and the theory of computation / John C. Martin.—4th ed. p. cm. Introduction to Languages and the Browse's Introduction to the Symptoms & Signs of Surgical . Introduction to Computer Theory by Daniel I. a. Cohen - Second Edition - Ebook download as PDF File .pdf), Text File .txt) or read book online. Introduction to. DANIEL. 1. A. OE. INTODSONT SOPUE. THEOR. INTRODUCTION TO COMPUTER THEORY. Daniel I. A. Cohen Hunter College City University of New York.
|Language:||English, French, Arabic|
|Genre:||Business & Career|
|ePub File Size:||18.41 MB|
|PDF File Size:||11.42 MB|
|Distribution:||Free* [*Registration needed]|
Library of Congress Cataloging-in-Publication Data: Cohen, Daniel I. A., Introduction to computer theory. Includes Index 1. Electronic digital computers. Where can I get "Intro to Computer Theory by Daniel I.A Cohen 2nd Edition" ebook for free and legally? akbdBt CGbQVyUjh Introduction to Computer Theory 2nd Edition What sites can I download free PDF and E-books?. Feb 18, Download Introduction to Computer Theory by Daniel I.A Cohen PDF free - ALL YOU NEED.
The best part? As a Chegg Study subscriber, you can view available interactive solutions manuals for each of your classes for one low monthly price. Why download extra books when you can get all the homework help you need in one place? Can I get help with questions outside of textbook solution manuals? You bet! Just post a question you need help with, and one of our experts will provide a custom solution. You can also find solutions immediately by searching the millions of fully answered study questions in our archive.
How do I view solution manuals on my smartphone? You can download our homework help app on iOS or Android to access solutions manuals on your mobile device.
Asking a study question in a snap - just take a pic. Is it bigger than S? Give an example of two sets of strings. If what remains is the string A. Step 1. If what remains is not A this means some letters are left but we cannot find a word in S at the beginning. Is abaabbabbaabb? Find a string that disproves this algorithm. Step 2. Give another description of this language. Give letter more such Let us take an example. A recursive definition is characteristically a three-step process.
One standard way of defining this set is: EVEN is the set of all positive whole numbers divisible by 2. Suppose that we are trying to define the set of positive even integers for someone who knows about arithmetic but has never heard of the even numbers.
To prove that 14 is even using the recursive definition is a lengthier process. We might use: The set EVEN is defined by these three rules.
We could proceed as below: By Rule 1. Rule 1 2 is in even. The third method we present is sneaky. To show this using the first definition we divide 14 by 2 and find that there is no remainder. The set EVEN is defined by these three rules: Rule 1 2 is in EVEN.. Pretty horrible.
There is a reason that the third definition is less popular than the others: It is much harder to use in most practical applications. To prove that 14 is in EVEN by the second definition we have to somehow come up with the number 7 and then.
Once more applying Rule 2. Whether or not we want a recursive definition depends on two things: For instance. In later chapters we shall be interested in certain sets that have no better definition than the recursive one.
It should be understood that we mean we can apply Rule 2 also to the case where x and y stand for the same number. Let us consider the way polynomials are usually defined: We can now prove that 14 is in EVEN in fewer steps: The set EVEN. Now let us consider a recursive definition that is designed for people who know algebraic notation but do not know what a polynomial is. Before leaving this example. This is a trivial conclusion from the second recursive definition.
This is a little more complicated to see if we had to provide a proof based on the classical definition. In this way we can have proven that we can differentiate all polynomials without giving the best algorithm to do it.
It is for these reasons that recursive definitions are important to us. We could include rules for making the notation prettier. This becomes a theorem that can be proven directly from the recursive definition. It is true that we do not then know that the derivative of x' is nx'-'.
On the plus side. What is even more astounding is that we shall be able to prove that certain tasks are theoretically impossible for any computer remember: As soon as we prove that the derivative of a number is 0 and that the derivative of x is 1 we have automatically shown that we can differentiate all polynomials.
Since the topic of this book is Computer Theory. There are some advantages to this definition as well as the evident disadvantages.
By Rule 3 In fact. Is it not: In computer languages.. If Q is any word in LI. Also in mathematics we often see the following definition of factorial: Rule 1 Rule 2 0! The reason that these definitions are called "recursive" is that one of the rules used to define the set mentions the set itself..
These definitions have the same self-referential sense. Rule 2 Rule 3 If x is in AE.
