Theory of Computation CSCI 3434, Spring 2010

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1 Homework 4: Solutions Theory of Computation CSCI 3434, Spring This was listed as a question to make it clear that you would lose points if you did not answer these questions or work with another student in the class. Some of the content in these solutions was taken from wikipedia. 2. A context-sensitive grammar (CSG) is an unrestricted grammar where every rule is of the form: α β where β α S ε is allowed only if S does not appear on the right-hand side of any rule A context-sensitive language (CSL) is a language that can be generated by a CSG. CSLs can be computed by a linearly-bounded Turing Machine, a TM that uses only the space used by the input. Construct CSGs for the following languages over Σ = {0, 1}: a. { 0 n 1 n 0 n 1 n n 0} S BCDT 0101 ε Generates strings of characters with equal numbers T ABCDT ABCD of B, C, and D, and 1 fewer A. ========== BA AB CA AC Rules to rearrange letters and CB BC DA AD put them in correct order DB BD DC CD ========= DT T1 CT U0 These rules turn the non-terminals A-D into CU U0 BU V1 the appropriate terminals. The end symbol BV V1 AV X0 goes from T to U to V to X to insure that we can t AX X0 X 0 change them when they re out of order. There was 1 fewer A, so turn end symbol to 0 when finished. Note this can be done a bit more simply if we start with the initial 0s at the beginning. b. { ww w {0, 1}* }

2 c. {0 n x 1 n x n } S ZLQ 01 ε Z is last 0 before start of x; Q is first 1 after x. L A0LB A1LB 1 0 1P 0P Generate x, along with an A and B for each char. P AB ABP Generate additional As and Bs beyond len of x. ========= 0A A0 1A A1 Rearrange non-terminals to push As to the left B0 0B B1 1B and Bs to the right. BA AB ========= ZA 0Z BQ Q1 When A passes Z or B passes Q, make into 0 or 1. Z 0 Q 1 Turn Z into 1 and Q and into Construct TMs for the following languages over Σ = {0, 1}: a. { 0 n 1 n 0 n n 0} Idea: Use a 2-tape TM (2TM). Copy the first group of 0s to tape 2. Then read these 0s (can read from right to left on tape 2) while reading 1s on tape 1 to verify there are the same number. Read tape 2 again (left to right this time) while reading the last group of 0s on tape 1 to verify there are the same number. b. { v#w w {0, 1}*, v is a substring of w } Idea: Use a 2TM. Read v and copy it to tape 1. Then nondeterministically determine (guess) when in w to begin comparing to v as follows: When # is read, move to beginning of w on tape 1 and back to the beginning of tape 2. Begin reading w, and at each character, nondeterministically begin reading v if you haven t already begun reading it. Once you begin reading v, verify that each character of v matches the current character of w. Continue reading both v and w until a character differs (reject), you reach the end of v (accept), or you reach the end of v without reaching the end of w (reject). 4. Construct TMs for the following functions with Σ = {0, 1}: a. f (x)=x+1, x is a binary integer Idea: Read to the end of the input. If the last character is a 0, change it to 1 and you re done. If it s a 1, change it to 0, then move one character to the left and repeat. If you get to a blank (in front of the first digit) without yet encountering a 0 to change to a 1, replace this leading blank with a 1.

3 b. f (x,y)=x+y, x,y are binary integers Idea: Use a 3TM. Copy x to tape 2 and y to tape 3, clearing tape 1 as it s read. Then begin at the end of the strings on tapes 2 and 3. Move from right to left, placing the correct sum on tape 1, and keeping track of carry bits using the states of the 3TM. c. f (0 i ) = 0 m, m = i 2 Idea: Use a 3TM. Copy all i 0s to tape 2, and i -1 0s to tape 3. Tape 2 will always contain i 0s, and tape 3 will be a counter showing how many more times i 0s (copied from tape 2) need to be appended to tape Describe the Kuroda normal form. Modify this definition to allow languages that include the empty string. Why is the Kuroda normal form useful to programmers? A grammar is in Kuroda normal form if all rules are of the following types: AB CD A BC A B A a Where A, B, C, and D are non-terminals and a is a terminal. We could modify Kuroda normal form to allow the empty string by adding a similar rule to that for CSGs, where we allow the rule S ε only if S does not appear on the right side of any rule. Kuroda normal form has many of the same advantages as Chomsky normal form: while any CSG that doesn t contain the empty string can be written in KNF, but derivation trees are all binary, allowing them to be more easily described than derivations in an arbitrary CSG. This allows a parsing algorithm similar to the CYK algorithm for CFGs. 6. Describe the Chomsky Hierarchy of languages. Give a Venn diagram that includes the following language types: context-free, context-sensitive, deterministic context-free, finite, recursively enumerable, regular. Give a table that gives a machine that can accept each of these languages (and only those languages e.g., do not say a Turing Machine accepts regular languages).

