x, y, z for variables;
E, M, N, P, Q for expressions
E ::= x | ( x. E1 ) | ( E1 E2 )
A -expression is one of
Variable: x
Abstraction (i.e., function definition): x. E1
Application: E1 E2
-calculus formulae (e.g., ( x. (x y) )) are called expressions or terms
( x. (x y) ) corresponds to (lambda (x) (x y)) in Scheme!
I use notation f, x, y, z for variables; E, M, N, P, Q for expressions
Fall 18 CSCI 4430, A Milanova/BG Ryder

7. Syntactic Conventions
May drop parenthesis from ( E1 E2 ) or ( x. E )
E.g., ( f x ) may be written as f x
Function application is left-associative
I.e., it groups from left-to-right
E.g., x y z abbreviates ( ( x y ) z )
E.g., E1 E2 E3 E4 abbreviates ( ( ( E1 E2 ) E3 ) E4 )
Application has higher precedence than abstraction
Another way to say this is that the scope of the dot extends as far to the right as possible
E.g., x. x y = x. ( x y ) = ( x. ( x y ) ) ≠ ( ( x. x ) y )

8. Free and Bound Variables
Abstraction ( x. E ) introduces a “binding”
Variable x is said to be bound in x. E
The set of free variables of E is the set of variables that are unbound in E
Defined by cases on E
Var x:
App E1 E2:
Abs x.E:
free(x) = {x}
free(E1 E2) = free(E1) U free(E2)
free(x.E) = free(E) - {x}
Fall 18 CSCI 4430, A Milanova

9. Free and Bound Variables
A variable x is bound if it is in the scope of a lambda abstraction: as in x. E
Variable is free otherwise
1. (x. x) y
2. (z. z z) (x. x)
3. x.y.z. x z (y (u. u))
y is free.
No free variables
Fall 18 CSCI 4430, A Milanova

10. Substitution, formally
(x. E) M  E[M/x] replaces all free occurrences of x in E by M
E[M/x] is defined by cases on E:
Var: y[M/x] =
y[M/x] =
App: (E1 E2)[M/x] =
Abs: (y. E1)[M/x] =
(y. E1)[M/x] =
M if x = y
y otherwise
(E1[M/x] E2[M/x])
y. E1 if x = y
z. ((E1[z/y])[M/x]) otherwise,
where z NOT in free(E1) U free(M) U {x}
Fall 18 CSCI 4430, A Milanova

11. Substitution, formally
(x.y. x y) (y w)
(y. x y)[(y w)/x]
1_. ( ((x y)[1_/y])[(y w)/x] )
1_. ( (x 1_)[(y w)/x] )
1_. ( (y w) 1_ )
1_. y w 1_ /* same as z. y w z */
You will be implementing this exact substitution algorithm in Haskell!
1_ is a “fresh” variable.
It is not free in either (x y), (y w) or x!
Fall 18 CSCI 4430, A Milanova/BG Ryder

12. Rules (Axioms) of Lambda Calculus
 rule (-conversion): renaming of parameter (choice of parameter name does not matter)
x.E  z.(E[z/x]) provided that z is not free in E
e.g., x. x x is the same as z. z z
 rule (-reduction): function application (substitutes argument for parameter)
(x. E) M  E[M/x]
Note: E[M/x] as defined on previous slide!
e.g., (x. x) z  z
Fall 18 CSCI 4430, A Milanova

13. Rules of Lambda Calculus: Exercises
Use -conversion and/or β-reduction:
(x. x) y  ?
(x. x) (y. y)  ?
(x.y.z. x z (y z)) (u. u) (v. v) 
Notation:  denotes that expression on the left reduces
to the expression on the right, through a sequence -conversions
and β-reductions.
Fall 18 CSCI 4430, A Milanova

14. Reductions
An expression ( x.E ) M is called a redex (for reducible expression)
An expression is in normal form if it cannot be β-reduced
The normal form is the meaning of the term, the “answer”
Fall 18 CSCI 4430, A Milanova

15. Questions Is z. z z in normal form?
Answer: yes, it cannot be beta-reduced
Is (z. z z) (x. x) in normal form?
Answer: no, it can be beta-reduced
Fall 18 CSCI 4430, A Milanova/BG Ryder

