2008-01-25 Bindings & Substitution

========================================================================
>>> Bindings & Substitution

We now get to an important concept: substitution.

Even in our simple language, we encounter repeated expressions.  For
example, if we want to compute the square of some expression:

  {* {+ 4 2} {+ 4 2}}

Why would we want to get rid of the repeated sub-expression?

* It introduces a redundant computation.  In this example, we want to
  avoid computing the same sub-expression a second time.

* It makes the computation more complicated than it could be without the
  repetition.  Compare the above with:

    with x = {+ 4 2},
      {* x x}

* This is related to a basic fact in programming that we have already
  discussed: duplicating information is always a bad thing.  Among other
  bad consequences, it can even lead to bugs that could not happen if we
  wouldn't duplicate code.  A toy example is "fixing" one of the numbers
  in one expression and forgetting to fix the corresponding one:

    {* {+ 4 2} {+ 4 1}}

  Real world examples involve much more code, which make such bugs very
  difficult to find, but they still follow the same principle.

* This gives us more expressive power -- we don't just say that we want
  to multiply two expressions that both happen to be {+ 4 2}, we say
  that we multiply the {+ 4 2} expression by *itself*.  It allows us to
  express identity of two values as well as using two values that happen
  to be the same.

So, the normal way to avoid redundancy is to introduce an identifier.
Even when we speak, we might say: "let x be 4 plus 2, multiply x by x".

(These are often called "variables", but we will try to avoid this name:
what if the identifier does not change (vary)?)

To get this, we introduce a new form into our language:

  {with {x {+ 4 2}}
    {* x x}}

We expect to be able to reduce this to:

  {* 6 6}

by substituting 6 for `x' in the body sub-expression of `with'.

A little more complicated example:

  {with {x {+ 4 2}}
    {with {y {* x x}}
      {+ y y}}}
  [add]  = {with {x 6} {with y {* x x} {+ y y}}}
  [subst]= {with {y {* 6 6}} {+ y y}}
  [mul]  = {with {y 36} {+ y y}}
  [subst]= {+ 36 36}
  [add]  = 72

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

To add this to our language, we start with the BNF.  We now call our
language `WAE' (With+AE):

  <WAE> ::= <num>
          | { + <WAE> <WAE> }
          | { - <WAE> <WAE> }
          | { * <WAE> <WAE> }
          | { with { <id> <WAE> } <WAE> }
          | <id>

Note that we had to introduce two new rules: one for introducing an
identifier, and one for using it.  This is common in many language
specifications, for example `define-type' introduced a new type, and it
comes with `cases' that allows us to destruct its instances.

For <id> we need to use some form of identifiers, the natural choice in
Scheme is to use symbols.  We can therefore write the corresponding type
definition:

  (define-type WAE
    [Num  (n Number)]
    [Add  (lhs WAE) (rhs WAE)]
    [Sub  (lhs WAE) (rhs WAE)]
    [Mul  (lhs WAE) (rhs WAE)]
    [Id   (name Symbol)]
    [With (name Symbol) (named WAE) (body WAE)])

The parser is easily extended to produce these syntax objects:

  (: parse-sexpr : (Sexpr -> WAE))
  ;; to convert s-expressions into WAEs
  (define (parse-sexpr sexpr)
    (match sexpr
      [(number: n)    (Num n)]
      [(symbol: name) (Id name)]
      [(cons 'with more)
       ;; go in here for all sexpr that begin with a 'with
       (match sexpr
         [(list 'with (list (symbol: name) named) body)
          (With name
                (parse-sexpr named)
                (parse-sexpr body))]
         [else (error 'parse-sexpr "bad `with' syntax in ~s" sexpr)])]
      [(list op left right)
       (let ([make-node
              (match op
                ['+ Add]
                ['- Sub]
                ['* Mul]
                [else (error 'parse-sexpr "don't know about ~s" op)])])
         (make-node (parse-sexpr left)
                    (parse-sexpr right)))]
      [else (error 'parse-sexpr "bad syntax in ~s" sexpr)]))

And we're done with the syntactic part of the `with' extension.

