Parsing techniques a practical guide

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Some of the more advanced techniques are used in code generation in compilers and in data compression. In linguistics, the importance of formal grammars was recognized early on, but only recently have the corresponding parsing techniques been applied. Also their importance as general pattern recognizers is slowly being acknowledged. This text Parsing Techniques explores new developments, such as generalized deterministic parsing, linear-time substring parsing, parallel parsing, parsing as intersection, non-canonical methods, and non-Chomsky systems.

To provide readers with low-threshold access to the full field of parsing techniques, this new edition uses a two-tiered structure. The basic ideas behind the dozen or so existing parsing techniques are explained in an intuitive and narrative style, and problems are presented at the conclusion of each chapter, allowing the reader to step outside the bounds of the covered material and explore parsing techniques at various levels.

The reader is also provided with an extensive annotated bibliography as well as hints and partial solutions to a number of problems. In the bibliography, hundreds of realizations and improvements of parsing techniques are explained in a much terser, yet still informal, style, improving its readability and usability. The reader should have an understanding of algorithmic thinking, especially recursion; however, knowledge of any particular programming language is not required.

Springer Shop Labirint Ozon. So while the XML Schema is simply a context-free grammar, the semantics of the data types imposes an additional layer of constraint on the XML instance. Still, this grammar allows the cats to bark … For a better way to handle context, see various sections in Chapter 15, especially Van Wijngaarden grammars Section Here is a sample: name: sentence: list: tom symbol; dick symbol; harry symbol.

For example, the comma binds tighter than the semicolon. Since the iterative implementation is complex, most practical parser generators use the recursive interpretation in some form or another, whereas most research has been done on the iterative interpretation. The XML instance is a sentence of the grammar.

The issue is: what is the parse tree DOM tree for this instance? While we are working on the Subject, the Verb and Object symbols remain queued at the right in the sentential form. This is a key concept Chomsky definition of Type 3 Our definition of Type 3: A non-terminal produces zero or more terminals.

A non-terminal produces zero or more terminals followed by one non-terminal. Chomsky definition of Type 3: A non-terminal produces one terminal. A non-terminal produces one terminal followed by one non-terminal. Our definition of Type 3: A non-terminal produces zero or more terminals.

Right-regular has nothing to do with recursion. See Section 5. A notational device has been invented to abate this nuisance. Depending on which definition you use, the grammars you create may vary in user friendliness and in ease of processing. The formal linguist is interested in and helped by minimally sufficient grammars.

A rule is context-sensitive if only one non-terminal symbol in its left-hand side gets replaced by other symbols, while we find the others back, undamaged and in the same order, in the right-hand side. Now continue doing the following: — Consider the first sentential form in the queue. There is no grammar for it, because this set cannot be generated you cannot tell whether a Type 0 grammar produces the empty string. If phrasestructure is not sufficient, only natural language description will do, as shown here.

That is, it is a languagegenerating procedure. The reason is that during the process more and more side-lines develop which all require equal attention. Computers are better at it than humans. Doing both substitutions replace A by b and AC by ac also leads to a blind alley, but there will be an output, ac. If you ignore a context you run the risk of creating false productions. Remember, this discussion is just for phrase-structure grammars. It generates no sentences.

Therefore, it produces the empty set. It means that the proof method used will not work for all PS grammars. In fact, the queue procedure answers Yes in finite time but takes an infinite time if the answer is No. The computer scientist is aware of but not daunted by the impossibilities from formal languages. The production rules must produce another sentential form with a non-terminal.

So the sentential forms never halt. Upon the first item generated that has no non-terminals is a sentence , return Yes the grammar does produce at least one sentence and stop. It turns out, there is no other algorithm. And you can trivially rewrite preserving the sets they generate any of these so they are no longer Type That is, it may take an infinite amount of time to determine that an item is not in the set. More formally, determining if a given item belongs to a set generated by a Type 0 grammar is undecidable.