Are there more rules? Q9 are also words in L 3. The definition can be written as: Rule 1 Any number positive. If x and y are in AE. Are there still more? The most natural way of defining a valid arithmetic expression. By applying Rule 2 we could always put in enough parentheses to avoid any confusion if we so desired.
This definition determines the set AE in a manner useful for proving many theorems about arithmetic expressions. We may never have seen a definition of "arithmetic expressions" before. So we usually adopt conventions of operator hierarchy and left to right execution.
We return to this point in Part 1I. There is no doubt that the string is a word in AE. Any AE formed by Rule 2 must begin and end with parentheses or begin with a minus sign. One can add only potatoes to the pot. Even without knowing exactly in what order the chefs visit the pot or how often. We shall show that it must have been Rule 3 iv. One can add only meat to the pot. If it were Rule 3 iii.
Let the shortest of these be a string called w. One can add only carrots to the pot. Our first question is: Which was the last rule used in the production of w? This is easy to answer. Similarly we can eliminate all the other possibilities. But since both x and y are AE's.
In computer science it is usually called a "slash. Now we know that w. This method of argument should sound familiar. The long-winded but careful proof of the last theorem is given to illustrate that recursive definitions can be conveniently employed in rigorous mathematical proofs.
Admittedly, this was a trival example of the application of this method. Most people would be just as convinced by the following "proof.
What would it mean? What are you, crazy or something? We should bear in mind that we are only on the threshold of investigating a very complex and profound subject and that in this early chapter we wish to introduce a feel for the techniques and viewpoints that will be relied on heavily later, under far less obvious circumstances.
We will use our learner's permit to spend a few hours driving around an empty parking lot before venturing onto the highway. Another common use for recursive definitions is to determine what expressions are valid in Symbolic Logic. We shall be interested in one particular branch of Symbolic Logic called the Sentential Calculus or the Propositional Calculus. The version we shall define here uses only negationand implication - along with the phrase variables, although conjunction and disjunction could easily be added to the system.
The valid expressions in this language are traditionally called WFF's for well-formed formulas. Rule 1 Rule 2 Rule 3.
Any single Latin letter is a WFF, a b c d If p is a WFF, then so are p and -- p. If p and q are WFF's, then so is p -- q. As a final note in this section, we should be wary that we have sometimes. Write another recursive definition for the language L, of Chapter 2. Describe a good method for showing that 2n is in EVEN. Show that the following is another recursive definition of the set EVEN. Rule 1 2 and 4 are in EVEN. Show that there are infinitely many different recursive definitions for the set EVEN.
Using any recursive definition of the set EVEN, show that all the numbers in it end in the digits 0, 2, 4, 6, or 8.
Write a recursive definition for the set of all polynomials in the two variables x and y. Define the set of valid algebraic expressions ALEX as follows:. What is the complete list of substrings of length 2 that cannot occur?
Are there any substrings of length 3 that cannot occur that do not contain forbidden substrings of length 2? What is the longest forbidden substring that does not contain a shorter forbidden substring?
The rules given above for the set AE, allow for the peculiar expressions 9. It is not really harmful to allow these in AE, but is there some modified definition of AE that eliminates this problem?
Write out the full recursive definition for the propositional calculus that contains the symbols V and A as well as i and What are all the.
Fix this problem. Use one of them to prove that the product of two powers of two is also a power of two. We defined L, in Chapter 2 by the symbols: We might define a language by the symbols: More precision and less guesswork is required, especially where computers are concerned. In this chapter we shall develop some new language-defining symbolism that will be much more precise than the ellipsis which is what the three dots. The language-defining symbols we are about to create are called regular expressions.
We will define the term regular expression itself recursively. The languages that are associated with these regular expressions are called regular languages and are also said to be defined by finite representation. These terms will make more sense when they are associated with concepts.
Let us reconsider the language L 4 of Chapter 2. In that chapter we presented one method for indicating this set as the closure of a smaller set. We now introduce the use of the Kleene star applied not to a set but directly to the letter x and written as a superscript as if it were an exponent. We can think of the star as an unknown power or undetermined power.
It stands for any string of x's in the language L 4. The star operator applied to a letter is analogous to the star operator applied to a set. It represents an arbituary concatenation of copies of that letter maybe none at all. This is why we use the word "language" in the equation. We could summarize this language by the English phrase "all words of the form one a followed by some number of b's maybe no b's at all.