(Quick question -- why is `With' indented like a normal function in this
code?)

========================================================================
Now, to make this work, we will need to do some substitutions.

We basically want to say that to evaluate:

  {with {id WAE1} WAE2}

we need to evaluate WAE2 with id substituted by WAE1.  Formally:

  eval( {with {id WAE1} WAE2} ) = eval( subst(WAE2,id,WAE1) )

There is a more common syntax for substitution (quick: what do I mean by
this "syntax"?):

  eval( {with {id WAE1} WAE2} ) = eval( WAE2[WAE1/id] )

Now all we need is an exact definition of substitution.  (Note that
substitution is not the same as evaluation, only part of the evaluation
process.  In the previous examples, when we evaluated the expression we
did substitutions as well as the usual arithmetic operations that were
already part of the AE evaluator.  In this last definition there is
still a missing evaluation step, see if you can find it.)

So let us try to define substitution now:

  [substitution, take 1]  e[v/i]
  To substitute an identifier `i' in an expression `e' with an
  expression `v', replace all identifiers in `e' that have the same name
  `i' by the expression `v'.

This seems to work with simple expressions, for example:

  {with {x 5} {+ x x}}   -->  {+ 5 5}
  {with {x 5} {+ 10 4}}  -->  {+ 10 4}

however, we crash with an invalid syntax if we try:

  {with {x 5} {+ x {with {x 3} 10}}}  -->  {+ 5 {with {5 3} 10}} ???

-- we got to an invalid expression.

To fix this, we need to distinguish "normal" occurrences of identifiers,
and ones that are used as new bindings.  We need a few new terms for
this:

1. Binding Instance: a binding instance of an identifier is one that is
   used to name it in a new binding.  In our <WAE> syntax, binding
   instances are only the <id> position of the `with' form.

2. Scope: the scope of a binding instance is the region of program text
   in which instances of the identifier refer to the value bound in the
   binding instance.  (Note that this definition actually relies on a
   definition of substitution, because that is what is used to specify
   how identifiers refer to values.)

3. Bound Instance: an instance of an identifier is bound if it is
   contained within the scope of a binding instance of its name.

4. Free Instance: An identifier that is not contained in any binding
   instance of its name is said to be free.

Using this we can say that the problem with the previous definition of
substitution is that it failed to distinguish between bound instances
(which should be substituted) and binding instances (which should not).
So we try to fix this:

  [substitution, take 2]  e[v/i]
  To substitute an identifier `i' in an expression `e' with an
  expression `v', replace all instances of `i' that are not themselves
  binding instances with the expression `v'.

First of all, check the previous examples:

  {with {x 5} {+ x x}}   -->  {+ 5 5}
  {with {x 5} {+ 10 4}}  -->  {+ 10 4}

still work, and

  {with {x 5} {+ x {with {x 3} 10}}}  -->  {+ 5 {with {x 3} 10}}
                                      -->  {+ 5 10}

also works.  However, if we try this:

  {with {x 5} {+ x {with {x 3} x}}}

we get:

  -->  {+ 5 {with {x 3} 5}}
  -->  {+ 5 5}
  -->  10

but we want that to be 8: the inner `x' should be bound by the closest
`with' that binds it.

The problem is that the new definition of substitution that we have
respects binding instances, but it fails to deal with their scope.  In
the above example, we want the inner `with' to *shadow* the outer
`with's binding for `x'.

  [substitution, take 3]  e[v/i]
  To substitute an identifier `i' in an expression `e' with an
  expression `v', replace all instances of `i' that are not themselves
  binding instances, and that are not in any nested scope, with the
  expression `v'.

This avoids bad substitution above, but it is now doing things too
carefully:

  {with {x 5} {+ x {with {y 3} x}}}

becomes

  -->  {+ 5 {with {y 3} x}}
  -->  {+ 5 x}

which is an error because `x' is unbound (and there is reasonable no
rule that we can specify to evaluate it).

