Computer Science Articles - Page 49 of 62

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

The Non-deterministic Polynomial (NP) problems were a little harder to understand. In terms of solving a NP problem, the run-time cannot be 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 the 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 ... Read More

Before we understand about the decidable and undecidable problems in the theory of computation (TOC), we must learn about the decidable and undecidable language. Hence, let us first see what do you mean by decidable language.Decidable LanguageA language L is called decidable if there is a decider M such that L( M) = L.Given a decider M, you can learn whether or not a string w ∈ L(M).Run M on w.Although it might take a long time, M will accept or reject w.The set R is the set of all decidable languages.L ∈ R if L is decidable.Undecidable LanguageA decision ... Read More

There are two methods for converting deterministic finite automata (DFA) to Regular expression (RE). These are as follows −Arden’s MethodState Elimination MethodLet us understand these methods in detail.Arden’s TheoremLet P and Q be the two regular expressions.If P does not contain null string, then following equation in R, viz R = Q + RP,Which has a unique solution by R = QP*Here,The finite Automata do not have epsilon movesIt must have only initial state q1It’s states are q1, q2, q3,…….qn. The final state may be some qi where i

The star height of Regular expression (RE) is nothing but the depth of Kleene stars in the theory of computation (TOC).For example, a+b the star height is 0(a+b)* the star height is 1(a*+b*)* the star height is 2 …….Star height is used to indicate the structural complexity of regular languages and expressions.The regular expressions may have different star height that depends on structural complexity or nesting.The star height of a regular language is a unique number and that is equal to the least star height of any regular expression which represents that language.The star height of regular expressions is a ... Read More

A grammar with at most one variable at the right side of production is called linear grammar.Following is an example of the linear grammar −S→aSb/εHere, if you observe, we can write the same production by dividing …..S→AbA→aAbA→εLeft Linear GrammarA grammar is left linear grammar where all non-terminals in the right hand sides are at the left end.For example, A→Sa/εSteps for conversionThe steps for the conversion of finite automata (FA) to the left linear grammar are as follows −Step 1 − Take reverse of the finite automataStep 2 − write right linear grammarStep 3 − Then take reverse of the right ... Read More

A grammar with at most one variable at the right side of production is called linear grammar.Example 1      S→aSb/εExample 2      S→Ab      A→aAb      A→εRight Linear GrammarA grammar is right linear grammar where all the non terminals in the right hand sides are at the right end.For example,       S->aS/εAlgorithm for conversionThe algorithm to convert the finite automata (FA) to the right linear grammar is as follows −Step 1 − Begin the process from the start state.Step 2 − Repeat the process for each state.Step 3 − Write the production as the ... Read More

To identify whether a language is regular or not, is based on Pigeon Hole Principle. This is generally called as the Pumping Lemma.Pumping lemma for Regular languagesIt gives a method for pumping (generating) many substrings from a given string.In other words, we say it provides means to break a given long input string into several substrings.It gives the necessary condition(s) to prove a set of strings is not regular.But the pumping lemma is a contradiction test. This means if any language does not satisfy pumping lemma, then we can say that it is not regular. However, if it satisfies the ... Read More

ProblemEliminate epsilon, unit and the useless symbols for the given grammar and rewrite it into CNF.S->0E0|1FF| εE->GF->S|EG->S| εSolutionIn the given grammar, we will first remove the null production. There are two null productions in the grammar, as given below −S ==> εG ==> εSo, remove null production and rewrite all the other rules containing G by epsilon there, along with old productions. We do not remove S ==> epsilon as it is the start symbol.Remove G ==> epsilon, we get the following −S ==> 0E0 | 1FF | εE ==> G | εF ==> S | EG ==> SNow remove ... Read More

ProblemGenerate a Chomsky normal form (CNF) for the following context free grammar (CFG).S->aAa|bBb|eA->C|aB->C|bC->CDE|eD->A|B|abSolutionFollow the steps mentioned below to generate a CNF for the given CFGStep 1 − Eliminate ∧ -productionsWe can delete, erase or ∧ -productions double time repeated.S --> aAa | bBb | ∧A --> a | ∧B --> b | ∧D --> A | B | abStep 2 − Eliminate unit productions in above grammarEliminate R.H.S one symbol productionsS --> aDa | bDbD --> a | b | abStep 3 − Eliminate useless symbolsE is a useless symbol from given grammar since it is not derivative in RHS.S ... Read More