I came across a statement postulating that it's undecidable whether a TM overwrites any of its own input.
What would be intuition as well as an actual proof for that?
PROOF:
Build a machine B that takes as input a machine A, and simulates it without disturbing the input string (the string describing A). This is not difficult.
Now build machine C, a modified version of B. If A ever halts, C will overwrite the input string; until then it will leave the input string untouched.
In order to decide whether C (acting on A) ever overwrites its input string, one must decide whether A ever halts. But "does A halt" is undecidable, therefore so is "does C overwrite its input".
INTUITION:
With Turing machines, almost anything that isn't easy is impossible.
Draw the diagram of a two tape Non deterministic Turing Machine M that decides the language
L={w∈Σ* | w=uuu ∈Σ* }
if i could get help explaining the steps how to construct the NDTM (linguistically), I believe I could draw the diagram but I couldnt come out with an answer..
thank you
By u*u*u (viewed in the edit history), I presume what you intend is the language of all words of the form u^3 (u repeated three times) where u is any string over the alphabet.
Our NDTM needs to accept strings in the language in at least one way, and it must never accept anything not in the language. In particular, the key is that an NDTM can reject strings in the language, as long as some path through the NDTM does accept every string in the language.
Given that, our first step can be do guess about the length of u. The NDTM can mark three tape symbols (say, by writing versions of the symbols that are underlined) by nondeterministically transitioning from state q0 to q1 then q2 at arbitrary points while scanning right. Then, we can reset the tape head and use a deterministic TM to answer the question: did the split we guessed in the first step result in a string of the form u^3?
This is deterministic since we know the delineation of parts. We can check the first two parts (say, by bouncing back ad forth and marking symbols we've already processed), and then the second two parts (using the same technique, but applied to the 2nd and 3rd parts).
We have reduced the problem to that of checking whether a string is of the form w|w where we know the split. This deterministic TM is easier to come up with. When we put it after the NDTM that guesses about how to split up the initial input, we get a NDTM that can (and for exactly one guess, does) accept any string of the form u^3, but cannot possibly accept anything else. This is what we were after and we are done.
I was looking at a problem in K&R (Exercise 1-18), which asked to remove any trailing blanks or tabs. That pushed me to think about text messengers like Whatsapp. The thing is lets say I am writing a word Parochial, then the moment I had just written paro, it shows parochial as options, I click on that replaces the entire word (even if the spelling is wrong written by me, it replaces when I chose an option).
What I am thinking is the pointer goes back to the starting of the word or say that with start of every new word when I am writing, the pointer gets fixed to the 1st letter & if I choose some option it replaces that entire word in the stream (don't know if I'm thinking in the right direction).
I can use getchar() to point at the next letter but how do I:
1: Go backward from the current position of the pointer pointing the stream?
(By using fseek())?
2: How to fix a pointer a position in an I/o stream, so that I can fix it at the beginning of a new word.
Please tell me my approach is correct or understanding of some different concept is needed. Thanks in advance
Standard streams are mainly for going forward*, minimizing the number of IO system calls, and for avoiding the need to keep large files in memory at once.
A GUI app is likely to want to keep all of its display output in memory, and when you have the whole thing in memory, going back and forth is just a simple mater of incrementing and decrementing pointers or indices.
*(random seeks aren't always optimal and they limit you from doing IO on nonseekable files such as pipes or sockets)
I feel like this is a pretty common problem but I wasn't really sure what to search for.
I have a large file (so I don't want to load it all into memory) that I need to parse control strings out of and then stream that data to another computer. I'm currently reading in the file in 1000 byte chunks.
So for example if I have a string that contains ASCII codes escaped with ('$' some number of digits ';') and the data looked like this... "quick $33;brown $126;fox $a $12a". The string going to the other computer would be "quick brown! ~fox $a $12a".
In my current approach I have the following problems:
What happens when the control strings falls on a buffer boundary?
If the string is '$' followed by anything but digits and a ';' I want to ignore it. So I need to read ahead until the full control string is found.
I'm writing this in straight C so I don't have streams to help me.
Would an alternating double buffer approach work and if so how does one manage the current locations etc.
If I've followed what you are asking about it is called lexical analysis or tokenization or regular expressions. For regular languages you can construct a finite state machine which will recognize your input. In practice you can use a tool that understands regular expressions to recognize and perform different actions for the input.
Depending on different requirements you might go about this differently. For more complicated languages you might want to use a tool like lex to help you generate an input processor, but for this, as I understand it, you can use a much more simple approach, after we fix your buffer problem.
You should use a circular buffer for your input, so that indexing off the end wraps around to the front again. Whenever half of the data that the buffer can hold has been processed you should do another read to refill that. Your buffer size should be at least twice as large as the largest "word" you need to recognize. The indexing into this buffer will use the modulus (remainder) operator % to perform the wrapping (if you choose a buffer size that is a power of 2, such as 4096, then you can use bitwise & instead).