The problem is that our substitution halts at every new scope, in this
case, it stopped at the new `y' scope, but it shouldn't have because it
uses a different name.  In fact, that last definition of substitution
cannot handle any nested scope.

Revise again:

  [substitution, take 4]  e[v/i]
  To substitute an identifier `i' in an expression `e' with an
  expression `v', replace all instances of `i' that are not themselves
  binding instances, and that are not in any nested scope of `i', with
  the expression `v'.

which, finally, is a good definition.  This is just a little too
mechanical.  Notice that we actually refer to all instances of `i' that
are not in a scope of a binding instance of `i', which simply means all
*free occurrences* of `i' -- free in `e' (why? -- remember the
definition of "free"?):

  [substitution, take 4b]  e[v/i]
  To substitute an identifier `i' in an expression `e' with an
  expression `v', replace all instances of `i' that are free in `e' with
  the expression `v'.

Based on this we can finally write the code for it:

  (: subst : (WAE Symbol WAE -> WAE))
  ;; substitutes the second argument with the third argument in the
  ;; first argument, as per the rules of substitution; the resulting
  ;; expression contains no free instances of the second argument
  (define (subst expr from to)
    (cases expr
      [(Num n) expr]
      [(Add l r) (Add (subst l from to) (subst r from to))]
      [(Sub l r) (Sub (subst l from to) (subst r from to))]
      [(Mul l r) (Mul (subst l from to) (subst r from to))]
      [(Id name) (if (eq? name from) to expr)]
      [(With bound-id named-expr bound-body)
       (if (eq? bound-id from)
         expr           ; <-- don't go in!
         (With bound-id
               named-expr
               (subst bound-body from to)))]))

... and this is just the same as writing a formal "paper version" of the
substitution rule.

... but we still have bugs!

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

Before we find the bugs, we need to see when and how substitution is
used in the evaluation process.

To modify our evaluator, we will need rules to deal with the new syntax
pieces -- `with' expressions and identifiers.

When we see an expression that looks like:

  {with {x E1} E2}

we continue by *evaluating* `E1' to get a value `V1', we then substitute
the identifier `x' with the expression `V1' in `E2', and continue by
evaluating this new expression.  In other words, we have the following
evaluation rule:

  eval( {with {x E1} E2} ) = eval( E2[eval(E1)/x] )

So we know what to do with `with' expressions.  How about identifiers?
The main feature of `subst', as said in the purpose statement, is that
it leaves no free instances of the substituted variable around.  This
means that if the initial expression is valid (did not contain any free
variables), then when we go from

  {with {x E1} E2}

to

  E2[E1/x]

the result is an expression that has *no* free instances of `x'.  So we
don't need to handle identifiers in the evaluator -- substitutions make
them all go away.

We can now extend the formal definition of AE to that of WAE:

    eval(...) = ... same as the AE rules ...
    eval({with {x E1} E2}) = eval(E2[eval(E1)/x])
    eval(id) = error!

If you're paying close attention, you might catch a potential problem in
this definition: we're substituting `eval(E1)' for `x' in `E2' -- an
operation that requires a WAE expression, but `eval(E1)' is a number.
(Look at the type of the `eval' definition we had for AE, then look at
the above definition of `subst'.)  This seems like being overly
pedantic, but we it will require some resolution when we get to the
code.  The above rules are easily coded as follows:

  (: eval : (WAE -> Number))
  ;; evaluates WAE expressions by reducing them to numbers
  (define (eval expr)
    (cases expr
      [(Num n) n]
      [(Add l r) (+ (eval l) (eval r))]
      [(Sub l r) (- (eval l) (eval r))]
      [(Mul l r) (* (eval l) (eval r))]
      [(With bound-id named-expr bound-body)
       (eval (subst bound-body
                    bound-id
                    (Num (eval named-expr))))] ; <-***
      [(Id name) (error 'eval "free identifier: ~s" name)]))

Note the `Num' expression in the marked line: evaluating the named
expression gives us back a number -- we need to convert this number into
a syntax to be able to use it with `subst'.  The solution is to use
`Num' to convert the resulting number into a numeral (the syntax of a
number).  It's not an elegant solution, but it will do for now.

