;;; 6.945 Problem Set #5 Comments
;;; 03/11/2009
;;; Bryan Newbold <bnewbold@mit.edu>

Note: Problem numbers were totally off! Jeez! ;)

Problem 5.1: Multiple Matches 
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The multiple suceeds are due to the looping let expression ("lp") in 
match:segment. The first segement match attempt is with n=0, which is why the
first succeed statement has x null. This is a valid match with y equaling
(b b b b b b), so the success procedure is called and this match printed, but
then the failure causes evaluation to drop back to "lp", with n now incremented
to 1, and the process continutes until n=4, which has no matches at all and
the #f thunks all the way down to the floor.

Problem 5.2: Multiple Matches 
--------------------------------
I allowed match:element declarations to have an optional predicate; the
procedure corresponding to the predicate symbol is pulled from the nearest repl
environment. A problem with my implementation is that the predicate is only
checked for the first reference to the variable; if the predicate is different
or only specified in one of the later references, the right thing doesn't
happen.  Eg,

((matcher '(a (? b) (? b integer?) c))
  '(a t t c))
; ((b t))
; should be #f?

Also the predicate has to be a simple symbol existing in the environmnet,
instead of a lambda expression, maybe I should have used apply?

[see code in bnewbold_ps05_work.scm]

Problem 5.3: Multiple Match Implementation
--------------------------------
Ugh, I should really read these problem sets through the whole way first. See
implementation above, a somewhat ugly method.

Problem 5.4: ?:choice
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[see code in bnewbold_ps05_work.scm]

Problem 5.5: ?:pletrec, ?:ref
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Problem 5.5: restrict
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