Description
Theoretical background
A derivation is a rule list that describes how to derive a phrase from a nonterminal symbol. For example,
suppose we have the following grammar with start symbol Expr:
Expr → Term Binop Expr
Expr → Term
Term → Num
Term → Lvalue
Term → Incrop Lvalue
Term → Lvalue Incrop
Term → “(” Expr “)”
Lvalue → $ Expr
Incrop → “++”
Incrop → “−−”
Binop → “+”
Binop → “−”
Num → “0”
Num → “1”
Num → “2”
Num → “3”
Num → “4”
Num → “5”
Num → “6”
Num → “7”
Num → “8”
Num → “9”
Then here is a derivation for the phrase “3” “+” “4” from the nonterminal Expr. After each rule is applied, the
resulting list of terminals and nonterminals is given.
rule after rule is applied
(at start) Expr
Expr → Term Binop Expr Term Binop Expr
Term → Num Num Binop Expr
Num → “3” “3” Binop Expr
Binop → “+” “3” “+” Expr
Expr → Term “3” “+” Term
Term → Num “3” “+” Num
Num → “4” “3” “+” “4”
In a leftmost derivation, the leftmost nonterminal is always the one that is expanded next. The above example is
a leftmost derivation.
Motivation
You’d like to test grammars that are being proposed as test cases for CS 132 projects. One way is to test it on
actual CS 132 projects, but those projects aren’t done yet and anyway you’d like a second opinion in case the
student projects are incorrect. So you decide to write a simple parser generator. Given a grammar in the style of
Homework 1, your program will generate a function that is a parser. When this parser is given a program to
parse, it produces a derivation for that program, or an error indication if the program contains a syntax error and
cannot be parsed.
The key notion of this assignment is that of a matcher. A matcher is a function that inspects a given string of
terminals to find a match for a prefix that corresponds to a nonterminal symbol of a grammar, and then checks
whether the match is acceptable by testing whether a given acceptor succeeds on the corresponding derivation
and suffix. For example, a matcher for awkish_grammar below might inspect the string [“3″;”+”;”4″;”-“] and
find two possible prefixes that match, namely [“3″;”+”;”4″] and [“3”]. The matcher will first apply the
acceptor to a derivation for the first prefix [“3″;”+”;”4″], along with the corresponding suffix [“-“]. If this is
accepted, the matcher will return whatever the acceptor returns. Otherwise, the matcher will apply the acceptor
to a derivation for the second prefix [“3”], along with the corresponding suffix [“+”;”4″;”-“], and will return
whatever the acceptor returns. If a matcher finds no matching prefixes, it returns the special value None.
As you can see by mentally executing the example, matchers sometimes need to try multiple alternatives and to
backtrack to a later alternative if an earlier one is a blind alley.
An acceptor is a function that accepts a rule list and a suffix by returning some value wrapped inside the Some
constructor. The acceptor rejects the rule list and suffix by returning None. For example, the acceptor (fun d ->
function | “+”::t -> Some (d,”+”::t) | _ -> None) accepts any rule list but accepts only suffixes
beginning with “+”. Such an acceptor would cause the example matcher to fail on the prefix [“3″;”+”;”4″]
(since the corresponding suffix begins with “-“, not “+”) but it would succeed on the prefix [“3”].
By convention, an acceptor that is successful returns Some (d,s), where d is a rule list that typically contains the
acceptor’s input rule list as a sublist (because the acceptor may do further parsing, and therefore has applied more
rules than before), and s is a tail of the input suffix (again, because the acceptor may have parsed more of the
input, and has therefore consumed some of the suffix). This allows the matcher’s caller to retrieve the derivation
for the matched prefix, along with an indication where the matched prefix ends (since it ends just before the
suffix starts). Although this behavior is crucial for the internal acceptors used by your code, it is not required for
top-level acceptors supplied by test programs: a top-level acceptor needs only to return a Some x value to
succeed.
Whenever there are several rules to try for a nonterminal, you should always try them left-to-right. For example,
awkish_grammar below contains this:
| Expr ->
[[N Term; N Binop; N Expr];
[N Term]]
and therefore, your matcher should attempt to use the rule “Expr → Term Binop Expr” before attempting to use
the simpler rule “Expr → Term”.
