CS221/321 Lecture 2, 9/30/2010
-------------------------------

[Note on terminology: we use the words "expression" and "term"
interchangeably.]

How do we precisely define how arithmetic expressions are evaluated?

1. By writing a program implementing evaluation (say in ML).

------------------------------------------------------------ 
Program 1.1: SAE eval  (SAE-interp.sml)
------------------------------------------------------------ 
datatype expr
  = Num of int
  | Plus of expr * expr
  | Times of expr * expr

(* eval : expr -> int *)
fun eval (Num n) = n
  | eval (Plus(e1,e2)) = eval e1 + eval e2
  | eval (Times(e1,e2)) = eval e1 * eval e2

val exp1 = Plus(Num 2, Times(Num 13, Num 4))

eval exp1  ==>  54
------------------------------------------------------------

2. But what if we want to define evaluation "mathematically"?

Define an abstract grammar of simple arithmetic expressions (SAEs)

   n    :: = <natural number>    (i.e. n ∈ Nat)
   expr :: = Num n | Plus(expr,expr) | Times(expr,expr)


We define a binary evaluation relation:


---------------------------------------------------------------------- 
Figure 1.1: SAE[BS] - Eval relation
----------------------------------------------------------------------

   Eval  ⊆ expr × Nat

   Eval ((Num n),n)
   Eval ((Plus(e1,e2)), p)  <=  Eval (e1, n1) & Eval (e2, n2) & p = n1 + n2
   Eval ((Plus(e1,e2)), p)  <=  Eval (e1, n1) & Eval (e2, n2) & p = n1 + n2
----------------------------------------------------------------------


Alternatively, we can use an infix down arrow symbol (⇓) to represent the
Eval relation:

---------------------------------------------------------------------- 
Figure 1.1a: SAE[BS] - ⇓ evaluation relation
----------------------------------------------------------------------
⇓  ⊆ expr × nat

Rules: 

(1)  (Num n) ⇓ n

(2)  Plus(e1,e2) ⇓ p  <=  e1 ⇓ n1  &  e2 ⇓ n2  &  p = n1 + n2 

(3)  Times(e1,e2) ⇓ p  <=  e1 ⇓ n1  &  e2 ⇓ n2  &  p = n1 * n2 
----------------------------------------------------------------------


Is evaluation deterministic?  I.e. For any e, is the n such that
e ⇓ n unique?  Intuitively, we certainly expect so.

Theorem 1.1:  ∀e ∀m ∀n. e ⇓ m & e ⇓ n => m = n

Exercise: How do we prove it?

-----------------------
This Eval (⇓) relation defines what is known as a "big-step" semantics,
because it relates an expression to its final value, ignoring how many
intermediate steps of computation are involved in the evaluation.


======================================================================
Note: On the inductive definition of expr

The BNF definition

  expr :: = Num n | Plus(expr,expr) | Times(expr,expr)

defines the set of SAE expressions recursively, or "inductively".
What is the meaning of such a definition?

We are working in basic set theory, augmented with a notion of
general expressions consisting of syntax constructors like Plus
and Times applied to tuples of expressions.  These expressions
could be themselves encoded by basic set constructions, but we
will just assume them as primitives.

It is assumed that syntax constructors are 1-1 functions, so, for
instance, Plus(e1,e2) = Plus(e1',e2') implies e1 = e1' and e2 = e2'.
Num is assumed to be a 1-1 function mapping Nat to expressions.
Furthermore the ranges of Nat, Plus, and Times are disjoint; 
expressions formed by Nat, Plus, and Times are always distinct.

We define a "set closure" function C as follows:

    C(S) = {Num(n) | n ∈ Nat}⋃ {Plus(s1,s2) | s1, s2 ∈ S}
           ⋃ {Times(s1,s2) | s1, s2 ∈ S}

Such a closure function maps sets to sets and is monotonic:
A ⊆ B  =>  C(A) ⊆ C(B).

Define a sequence of sets E(i) as follows:

   E(0) = ∅ 
   E(n+1) = C(E(n))

For instance, 

   E(1) = {Num(n) | n ∈ Nat}
   E(2) = {Num(n) | n ∈ Nat} ⋃ {Plus(Num(n1),Num(n2)) | n1, n2 ∈ Nat}
          ⋃ {Times(Num(n1),Num(n2)) | n1, n2 ∈ Nat}

and in general E(n) is the set of terms built from Num, Plus and Times
of depth less than or equal to n.

Note that E(n) ⊆ E(n+1), and each finite expression e in SAE will 
appear in E(n), where n = depth(e), defined by:

    depth(Num(n)) = 1
    depth(Plus(e1,e2)) = 1 + max(depth(e1), depth(e2))
    depth(Times(e1,e2)) = 1 + max(depth(e1), depth(e2))

Now define the limit set

    E = ⋃{E(n) | n ∈ Nat}

Prop: E is the least fixed point of C, i.e.
      (i) E = C(E).
      (ii) If S = C(S), then E ⊆ S.

Proof: informal homework.

We take E, the least fixed point of C to be the meaning of the
inductive definition of expr, i.e. expr = E.

Note that all the expressions in E are finite, because all the
expressions in each E(n) are finite.  This definition excludes
"infinite expressions" such as the expression defined by the
recursive equation

    e = Plus(e,e)

which looks like an infinite binary tree with each node labeled
by the Plus constructor.

    e  =   Plus
          /     \
        Plus    Plus
       /    \  /    \
     ...   ... ...  ...

As a consequence, the relation of subexpression is a well-founded
partial order on expr.  This is the basis for structural induction
on expressions.

========================================================================

Small-step semantics
--------------------

A more fine-grained definition of evaluation takes into account all
the intermediate steps involved.

