//the exercice is : //longage 2-------------------- //Language L2 = {(ab) n | n ≥0} words formed from a succession //of a and b so that each a is followed by b. By //example, the words {ε, ab, abab, abababab} must be recognized //but the words {abaab, baba, ababbab, bbabab, bababab} must not //not be recognized. //longage3--------------------------------------- //L3 language = {a2nb //m / n, m> 0} words formed from a //succession of an even number of a followed by a number //any of b. For example, the words {aab, aabbb, aaaabb, //aaaab, aabbbbb} must be recognized but the words {ε, a, b, //ab, ba, aba, bab, abab, baba} must not be recognized. //longage 4------------------------------------------ //Langage L4=L2.L3, les mots appartenant au langage L4 qui est //la concaténation des langages L2 et L3. exp: abaab or ababaaaab #include #include #include int longer(char phr[50],int i){ //calculates the length of word like strlen() if(phr[i]=='\0'){ return 0; }else{ return longer(phr,i+1)+1;}} int occurrence(char phe[],char a){ ////counts how many times the letter is repeated in a word exp: phe =aba occurrence(phe,'a')=2 whe have two a int k=0; for(int i=0;i<=longer(phe,0);i++){ if(phe[i]==a){ k=k+1; }}return k;} //--------------------------------------------------------------------longage 2 //---------------------------monitors whether the word belongs to the longage 2 or no //--------longage 2 is (ab)^(2n) n>1 aand b caractare and n is nember ////////--------exp:ab or abab int longage2(char phe[ ]){ if (strlen(phe)==1||strlen(phe)%2!=0||phe[strlen(phe)]=='a'){ return 0; } for(int i=0;i

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