Definitions
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symbol, right hand side, rule
same as in Homework 1.
alternative list
A list of right hand sides. It corresponds to all of a grammar’s rules for a given nonterminal symbol. By
convention, an empty alternative list [] is treated as if it were a singleton list [[]] containing the empty
symbol string.
production function
A function whose argument is a nonterminal value. It returns a grammar’s alternative list for that
nonterminal.
grammar
A pair, consisting of a start symbol and a production function. The start symbol is a nonterminal value.
derivation
a list of rules used to derive a phrase from a nonterminal. For example, the OCaml representation of the
example derivation shown above is as follows:
[Expr, [N Term; N Binop; N Expr];
Term, [N Num];
Num, [T “3”];
Binop, [T “+”];
Expr, [N Term];
Term, [N Num];
Num, [T “4”]]
fragment
a list of terminal symbols, e.g., [“3”; “+”; “4”; “xyzzy”].
acceptor
a curried function with two arguments: a derivation d and a fragment frag. If the fragment is not
acceptable, it returns None; otherwise it returns Some x for some value x.
matcher
a curried function with two arguments: an acceptor accept and a fragment frag. A matcher matches a
prefix p of frag such that accept (when passed a derivation and the corresponding suffix) accepts the
corresponding suffix (i.e., the suffix of frag that remains after p is removed). If there is such a match, the
matcher returns whatever accept returns; otherwise it returns None.
Assignment
1. To warm up, notice that the format of grammars is different in this assignment, versus Homework 1. Write
a function convert_grammar gram1 that returns a Homework 2-style grammar, which is converted from
the Homework 1-style grammar gram1. Test your implementation of convert_grammar on the test
grammars given in Homework 1. For example, the top-level definition let awksub_grammar_2 =
convert_grammar awksub_grammar should bind awksub_grammar_2 to a Homework 2-style grammar that
is equivalent to the Homework 1-style grammar awksub_grammar.
2. Write a function parse_prefix gram that returns a matcher for the grammar gram. When applied to an
acceptor accept and a fragment frag, the matcher must return the first acceptable match of a prefix of frag,
by trying the grammar rules in order; this is not necessarily the shortest nor the longest acceptable match.
A match is considered to be acceptable if accept succeeds when given a derivation and the suffix fragment
that immediately follows the matching prefix. When this happens, the matcher returns whatever the
acceptor returned. If no acceptable match is found, the matcher returns None.
3. Write two good, nontrivial test cases for your parse_prefix function. These test cases should all be in the
style of the test cases given below, but should cover different problem areas. Your test cases should be
named test_1 and test_2 (note the underscores; this distinguishes your test cases from the standard ones
given below). Your test cases should test at least one grammar of your own. You may reuse your test cases
for Homework 1 as part of test_1, but test_2 should be new.
4. Assess your work by writing an after-action report that summarizes why you solved the problem the way
you did, other approaches that you considered and rejected (and why you rejected them), and any
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weaknesses in your solution in the context of its intended application. If possible, illustrate weaknesses by
test cases that fail with your implementation. This report should be a simple ASCII plain text file that
consumes a page or so (at most 100 lines and 80 columns per line, and at least 50 lines, please). See
Resources for oral presentations and written reports for advice on how to write assessments; admittedly
much of the advice there is overkill for the simple kind of report we’re looking for here.
Unlike Homework 1, we are expecting some weaknesses here, so your assessment should talk about them. For
example, we don’t expect that your implementation will work with all possible grammars, but we would like to
know which sort of grammars it will have trouble with.
As with Homework 1, your code may use the Pervasives and List modules, but it should use no other
modules. Your code should be free of side effects. Simplicity is more important than efficiency, but your code
should avoid using unnecessary time and space when it is easy to do so.
Submit
We will test your program on the SEASnet Linux servers as before, so make sure that /usr/local/cs/bin is at
the start of your path, using the same technique as in Homework 1.
Submit three files:
hw2.ml should define parse_prefix along with any auxiliary types and functions needed to define
parse_prefix.
hw2test.ml should contain your test cases.
hw2.txt should hold your assessment.
Please do not put your name, student ID, or other personally identifying information in your files.
Sample test cases
let accept_all derivation string = Some (derivation, string)
let accept_empty_suffix derivation = function
| [] -> Some (derivation, [])
| _ -> None
(* An example grammar for a small subset of Awk.