16. Definitions of Normal Form
Normal form (NF): a term without redexes
Head normal form (HNF)
x is in HNF
(x. E) is in HNF if E is in HNF
(x E1 E2 … En) is in HNF
Weak head normal form (WHNF)
x is in WHNF
(x. E) is in WHNF
(x E1 E2 … En) is in WHNF
Fall 18 CSCI 4430, A Milanova (from MIT’s 2015 Program Analysis OCW)

17. Questions z. z z is in NF, HNF, or WHNF? (z. z z) (x. x) is in?
x.y.z. x z (y (u. u)) is in?
(x.y. x) z ((x. z x) (x. z x)) is in?
z ((x. z x) (x. z x)) is in?
z.(x.y. x) z ((x. z x) (x. z x)) is in?
(We will be reducing to NF, mostly)

18. More Reduction Exercises
C = x.y.f. f x y
H = f. f (x.y. x) T = f. f (x.y. y)
What is H (C a b)?
(f. f (x.y. x)) (C a b)
(C a b) (x.y. x)
((x.y.f. f x y) a b) (x.y. x)
(f. f a b) (x.y. x)
(x.y. x) a b a
Fall 18 CSCI 4430, A Milanova (from MIT 2015 Program Analysis OCW)

19. Exercise S = x.y.z. x z (y z) I = x. x What is S I I I?
An expression with no free
variables is called combinator.
S, I, C, H, T are combinators.
S = x.y.z. x z (y z)
I = x. x
What is S I I I?
( x.y.z. x z (y z) ) I I I
(y.z. I z (y z) ) I I
(z. I z (I z) ) I
I I (I I) = (x. x) I (I I)
I (I I) = (x. x) (I I)
I I = (x. x) I  I
Reducible expression is underlined
at each step.
Fall 18 CSCI 4430, A Milanova

20. Today’s Lecture Outline
Lambda calculus, continued
Substitution, review
Rules of the lambda calculus
Normal forms
Reduction strategies
Review for test
Fall 18 CSCI 4430, A Milanova/BG Ryder

21. Reduction Strategy Look again at (x.y.z. x z (y z)) (u. u) (v. v)
Actually, there are (at least) two “reduction paths”:
Path 1: (x.y.z. x z (y z)) (u. u) (v. v) β (y.z. (u. u) z (y z)) (v. v) β
(z. (u. u) z ((v. v) z)) β (z. z ((v. v) z)) β
z. z z
Path 2: (x.y.z. x z (y z)) (u. u) (v. v) β (y.z. (u. u) z (y z)) (v. v) β
(y.z. z (y z)) (v. v) β (z. z ((v. v) z)) β

22. Reduction Strategy
A reduction strategy (also called evaluation order) is a strategy for choosing redexes
How do we arrive at a normal form (answer)?
Applicative order reduction chooses the leftmost-innermost redex in an expression
Also referred to as call-by-value reduction
Normal order reduction chooses the leftmost-outermost redex in an expression
Also referred to as call-by-name reduction
Fall 18 CSCI 4430, A Milanova

23. Reduction Strategy: Examples
Evaluate (x. x x) ( (y. y) (z. z) )
Using applicative order reduction:
(x. x x) ( (y. y) (z. z) )
(x. x x) (z. z)
(z. z) (z. z)  (z. z)
Using normal order reduction
(y. y) (z. z) ( (y. y) (z. z) )
(z. z) ( (y. y) (z. z) )
(y. y) (z. z)  (z. z)
Both orders arrived at the same normal form (generally, but not always true!)
Normal order took more work (generally true).

24. Reduction Strategy
In our examples, both strategies produced the same result. This is not always the case
First, look at expression (x. x x) (x. x x). What happens when we apply β-reduction to this expression?
Then look at (z. y) ((x. x x) (x. x x))
Applicative order reduction – what happens?
Normal order reduction – what happens?
Applicative order never terminates!
Normal order reduction terminates. It evaluates to y.
Fall 18 CSCI 4430, A Milanova

25. Church-Rosser Theorem
Normal form implies that there are no more reductions possible
Church-Rosser Theorem, informally
If normal form exists, then it is unique (i.e., result of computation does not depend on the order that reductions are applied; i.e., no expression can have two distinct normal forms)
If normal form exists, then normal order will find it
Church-Rosser Theorem, more formally:
For all pure -expressions M, P and Q, if M * P and M * Q, then there must exist an expression R such that P * R and Q * R