We noted that PS sentential forms can expand and shrink during a production process. Thus a production can grow. Thus a production cannot shrink. A production is either of the same length or longer. The halting problem is undecidable for CS grammars. These terminals are guaranteed to produce something. Now scan again to find non-terminals which have a right-hand side that consists of only terminals and non-terminals that are guaranteed to produce something.

This will give us new nonterminals that are guaranteed to produce something. Repeat 2 until we find no more new non-terminals. If we have not met the start symbol this way, the grammar will not produce anything. The sequences of production rules are not as similar as we would expect. In grand total the same rules and alternatives are used. But the order of rewriting differs. The left-hand side of this rule, the parent of our symbol, was produced as the 1st member of rule 3.

And so on, until we reach the start symbol. We can, in a sense, trace the lineage of the symbol in this way. Therefore the number of original symbols is finite and the number of original sentences is finite. We arrive at the surprising conclusion that any CF grammar produces a finite-size kernel of original sentences and probably an infinite number of unoriginal sentences. Unoriginal sentence 1 1,2 b b 2 2,2 a 1. Consequently, there is no CF grammar for it.

How we can make a CF grammar infinitely complicated is described in the section on two-level grammars, When leaving the transition graph for N, pop n2 from the stack and continue at node n2. No need for stacking: interpret an arrow marked with a non-terminal N as a jump to the transition graph for N. So a regular grammar corresponds to a nonrecursive transition network.

If we start with the second rule for S, we get stuck. It is a useless rule. If we start with the third rule for S, we get ourselves into an infinite loop, producing more and more Cs. Rules 2, 3, and 5 can never occur in a successful production process; they are useless rules and can be removed from the grammar without affecting the language produced. If you know that it is Type 2, then you can easily build the program. It may contain an undefined non-terminal. It may not be reachable from the start symbol.

It may fail to produce anything. But now B may be undefined, so remove rules with B on the righthand side, etc. This happens when all right-hand sides in the grammar contain at least one non-terminal. Then there is no way to get rid of the non-terminals, and the grammar itself is non-productive. Remove non-productive rules 2.

After finding all productive rules, the other, remaining rules are the nonproductive rules. Apply this knowledge in a second round through the grammar. That means they are non-productive and can be removed from the grammar. And so forth. Initialization: an assessment of what we know initially. For our problem we knew: The grammar rules Terminals and empty are productive 2.

Inference rule: a rule telling how knowledge from several places is to be combined. The inference rule is repeated until nothing changes any more. So you will have to redo the algorithm for removing unreachable nonterminals. Will we have to run the algorithm for removing non-productive rules again?

In the process of removing non-productive rules we determined that all symbols on the right-hand side of X are productive. That means that N is productive N is defined. A grammar describes a language. So the intersection language consists of strings of the form anbncn and we know that language is not context-free.

The queue algorithm outputs the strings in order of increasing length. Suppose grammar 1 generates string abc. We can determine if grammar 2 generates abc by running the queue algorithm on grammar 2 until a it outputs abc, or b it outputs a string with length greater than 3 the length of abc. After erasing the mirrors we have abaaba, which is aba 2 Create Type 0 from the intersection of context-free languages Using a massive application of the mirror-mirror trick, one can relatively easily prove that any Type 0 language can be constructed as the intersection of two CF languages, plus a set of erasable symbols.

But like all text in this book the explanation is very dense and is severely complicated by the fact that the author inside the explanation wants to prove that the two CF languages you need are deterministic. It is easy to produce S x -S by CF grammar, so apply twice, once for the inner Xnr, once for the outer Xnl, and you're done. This is where L2 comes in. This makes the intersection of string 1 and string 2 a representation of a valid Type 0 production process. There are two more details to care for.

One is the start-up, which is next to trivial. The second is the close-down and the harvesting of the result. This is where the homomorphism the erasing of the erasable symbols comes in. Before we start the whole construction we replace all terminals in G by non-terminals with similar names, and declare all symbols in G erasable.