Finally, here are a few test cases.  We use a new `test' special form
which is part of the course plugin.  The way to use `test' is with two
expressions and an `=>' arrow -- DrScheme evaluates both, and nothing
will happen if the results are equal.  If the results are different, you
will get a warning line, but evaluation will continue so you can try
additional tests.  You can also use an `=error>' arrow to test an error
message -- use it with some text from the expected error, `?' stands for
any single character, and `*' is a sequence of zero or more characters.
(When you use `test' in your homework, the handin server will abort when
tests fail.)  We expect these tests to succeed (make sure that you
understand *why* they should succeed).

  ;; tests
  (test (run "5") => 5)
  (test (run "{+ 5 5}") => 10)
  (test (run "{with {x {+ 5 5}} {+ x x}}") => 20)
  (test (run "{with {x 5} {+ x x}}") => 10)
  (test (run "{with {x {+ 5 5}} {with {y {- x 3}} {+ y y}}}") => 14)
  (test (run "{with {x 5} {with {y {- x 3}} {+ y y}}}") => 4)
  (test (run "{with {x 5} {+ x {with {x 3} 10}}}") => 15)
  (test (run "{with {x 5} {+ x {with {x 3} x}}}") => 8)
  (test (run "{with {x 5} {+ x {with {y 3} x}}}") => 10)
  (test (run "{with {x 5} {with {y x} y}}") => 5)
  (test (run "{with {x 5} {with {x x} x}}") => 5)
  (test (run "{with {x 1} y}") =error> "free identifier")

Putting this all together, we get the following code; trying to run this
code will raise an unexpected error...

  --------------------------------------------------------------------
  #| BNF for the WAE language:
       <WAE> ::= <num>
               | { + <WAE> <WAE> }
               | { - <WAE> <WAE> }
               | { * <WAE> <WAE> }
               | { with { <id> <WAE> } <WAE> }
               | <id>
  |#

  ;; WAE abstract syntax trees
  (define-type WAE
    [Num  (n Number)]
    [Add  (lhs WAE) (rhs WAE)]
    [Sub  (lhs WAE) (rhs WAE)]
    [Mul  (lhs WAE) (rhs WAE)]
    [Id   (name Symbol)]
    [With (name Symbol) (named WAE) (body WAE)])

  (: parse-sexpr : (Sexpr -> WAE))
  ;; to convert s-expressions into WAEs
  (define (parse-sexpr sexpr)
    (match sexpr
      [(number: n)    (Num n)]
      [(symbol: name) (Id name)]
      [(cons 'with more)
       (match sexpr
         [(list 'with (list (symbol: name) named) body)
          (With name (parse-sexpr named) (parse-sexpr body))]
         [else (error 'parse-sexpr "bad `with' syntax in ~s" sexpr)])]
      [(list op left right)
       (let ([make-node
              (match op
                ['+ Add]
                ['- Sub]
                ['* Mul]
                [else (error 'parse-sexpr "don't know about ~s" op)])])
         (make-node (parse-sexpr left)
                    (parse-sexpr right)))]
      [else (error 'parse-sexpr "bad syntax in ~s" sexpr)]))

  (: parse : (String -> WAE))
  ;; parses a string containing a WAE expression to a WAE AST
  (define (parse str)
    (parse-sexpr (string->sexpr str)))

  (: subst : (WAE Symbol WAE -> WAE))
  ;; substitutes the second argument with the third argument in the
  ;; first argument, as per the rules of substitution; the resulting
  ;; expression contains no free instances of the second argument
  (define (subst expr from to)
    (cases expr
      [(Num n) expr]
      [(Add l r) (Add (subst l from to) (subst r from to))]
      [(Sub l r) (Sub (subst l from to) (subst r from to))]
      [(Mul l r) (Mul (subst l from to) (subst r from to))]
      [(Id name) (if (eq? name from) to expr)]
      [(With bound-id named-expr bound-body)
       (if (eq? bound-id from)
         expr
         (With bound-id
               named-expr
               (subst bound-body from to)))]))

