Related
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´m searching information about how to compare two codes and decide if the code submitted by someone is correct or not (based on a solution code defined before).
I could compare the output but many codes may have the same output. Then I think I must compare someway the codes and give a percentage of similitude.
Anybody can help me?
(the language code is C but I think this isn´t important)
Some of my teachers used online automated program grading systems like http://web-cat.org/
In the assignment they would specify a public api you must provide, and then they would just write tests against your functions, much like unit tests. They would intentionally pick tests that would exploit boundary conditions and other things students are notorious for not thinking about, and just call your code with many different inputs to try to get your code to fail.
Sometimes they would hardcode the expected values, other times they would allow values within a range, and other times they just did the assignment themselves and made it so your own code has to match the results produced by their code.
Obviously, not all programs can be effectively graded this way. It's also kinda error prone in that sometimes even the teacher made a mistake and overflowed an int or something, then the correct student submissions wouldn't match the teachers incorrect results. But, a system doesn't need to be perfect to be useful. But I think this raises an important point in that manually grading by reading the code won't necessarily reveal all mistakes either.
Another possibility is copy the submitted code, strip out all of the white space and search for substrings that must exist for the code to be correct and/or substrings that cannot exist for the code to be considered correct. The troublesome bit might be setting up to allow for some of the more tricky requirements such as [(a or c),((a or b) and c),((a or b) and c)], where the variables are the result of a boolean check as to if the substring related to the variable exists within the code.
For example, [("printf"),("for"), (not "1,2,3,4,5,6,7,9,10")], would require that "printf" and "for" be substrings in the code, while "1,2,3,4,5,6,7,9,10" i I'm not familiar with C, so I'm I'm assuming here that "printf" is required to be able to print anything without involving output streams, which could be accounted for by something like [("printf" or "out"),("for"), (not "1,2,3,4,5,6,7,9,10")], where "out" is part of C code required to make use of output streams.
It might be possible to automatically find required substrings based on a "correct" code, but as others have mentioned, there are alternative ways to do things. Which is why hard-coding the "solution" is probably required. Even so, it's quite possible that you'll miss a required substring, and it'll be marked as wrong, but it's probably the only way you can do what you ask with some degree of success.
Regular expressions might be useful here.
Say I have a C program which wants to call a very simple Lua function with two strings (let's say two comma separated lists, returning true if the lists intersect at all, false if not).
The obvious way to do this is to push them onto the stack with lua_pushstring, which works fine, however, from the doc it looks like lua_pushstring but makes a copy of the string for Lua to work with.
That means that to cross over to the Lua function is going to require two string copies which I could avoid by rewriting the Lua function in C. Is there any way to arrange things so that the existing C strings could be reused on the Lua side for the sake of performance (or would the strcpy cost pale into insignificance anyway)?
From my investigation so far (my first couple of hours looking seriously at Lua), lite userdata seems like the sort of thing I want, but in the form of a string.
No. You cannot forbid Lua making a copy of the string when you call lua_pushstring().
Reason: Unless, the internal garbage collector would not be able to free unused memory (like your 2 input strings).
Even if you use the light user data functionality (which would be an overkill in this case), you would have to use lua_pushstring() later, when the Lua program asks for the string.
Hmm.. You could certainly write some C functions so that the work is being done on the C side, but as the other answer points out, you might get stuck pushing the string or sections of it in anyways.
Of note: Lua only stores strings once when they are brought in. i.e.: if I have 1 string containing "The quick brown fox jumps over the lazy dog" and I push it into Lua, and there are no other string objects that contain that string, it will make a new copy of it. If, on the other hand, I've already inserted it, you'll just get a pointer to that first string so equality checks are pretty cheap. Importing those strings can be a little expensive, I would guess, if this is done at a high frequency, however comparisons, again, are cheap.
I would try profiling what you're implementing and see if the performance is up to your expectations or not.
As the Lua Performance Tips document (which I recommend reading if you are thinking about maximizing performance with Lua), the two programming maxims related to optimizing are:
Rule #1: Don’t do it.
Rule #2: Don’t do it yet. (for experts only)
(in c90) (linux)
input:
sqrt(2 - sin(3*A/B)^2.5) + 0.5*(C*~(D) + 3.11 +B)
a
b /*there are values for a,b,c,d */
c
d
input:
cos(2 - asin(3*A/B)^2.5) +cos(0.5*(C*~(D)) + 3.11 +B)
a
b /*there are values for a,b,c,d */
c
d
input:
sqrt(2 - sin(3*A/B)^2.5)/(0.5*(C*~(D)) + sin(3.11) +ln(B))
/*max lenght of formula is 250 characters*/
a
b /*there are values for a,b,c,d */
c /*each variable with set of floating numbers*/
d
As you can see infix formula in the input depends on user.
My program will take a formula and n-tuples value.
Then it calculate the results for each value of a,b,c and d.
If you wonder I am saying ;outcome of program is graph.
/sometimes,I think i will take input and store in string.
then another idea is arise " I should store formula in the struct"
but ı don't know how I can construct
the code on the base of structure./
really, I don't know way how to store the formula in program code so that
I can do my job.
can you show me?