26. Reduction Strategy Intuitively:
Applicative order (call-by-value) is an eager evaluation strategy. Also known as strict
Normal order (call-by-name) is a lazy evaluation strategy
What order of evaluation do most programming languages use?
Fall 18 CSCI 4430, A Milanova/BG Ryder

27. Exercises Evaluate (x.y. x y) ((z. z) w)
Using applicative order reduction
Using normal order reduction
Fall 18 CSCI 4430, A Milanova

28. Exercise Evaluate (x.y. x y) ((z. z) w)
Using applicative order reduction
Using normal order reduction
Let S = xyz. x z (y z) and let I = x. x
Evaluate S I I I
Remember function application is left-associative, S I I I stands for ((S I) I) I

29. Today’s Lecture Outline
Lambda calculus, continued
Substitution, review
Rules of the lambda calculus
Normal forms
Reduction strategies
Review for test
Fall 18 CSCI 4430, A Milanova/BG Ryder

30. Exam 2 Practice (Quiz 4) Under static scoping, where does x in A bind?
Answer: global x
2. Under static scoping, what gets printed?
Answer: 101
3. Under dynamic scoping (with shallow binding), where does x in A bind?
Answer: global x and B’s x
Binds to global x.
101
Sometimes it binds to global x sometimes it binds to B’s x.
4. Under dynamic scoping (with shallow binding), what gets printed?
Answer: 0
Fall 18 CSCI 4430, A Milanova

31. Exam 2 Practice (Quiz 4)
Willy Wazoo wants to write a verifying compiler, which
(among other things) would guarantee that if a program compiled successfully, then when it executed all loops would terminate. Willy's compiler would be an example of
(a) static syntax analysis
(b) dynamic syntax analysis
(c) static semantic analysis
(d) dynamic semantic analysis
(c) Static semantic analysis
Spring 18 CSCI 4430, A Milanova

32. Exam 2 Practice (Quiz 6)
1. What gets printed under dynamic scoping with shallow binding?
Answer: 100
2. What gets printed under dynamic scoping with deep binding?
Answer: 101
100
101
Spring 18 CSCI 4430, A Milanova

33. Exam 2 Practice (Quiz 6) What does f compute? (define (f lis)
(foldl (lambda (x y) (if (> x y) x y)) lis (car lis)) )
max element in lis
Answer: maximal element in lis
Spring 18 CSCI 4430, A Milanova

34. Exam 2 Practice (Quiz 6)
Consider the problem of figuring whether two trees
(lists in Scheme) have the same fringe, that is, the
same leaves, in the same order, regardless of structure.
E.g., ((1 2) 3) and (1 (2 3)) have the same fringe.
What is the obvious way to solve this problem?
Flatten then compare.
Answer: flatten then compare:
(define (same-fringe? lis1 lis2)
(equal? (flatten lis1) (flatten lis2)))
Spring 18 CSCI 4430, A Milanova

35. Exam 2 Practice (Quiz 6) (____ ((is_even? (lambda (n) (or (zero? n)
Which let term produces the expected result, #t: let, let*, letrec
(____ ((is_even?
(lambda (n) (or (zero? n)
(is_odd? (- n 1)) ) )
(is_odd?
(lambda (n) (and (not (zero? n))
(is_even? (- n 1)) ) ))
(is_even? 10)
) Answer: letrec

36. Exam 2 Practice (Quiz 5) Scheme’s scoping discipline is?
Scheme’s typing discipline is?
What does f do? Is f tail-recursive?
(define (f a b)
(cond ((= a b) a)
((> a b) (f (- a b) b))
(else (f a (- b a))) )
Answer: GCD, it is tail-recursive
Fall 18 CSCI 4430, A Milanova

37. Exam 2 Practice (Quiz 5) (define (atom? obj) (not (pair? obj)) )
Answer: 6
Fall 18 CSCI 4430, A Milanova

38. Attribute Grammars Syntax of the Lambda calculus E  x E  ( x. E )
E  ( E E )
Give an attribute grammar that associates attribute free with each parse tree node M such that free contains the set of free variables in the expression represented by M.
Is your grammar S-attributed?
Fall 18 CSCI 4430, A Milanova

39. Attribute Grammars Syntax of the Lambda calculus E  x E  ( x. E )
E  ( E E )
Write an attribute grammar that translates lambda expressions into properly parenthesized expressions without redundant parentheses. (Use syntactic conventions we defined in class.)
Fall 18 CSCI 4430, A Milanova

40. Fall 18 CSCI 4430, A Milanova/BG Ryder