The meaning is clear: No string can contain a blank unless a blank is a character in the alphabet 1. If we want blanks to be in the alphabet, we normally introduce some special symbol to stand for them, as blanks themselves are invisible to the naked eye.
We have now used a boldface letter without a star as well as with a star. We can apply the Kleene star to the string ab if we want, as follows: Parentheses are not letters in the alphabet of this language, so they can be used to indicate factoring without accidently changing the words.
Since the star represents some kind of exponentiation, we use it as powers are used in algebra, where by universal understanding the expression xy2 means x y2 , not Xy 2.
It would be a subtle mistake to say only that this language is the set of all words that begin and end with an a and have only b's in between, because this description may also apply to the word "a," depending on how it is interpreted.
Our symbolism eliminates this ambiguity. Notice that ba and aba are not in this language. Notice also that there need not be the same number of a's and b's.
We now introduce another use for the where x and y are strings of characters x or y". Care should be taken so as not to. All the words in T begin with an a or a c and then are followed by some number of b's.
We should, of course, have said "some or no b's". We often drop the zerooption because it is tiresome. We let the word "some" always mean "some or no," and when we mean "some positive number of' we say that. For every set of choices we have generated a particular string. The set of all strings produceable by this method is the language of the expression.
Each set of choices is a word. The first letter of each word in L is either an a or a b. The second letter of each word in L is either an a or a b. The third letter of each word in L is either an a or a b. This is a very important regular expression and we use it often.
Again this expression represents a language. This is a very powerful notation. We can describe all words that begin with the letter a simply as:. The only words left out are those that have only b's and the word A. For example, the word abbaab can be considered to be of this form in three ways: Another expression that denotes all the words with at least two a's is: We scan through some jungle of b's or no b's until we find the first a, then more b's or no b's , then the second a, then we finish up with anything.
In this set are abbbabb and aaaaa.
We can write: To be careful about this point, we say that two regular expressions are equivalent if they describe the same language. The expressions below also describe the language of words with at least two a's. If we write.
Such words as ba and bbaaaa are not included in this set. Since, however, we know that either the a comes before the b or the b comes before the a, we could define this set by the expression: There is a simpler expression that defines the same language.
These exceptions are all defined by the regular expression: U These language-defining expressions cannot be treated like algebraic symbols. The only words that do not contain both an a and a b in them somewhere are the words of all a's, all b's, or A. When these are added into the language, we get everything. Therefore, the regular expression: We can then write: EXAMPLE All temptation to treat these language-defining expressions as if they were algebraic polynomials should be dispelled by these equivalences: It means that all the words that do not contain the substring ab which are accounted for in the first term are all a's, all b's, A, or some b's followed by some a's.
N Usually when we employ the star operator, we are defining an infinite language. We can represent a finite language by using the plus union sign alone. If L is a finite language that includes the null word A, then the expression that defines L must also employ the symbol A. Let V also contain the word A. Alternatively, we could define V by the expression: It is because of this analogy to algebra that we have denoted our disjunction by the plus sign instead of the union sign U or the symbolic logic sign V.
We have a hybrid system: But the analogies to algebra should be approached very suspiciously, since addition in algebra never means choice and algebraic multiplication has properties different from concatenation even though we sometimes conventionally refer to it as product: Rule 2 If r.
Rule 1 Every letter of 1 can be made into a regular expression by writing it in boldface. As with the recursive definition of arithmetic expressions. A is a regular expression. The same confusion occurs in everyday speech. The string "French" is both a word an adjective and a language-defining name a noun. English is much harder. Both are regular expressions and both can be generated. Every word in this book is a book-definer. We have all the parts we need in order to define regular expressions recursively.
It is analogous to a book that lists all the books in print. The symbols that appear in regular expressions are: Context and typography will guide us. This is a language of language definers. Because of Rule 1 we may have trouble in distinguishing when we write an a whether we mean a the letter in Y.
However difficult Computer Theory may seem. The set of words in this language is the null set. Another example of excessive care is the worry about the language that contains no words at all.
Care should always be taken to produce the expression we actually want. The null word is a word. To make the identification between the regular expressions and their associated languages more explicit. The language of no words cannot technically be defined by a regular expression since Rule 1 starts by putting something into the language.
We finesse this point by saying that the language of no words is defined by the regular expression of no symbols. It might not be clear why we can not just leave the rules for associating a language with a regular expression on the informal level..