Now you just look at the characters until you read a $, output what you've looked at up until that point, and then knowing that you are in a different state because you saw a $ you look at more characters until you see another character that ends the current state (the ;) and perform some other action on the data that you had read in. How to handle the case where the $ is seen without a well formatted number followed by an ; wasn't entirely clear in your question -- what to do if there are a million numbers before you see ;, for instance.
The regular expressions would be:
[^$]
Any non-dollar sign character. This could be augmented with a closure ([^$]* or [^$]+) to recognize a string of non$ characters at a time, but that could get very long.
$[0-9]{1,3};
This would recognize a dollar sign followed by up 1 to 3 digits followed by a semicolon.
[$]
This would recognize just a dollar sign. It is in the brackets because $ is special in many regular expression representations when it is at the end of a symbol (which it is in this case) and means "match only if at the end of line".
Anyway, in this case it would recognize a dollar sign in the case where it is not recognized by the other, longer, pattern that recognizes dollar signs.
In lex you might have
[^$]{1,1024} { write_string(yytext); }
$[0-9]{1,3}; { write_char(atoi(yytext)); }
[$] { write_char(*yytext); }
and it would generate a .c file that will function as a filter similar to what you are asking for. You will need to read up a little more on how to use lex though.
The "f" family of functions in <stdio.h> can take care of the streaming for you. Specifically, you're looking for fopen(), fgets(), fread(), etc.
Nategoose's answer about using lex (and I'll add yacc, depending on the complexity of your input) is also worth considering. They generate lexers and parsers that work, and after you've used them you'll never write one by hand again.
I saw this question, and was curious as to what the pumping lemma was (Wikipedia didn't help much).
I understand that it's basically a theoretical proof that must be true in order for a language to be in a certain class, but beyond that I don't really get it.
Anyone care to try to explain it at a fairly granular level in a way understandable by non mathematicians/comp sci doctorates?
The pumping lemma is a simple proof to show that a language is not regular, meaning that a Finite State Machine cannot be built for it. The canonical example is the language (a^n)(b^n). This is the simple language which is just any number of as, followed by the same number of bs. So the strings
ab
aabb
aaabbb
aaaabbbb
etc. are in the language, but
aab
bab
aaabbbbbb
etc. are not.
It's simple enough to build a FSM for these examples:
This one will work all the way up to n=4. The problem is that our language didn't put any constraint on n, and Finite State Machines have to be, well, finite. No matter how many states I add to this machine, someone can give me an input where n equals the number of states plus one and my machine will fail. So if there can be a machine built to read this language, there must be a loop somewhere in there to keep the number of states finite. With these loops added:
all of the strings in our language will be accepted, but there is a problem. After the first four as, the machine loses count of how many as have been input because it stays in the same state. That means that after four, I can add as many as as I want to the string, without adding any bs, and still get the same return value. This means that the strings:
aaaa(a*)bbbb
with (a*) representing any number of as, will all be accepted by the machine even though they obviously aren't all in the language. In this context, we would say that the part of the string (a*) can be pumped. The fact that the Finite State Machine is finite and n is not bounded, guarantees that any machine which accepts all strings in the language MUST have this property. The machine must loop at some point, and at the point that it loops the language can be pumped. Therefore no Finite State Machine can be built for this language, and the language is not regular.
Remember that Regular Expressions and Finite State Machines are equivalent, then replace a and b with opening and closing Html tags which can be embedded within each other, and you can see why it is not possible to use regular expressions to parse Html
It's a device intended to prove that a given language cannot be of a certain class.
Let's consider the language of balanced parentheses (meaning symbols '(' and ')', and including all strings that are balanced in the usual meaning, and none that aren't). We can use the pumping lemma to show this isn't regular.
(A language is a set of possible strings. A parser is some sort of mechanism we can use to see if a string is in the language, so it has to be able to tell the difference between a string in the language or a string outside the language. A language is "regular" (or "context-free" or "context-sensitive" or whatever) if there is a regular (or whatever) parser that can recognize it, distinguishing between strings in the language and strings not in the language.)
LFSR Consulting has provided a good description. We can draw a parser for a regular language as a finite collection of boxes and arrows, with the arrows representing characters and the boxes connecting them (acting as "states"). (If it's more complicated than that, it isn't a regular language.) If we can get a string longer than the number of boxes, it means we went through one box more than once. That means we had a loop, and we can go through the loop as many times as we want.
Therefore, for a regular language, if we can create an arbitrarily long string, we can divide it into xyz, where x is the characters we need to get to the start of the loop, y is the actual loop, and z is whatever we need to make the string valid after the loop. The important thing is that the total lengths of x and y are limited. After all, if the length is greater than the number of boxes, we've obviously gone through another box while doing this, and so there's a loop.
So, in our balanced language, we can start by writing any number of left parentheses. In particular, for any given parser, we can write more left parens than there are boxes, and so the parser can't tell how many left parens there are. Therefore, x is some amount of left parens, and this is fixed. y is also some number of left parens, and this can increase indefinitely. We can say that z is some number of right parens.
