cse130

Your Favorite Language

Probably has lots of features:

Assignment (x = x + 1)
Booleans, integers, characters, strings, …
Conditionals
Loops
return, break, continue
Functions
Recursion
References / pointers
Objects and classes
Inheritance
…

Which ones can we do without?

What is the smallest universal language?

What is computable?

Before 1930s

Informal notion of an effectively calculable function:

can be computed by a human with pen and paper, following an algorithm

1936: Formalization

What is the smallest universal language?

The Next 700 Languages

Whatever the next 700 languages turn out to be, they will surely be variants of lambda calculus.

Peter Landin, 1966

The Lambda Calculus

Has one feature:

Functions

No, really

~~Assignment (x = x + 1)~~
~~Booleans, integers, characters, strings, …~~
~~Conditionals~~
~~Loops~~
~~return, break, continue~~
Functions
~~Recursion~~
~~References / pointers~~
~~Objects and classes~~
~~Inheritance~~
~~Reflection~~

More precisely, only thing you can do is:

Define a function
Call a function

Describing a Programming Language

Syntax: what do programs look like?
Semantics: what do programs mean?
- Operational semantics: how do programs execute step-by-step?

Syntax: What Programs Look Like

e ::= x
    | \x -> e
    | e1 e2

Programs are expressions e (also called λ-terms) of one of three kinds:

Variable
- x, y, z
Abstraction (aka nameless function definition)
- \x -> e
- x is the formal parameter, e is the body
- “for any x compute e”
Application (aka function call)
- e1 e2
- e1 is the function, e2 is the argument
- in your favorite language: e1(e2)

(Here each of e, e1, e2 can itself be a variable, abstraction, or application)

Examples

\x -> x             -- The identity function
                    -- ("for any x compute x")

\x -> (\y -> y)     -- A function that returns the identity function

\f -> f (\x -> x)   -- A function that applies its argument 
                    -- to the identity function

QUIZ

Which of the following terms are syntactically incorrect?

A. \(\x -> x) -> y

B. \x -> x x

C. \x -> x (y x)

D. A and C

E. all of the above

Examples

\x -> x             -- The identity function
                    -- ("for any x compute x")

\x -> (\y -> y)     -- A function that returns the identity function
 
\f -> f (\x -> x)   -- A function that applies its argument 
                    -- to the identity function

How do I define a function with two arguments?

e.g. a function that takes x and y and returns y?

\x -> (\y -> y)     -- A function that returns the identity function
                    -- OR: a function that takes two arguments
                    -- and returns the second one!

How do I apply a function to two arguments?

e.g. apply \x -> (\y -> y) to apple and banana?

(((\x -> (\y -> y)) apple) banana) -- first apply to apple,
                                   -- then apply the result to banana

Syntactic Sugar

instead of	we write
`\x -> (\y -> (\z -> e))`	`\x -> \y -> \z -> e`
`\x -> \y -> \z -> e`	`\x y z -> e`
`(((e1 e2) e3) e4)`	`e1 e2 e3 e4`

\x y -> y     -- A function that that takes two arguments
              -- and returns the second one...
              
(\x y -> y) apple banana -- ... applied to two arguments

Semantics : What Programs Mean

How do I “run” / “execute” a λ-term?

Think of middle-school algebra:

-- Simplify expression:

  (x + 2)*(3*x - 1)
 =
  ???

Execute = rewrite step-by-step following simple rules, until no more rules apply

Rewrite Rules of Lambda Calculus

α-step (aka renaming formals)
β-step (aka function call)

But first we have to talk about scope

Semantics: Scope of a Variable

The part of a program where a variable is visible

In the expression \x -> e

x is the newly introduced variable
e is the scope of x
any occurrence of x in \x -> e is bound (by the binder \x)

For example, x is bound in:

  \x -> x
  \x -> (\y -> x)

An occurrence of x in e is free if it’s not bound by an enclosing abstraction

For example, x is free in:

  x y                -- no binders at all!  
  \y -> x y          -- no \x binder
  (\x -> \y -> y) x  -- x is outside the scope of the \x binder;
                     -- intuition: it's not "the same" x

QUIZ

In the expression (\x -> x) x, is x bound or free?

A. bound

B. free

C. first occurrence is bound, second is free

D. first occurrence is bound, second and third are free

E. first two occurrences are bound, third is free

Free Variables

An variable x is free in e if there exists a free occurrence of x in e

We can formally define the set of all free variables in a term like so:

FV(x)       = ???
FV(\x -> e) = ???
FV(e1 e2)   = ???

Closed Expressions

If e has no free variables it is said to be closed

Closed expressions are also called combinators

What is the shortest closed expression?