This ensures that when we in the end apply the homomorphism the erasure act the whole production process disappears. But of course we want to keep the final product which consists exclusively of those non-terminals that represent terminals. We harvest them by letting the productions of L1 and L2 end in the language T -T, where T is any string of the non-terminals created for the original terminals of G, and replaces each of the non-terminals by its corresponding terminal.

Again this is easy to do since its structure is again essentially S x -S. Now when we erase the erasable symbols, everything disappears except the final string of terminals, a production of G. The negation of it then produces a CF language. If you insist on having set intersection which is very tempting and convenient, see for example the ease with which you can construct anbncn by intersection , you'll never invent CF languages.

So you create a grammar to describe the pattern. Now you want to check that your grammar correctly describes the pattern. What is required of a parser? If it is not, we have the wrong grammar. The semantics of the members of the right-hand side of each rule is used to define the semantics of the left-hand side. The semantics of the left-hand side of each rule is used to define the semantics of the members of the right-hand side.

Attribute grammars 2. The overall semantics is composed as the result of all the local computations. If it computes a value of one of the non-terminals in the right-hand side of R, say A, then that value is inherited by A. The attribute for the symbol on the left-hand side, Sum, is named A0 Each symbol, including terminals, are indexed. So the attribute for the right-side Sum is A1, the attribute for Digit is A3. Initially only the attributes of the leaves are known, but as soon as all attributes in a right-hand side of a production rule are known, we can use its semantic clause to compute the attribute of its left-hand side.

This way the attribute values semantics percolate up the tree, finally reaching the start symbol and providing us with the semantics of the whole string. This is another example of a closure algorithm! Example: a Type 1 context-sensitive grammar can define sentences with the same number of as, bs, and cs i. It is approximated by increasingly finer outlines. In this metaphor, the rose corresponds to the language imagine the strings of the language as molecules in the rose ; the grammar serves to delineate its silhouette.

A regular grammar only allows us straight horizontal and vertical line segments to describe the flower. Ruler and T-square suffice, but the result is a course and mechanical looking picture. A CF grammar would approximate the outline by straight lines at any angle and by circle segments. The drawing could still be made using the classical tools of compass and ruler. The result is stilted but recognizable. A CS grammar would present us with a smooth curve tightly enveloping the flower, but the curve is too smooth: it cannot follow all the sharp turns, and it deviates slightly at complicated points.

Still, a very realistic picture results. An unrestricted phrase structure grammar can represent the outline perfectly.

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Parsing, also referred to as syntax analysis, has been and continues to be an essential part of computer science and linguistics. Today, parsing techniques are also implemented in a number of other disciplines, including but not limited to, document preparation and conversion, typesetting chemical formulae, and chromosome recognition. This second edition presents new developments and discoveries that have been made in the field.

Parsing techniques have grown considerably in importance, both in computational linguistics where such parsers are the only option, and computer science, where advanced compilers often use general CF parsers. Parsing techniques provide a solid basis for compiler construction and contribute to all existing software: enabling Web browsers to analyze HTML pages and PostScript printers to analyze PostScript.

Some of the more advanced techniques are used in code generation in compilers and in data compression. In linguistics, the importance of formal grammars was recognized early on, but only recently have the corresponding parsing techniques been applied. Also their importance as general pattern recognizers is slowly being acknowledged.

This text Parsing Techniques explores new developments, such as generalized deterministic parsing, linear-time substring parsing, parallel parsing, parsing as intersection, non-canonical methods, and non-Chomsky systems. To provide readers with low-threshold access to the full field of parsing techniques, this new edition uses a two-tiered structure.

A sentence is finished only when it no longer contains non-terminals 6. We start our replacement procedure with Sentence 66 Different types of clauses 1. It is valid generally and it is one of the rules of the game.

This name is called the start symbol, and it is required for every grammar. His analysis has been the foundation for almost all research and progress in formal languages, parsers, and a considerable part of compiler construction and linguistics. It allows a very concise expression of what and how but gives very little information on why. This tutorial gives the why. The right-hand side separated by vertical bars are also called alternatives.