This grammar is not the same as Homework 1; it is
instead the same as the grammar under
“Theoretical background” above. *)
type awksub_nonterminals =
| Expr | Term | Lvalue | Incrop | Binop | Num
let awkish_grammar =
(Expr,
function
| Expr ->
[[N Term; N Binop; N Expr];
[N Term]]
| Term ->
[[N Num];
[N Lvalue];
[N Incrop; N Lvalue];
[N Lvalue; N Incrop];
[T”(“; N Expr; T”)”]]
| Lvalue ->
[[T”$”; N Expr]]
| Incrop ->
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[[T”++”];
[T”–“]]
| Binop ->
[[T”+”];
[T”-“]]
| Num ->
[[T”0″]; [T”1″]; [T”2″]; [T”3″]; [T”4″];
[T”5″]; [T”6″]; [T”7″]; [T”8″]; [T”9”]])
let test0 =
((parse_prefix awkish_grammar accept_all [“ouch”]) = None)
let test1 =
((parse_prefix awkish_grammar accept_all [“9”])
= Some ([(Expr, [N Term]); (Term, [N Num]); (Num, [T “9”])], []))
let test2 =
((parse_prefix awkish_grammar accept_all [“9”; “+”; “$”; “1”; “+”])
= Some
([(Expr, [N Term; N Binop; N Expr]); (Term, [N Num]); (Num, [T “9”]);
(Binop, [T “+”]); (Expr, [N Term]); (Term, [N Lvalue]);
(Lvalue, [T “$”; N Expr]); (Expr, [N Term]); (Term, [N Num]);
(Num, [T “1”])],
[“+”]))
let test3 =
((parse_prefix awkish_grammar accept_empty_suffix [“9”; “+”; “$”; “1”; “+”])
= None)
(* This one might take a bit longer…. *)
let test4 =
((parse_prefix awkish_grammar accept_all
[“(“; “$”; “8”; “)”; “-“; “$”; “++”; “$”; “–“; “$”; “9”; “+”;
“(“; “$”; “++”; “$”; “2”; “+”; “(“; “8”; “)”; “-“; “9”; “)”;
“-“; “(“; “$”; “$”; “$”; “$”; “$”; “++”; “$”; “$”; “5”; “++”;
“++”; “–“; “)”; “-“; “++”; “$”; “$”; “(“; “$”; “8”; “++”; “)”;
“++”; “+”; “0”])
= Some
([(Expr, [N Term; N Binop; N Expr]); (Term, [T “(“; N Expr; T “)”]);
(Expr, [N Term]); (Term, [N Lvalue]); (Lvalue, [T “$”; N Expr]);
(Expr, [N Term]); (Term, [N Num]); (Num, [T “8”]); (Binop, [T “-“]);
(Expr, [N Term; N Binop; N Expr]); (Term, [N Lvalue]);
(Lvalue, [T “$”; N Expr]); (Expr, [N Term; N Binop; N Expr]);
(Term, [N Incrop; N Lvalue]); (Incrop, [T “++”]);
(Lvalue, [T “$”; N Expr]); (Expr, [N Term; N Binop; N Expr]);
(Term, [N Incrop; N Lvalue]); (Incrop, [T “–“]);
(Lvalue, [T “$”; N Expr]); (Expr, [N Term; N Binop; N Expr]);
(Term, [N Num]); (Num, [T “9”]); (Binop, [T “+”]); (Expr, [N Term]);
(Term, [T “(“; N Expr; T “)”]); (Expr, [N Term; N Binop; N Expr]);
(Term, [N Lvalue]); (Lvalue, [T “$”; N Expr]);
(Expr, [N Term; N Binop; N Expr]); (Term, [N Incrop; N Lvalue]);
(Incrop, [T “++”]); (Lvalue, [T “$”; N Expr]); (Expr, [N Term]);
(Term, [N Num]); (Num, [T “2”]); (Binop, [T “+”]); (Expr, [N Term]);
(Term, [T “(“; N Expr; T “)”]); (Expr, [N Term]); (Term, [N Num]);
(Num, [T “8”]); (Binop, [T “-“]); (Expr, [N Term]); (Term, [N Num]);
(Num, [T “9”]); (Binop, [T “-“]); (Expr, [N Term]);
(Term, [T “(“; N Expr; T “)”]); (Expr, [N Term]); (Term, [N Lvalue]);
(Lvalue, [T “$”; N Expr]); (Expr, [N Term]); (Term, [N Lvalue]);
(Lvalue, [T “$”; N Expr]); (Expr, [N Term]); (Term, [N Lvalue]);