  (: eval : (WAE -> Number))
  ;; evaluates WAE expressions by reducing them to numbers
  (define (eval expr)
    (cases expr
      [(Num n) n]
      [(Add l r) (+ (eval l) (eval r))]
      [(Sub l r) (- (eval l) (eval r))]
      [(Mul l r) (* (eval l) (eval r))]
      [(With bound-id named-expr bound-body)
       (eval (subst bound-body
                    bound-id
                    (Num (eval named-expr))))]
      [(Id name) (error 'eval "free identifier: ~s" name)]))

  (: run : (String -> Number))
  ;; evaluate a WAE program contained in a string
  (define (run str)
    (eval (parse str)))

  ;; tests
  (test (run "5") => 5)
  (test (run "{+ 5 5}") => 10)
  (test (run "{with {x {+ 5 5}} {+ x x}}") => 20)
  (test (run "{with {x 5} {+ x x}}") => 10)
  (test (run "{with {x {+ 5 5}} {with {y {- x 3}} {+ y y}}}") => 14)
  (test (run "{with {x 5} {with {y {- x 3}} {+ y y}}}") => 4)
  (test (run "{with {x 5} {+ x {with {x 3} 10}}}") => 15)
  (test (run "{with {x 5} {+ x {with {x 3} x}}}") => 8)
  (test (run "{with {x 5} {+ x {with {y 3} x}}}") => 10)
  (test (run "{with {x 5} {with {y x} y}}") => 5)
  (test (run "{with {x 5} {with {x x} x}}") => 5)
  (test (run "{with {x 1} y}") =error> "free identifier")
  --------------------------------------------------------------------

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

Oops, this program still has problems that were caught by the tests --
we encounter unexpected free identifier errors.  What's the problem now?
In expressions like:

  {with {x 5}
    {with {y x}
      y}}

we forgot to substitute `x' in the expression that `y' is bound to.  We
need to the recursive substitute in both the with's body expression as
well as its named expression:

  (: subst : (WAE Symbol WAE -> WAE))
  ;; substitutes the second argument with the third argument in the
  ;; first argument, as per the rules of substitution; the resulting
  ;; expression contains no free instances of the second argument
  (define (subst expr from to)
    (cases expr
      [(Num n) expr]
      [(Add l r) (Add (subst l from to) (subst r from to))]
      [(Sub l r) (Sub (subst l from to) (subst r from to))]
      [(Mul l r) (Mul (subst l from to) (subst r from to))]
      [(Id name) (if (eq? name from) to expr)]
      [(With bound-id named-expr bound-body)
       (if (eq? bound-id from)
         expr
         (With bound-id
               (subst named-expr from to)       ; <-- new
               (subst bound-body from to)))]))

And *still* we have a problem...  Now it's

  {with {x 5}
    {with {x x}
      x}}

that halts with an error, but we want it to evaluate to 5!  Carefully
trying out our substitution code reveals the problem: when we substitute
`5' for the outer `x', we don't go inside the inner `with' because it
has the same name -- but we *do* need to go into its named expression.
We need to substitute in the named expression even if the identifier is
the *same* one we substituting:

  (: subst : (WAE Symbol WAE -> WAE))
  ;; substitutes the second argument with the third argument in the
  ;; first argument, as per the rules of substitution; the resulting
  ;; expression contains no free instances of the second argument
  (define (subst expr from to)
    (cases expr
      [(Num n) expr]
      [(Add l r) (Add (subst l from to) (subst r from to))]
      [(Sub l r) (Sub (subst l from to) (subst r from to))]
      [(Mul l r) (Mul (subst l from to) (subst r from to))]
      [(Id name) (if (eq? name from) to expr)]
      [(With bound-id named-expr bound-body)
       (With bound-id
             (subst named-expr from to)
             (if (eq? bound-id from)
               bound-body
               (subst bound-body from to)))]))

The complete (and, finally, correct) version of the code is now:

