## Enumerate Turing recognizable language in lexicographic order - turing-machines

So a Turing recognizable language is also enumerable.
But is it possible for an enumerator to "print" a Turing recognizable language in lexicographic order?

No. Suppose it were possible. Then the language could be decided as follows:
Begin enumerating all accepted strings in order.
If you list a string whose lexicographic order is greater than the string you're looking for, halt reject.
If you list the string you're looking for, halt accept.
This will eventually terminate for any input string since for any given input string only finitely many enumerated strings exist with smaller lexicographic order. It will always halt reject if it doesn't find your target string and it will always halt accept if it does. It doesn't have to worry about missing your string since it knows where your string should be, thanks to having listed strings in order.
Since we know there are enumerable languages that aren't decidable, it must be that we can't enumerate enumerable languages in order.

## Related

### Algorithm to replace matched patterns in string

What is a simple way to implement a find/replace algorithm on a string? I would like to transform a string using dictionary which defines replacement rules. The problem is that after each replacement, I must make sure that subsequent replacements operate on the original string. For example: My string is: ABCABCDEFDEF My rules are: ABC -> DEF and DEF -> XXX So my result should be: DEFDEFXXXXXX and not XXXXXXXXXXXX (which would be the result if I were to first apply rule one and then apply rule two).

Simple way: Starting at the first character, try each key if it occurs at that position. If you find a match, replace and continue with the character after the replacement Otherwise, continue at the next character To kep in mind: Ambiguities: If you have both "AB" and "ABC" as keys, you need to decide which should match "ABCD". Usually you want the longer string to match (otherwise, it would never match) Unicode: Normalize keys and original string first. That is certainly sufficient for a handful of keys. However, it's O(N*M) where N is the string length and M is the number of replacements. Improvements: do not search linearly for a match; instead use a sorted list of keys and do a binary search for the character in the original string, then the next etc. Indeed, it might be beneficial to only remember the position and key of found matches in the first pass, and do the replacements in a second pass for large strings with many replacements, it is usually better to build a new string Use Aho-Corasick for search. This exploits the limited search spacce (i.e knowledge derived from the list of keywords) to avoid probing every character of the source string

Depending on the language you are using there might be so predesigned functions. String.Replace might be helpful if you are using C#. This would save you very much time. If you are still looking for algorithms which can find patterns within other strings, the Horspool-algorithm might be what you are looking for. You would still have to implement the logic for subsequent replacements to operate on the original string. But this sounds like not a hard thing to do.

### What is the Pumping Lemma in Layman's terms?

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.

### Turing machine decidability ambiguous cases

1) Is a Turing machine M that accepts the language L = {ε}, accepting no entry? In one hand, I think it can be false because the empty word could be an entry, but in another i think this could possibly be an indecidable problem. 2) Is every Turing machine whose language is decidable stops on any input ? Same idea, intuitively I would have say yes, due to the definition of decidable, but I don't know, something trouble me. 3) Is the language of the palindromes decidable whatever the aphabet ? For this one, I have almost no doubt that it's False, because with Rice's Theorem we can prove that, this probleme is indecidable.

1) I am not sure how to parse this but if a TM accepts the language consisting only of the empty set, it will eventually halt-accept on a blank tape. Whether that counts as an entry or not depends on your definition of "entry". I would count it as an entry, so I would answer "no". 2) The language consisting of only the empty string is decidable. However, we can write a TM that halt-accepts the empty string only and goes into an infinite loop for all other inputs. What is meant by "whose language" is debatable but for TMs that encode partial functions I would call the language of that TM the set of strings which it halt-accepts on, so I would answer "no". 3) It seems to me that, given an alphabet with n symbols, you can always construct a single-tape deterministic TM with O(n) states which halt-accepts on palindromes over that alphabet and halt-rejects other strings, thus deciding the language of palindromes over the alphabet. I would answer "yes", as long as the terms have their usual meanings. Note that Rice's theorem does not apply; it would apply to the problem of deciding whether a TM accepts the language of palindromes over an alphabet, but actually deciding whether something is a palindrome is of course possible (PDAs do it).

### Proving a language is in RE/R/coRE

Lets say we have a function for a Turing Machine like this one: 𝑓(𝑀) = { 1, for any 𝑤 where 𝑀(𝑤) halts only if w is a palindrome of even length 0, otherwise How can one prove that it belongs (or not) to RE, R, coRE. I mean, I know we can use a turing reduction using the f-halt to prove that it is not belonging to R. But what about RE/coRE?

A language is RE if we can halt-accept for any string in the language. A language is coRE if we can halt-reject for any string not in the language. R is the intersection of RE and coRE; a language is R if we can halt-accept on strings in the language and halt-reject on strings not in the language. You already know that the language isn't R. This can also be seen by Rice's theorem: halting only on palindromes of even length is a semantic property of the accepted language (subset of EPAL), so the inclusion problem isn't decidable. This tells you that the language cannot be both RE and coRE, though it might be neither. Given a machine M, can we determine that it does only accept strings which are even-length palindromes? This seems unlikely. We would need a way to be sure that all strings - maybe infinitely many - are even-length palindromes. We can't just find a counterexample and be done. Given a machine M, can we determine that it doesn't only accept strings which are even-length palindromes? We sure can! We can interleave executions of copies of this machine such that arbitrarily many possible input strings get arbitrarily much computing time; if M accepts any particular string, we can eventually find out, and if it accepts one that isn't an even-length palindrome, we can eventually tell. So, this language: is coRE is not RE is not R

### caesar cipher check in ocaml

I want to implement a check function that given two strings s1 and s2 will check if s2 is the caesar cipher of s1 or not. the inter face needs to be looked like string->string->bool. the problem is that I am not allowed to use any string functions other than String.length, so how can I solve it? i am not permitted any list array, iterations. Only recursions and pattern matching. Please help me. And also can you tell me how I can write a substring function in ocaml other than the module function with the above restrictions?

My guess is that you are probably allowed to use s.[i] to get the ith character of string s. This is the same as String.get, but the instructor may not think of it in those terms. Without some form of getting the individual characters for the string, I believe that this is impossible. You should probably double check with your instructor to be sure, but I would be surprised if he had meant for you to be unable to separate a string into characters (which is something that you cannot do with pattern-matching alone in Ocaml). Once you can get individual characters, the way to do it should be pretty clear (you do not need substring to traverse each string recursively). If you still want to write substring, creating it would be complex since you don't have access to String.create or other similar functions. But you can write your own version of String.create using recursion, one character string literals (like "x"), the ability to set a character in a string to another (like s.[0] <- c), and string concatenation (s1 ^ s2). Again, of course, all of this is assuming that those operators are allowed to be used.