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Dvinal [7]
3 years ago
5

Which of the following statement is False? 1 point Context free language is the subset of context sensitive language Regular lan

guage is the subset of context sensitive language Recursively enumerable language is the super set of regular language Context sensitive language is a subset of context free language
Computers and Technology
1 answer:
alisha [4.7K]3 years ago
3 0

Answer:

Context-sensitive language is a subset of context-free language

Explanation:

Considering the available options, the statement that is considered wrong is "Context-sensitive language is a subset of context-free language."

This is because generally every regular language can be produced through the means of context-free grammar, while context-free language can be produced through the means of context-sensitive grammar, and at the same time, context-sensitive grammars are produced through the means of Recursively innumerable.

Hence, the correct answer in this correct answer to the question is the last option *Context-sensitive language is a subset of context-free langage

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Write a C++ program that searches for anagrams in a dictionary. An anagram is a word obtained by scrambling the letters of some
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Answer:

d, avt, car, feet, more, pitch, rome, tac, teef,

Anagrams of b in dictionary

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Explanation:

// FindAnagrams.cpp : it is the starting point of application console.

//

#include <vector>

#include <hash_map>

#include <iostream>

#include <string>

#include <algorithm>

#include <map>

#include <set>

using namespace std;

using namespace stdext;

bool IsAnagrams(string& strA, string& strB)

{

   list<char> listA(strA.begin(), strA.end());

   list<char> listB(strB.begin(), strB.end());

   listA.sort();

   listB.sort();

   return listA == listB;

   // return equal(listA.begin(), listA.end(), listB.begin());

}

string SortChars(string str)

{

   list<char> l(str.begin(), str.end());

   l.sort();

   return string(l.begin(), l.end());

}

set<string> FindAnagrams(list<string>& dict, string findStr)

{

   map<string, set<string>> d;

   typedef pair<string, set<string>> MyPair;

   for(list<string>::const_iterator it = dict.begin(); it != dict.end(); ++it){

       string str(SortChars(*it));

       if(d.find(str) == d.end()){

           set<string> s;

           s.insert(*it);

           d.insert(MyPair(str, s));

       }

       else{

           d[str].insert(*it);

       }

   }

   string sortedStr(SortChars(findStr));

   return d[sortedStr];

}

int main(int argc, char* argv[])

{

   list<string> dict;

   dict.push_back("c");

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   dict.sort();

   cout << "The dictionary: " << endl;

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   testCases.push_back("rome");

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   for(list<string>::iterator it = testCases.begin(); it != testCases.end(); ++it)

   {

       cout << endl << "Anagrams of " << *it << " in dictionary" << endl;

       set<string> output = FindAnagrams(dict, *it);

       copy(output.begin(), output.end(), ostream_iterator<string>(cout, ", "));

       cout << endl;

   }

   return 0;

}

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