## Construct a Deterministic Finite Automata (DFA) for the language 1*01 (11)*(0 U 1)*, - dfa

Construct a Deterministic Finite Automata (DFA) for the language where the set of all strings are of the form 1*01 (11)*(0 U 1)*, that contain 01 as a substring

Without providing the direct answer, you should know the building blocks to be able to arrive there. Given that you know how finite automata work (otherwise read 'Languages and Machines' by Sudkamp), the DFA has a transitation for every symbol in every state:
Unlike for non-deterministic finite automata or NFA that we encounter in the next section, for a DFA in each state q ∈ Q and for every symbol a ∈ Σ the next state,
which is the state δ(q, a), is determined by the transition function δ. 1
Note: if you are a visual thinker and you wonder how the diagrams in these books are constructed, here is an example visualisation.

## Related

### Make DFA of the set of all strings from {0,1} whose tenth symbol from the right end is a 1

What will be the transition diagram of the set of all strings from {0,1} whose tenth symbol from the right end is a 1 ? I know the regular expression is (0+1)*1(0+1)(0+1)(0+1)(0+1)(0+1)(0+1)(0+1)(0+1)(0+1) But I couldn't turn it into DFA.

I encountered this question in Introduction to Automata Theory, Languages, and Computations by John E. Hopcroft, Rajeev Motwani, Jeffrey D. Ullman. First, we need to get hold of the NFA here. This would have 11 states in linear formation with the start having a self loop with 0,1 and a transition to the next state with 1 (not 0). All the other 9 transitions from qi -> qi+1 would have transition for 0,1 both. The last state, q10 will be our accept state. That was fairly a straightforward approach for the NFA. Now if we convert this NFA to a DFA considering all 211 subsets of set of states of the NFA, we would get the solution. I will add the NFA Diagram when I get some time. And try to figure out an easy trick to solve the DFA without considering all the possibilities. You can look at Example 1.30 from Introduction to the Theory of Computation - M. Sipser - 3rd Edition at page 51. There they showed an NFA for strings having 1 at the third position from the end would have 4 states but that corresponding DFA would have 8 states. Not posting the screenshots for copyright issues.

DFA of the set of all strings from {0,1} whose tenth symbol from the right end is a ... the regular expression for given language is : (0+1)*1(0+1)(0+1)(0+1)(0+1)(0+1)(0+1)(0+1)(0+1)(0+1) Hope you got it..!!

### Partially Observable Markov Decision Process Optimal Value function

I understood how belief states are updated in POMDP. But in Policy and Value function section, in http://en.wikipedia.org/wiki/Partially_observable_Markov_decision_process I could not figure out how to calculate value of V*(T(b,a,o)) for finding optimal value function V*(b). I have read a lot of resources on the internet but none explain how to calculate this clearly. Can some one provide me with a mathematically solved example with all the calculations or provide me with a mathematically clear explanation.

You should check out this tutorial on POMDPs: http://cs.brown.edu/research/ai/pomdp/tutorial/index.html It includes a section about Value Iteration, which can be used to find an optimal policy/value function.

I try to use the same notation in this answer as Wikipedia. First I repeat the Value Function as stated on Wikipedia: V*(b) is the value function with the belief b as parameter. b contains the probability of all states s, which sum up to 1: r(b,a) is the reward for belief b and action a which has to be calculated using the belief over each state given the original reward function R(s,a): the reward for being in state s and having done action a. We can also write the function O in terms of states instead of belief b: this is the probability of having observation o given a belief b and action a. Note that O and T are probability functions. Finally the function τ(b,a,o) gives the new belief state b'=τ(b,a,o) given the previous belief b, action a and observation o. Per state we can calculate the new probability: Now the new belief b' can be used to calculate iteratively: V(τ(b,a,o)). The optimal value function can be approached by using for example Value Iteration which applies dynamic programming. Then the function is iteratively updated until the difference is smaller then a small value ε. There is a lot more information on POMDPs, for example: Sebastian Thrun, Wolfram Burgard, and Dieter Fox. 2005. Probabilistic Robotics (Intelligent Robotics and Autonomous Agents). The MIT Press. A brief introduction to reinforcement learning A POMDP Tutorial Reinforcement Learning and Markov Decision Processes

### NFA to DFA conversion = deterministic?

I am struggling a bit with the meaning of determinism and nondeterminism. I get the difference when it comes to automata, but I can't seem to find an answer for the following: Is a NFA to DFA transformation deterministic? If multiple DFAs can be constructed for the same regular language, does that mean that the result of a NFA to DFA transformation is not unique? And thus a nondeterministic algorithm? I'm happy with any information you guys might be able to provide. Thanks in advance!

There are two different concepts at play here. First, you are correct that there can be many different DFAs equivalent to the same NFA, just as there can be many NFAs that are all equivalent to one another. Independently, there are several algorithms for converting an NFA into a DFA. The standard algorithm taught in most introductory classes on formal languages is the subset construction (also called the powerset construction). That algorithm is deterministic - there's a specific sequence of steps to follow to convert an NFA to a DFA, and accordingly you'll always get back the same DFA whenever you feed in the same NFA. You could conceivably have a nondeterministic algorithm for converting an NFA to a DFA, where the algorithm might produce one of many different DFAs as output, but to the best of my knowledge there aren't any famous algorithms of this sort. Hope this helps!

