Discrete Math Academic Essay

Discrete Math 1. [2 points] Let L = {hMi : M has an even number of states}. Is L decidable? Give a brief explanation for your answer. 2. [2 points] Let L = {hMi : L(M) has an even number of elements}. Is L decidable? Give a brief explanation for your answer. 3. [4 points] Consider two languages L1 and L2 such that L1 L2 = !. Show that if L1 and L2 are both recursively enumerable, then there exists a decidable language L such that L1 ? L and L2 ? L. 4. [3 points] Let 3-SAT-BOUNDED be defined as 3-SAT-BOUNDED = {! : ! is a satisfiable 3-CNF and every variable appears in at most 100 clauses}. Show that 3-SAT-BOUNDED is NP-complete. Hint: Consider the CNF ! = (x1 x2 x3) ^ (x1 x2 x4). In this 3-CNF, the variable x1 and x2 appear in two clauses whereas x3 and x4 in one of the clauses. For the version of 3-SAT we showed to be NP-complete in the class, there might be a variable which can appear in say all the clauses, or half of the clauses. In order to prove 3-SAT-BOUNDED is NP-complete, show that given any 3-CNF !, there is an equivalent 3-CNF !0 in which every variable appears in at most 100 clauses and !0 is satisfiable if and only if ! is satisfiable. Also, the number 100 is irrelevant. If you can do the reduction with a larger number (say 1000), I will accept that. 5.(a) [2 points] Show that L = {hMi||L(M)| 2} is recursively enumerable. 5.(b)[2 points] Show that L = {hMi||L(M)| = 2} is not recursively enumerable. 6. [5 points] A function f : {0, 1}?? ! {0, 1}?? is said to be length-preserving if for all x 2 {0, 1}??, |f(x)| = |x|. In other words, for any n, strings of length n get mapped to strings of length n under the map f. Further, assume that f is one-one. 1 f is onto. f is polynomial time computable. Note that the first two bullets ensure that f #1 : {0, 1}?? ! {0, 1}?? is a well-defined function which is also one-one, onto and length-preserving. However, it need not be polynomial time computable. Further, let g : {0, 1}?? ! {0, 1} be another polynomial time computable function. Prove that L = {y|g(f #1(y)) = 1} is in NP. Prove that L = {y|g(f #1(y)) 6= 1} is in NP. Hint: Please make sure that your proof does not somehow require f #1 to be e!ciently computable. 7.(a) [1 point] Suppose you come up with an algorithm running in time 2O( pn) for 3-SAT (where the input length is n). Will it necessarily imply that for all L 2 NP, L admits an algorithm of running time 2O( pn) ? (where n is the length of the instance in L). Reason briefly. 7. (b) [1 point] Suppose you come up with an algorithm running in time nO(log n) for 3-SAT (where the input length is n). Will it necessarily imply that for all L 2 NP, L admits an algorithm of running time nO(log n) ? (where n is the length of the instance in L). Reason briefly. 8. [2 points] Let B = {hM1, M2i|L(M1) = L(M2)}. Show that B is undecidable. 9. [1 point] Let L 2 NP. If P = NP, is L 2 P? If P 6= NP, is L 2 P? (justify briefly). PLACE THIS ORDER OR A SIMILAR ORDER WITH US TODAY AND GET A GOOD DISCOUNT

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