Rewrite Rules of Lambda Calculus

α-step (aka renaming formals)
β-step (aka function call)

Semantics: β-Reduction

  (\x -> e1) e2   =b>   e1[x := e2]

where e1[x := e2] means “e1 with all free occurrences of x replaced with e2”

Computation by search-and-replace:

If you see an abstraction applied to an argument, take the body of the abstraction and replace all free occurrences of the formal by that argument
We say that (\x -> e1) e2 β-steps to e1[x := e2]

Examples

(\x -> x) apple     
=b> apple

Is this right? Ask Elsa!

(\f -> f (\x -> x)) (give apple)
=b> ???

QUIZ

(\x -> (\y -> y)) apple
=b> ???

A. apple

B. \y -> apple

C. \x -> apple

D. \y -> y

E. \x -> y

QUIZ

(\x -> x (\x -> x)) apple
=b> ???

A. apple (\x -> x)

B. apple (\apple -> apple)

C. apple (\x -> apple)

D. apple

E. \x -> x

A Tricky One

(\x -> (\y -> x)) y
=b> \y -> y

Is this right?

Something is Fishy

(\x -> (\y -> x)) y
=b> \y -> y

Is this right?

Problem: the free y in the argument has been captured by \y!

Solution: make sure that all free variables of the argument are different from the binders in the body.

Capture-Avoiding Substitution

We have to fix our definition of β-reduction:

  (\x -> e1) e2   =b>   e1[x := e2]

where e1[x := e2] means ~~“e1 with all free occurrences of x replaced with e2”~~

e1 with all free occurrences of x replaced with e2, as long as no free variables of e2 get captured
undefined otherwise

Formally:

x[x := e]            = e
y[x := e]            = y            -- assuming x /= y
(e1 e2)[x := e]      = (e1[x := e]) (e2[x := e])
(\x -> e1)[x := e]   = \x -> e1     -- why do we leave `e1` alone?
(\y -> e1)[x := e] 
  | not (y in FV(e)) = \y -> e1[x := e]
  | otherise         = undefined    -- wait, but what do we do then???

Rewrite Rules of Lambda Calculus

α-step (aka renaming formals)
β-step (aka function call)

Semantics: α-Renaming

  \x -> e   =a>   \y -> e[x := y]
    where not (y in FV(e))

We can rename a formal parameter and replace all its occurrences in the body
We say that \x -> e α-steps to \y -> e[x := y]

Example:

\x -> x   =a>   \y -> y   =a>    \z -> z

All these expressions are α-equivalent

What’s wrong with these?

-- (A)
\f -> f x    =a>   \x -> x x

-- (B)
(\x -> \y -> y) y   =a>   (\x -> \z -> z) z

-- (C)
\x -> \y -> x y   =a>    \apple -> \orange -> apple orange

The Tricky One

(\x -> (\y -> x)) y
=a> ???

To avoid getting confused, you can always rename formals, so that different variables have different names!

Normal Forms

A redex is a λ-term of the form

(\x -> e1) e2

A λ-term is in normal form if it contains no redexes.

QUIZ

Which of the following term are not in normal form ?

A. x

B. x y

C. (\x -> x) y

D. x (\y -> y)

E. C and D

Semantics: Evaluation

A λ-term e evaluates to e' if

There is a sequence of steps

e =?> e_1 =?> ... =?> e_N =?> e'

where each =?> is either =a> or =b> and N >= 0

e' is in normal form

Examples of Evaluation

(\x -> x) apple
  =b> apple

(\f -> f (\x -> x)) (\x -> x)
  =?> ???

(\x -> x x) (\x -> x)
  =?> ???

Elsa shortcuts

Named λ-terms:

let ID = \x -> x  -- abbreviation for \x -> x

To substitute name with its definition, use a =d> step:

ID apple
  =d> (\x -> x x) apple  -- expand definition
  =b> apple              -- beta-reduce

Evaluation:

e1 =*> e2: e1 reduces to e2 in 0 or more steps
- where each step is =a>, =b>, or =d>
e1 =~> e2: e1 evaluates to e2

What is the difference?

Non-Terminating Evaluation

(\x -> x x) (\x -> x x)
  =b> (\x -> x x) (\x -> x x)

Oops, we can write programs that loop back to themselves…

and never reduce to a normal form!

This combinator is called Ω

What if we pass Ω as an argument to another function?

let OMEGA = (\x -> x x) (\x -> x x)

(\x -> \y -> y) OMEGA

Does this reduce to a normal form? Try it at home!

Programming in λ-calculus

Real languages have lots of features

Booleans
Records (structs, tuples)
Numbers
Functions [we got those]
Recursion

Lets see how to encode all of these features with the λ-calculus.

λ-calculus: Booleans

How can we encode Boolean values (TRUE and FALSE) as functions?

Well, what do we do with a Boolean b?