  ----------------------------------------------------------------------
  #| BNF for the WAE language:
       <WAE> ::= <num>
               | { + <WAE> <WAE> }
               | { - <WAE> <WAE> }
               | { * <WAE> <WAE> }
               | { with { <id> <WAE> } <WAE> }
               | <id>
  |#

  ;; WAE abstract syntax trees
  (define-type WAE
    [Num  (n Number)]
    [Add  (lhs WAE) (rhs WAE)]
    [Sub  (lhs WAE) (rhs WAE)]
    [Mul  (lhs WAE) (rhs WAE)]
    [Id   (name Symbol)]
    [With (name Symbol) (named WAE) (body WAE)])

  (: parse-sexpr : (Sexpr -> WAE))
  ;; to convert s-expressions into WAEs
  (define (parse-sexpr sexpr)
    (match sexpr
      [(number: n)    (Num n)]
      [(symbol: name) (Id name)]
      [(cons 'with more)
       (match sexpr
         [(list 'with (list (symbol: name) named) body)
          (With name (parse-sexpr named) (parse-sexpr body))]
         [else (error 'parse-sexpr "bad `with' syntax in ~s" sexpr)])]
      [(list op left right)
       (let ([make-node
              (match op
                ['+ Add]
                ['- Sub]
                ['* Mul]
                [else (error 'parse-sexpr "don't know about ~s" op)])])
         (make-node (parse-sexpr left)
                    (parse-sexpr right)))]
      [else (error 'parse-sexpr "bad syntax in ~s" sexpr)]))

  (: parse : (String -> WAE))
  ;; parses a string containing a WAE expression to a WAE AST
  (define (parse str)
    (parse-sexpr (string->sexpr str)))

  (: subst : (WAE Symbol WAE -> WAE))
  ;; substitutes the second argument with the third argument in the
  ;; first argument, as per the rules of substitution; the resulting
  ;; expression contains no free instances of the second argument
  (define (subst expr from to)
    (cases expr
      [(Num n) expr]
      [(Add l r) (Add (subst l from to) (subst r from to))]
      [(Sub l r) (Sub (subst l from to) (subst r from to))]
      [(Mul l r) (Mul (subst l from to) (subst r from to))]
      [(Id name) (if (eq? name from) to expr)]
      [(With bound-id named-expr bound-body)
       (With bound-id
             (subst named-expr from to)
             (if (eq? bound-id from)
               bound-body
               (subst bound-body from to)))]))

  (: eval : (WAE -> Number))
  ;; evaluates WAE expressions by reducing them to numbers
  (define (eval expr)
    (cases expr
      [(Num n) n]
      [(Add l r) (+ (eval l) (eval r))]
      [(Sub l r) (- (eval l) (eval r))]
      [(Mul l r) (* (eval l) (eval r))]
      [(With bound-id named-expr bound-body)
       (eval (subst bound-body
                    bound-id
                    (Num (eval named-expr))))]
      [(Id name) (error 'eval "free identifier: ~s" name)]))

  (: run : (String -> Number))
  ;; evaluate a WAE program contained in a string
  (define (run str)
    (eval (parse str)))

  ;; tests
  (test (run "5") => 5)
  (test (run "{+ 5 5}") => 10)
  (test (run "{with {x {+ 5 5}} {+ x x}}") => 20)
  (test (run "{with {x 5} {+ x x}}") => 10)
  (test (run "{with {x {+ 5 5}} {with {y {- x 3}} {+ y y}}}") => 14)
  (test (run "{with {x 5} {with {y {- x 3}} {+ y y}}}") => 4)
  (test (run "{with {x 5} {+ x {with {x 3} 10}}}") => 15)
  (test (run "{with {x 5} {+ x {with {x 3} x}}}") => 8)
  (test (run "{with {x 5} {+ x {with {y 3} x}}}") => 10)
  (test (run "{with {x 5} {with {y x} y}}") => 5)
  (test (run "{with {x 5} {with {x x} x}}") => 5)
  (test (run "{with {x 1} y}") =error> "free identifier")
  ----------------------------------------------------------------------

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