We define a "transition relation" on expressions (which is a binary
relation on expressions):

   e ↦ e'    ( ↦ ⊆ expr × expr )

where expression e' is obtained from expression e by one basic
step of evaluation (i.e. by a single "reduction" of a redex
subexpression).

[For definitions and notation for transition systems, see Chapter 2,
"Transistion Systems" in the Harper book. The expressions here are also
known as "states" in a transition system. In this case, the initial
states I = expr (i.e. all expressions are initial states), and the
final states F = {Num(n) | n ∈ Nat}, i.e. the atomic expressions
representing numbers.]

We have to account for the fact that the place that the reduction
takes place will generally be a subterm (possibly deeply nested)
of the main expression we are evaluating.

Here are some rules defining the small-step transition relation
for evaluating SAEs.


----------------------------------------------------------------------
Figure 1.2: SAE[SS]
----------------------------------------------------------------------

(1)  Plus(Num n1, Num n2)  ↦  Num p   where p = n1 + n2

(2)  Plus(e1, e2)  ↦  Plus(e1', e2)
      <=  e1 ↦ e1'

(3)  Plus(Num n1, e2)  ↦  Plus(Num n1, e2')
      <=  e2 ↦ e2'

(4)  Times(Num n1, Num n2)  ↦  Num p   where p = n1 * n2
 
(5)  Times(e1, e2)  ↦  Times(e1', e2)
      <=  e1 ↦ e1'

(6)  Times(Num n1, e2)  ↦  Times(Num n1, e2')
      <=  e2 ↦ e2'
----------------------------------------------------------------------


Rules (1) and (4) are called "instructions" or axioms. They specify
the basic reductions. A term matching one of the patterns on the left
is called a redex, and the reduced term on the right is called the
"contractum".

Rules (2), (3), (5), and (6) are called "search rules".  They are used
to isolate the site of a redex and propagate the reduction of the
redex to the larger enclosing expression.

See lecture2.pdf for the conventional presentation this set of
"inference" rules, with a line separating premisses above the line
and a conclusion below the line.


How do we use these rules to justify a reduction like:

    2 + (13 * 4)  ↦  2 + 52   ?

We have to "prove" this transition using a derivation from the
rules. (To make terms and derivations more concise, we'll leave
out the Num constructors and just treat the numbers themselves
as basic expressions.)

----------------------------------------------------------------------
Figure 1.3: Example: 2 step derivation
----------------------------------------------------------------------

         ------------------ (2)
          Times(13,4) ↦ 52
   ------------------------------------ (5)
    Plus(2, Times(13,4)) ↦ Plus(2, 52)

----------------------------------------------------------------------

This is called a proof tree or a derivation tree.  Each node is
labeled by the rule that justifies the inference. A search rule
like (5) promotes a transition on a subexpression to a transition on
the (immediate) containing expression.

If the redex is nested more deeply, the transition derivation will
also be deeper.  In fact, its shape exactly corresponds to the
form of the expression in the conclusion.

The transition

    2 + ((5 + 8) * 4)  ↦  2 + (13 * 4)
          -----

has the derivation

----------------------------------------------------------------------
Figure 1.4: Example: 3 step derivation
----------------------------------------------------------------------

              ---------------- (1)
               Plus(5,8) ↦ 13
        ----------------------------------- (4)
         Times(Plus(5,8),4)) ↦ Times(13,4)
   --------------------------------------------------- (5)
    Plus(2,Times(Plus(5,8),4)) ↦ Plus(2,Times(13,4))

----------------------------------------------------------------------

Observe that the left-hand-sides of the transitions in this 
derivation are just the nested sequence of subexpressions along
the path from the root of the full term to the redex subexpression.
This will be true in general, so the structure of any reduction
derivation reflects the path from the root to the redex subexpression
of the expression being reduced.

How do we get full evaluation from the transition relation ↦ ?

We construct sequences of chained transitions:

    e1 ↦ e2 ↦ e3 ↦ ... ↦ e

How do we know we are done?  When there are no further transitions
possible.  I.e when there does not exist an e' such that e ↦ e'. Such
an expression/state is known as a final state.  In this case, the
final states are exactly the simple Num terms -- they are already
fully evaluated.  The set of final states is F = {Num(n) | n ∈ Nat}

There are a number of natural questions to ask about this transition
system.

Question 1. Does evaluation always terminate?  That is, for any 
given initial expression e1, is there always a complete transition
sequence ending in a final state, that is, in a number expression?

Question 2. Is evaluation deterministic? That is, does the evaluation
of an expression, assuming it terminates, always yield the same,
uniquely determined number?

A question related to Question 2 is:

Question 3. Is the transition relation deterministic, in the sense
that for a given nonfinal expression (i.e. a compound expression)
e1, there is a unique expression e2 such that e1 ↦ e2.

An affirmative answer to Question 3 obviously implies an affirmative
answer to Question 3.

In fact, the answer for all three questions is yes.


======================================================================
Homework 1.2
------------
Prove that small-step (transition) evaluation for SAE always terminates.

Homework 1.3
------------
Prove that the transistion relation is deterministic: for any 
e ∈ expr, there exists at most one e' such that e ↦ e'.

======================================================================


Comparing Big-Step and Small-Step evaluation

Now we have two alternate explanations of evaluation of SAE, using
either the big-step semantics ⇓	or the small-step semantics ↦.

We would hope that they are consistent with one another, and this
is indeed the case.

Theorem 1.2. For all e ∈ expr and n ∈ Nat,
    e ⇓ n  <=>  e ↦! Num(n).


======================================================================
Homework 1.4
------------
Prove Theorem 1.2.

======================================================================