Make a binary choice

if b then e1 else e2

Booleans: API

We need to define three functions

let TRUE  = ???
let FALSE = ???
let ITE   = \b x y -> ???  -- if b then x else y

such that

ITE TRUE apple banana =~> apple
ITE FALSE apple banana =~> banana

(Here, let NAME = e means NAME is an abbreviation for e)

Booleans: Implementation

let TRUE  = \x y -> x        -- Returns its first argument
let FALSE = \x y -> y        -- Returns its second argument
let ITE   = \b x y -> b x y  -- Applies condition to branches
                             -- (redundant, but improves readability)

Example: Branches step-by-step

eval ite_true:
  ITE TRUE e1 e2
  =d> (\b x y -> b    x  y) TRUE e1 e2    -- expand def ITE  
  =b>   (\x y -> TRUE x  y)      e1 e2    -- beta-step
  =b>     (\y -> TRUE e1 y)         e2    -- beta-step
  =b>            TRUE e1 e2               -- expand def TRUE
  =d>     (\x y -> x) e1 e2               -- beta-step
  =b>       (\y -> e1)   e2               -- beta-step
  =b> e1

Example: Branches step-by-step

Now you try it!

Can you fill in the blanks to make it happen?

eval ite_false:
  ITE FALSE e1 e2

  -- fill the steps in!

  =b> e2

Boolean Operators

Now that we have ITE it’s easy to define other Boolean operators:

let NOT = \b     -> ???

let AND = \b1 b2 -> ???

let OR  = \b1 b2 -> ???

let NOT = \b     -> ITE b FALSE TRUE 

let AND = \b1 b2 -> ITE b1 b2 FALSE

let OR  = \b1 b2 -> ITE b1 TRUE b2

Or, since ITE is redundant:

let NOT = \b     -> b FALSE TRUE 

let AND = \b1 b2 -> b1 b2 FALSE

let OR  = \b1 b2 -> b1 TRUE b2

Which definition to do you prefer and why?

Programming in λ-calculus

Booleans [done]
Records (structs, tuples)
Numbers
Functions [we got those]
Recursion

λ-calculus: Records

Let’s start with records with two fields (aka pairs)

What do we do with a pair?

Pack two items into a pair, then
Get first item, or
Get second item.

Pairs : API

We need to define three functions

let PAIR = \x y -> ???    -- Make a pair with elements x and y 
                          -- { fst : x, snd : y }
let FST  = \p -> ???      -- Return first element 
                          -- p.fst
let SND  = \p -> ???      -- Return second element
                          -- p.snd

such that

FST (PAIR apple banana) =~> apple
SND (PAIR apple banana) =~> banana

Pairs: Implementation

A pair of x and y is just something that lets you pick between x and y! (I.e. a function that takes a boolean and returns either x or y)

let PAIR = \x y -> (\b -> ITE b x y)
let FST  = \p -> p TRUE   -- call w/ TRUE, get first value
let SND  = \p -> p FALSE  -- call w/ FALSE, get second value

Exercise: Triples?

How can we implement a record that contains three values?

let TRIPLE = \x y z -> ???
let FST3  = \t -> ???
let SND3  = \t -> ???
let TRD3  = \t -> ???

Programming in λ-calculus

Booleans [done]
Records (structs, tuples) [done]
Numbers
Functions [we got those]
Recursion

λ-calculus: Numbers

Let’s start with natural numbers (0, 1, 2, …)

What do we do with natural numbers?

Count: 0, inc
Arithmetic: dec, +, -, *
Comparisons: ==, <=, etc

Natural Numbers: API

We need to define:

A family of numerals: ZERO, ONE, TWO, THREE, …
Arithmetic functions: INC, DEC, ADD, SUB, MULT
Comparisons: IS_ZERO, EQ

Such that they respect all regular laws of arithmetic, e.g.

IS_ZERO ZERO       =~> TRUE
IS_ZERO (INC ZERO) =~> FALSE
INC ONE            =~> TWO
...

Natural Numbers: Implementation

Church numerals: a number N is encoded as a combinator that calls a function on an argument N times

let ONE   = \f x -> f x
let TWO   = \f x -> f (f x)
let THREE = \f x -> f (f (f x))
let FOUR  = \f x -> f (f (f (f x)))
let FIVE  = \f x -> f (f (f (f (f x))))
let SIX   = \f x -> f (f (f (f (f (f x)))))
...

QUIZ: Church Numerals

Which of these is a valid encoding of ZERO ?

A: let ZERO = \f x -> x
B: let ZERO = \f x -> f
C: let ZERO = \f x -> f x
D: let ZERO = \x -> x
E: None of the above

Does this function look familiar?

λ-calculus: Increment

-- Call `f` on `x` one more time than `n` does
let INC   = \n -> (\f x -> ???)

Example:

eval inc_zero :
  INC ZERO
  =d> (\n f x -> f (n f x)) ZERO
  =b> \f x -> f (ZERO f x)
  =*> \f x -> f x
  =d> ONE

QUIZ

How shall we implement ADD?