Some words in this language are ennuiverboten and souffl6Gesundheit Using regular expressions. The rules seem to show us how we can interpret the regular expression as a language.
As we build up a regular expression from the rules. As we might suspect. Rule 2 i The regular expression rl r 2 is associated with the language L. This is a question of meaning. We are now ready to give the rules for associating a language with every regular expression. In our next theorem we show that every finite language can be defined by a regular expression.
The situation for languages with infinitely many words is different. Is there some way of telling when this happens? By "way" we mean. We have already seen examples where completely different regular expressions end up describing the same language.
As to the first and perhaps most important question. We may be centuries away from being able to do that. We present an algorithmic procedure in Chapter 12 to determine whether or not two regular expressions define the same language. Before we can construct an algorithm for obtaining understanding we must have some good definition of what it means to understand. We prove in Chapter 11 that there are some languages that cannot be defined by any regular expression.
Another fundamental question is this: We have seen that every regular expression is associated with some language. The reason this trick only works for finite languages is that an infinite language would become a regular expression that is infinitely long. We can think of it as arbitrary double letter arbitrary Let us now ask.
A a b ab ba aba bab abab baba. The first term above clearly represents all words that have at least three a's in them. One obvious fact is that all the words in the language of E must have at least two a's in them.
But since all strings with three or more a's are themselves already strings with two or more a's. Let us break up the middle plus sign into its two cases: U It is possible by repeated application of the rules for forming regular expressions to produce an expression in which the star operator is applied to a subexpression that already has a star in it.
Some examples are: It cannot contain more than everything. The language defined by the regular expression on the right cannot contain any word with a double b.
It starts with an undoubled pair either ab or ba. All words with an even number of a's and an even number of b's belong to the language of E. If this were all we wanted to conclude. If the word ends before we find such a pair. The proof of this parallels our argument above. After this section of type 3 we could proceed with more sections of type. Continue scanning by skipping over the double a's and double b's that get in the way until we find the balancing unmatched pair ab or ba to even off the count of a's and b's.
We know that another undoubled pair will be coming up to balance off the initial one. If the first two letters are the same. All words in the language of E are made up of these three types of substrings and. If the section started with a ba. Scan over the doubled letter pairs until we come to an unmatched pair such as ab or ba. One property of this section of the word is that it has an even number of a's and an even number of b's.
Consider a word w with even a's and even b's. The total effect is that every word of the language of E contains an even number of a's and an even number of b's.
This must begin a substring of type 3.
If it started with a ba and ended with an ab. First we come to a double a type. Then perhaps we come upon a pair of letters that are not the same. Let us have only one flag called the type 3-flag.
This is another starred expression with a star inside. Once we have found the balancing unmatched pair. Consider them bookends or open and close parentheses. We start both flags at 0 and check to be sure they are both 0 at the end. If the flag starts at 0. The typical word here starts with some b's.
But there is another method that also works that uses only one flag-the method that corresponds to the discussion above. We will refer to this language again later. Every time an a is read. We are about to feed in a long string of a's and b's. We read the letters in two at a time. If they are the same. By "balancing" we do not mean it has to be the same unmatched pair: If it is 0 at the end.
Solutions by Chapter
Then we finish up with a final a or we leave the last b's as they are. Let us consider this as a computer algorithm. This method will work. One method is to keep two binary flags.
E represents the language of all strings with even a's and even b's. This means each word contains the substring aaa or the substring bbb but not both. All strings that have an even number of a's and an odd number of b's. This means that every clump of a's contains 3 or 6 or 9 or All strings in which the total number of a's is divisible by three.
All strings in which the letter b is never tripled. All strings that do not have both the substrings bba and abb. All strings that have an odd number of a's and an odd number of b's. This means that no word contains the substring bbb.
All words in which a appears tripled. Let ri. All words that contain exactly three b's in total. All words that contain exactly two b's or exactly three b's. This will be the same as "proving a distributive law" for regular expressions. Show that the language associated with r. All words that contain at least one of the strings s. All words in which a is tripled or b is tripled.
We have defined the product of two sets of strings in general. Similarly we can define S3. If we apply this to the case where both factors are the same set.
Arden Let R. Explain why we can take any pair of equivalent regular expressions and replace the letter a in both with any regular expression R and the letter b with any regular expression S and the resulting regular expressions will have the same language. Prove the following statements. After a certain number of rolls. Usually it is then some other child's turn to throw the dice and make his or her move.