(Lvalue, [T “$”; N Expr]); (Expr, [N Term]); (Term, [N Lvalue; N Incrop]);
(Lvalue, [T “$”; N Expr]); (Expr, [N Term]); (Term, [N Lvalue; N Incrop]);
(Lvalue, [T “$”; N Expr]); (Expr, [N Term]); (Term, [N Incrop; N Lvalue]);
(Incrop, [T “++”]); (Lvalue, [T “$”; N Expr]); (Expr, [N Term]);
(Term, [N Lvalue; N Incrop]); (Lvalue, [T “$”; N Expr]); (Expr, [N Term]);
(Term, [N Num]); (Num, [T “5”]); (Incrop, [T “++”]); (Incrop, [T “++”]);
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(Incrop, [T “–“]); (Binop, [T “-“]); (Expr, [N Term]);
(Term, [N Incrop; N Lvalue]); (Incrop, [T “++”]);
(Lvalue, [T “$”; N Expr]); (Expr, [N Term]); (Term, [N Lvalue; N Incrop]);
(Lvalue, [T “$”; N Expr]); (Expr, [N Term]);
(Term, [T “(“; N Expr; T “)”]); (Expr, [N Term]);
(Term, [N Lvalue; N Incrop]); (Lvalue, [T “$”; N Expr]); (Expr, [N Term]);
(Term, [N Num]); (Num, [T “8”]); (Incrop, [T “++”]); (Incrop, [T “++”]);
(Binop, [T “+”]); (Expr, [N Term]); (Term, [N Num]); (Num, [T “0”])],
[]))
let rec contains_lvalue = function
| [] -> false
| (Lvalue,_)::_ -> true
| _::rules -> contains_lvalue rules
let accept_only_non_lvalues rules frag =
if contains_lvalue rules
then None
else Some (rules, frag)
let test5 =
((parse_prefix awkish_grammar accept_only_non_lvalues
[“3”; “-“; “4”; “+”; “$”; “5”; “-“; “6”])
= Some
([(Expr, [N Term; N Binop; N Expr]); (Term, [N Num]); (Num, [T “3”]);
(Binop, [T “-“]); (Expr, [N Term]); (Term, [N Num]); (Num, [T “4”])],
[“+”; “$”; “5”; “-“; “6”]))
Sample use of test cases
If you put the sample test cases into a file hw2sample.ml, you should be able to use it as follows to test your
hw2.ml solution on the SEASnet implementation of OCaml. Similarly, the command #use “hw2test.ml”;;
should run your own test cases on your solution.
$ ocaml
OCaml version 4.04.0
# #use “hw2.ml”;;
…
val parse_prefix :
‘a * (‘a -> (‘a, ‘b) symbol list list) ->
((‘a * (‘a, ‘b) symbol list) list -> ‘b list -> (‘c list * ‘d) option) ->
‘b list -> (‘c list * ‘d) option =
…
# #use “hw2sample.ml”;;
val accept_all : ‘a -> ‘b -> (‘a * ‘b) option =
val accept_empty_suffix : ‘a -> ‘b list -> (‘a * ‘c list) option =
type awksub_nonterminals = …
val awkish_grammar :
awksub_nonterminals *
(awksub_nonterminals -> (awksub_nonterminals, string) symbol list list) =
(Expr,
val test0 : bool = true
val test1 : bool = true
val test2 : bool = true
val test3 : bool = true
val test4 : bool = true
val contains_lvalue : (awksub_nonterminals * ‘a) list -> bool =
val accept_only_non_lvalues :
(awksub_nonterminals * ‘a) list ->
‘b -> ((awksub_nonterminals * ‘a) list * ‘b) option =
val test5 : bool = true
#
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Hint
You can use a previous Homework 2 as a hint. It is a tough homework and is not the same problem but there are
some common ideas. Look for the sample solution at the end.
© 2003, 2004, 2006–2012, 2014–2017 Paul Eggert. See copying rules.
$Id: hw2.html,v 1.65 2017/10/23 00:49:12 eggert Exp $