DFA- means deterministic finite automata Where as NFA- means non deterministic finite automata.. In dfa for every state there is a transition for both the inputs... I we have...{a, b} are the inputs for the given question.. For.. Every state there is a transition for both a and b... That automata is known as deterministic finite automata.. Where as in NDA we need not to have both input transitions for every state... At least one transition... is sufficient... In NFA Epsilon transition is also accepted.. And dead state is also accepted... In nfa... No of states required is less.. When compare to dfa.. Every dfa is equivalent to nfa... But every dfa is not equivalent to nfa...

### How can I quickly do a string matching on many regex keys?

I have an array of elements whose key is a regex. I would like to come up with a fast algorithm that given a string (not a regex) will return in less than O(N) time what are the matching array values based on execution of the key regex. Currently I do a linear scan of the array, for each element I execute the respective regex with posix regexec API, but this means that to find the matching elements I have to search across the whole array. I understand if the aray was composed by only simple strings as key, I could have kept it orderer and use a bsearch style API, but with regex looks like is not so easy. Am I missing something here? Example follows // this is mainly to be considered // as pseudocode typedef struct { regex_t r; ... some other data } value; const char *key = "some/key"; value my_array[1024]; bool my_matches[1024]; for(int i =0; i < 1024; ++i) { if(!regexec(&my_array[i].r, key, 0, 0, REG_EXTENDED)) my_matches[i] = 1; else my_matches[i] = 0; } But the above, as you can see, is linear. Thanks Addendum: I've put together a simple executable which executed above algorithm and something proposed in below answer, where form a large regex it build a binary tree of sub-regex and navigates it to find all the matches. Source code is here (GPLv3): http://qpsnr.youlink.org/data/regex_search.cpp Compile with: g++ -O3 -o regex_search ./regex_search.cpp -lrt And run with: ./regex_search "a/b" (or use --help flag for options) Interestingly (and I would say as expected) when searching in the tree, it takes less number of regex to execute, but these are far more complex to run for each comparison, so eventually the time it takes balances out with the linear scan of vectors. Results are printed on std::cerr so you can see that those are the same. When running with long strings and/or many token, watch out for memory usage; be ready to hit Ctrl-C to stop it from preventing your system to crash.

This is possible but I think you would need to write your own regex library to achieve it. Since you're using posix regexen, I'm going to assume that you intend to actually use regular expressions, as opposed to the random collection of computational features which modern regex libraries tend to implement. Regular expressions are closed under union (and many other operations), so you can construct a single regular expression from your array of regular expressions. Every regular expression can be recognized by a DFA (deterministic finite-state automaton), and a DFA -- regardless of how complex -- recognizes (or fails to recognize) a string in time linear to the length of the string. Given a set of DFAs, you can construct a union DFA which recognizes the languages of all DFAs, and furthermore (with a slight modification of what it means for a DFA to accept a string), you can recover the information about which subset of the DFAs matched the string. I'm going to try to use the same terminology as the Wikipedia article on DFAs. Let's suppose we have a set of DFAs M = {M1...Mn} which share a single alphabet Σ. So we have: Mi = (Qi, Σ, δi, qi0, Fi) where Qi = {qij} for 0 ≤ j < |Qi|, and Qi ⊂ Fi. We construct the union-DFA M⋃ = (Q⋃, Σ, δ⋃, q⋃0) (yes, no F; I'll get to that) as follows: q⋃0 = <q10,...,qn0> δ⋃(<q1j1,...,qnjn>, α) = <δ1(q1j1, α),... , δn(qnjn, α)> for each α ∈ Σ Q⋃ consists of all states reachable through δ⋃ starting from q⋃0. We can compute this using a standard closure algorithm in time proportional to the product of the sizes of the δi transition functions. Now to do a union match on a string α1...αm, we run the union DFA in the usual fashion, starting with its start symbol and applying its transition function to each α in turn. Once we've read the last symbol in the string, the DFA will be in some state <q1j1,...,qnjn>. From that state, we can extract the set of Mi which would have matched the string as: {Mi | qiji ∈ Fi}. In order for this to work, we need the individual DFAs to be complete (i.e., they have a transition from every state on every symbol). Some DFA construction algorithms produce DFAs which are lacking transitions on some symbols (indicating that no string with that prefix is in the language); such DFAs must be augmented with a non-accepting "sink" state which has a transition to itself on every symbol. I don't know of any regex library which exposes its DFAs sufficiently to implement the above algorithm, but it's not too much work to write a simple regex library which does not attempt to implement any non-regular features. You might also be able to find a DFA library. Constructing a DFA from a regular expression is potentially exponential in the size of the expression, although such cases are rare. (The non-deterministic FA can be constructed in linear time, but in some cases, the powerset construction on the NFA will require exponential time and space. See the Wikipedia article.) Once the DFAs are constructed, however, the union FA can be constructed in time proportional to the product of the sizes of the DFAs. So it should be easy enough to allow dynamic modification to the set of regular expressions, by compiling each regex to a DFA once, and maintaining the set of DFAs. When the set of regular expressions changes, it is only necessary to regenerate the union DFA. Hope that all helps.

### NFA to DFA conversion whose language is the complement of L(A)

Can someone please help me with this question? Describe an algorithm that converts an NFA into a DFA whose language is the complement of L(A). The complement should be taken with respect to the alphabet of A. Given an informal argument for why your construction works. You need not give a formal proof. Any kind of guidance is appreciated...

You can convert a FA to its complement by turning its accept states into non-accept states, and turning its non-accept states into accept states. Easy! You can convert a NFA to a DFA by considering that any NFA state is a power of the states: that is, for each state in the NFA, it is either active or not active. You can map each of these states to a state in a DFA, so you end up with at most 2|Q| states for your DFA which represents your NFA. Edit: this algorithm and its proof do not actually need the details of A, so long as it is a valid Finite State Automaton.