Pieces are set up on a playing board. Depending on the number. The child has no options about changing the board. Everything is determined by the dice. The game changes from one state to another in a fashion determined by the input of a certain number. We call this a final state. We could eliminate the opponent and have the one child move first the white pieces and then the black. We should allow for the possibility that after a number is entered the game is still in the same state as it was before.
Let us look at all possible positions of the pieces on the board and call them states. For each possible number there is one and only one resulting state. Dice are thrown or a wheel is spun. There might be Whether or not the white pieces win the game is dependent entirely on what sequence of numbers is generated by the dice. A child has a simple computer input device.
Two machines are in the same state if their output pages look the same and their memories look the same cell by cell. Some sequences of input instructions may lead to success printing the 7 and some may not. In the first example. Nothing else. Printing the 7 is what is. Success is entirely determined by the sequence of inputs. If all goes well. Here the board is the computer and the different arrangements of pieces on the board correspond to the different arrangements of O's and l's in the cells of memory.
Let us put this game back on the shelf and take another example. No choice is involved. As in the case of the board-game. We can consider this process to be similar to the boardgame. We can then define a certain language as the set of strings of those letters that lead to success.
This is the language with words that are all programs that print a 7. The most general model. Each instruction is executed as soon as it is read. No knowledge is required of the state the machine was in six instructions ago. The computer is also deterministic. Beginning with the initial state which we presume to be unique some input sequences of numbers lead to victory for the first child and some do not.
One small difference between these two situations is that in the child's game the number of pieces of input is determined by whether either player has yet reached a final state whereas with the computer the number of pieces of input is a matter of choice made before run time.
Either the program will work or it won't. It is automatic involuntary and mechanical not willful.
The child writes a program. In Computer Theory these are also called halting states or terminal states or accepting states. We can then define a language to be the set of all words over this alphabet that lead to success. Instead of writing out the whole phrase "finite automaton" it is customary to refer to one by its initials. From state y and input a go to state x. Suppose that the input alphabet has only the two letters a and b.
The term FA is read by naming its letters. Even so. From state x and input a go to state y. It does not do anything like print output or play music. A finite set of transitions that tell for each state and for each letter of the input alphabet which state to go to next.
It works by being presented with an input string of letters that it reads letter by letter starting at the leftmost letter. Let the following be the rules of transition: From state x and input b go to state z. U The definition above is incomplete in the sense that it describes what a finite automation is but not how it works. Computer theory is rife with acronyms. The sequence ends when the last input letter has been read.
Throughout this chapter we use only this alphabet except for a couple of problems at the end. Beginning at the start state the letters determine a sequence of states. We present the precise definition below. Let us also assume that there are only three states.
An alphabet I of possible input letters. Some people prefer to call the object we have just defined a finite acceptor because its sole job is to accept certain input strings and run on them. A finite set of states. From state y and input b go to state z. State z is designated a final state. We use this expression often. Rule 4 now tells us to move to state z. The string aaa is not in the language of all strings that leave this FA in state z.
The fourth input letter is an a. Using other terminology. The third input is another a. There are no more input letters in the input string. This is solely a matter of standard usage. At the moment.
When we wish to be anthropomorphic. We may also say "aaa is rejected by this FA. Let the input be abba. This last term is good because the FA merely recognizes whether the input string is in its language much the same way we might recognize when we hear someone speak Russian without necessarily understanding what it means.
We now have a perfectly defined finite automaton. Let us examine what happens to various input strings when presented to this FA. We did not finish up in the final state state z. If language L1 is contained in language L 2 and a certain FA accepts L 2 all the words in L 2 are accepted and all the inputs accepted are words in L 2. We begin. Once we are in state y we read the second input letter. Rule 1 tells us that the first input letter. From state z and any input stay at state z.
The first letter of the string is an a. Rule 5 tells us to stay there. We say. The input string abba. Let us examine a different input string for this same FA.
The input string aaa is not in the language defined by this FA. Let us start with the string aaa. The set of all strings that do leave us in a final state is called the language defined by the finite automaton.
The third input letter is a b. Let us also designate state x as the starting state and state z as the only final state. The next input instruction is also an a. We may imagine that the state we are in at any given time is lit up and the others are dark. The string abba is therefore a word in the language associated with this FA. The FA above will accept all strings that have the letter b in them and no other strings. An FA running on an input string then looks like a pinball machine.