So this is a production graph, not a production tree. But since the production process always makes new copies for the nodes it produces, it cannot produce an already existing node. Directed acyclic graphs are called dags. You draw it for the process taken to generate a sentence. It loops. So it cannot generate a sentence. With the path we have taken, we have arrived at a blind alley. All other methods known to mankind for generating sets have been proved to be equivalent-to, or less powerful than a phrase structure grammar.

That result is called a sentence in formal language theory. It has been proved that any set language that can be generated by a program can be generated by a phrase structure grammar. A Manhattan turtle moves in a plane and can only move north, east, south or west in distances of one block. The grammar below produces all paths that return to their own starting point. Thus, the replacement may be more than one symbol.

In this case the replacement is two symbols. Because only a non-terminal may be replaced. EndName context is and. We had to introduce a new non-terminal, Comma. There are languages that can be generated by Type 0 grammars that cannot be generated by any Type 1 or Type 2, 3, 4 grammar. Type 1 context-sensitive grammars Type 2 context-free grammars Type 3 regular grammars Type 4 finite-choice grammars Roger, not sure this is true Contradiction?

If we want to remember how many there were, we shall have to append something to the end as well, and it cannot be a b or c. We shall use a yet unknown symbol Q. The following rule both prepends and appends: 1. There the newly inserted bc will do no harm: 2. This can be remedied by allowing Q to hop left over c: 3.

The last rule is not contextsensitive since it does not conform to: only one nonterminal symbol in its left-hand side gets replaced by other symbols, while we find the others back, undamaged and in the same order, in the right-hand side.

Each node fans out, no nodes come together. That XML Schema is a context-free grammar. But the sentences that are generated i. So while the XML Schema is simply a context-free grammar, the semantics of the data types imposes an additional layer of constraint on the XML instance.

Still, this grammar allows the cats to bark … For a better way to handle context, see various sections in Chapter 15, especially Van Wijngaarden grammars Section Here is a sample: name: sentence: list: tom symbol; dick symbol; harry symbol. For example, the comma binds tighter than the semicolon. Since the iterative implementation is complex, most practical parser generators use the recursive interpretation in some form or another, whereas most research has been done on the iterative interpretation.

The XML instance is a sentence of the grammar. The issue is: what is the parse tree DOM tree for this instance? While we are working on the Subject, the Verb and Object symbols remain queued at the right in the sentential form. This is a key concept Chomsky definition of Type 3 Our definition of Type 3: A non-terminal produces zero or more terminals. A non-terminal produces zero or more terminals followed by one non-terminal.

Chomsky definition of Type 3: A non-terminal produces one terminal. A non-terminal produces one terminal followed by one non-terminal. Our definition of Type 3: A non-terminal produces zero or more terminals. Right-regular has nothing to do with recursion. See Section 5. A notational device has been invented to abate this nuisance.

Depending on which definition you use, the grammars you create may vary in user friendliness and in ease of processing. The formal linguist is interested in and helped by minimally sufficient grammars. A rule is context-sensitive if only one non-terminal symbol in its left-hand side gets replaced by other symbols, while we find the others back, undamaged and in the same order, in the right-hand side.

Now continue doing the following: — Consider the first sentential form in the queue. There is no grammar for it, because this set cannot be generated you cannot tell whether a Type 0 grammar produces the empty string. If phrasestructure is not sufficient, only natural language description will do, as shown here. That is, it is a languagegenerating procedure. The reason is that during the process more and more side-lines develop which all require equal attention.

Computers are better at it than humans. Doing both substitutions replace A by b and AC by ac also leads to a blind alley, but there will be an output, ac. If you ignore a context you run the risk of creating false productions. Remember, this discussion is just for phrase-structure grammars. It generates no sentences. Therefore, it produces the empty set. It means that the proof method used will not work for all PS grammars.

In fact, the queue procedure answers Yes in finite time but takes an infinite time if the answer is No. The computer scientist is aware of but not daunted by the impossibilities from formal languages. The production rules must produce another sentential form with a non-terminal. So the sentential forms never halt. Upon the first item generated that has no non-terminals is a sentence , return Yes the grammar does produce at least one sentence and stop.