This means that we might have a string of 43 left parens and 43 right parens recognized by our parser, but the parser can't tell that from a string of 44 left parens and 43 right parens, which isn't in our language, so the parser can't parse our language.
Since any possible regular parser has a fixed number of boxes, we can always write more left parens than that, and by the pumping lemma we can then add more left parens in a way that the parser can't tell. Therefore, the balanced parenthesis language can't be parsed by a regular parser, and therefore isn't a regular expression.
Its a difficult thing to get in layman's terms, but basically regular expressions should have a non-empty substring within it that can be repeated as many times as you wish while the entire new word remains valid for the language.
In practice, pumping lemmas are not sufficient to PROVE a language correct, but rather as a way to do a proof by contradiction and show a language does not fit in the class of languages (Regular or Context-Free) by showing the pumping lemma does not work for it.
Basically, you have a definition of a language (like XML), which is a way to tell whether a given string of characters (a "word") is a member of that language or not.
The pumping lemma establishes a method by which you can pick a "word" from the language, and then apply some changes to it. The theorem states that if the language is regular, these changes should yield a "word" that is still from the same language. If the word you come up with isn't in the language, then the language could not have been regular in the first place.
The simple pumping lemma is the one for regular languages, which are the sets of strings described by finite automata, among other things. The main characteristic of a finite automation is that it only has a finite amount of memory, described by its states.
Now suppose you have a string, which is recognized by a finite automaton, and which is long enough to "exceed" the memory of the automation, i.e. in which states must repeat. Then there is a substring where the state of the automaton at the beginning of the substring is the same as the state at the end of the substring. Since reading the substring doesn't change the state it may be removed or duplicated an arbitrary number of times, without the automaton being the wiser. So these modified strings must also be accepted.
There is also a somewhat more complicated pumping lemma for context-free languages, where you can remove/insert what may intuitively be viewed as matching parentheses at two places in the string.
By definition regular languages are those recognized by a finite state automaton. Think of it as a labyrinth : states are rooms, transitions are one-way corridors between rooms, there's an initial room, and an exit (final) room. As the name 'finite state automaton' says, there is a finite number of rooms. Each time you travel along a corridor, you jot down the letter written on its wall. A word can be recognized if you can find a path from the initial to the final room, going through corridors labelled with its letters, in the correct order.
The pumping lemma says that there is a maximum length (the pumping length) for which you can wander through the labyrinth without ever going back to a room through which you have gone before. The idea is that since there are only so many distinct rooms you can walk in, past a certain point, you have to either exit the labyrinth or cross over your tracks. If you manage to walk a longer path than this pumping length in the labyrinth, then you are taking a detour : you are inserting a(t least one) cycle in your path that could be removed (if you want your crossing of the labyrinth to recognize a smaller word) or repeated (pumped) indefinitely (allowing to recognize a super-long word).
There is a similar lemma for context-free languages. Those languages can be represented as word accepted by pushdown automata, which are finite state automata that can make use of a stack to decide which transitions to perform. Nonetheless, since there is stilla finite number of states, the intuition explained above carries over, even through the formal expression of the property may be slightly more complex.
In laymans terms, I think you have it almost right. It's a proof technique (two actually) for proving that a language is NOT in a certain class.
Fer example, consider a regular language (regexp, automata, etc) with an infinite number of strings in it. At a certain point, as starblue said, you run out of memory because the string is too long for the automaton. This means that there has to be a chunk of the string that the automaton can't tell how many copies of it you have (you're in a loop). So, any number of copies of that substring in the middle of the string, and you still are in the language.
This means that if you have a language that does NOT have this property, ie, there is a sufficiently long string with NO substring that you can repeat any number of times and still be in the language, then the language isn't regular.
For example, take this language L = anbn.
Now try to visualize finite automaton for the above language for some n's.
if n = 1, the string w = ab. Here we can make a finite automaton with out looping
if n = 2, the string w = a2b2. Here we can make a finite automaton with out looping
if n = p, the string w = apbp. Essentially a finite automaton can be assumed with 3 stages.
First stage, it takes a series of inputs and enter second stage. Similarly from stage 2 to stage 3. Let us call these stages as x, y and z.
There are some observations
Definitely x will contain 'a' and z will contain 'b'.
Now we have to be clear about y:
case a: y may contain 'a' only
case b: y may contain 'b' only
case c: y may contain a combination of 'a' and 'b'
So the finite automaton states for stage y should be able to take inputs 'a' and 'b' and also it should not take more a's and b's which cannot be countable.
If stage y is taking only one 'a' and one 'b', then there are two states required
If it is taking two 'a' and one 'b', three states are required with out loops
and so on....
So the design of stage y is purely infinite. We can only make it finite by putting some loops and if we put loops, the finite automaton can accept languages beyond L = anbn. So for this language we can't construct a finite automaton. Hence it is not regular.
This is not an explanation as such but it is simple.
For a^n b^n our FSM should be built in such a way that b must know the number of a's already parsed and will accept the same n number of b's. A FSM can not simply do stuff like that.