Once in state z. The entries inside the table are the new states that the FA moves into-the transition states. Each row of the table is the name of one of the states in the FA. It is much simpler to summarize them in a table format.
The transition table for the FA we have described is: If an input string is made up of only the letter a repeated some number of times. As soon as a b is encountered in the input string. The word abba is accepted by this FA. Even though it is no more than a table of symbols. There is a pictorial representation of an FA that gives us more of a feel for the motion.
We label these arrows with the corresponding alphabet letters. To get into state z. From each state we draw arrows showing to which other states the different letters of the input alphabet will lead us. From the table format it is hard to see the moving parts. It moves. When the input string runs out.
We begin by representing each state by a small circle drawn on a sheet of paper. It processes input. No such word can ever be accepted. This table has all the information necessary to define an FA.
Something goes from state to state as the input is read in and executed. It is not hard for us to predict which strings will be accepted by this FA.
The machine we have already defined by the transition list and the transition table can be depicted by the transition diagram: They are the map and the fuel needed for motion. This machine can also be depicted as: When we are out of letters we must stop. The letters of the input string dictate the directions of travel. The minus and plus signs.
If it is a final state. Let us look at this machine again and at the paths generated by the input strings aaaabba and bbaabbbb. We can indicate the start state by labeling it with the word "start" or by a minus sign.
Every state has as many outgoing edges as there are letters in the alphabet. Graph Theory is an exciting subject in its own right. We borrow from Graph Theory the name directed edge or simply edge for the arrow between states.
It is possible for a state to have no incoming edges. An edge comes from one state and leads to another or the same. There are machines for which it is not necessary to give the states specific names. When we depict an FA as circles and arrows.
Introduction to Computer Theory by Daniel I. a. Cohen - Second Edition
The language accepted by this machine is the set of all strings except A. By convention we shall say that A starts in the start state and then ends right there on all FA's.
We save ourselves from drawing a second loop edge. This description. At first glance it looks as if this machine accepts everything. Let us consider some more simple examples of FA's. We start at the minus sign and proceed along the indicated edges until we are out of input letters. If the input string is the null string.
There is a small problem about understanding how it is possible for A ever to be an input string to an. Even though we do not have names for the states. The first letter of the input takes us to the right-hand state and.
If we are then at a plus sign. This may be either because the picture is in two separate components as with a. These are of two types: FA's that have no final states. We start at state x and. If the first letter is an a we go to the dead-end state z. The machine looks like this: This idea can be carried further to a five-state FA as below: A "dead-end state" is an informal way of describing a state that no string can leave once it has entered. Let us trace the word babbb.
The first letter b takes us to state 3. The third letter b takes us back to state 3. The last letter is an a that returns us to state 2 where we end. There are two ways to get to state 4 in this FA. If we read another a immediately. So when we are in state 2 we know we have just.
The first letter is an a.
Introduction to Computer Theory by Cohen
From them we can see that there is not a unique machine for a given language. From there the next letter. Once in state 4 we must stay in state 4. An a then takes us to state 2. The fourth letter is a b and that takes us to state 3 again. Now another b takes us to state 4. Let us input the string ababa.
The next letter. We may then ask the question. We begin at the start state 1. Let us consider some more examples of FA's. The only way to get to state 2 is by reading the input letter a either while. One is from state 2. Once in state 4. More precisely. The situation with bb is analogous. Once in state 3. This language.
Similarly with state 3. The first couple of states are only waiting states first two letters of input. Either way. To get to state 3 we need to read a b. In summary.Show More. Alternatively, we could define V by the expression: Let us look at this machine again and at the paths generated by the input strings aaaabba and bbaabbbb. The machine must have a start state. There are differences between the two machines such as the number of total states they have, but as language acceptors they are equivalent.
- NEW ENGLISH FILE INTERMEDIATE PLUS TEACHERS BOOK PDF
- ENGLISH FOR NEGOTIATING PDF
- THE PRINCESS BY JUDE DEVERAUX PDF
- ADVANCED JAVA BLACK BOOK PDF
- DICCIONARIO BIBLICO VINE EPUB
- WORLD BOOK DAY S
- BLACK SWAN SCRIPT PDF
- STEPHENIE MEYER TWILIGHT NEW MOON PDF
- EKA TELIYANE PDF
- ADVANCED JAVA BLACK BOOK BY SANTOSH KUMAR