It turns out, there is no other algorithm. And you can trivially rewrite preserving the sets they generate any of these so they are no longer Type That is, it may take an infinite amount of time to determine that an item is not in the set. More formally, determining if a given item belongs to a set generated by a Type 0 grammar is undecidable.

We noted that PS sentential forms can expand and shrink during a production process. Thus a production can grow. Thus a production cannot shrink. A production is either of the same length or longer. The halting problem is undecidable for CS grammars. These terminals are guaranteed to produce something. Now scan again to find non-terminals which have a right-hand side that consists of only terminals and non-terminals that are guaranteed to produce something. This will give us new nonterminals that are guaranteed to produce something.

Repeat 2 until we find no more new non-terminals. If we have not met the start symbol this way, the grammar will not produce anything. The sequences of production rules are not as similar as we would expect. In grand total the same rules and alternatives are used. But the order of rewriting differs. The left-hand side of this rule, the parent of our symbol, was produced as the 1st member of rule 3.

And so on, until we reach the start symbol. We can, in a sense, trace the lineage of the symbol in this way. Therefore the number of original symbols is finite and the number of original sentences is finite. We arrive at the surprising conclusion that any CF grammar produces a finite-size kernel of original sentences and probably an infinite number of unoriginal sentences.

Unoriginal sentence 1 1,2 b b 2 2,2 a 1. Consequently, there is no CF grammar for it. How we can make a CF grammar infinitely complicated is described in the section on two-level grammars, When leaving the transition graph for N, pop n2 from the stack and continue at node n2. No need for stacking: interpret an arrow marked with a non-terminal N as a jump to the transition graph for N.

So a regular grammar corresponds to a nonrecursive transition network. If we start with the second rule for S, we get stuck. It is a useless rule. If we start with the third rule for S, we get ourselves into an infinite loop, producing more and more Cs.

Rules 2, 3, and 5 can never occur in a successful production process; they are useless rules and can be removed from the grammar without affecting the language produced. If you know that it is Type 2, then you can easily build the program. It may contain an undefined non-terminal. It may not be reachable from the start symbol. It may fail to produce anything. But now B may be undefined, so remove rules with B on the righthand side, etc.

This happens when all right-hand sides in the grammar contain at least one non-terminal. Then there is no way to get rid of the non-terminals, and the grammar itself is non-productive. Remove non-productive rules 2. After finding all productive rules, the other, remaining rules are the nonproductive rules. Apply this knowledge in a second round through the grammar. That means they are non-productive and can be removed from the grammar.

And so forth. Initialization: an assessment of what we know initially. For our problem we knew: The grammar rules Terminals and empty are productive 2. Inference rule: a rule telling how knowledge from several places is to be combined. The inference rule is repeated until nothing changes any more. So you will have to redo the algorithm for removing unreachable nonterminals. Will we have to run the algorithm for removing non-productive rules again?

In the process of removing non-productive rules we determined that all symbols on the right-hand side of X are productive. That means that N is productive N is defined. A grammar describes a language. So the intersection language consists of strings of the form anbncn and we know that language is not context-free.

The queue algorithm outputs the strings in order of increasing length. Suppose grammar 1 generates string abc. We can determine if grammar 2 generates abc by running the queue algorithm on grammar 2 until a it outputs abc, or b it outputs a string with length greater than 3 the length of abc. After erasing the mirrors we have abaaba, which is aba 2 Create Type 0 from the intersection of context-free languages Using a massive application of the mirror-mirror trick, one can relatively easily prove that any Type 0 language can be constructed as the intersection of two CF languages, plus a set of erasable symbols.

But like all text in this book the explanation is very dense and is severely complicated by the fact that the author inside the explanation wants to prove that the two CF languages you need are deterministic. It is easy to produce S x -S by CF grammar, so apply twice, once for the inner Xnr, once for the outer Xnl, and you're done.

This is where L2 comes in. This makes the intersection of string 1 and string 2 a representation of a valid Type 0 production process. There are two more details to care for.