Article Categories
- All Categories
-
Data Structure
-
Networking
-
RDBMS
-
Operating System
-
Java
-
MS Excel
-
iOS
-
HTML
-
CSS
-
Android
-
Python
-
C Programming
-
C++
-
C#
-
MongoDB
-
MySQL
-
Javascript
-
PHP
Articles by Bhanu Priya
Page 84 of 106
C program for DFA accepting all strings over w ∈(a,b)* containing “aba” as a substring
ProblemDesign a DFA for the language L={w1abaw2 | w1, w2 Є(a, b)*}, which means the DFA accepts all strings which contain “aba” as a substring.SolutionThe strings that are accepted by language L= {aba, aabaa, aabab, babab, ababa, …….}Step 1 − Transition diagram for minimal string (starting string) −If w1 and w2 are null then the string it generates is “aba” because w1, w2 ε(a, b)*q0 is the initial state and q3 is the final state.Step 2 − The final DFA for the given language is as follows −Explanationqo is the initial state q0 on ‘a’ goes to q1 and on ...
Read MoreConstruct ∈-NFA of Regular Language L = (0+1)*(00+ 11)
The ε transitions in Non-deterministic finite automata (NFA) are used to move from one state to another without having any symbol from input set Σε-NFA is defined in five tuple{Q, q0, Σ, δ, F}Where, δ − Q × (Σ∪ε)→2QQ − Finite set of statesΣ − Finite set of the input symbolq0 − Initial stateF − Final stateδ − Transition functionNFA without ε transitionNFA is defined in 5 tuple representation{Q, q0, Σ, δ, F}Where, δ − Q X Σ→ 2QQ − Finite set of statesΣ, − Finite set of the input symbolq0 − Initial stateF − Final stateδ − Transition functionNFA ...
Read MoreC Program to construct a DFA which accepts L = {aN | N ≥ 1}
Let us take a string S of size N, we have to design a Deterministic Finite Automata (DFA) for accepting the language L = {aN | N ≥ 1}.The string accepting the language L is {a, aa, aaa, aaaaaaa…, }.Now the user has to enter a string, if that string is present in the given language, then print “entered string is Accepted”. Otherwise, print “entered string is not Accepted”.DFA transition diagram for the given language is −ExampleFollowing is the C program to construct DFA which accepts the language L = {aN | N ≥ 1} −#include int main() { char S[30]; ...
Read MoreC Program to construct DFA for Regular Expression (a+aa*b)*
Design a Deterministic Finite Automata (DFA) for accepting the language L = (a+aa*b)* If the given string is accepted by DFA, then print “string is accepted”. Otherwise, print “string is rejected”.Example 1Input: Enter Input String aaaba Output: String Accepted.Explanation − The given string is of the form (a+aa*b)* as the first character is a and it is followed by a or ab.Example 2Input: Enter Input String baabaab Output: String not Accepted.The DFA for the given regular expression (a+aa*b) is −Explanation −If the first character is always a, then traverse the remaining string and check ...
Read MoreHow to convert right linear grammar to left linear grammar?
For every finite automata (FA) there exists a regular grammar and for every regular grammar there is a left linear and right linear regular grammar.Example 1Consider a regular grammar − a(a+b)* A → aB B → aB|bB|eFor the given regular expression, the above grammar is right linear grammar.Now, convert the above right linear grammar to left linear grammar.The rule to follow for conversion is, Finite Automata → Right linearThe reverse of right linear →left linear grammar.So, A → BaB → Ba|Bb|eFinally for every right linear there is aExampleConsider a language {bnabma| n>=2, m>=2}The right linear grammar for the given language ...
Read MoreExplain about left linear regular grammar in TOC
Regular grammar describes a regular language. It consists of four components, which are as follows −G = (N, E, P, S)Where, N − finite set of non-terminal symbols, E − a finite set of terminal symbols, P − a set of production rules, each of one is in the formsS → aBS → aS → ∈, S ∈ N is the start symbol.The above grammar can be of two forms −Right Linear Regular GrammarLeft Linear Regular GrammarLinear GrammarWhen the right side of the Grammar part has only one terminal then it's linear else nonv linear.Left linear grammarIn a left-regular grammar ...
Read MoreExplain about right linear regular grammars in TOC
Regular grammar describes a regular language. It consists of four components, which are as follows −G = (N, E, P, S)Where, N − finite set of non-terminal symbols, E − a finite set of terminal symbols, P − a set of production rules, each of one is in the formsS → aBS → aS → ∈, S ∈ N is the start symbol.The above grammar can be of two forms −Right Linear Regular GrammarLeft Linear Regular GrammarLinear GrammarWhen the right side of the Grammar part has only one terminal then it's linear else non linear.Let’s discuss about right linear grammar ...
Read MoreProve that the vertex cover is NP complete in TOC
It is the subset(minimum size) of vertices of a graph G such that every edge in G incident to at least one vertex in G.Vertex Cover (VC) ProblemTo prove VC is NP-complete we have to prove the following −VC is Non-deterministic Polynomial (NP).A NPC problem can be reduced into VC.To prove VC is NP, find a verifier which is a subset of vertices which is VC and that can be verified in polynomial time. For a graph of n vertices it can be proved in O(n2). Thus, VC is NP.Now consider the “clique” problem which is NPC and reduce it ...
Read MoreProve that the Hamiltonian Path is NP-Complete in TOC
A Hamilton cycle is a round trip path along n edges of graph G which visits every vertex once and returns to its starting vertexExampleGiven below is an example of the Hamilton cycle path −Hamilton cycle path: 1, 2, 8, 7, 6, 5, 4, 3, 1TSP is NP-CompleteThe travelling salesman problem (TSP) is having a salesman and a set of cities. The salesman needs to visit each one of the cities starting from a certain one and returning to the same city i.e. back to starting position. The challenge of this problem is that the travelling salesman wants to minimise ...
Read MoreWhat is NP-completeness in TOC?
The Non-deterministic Polynomial (NP) problems were a little harder to understand. In terms of solving a NP problem, the run-time is not polynomial. It would be something like O(n!) or something larger.However, this class of problems are given a specific solution, and checking the solution would have a polynomial run-time.For example, the Sudoku game.NP-Hard ProblemsA problem is said to be NP-Hard when an algorithm for solving NP Hard can be translated to solve any NP problem. Then we can say, this problem is at least as hard as any NP problem, but it could be much harder or more complex.NP-Complete ...
Read More