/* a,b,c,d is letters
cos,sin,sqrt,ln is function*/
You need to write a lexical analyzer to tokenize the input (break it into its component parts--operators, punctuators, identifiers, etc.). Inevitably, you'll end up with some sequence of tokens.
After that, there are a number of ways to evaluate the input. One of the easiest ways to do this is to convert the expression to postfix using the shunting yard algorithm (evaluation of a postfix expression is Easy with a capital E).
You should look up "abstract syntax trees" and "expression trees" as well as "lexical analysis", "syntax", "parse", and "compiler theory". Reading text input and getting meaning from it is quite difficult for most things (though we often try to make sure we have simple input).
The first step in generating a parser is to write down the grammar for your input language. In this case your input language is some Mathematical expressions, so you would do something like:
expr => <function_identifier> ( stmt )
( stmt )
<variable_identifier>
<numerical_constant>
stmt => expr <operator> stmt
(I haven't written a grammar like this {look up BNF and EBNF} in a few years so I've probably made some glaring errors that someone else will kindly point out)
This can get a lot more complicated depending on how you handle operator precedence (multiply and device before add and subtract type stuff), but the point of the grammar in this case is to help you to write a parser.
There are tools that will help you do this (yacc, bison, antlr, and others) but you can do it by hand as well. There are many many ways to go about doing this, but they all have one thing in common -- a stack. Processing a language such as this requires something called a push down automaton, which is just a fancy way of saying something that can make decisions based on new input, a current state, and the top item of the stack. The decisions that it can make include pushing, popping, changing state, and combining (turning 2+3 into 5 is a form of combining). Combining is usually referred to as a production because it produces a result.
Of the various common types of parsers you will almost certainly start out with a recursive decent parser. They are usually written directly in a general purpose programming language, such as C. This type of parser is made up of several (often many) functions that call each other, and they end up using the system stack as the push down automaton stack.
Another thing you will need to do is to write down the different types of words and operators that make up your language. These words and operators are called lexemes and represent the tokens of your language. I represented these tokens in the grammar <like_this>, except for the parenthesis which represented themselves.
You will most likely want to describe your lexemes with a set of regular expressions. You should be familiar with these if you use grep, sed, awk, or perl. They are a way of describing what is known as a regular language which can be processed by something known as a Finite State Automaton. That is just a fancy way of saying that it is a program that can make a decision about changing state by considering only its current state and the next input (the next character of input). For example part of your lexical description might be:
[A-Z] variable-identifier
sqrt function-identifier
log function-identifier
[0-9]+ unsigned-literal
+ operator
- operator
There are also tools which can generate code for this. lex which is one of these is highly integrated with the parser generating program yacc, but since you are trying to learn you can also write your own tokenizer/lexical analysis code in C.
After you have done all of this (it will probably take you quite a while) you will need to have your parser build a tree to represent the expressions and grammar of the input. In the simple case of expression evaluation (like writing a simple command line calculator program) you could have your parser evaluate the formula as it processed the input, but for your case, as I understand it, you will need to make a tree (or Reverse Polish representation, but trees are easier in my opinion).
Then after you have read the values for the variables you can traverse the tree and calculate an actual number.
Possibly the easiest thing to do is use an embedded language like Lua or Python, for both of which the interpreter is written in C. Unfortunately, if you go the Lua route you'll have to convert the binary operations to function calls, in which case it's likely easier to use Python. So I'll go down that path.
If you just want to output the result to the console this is really easy and you won't even have to delve too deep in Python embedding. Since, then you only have to write a single line program in Python to output the value.
Here is the Python code you could use:
exec "import math;A=<vala>;B=<valb>;C=<valc>;D=<vald>;print <formula>".replace("^", "**").replace("log","math.log").replace("ln", "math.log").replace("sin","math.sin").replace("sqrt", "math.sqrt").replace("cos","math.cos")
Note the replaces are done in Python, since I'm quite sure it's easier to do this in Python and not C. Also note, that if you want to use xor('^') you'll have to remove .replace("^","**") and use ** for powering.
I don't know enough C to be able to tell you how to generate this string in C, but after you have, you can use the following program to run it:
#include <Python.h>
int main(int argc, char* argv[])
{
char* progstr = "...";
Py_Initialize();
PyRun_SimpleString(progstr);
Py_Finalize();
return 0;
}
You can look up more information about embedding Python in C here: Python Extension and Embedding Documentation
If you need to use the result of the calculation in your program there are ways to read this value from Python, but you'll have to read up on them yourself.
Also, you should review your posts to SO and other posts regarding Binary Trees. Implement this using a tree structure. Traverse as infix to evaluate. There have been some excellent answers to tree questions.
If you need to store this (for persistance as in a file), I suggest XML. Parsing XML should make you really appreciate how easy your assignment is.
Check out this post:
http://blog.barvinograd.com/2011/03/online-function-grapher-formula-parser-part-2/
It uses ANTLR library for parsing math expression, this one specifically uses JavaScript output but ANTLR has many outputs such as Java, Ruby, C++, C# and you should be able to use the grammar in the post for any output language.
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.