A. let ADD = \n m -> n INC m

B. let ADD = \n m -> INC n m

C. let ADD = \n m -> n m INC

D. let ADD = \n m -> n (m INC)

E. let ADD = \n m -> n (INC m)

λ-calculus: Addition

--  Call `f` on `x` exactly `n + m` times
let ADD = \n m -> n INC m

Example:

eval add_one_zero :
  ADD ONE ZERO
  =~> ONE

QUIZ

How shall we implement MULT?

A. let MULT = \n m -> n ADD m

B. let MULT = \n m -> n (ADD m) ZERO

C. let MULT = \n m -> m (ADD n) ZERO

D. let MULT = \n m -> n (ADD m ZERO)

E. let MULT = \n m -> (n ADD m) ZERO

λ-calculus: Multiplication

--  Call `f` on `x` exactly `n * m` times
let MULT = \n m -> n (ADD m) ZERO

Example:

eval two_times_three :
  MULT TWO ONE
  =~> TWO

Programming in λ-calculus

Booleans [done]
Records (structs, tuples) [done]
Numbers [done]
Functions [we got those]
Recursion

λ-calculus: Recursion

I want to write a function that sums up natural numbers up to n:

\n -> ...          -- 1 + 2 + ... + n

QUIZ

Is this a correct implementation of SUM?

let SUM = \n -> ITE (ISZ n) 
            ZERO 
            (ADD n (SUM (DEC n)))

A. Yes

B. No

No!

Named terms in Elsa are just syntactic sugar
To translate an Elsa term to λ-calculus: replace each name with its definition

\n -> ITE (ISZ n) 
        ZERO 
        (ADD n (SUM (DEC n))) -- But SUM is not a thing!

Recursion:

Inside this function I want to call the same function on DEC n

Looks like we can’t do recursion, because it requires being able to refer to functions by name, but in λ-calculus functions are anonymous.

Right?

λ-calculus: Recursion

Think again!

Recursion:

~~Inside this function I want to call the same function on DEC n~~
Inside this function I want to call a function on DEC n
And BTW, I want it to be the same function

Step 1: Pass in the function to call “recursively”

let STEP = 
  \rec -> \n -> ITE (ISZ n) 
                  ZERO 
                  (ADD n (rec (DEC n))) -- Call some rec

Step 2: Do something clever to STEP, so that the function passed as rec itself becomes

\n -> ITE (ISZ n) ZERO (ADD n (rec (DEC n)))

λ-calculus: Fixpoint Combinator

Wanted: a combinator FIX such that FIX STEP calls STEP with itself as the first argument:

FIX STEP
=*> STEP (FIX STEP)

(In math: a fixpoint of a function f(x) is a point x, such that f(x) = x)

Once we have it, we can define:

let SUM = FIX STEP

Then by property of FIX we have:

SUM =*> STEP SUM -- (1)

eval sum_one:
  SUM ONE
  =*> STEP SUM ONE                 -- (1)
  =d> (\rec n -> ITE (ISZ n) ZERO (ADD n (rec (DEC n)))) SUM ONE
  =b> (\n -> ITE (ISZ n) ZERO (ADD n (SUM (DEC n)))) ONE 
                                   -- ^^^ the magic happened!
  =b> ITE (ISZ ONE) ZERO (ADD ONE (SUM (DEC ONE)))
  =*> ADD ONE (SUM ZERO)           -- def of ISZ, ITE, DEC, ...
  =*> ADD ONE (STEP SUM ZERO)      -- (1)
  =d> ADD ONE 
        ((\rec n -> ITE (ISZ n) ZERO (ADD n (rec (DEC n)))) SUM ZERO)
  =b> ADD ONE ((\n -> ITE (ISZ n) ZERO (ADD n (SUM (DEC n)))) ZERO)
  =b> ADD ONE (ITE (ISZ ZERO) ZERO (ADD ZERO (SUM (DEC ZERO))))
  =b> ADD ONE ZERO
  =~> ONE

How should we define FIX???

The Y combinator

Remember Ω?

(\x -> x x) (\x -> x x)
=b> (\x -> x x) (\x -> x x)

This is self-replcating code! We need something like this but a bit more involved…

The Y combinator discovered by Haskell Curry:

let FIX   = \stp -> (\x -> stp (x x)) (\x -> stp (x x))

How does it work?

eval fix_step:
  FIX STEP
  =d> (\stp -> (\x -> stp (x x)) (\x -> stp (x x))) STEP
  =b> (\x -> STEP (x x)) (\x -> STEP (x x))
  =b> STEP ((\x -> STEP (x x)) (\x -> STEP (x x)))
  --       ^^^^^^^^^^ this is FIX STEP ^^^^^^^^^^^

That’s all folks!