4 Recursively Context-Sensitive Context-Free Deterministic Context-Free Regular Finite context-free context-sensitive deterministic context-free finite recursively enumerable regular PDA linear-bounded TM DPDA DFM or NFM without any loops TM DFM, NFM 7. What is the Church-Turing thesis? 8. Show that the following problems are undecidable: a. P Accept : Given a TM M and input x, is x L(M) (does M accept x)? Note that this differs from P Halt, because a TM may halt simply because there are no allowed transitions from the current configuration (state and tape symbol) without transitioning to the accept state. P Halt P Accept : Given any P Halt instance, with a TM M and input x, construct another TM M' that simulates M on any input, then if M would halt, M' accepts. M' will accept anything on which M would halt, and if M would halt, then M' accepts, so we can determine the answer to the Halting problem if we could answer P Accept. So P Accept must also be undecidable. b. P 2Nonblank : Given a 2-tape TM M and input x, does M ever write a non-blank symbol on tape 2 when run on x? P Halt P 2Nonblank : Given any P Halt instance, with a TM M (assumed to have 1 tape) and input x, construct another TM M' that simulates M on any input, then if M would halt, M' writes a non-blank symbol on tape 2. M' will write a nonblank symbol on tape 2 if M would halt, and if M would halt, then M' writes

5 a nonblank symbol on tape 2, so we can determine the answer to the Halting problem if we could answer P 2Nonblank. So P 2Nonblank must also be undecidable. 9. Ex on p. 28. a. This modification says that each vertex must have at least one color instead of exactly one color. So, this transformation to SAT differs in semantics from the 3-Color problem. However, to be a transformation means that the Boolean formula is satisfiable if and only if the graph can be 3-colored, and semantic details are irrelevant. Furthermore, it is not necessary to find a satisfying 3-coloring of the graph from the satisfiable Boolean formula, one merely needs to be able to state correctly whether the graph is 3-colorable or not. In this case, since there are clauses requiring that no color can be true for both vertices connected by edge, this is a valid transformation. If there is not a satisfying truth assignment to this relaxed (more permissive) mapping to SAT, then there is no way to 3-color the graph. If there is a satisfying truth assignment, see part b. b. If there is a satisfying truth assignment, and if a vertex is assigned all 3 colors, then (by the edge clauses) that vertex must not have any neighbors (if it had a neighbor, it could not be any of the 3 colors by the edge clauses, but it must have at least one color by the vertex clause, a contradiction), so this node can be given any color. If a vertex is assigned 2 colors, then all neighbors must be assigned only the 3 rd color, so either of the 2 colors can be chosen to give a valid 3-coloring of the graph. 10. Ex. 4.4 on p i. If A and B are two problems in NP, A is NP-complete, and there is a polynomial-time transformation from A to B then B is NP-complete as well. TRUE ii. iii. iv. If A and B are two problems in NP, A is NP-complete, and there is a polynomial-time transformation from B to A then B is NP-complete as well. FALSE: Any problem in P (which is contained in NP) can be transformed to any NP-complete problem. If A and B are two NP-complete problems then there is a polynomial-time transformation from A to B and also a polynomial-time transformation from B to A. TRUE: By the definition of an NP-complete problem, any problem in NP (including all NP-complete problems) can be polynomial-time reduced to the given NP-complete problem. If A is a problem that cannot be solved in polynomial time and there is a polynomial-time transformation from A to B then B cannot be solved in polynomial time either. TRUE: If B could be solved in polynomial time, so could A.

6 v. Any problem A that can be transformed into Circuit Satisfiability by a polynomial-time transformation is NP-complete. FALSE: See part Ex. 4.6 on p Show that the following problems are NP-complete: a. Longest Common Subsequence: Given an alphabet, a finite set of strings on the alphabet R *, and a natural number k, does there exist a string w *, of length at least k, that is a subsequence of each string in R? Note that a valid subsequence does not need to contain consecutive characters as in the original string, but it does need to retain order. (Hint: transform from Independent Set or Vertex Cover.) This problem has applications to genome sequencing and gene function prediction using sequence similarity. 1) The problem LCS is in NP, as we can use a NTM to guess a substring and check (in polynomial time) to see whether it is a common subsequence. Using the validation definition of NP, we can verify an input solution to LCS in polynomial time. 2). LCS is NP-hard: We can show that LCS can be reduced in polynomial time from VC, which is proven to be NP-complete. The proof is actually a reduction from Independent- Set (find a set of vertices with no edges amongst them), another NP-complete problem, but since we didn t discuss this problem, we ll sketch a reduction from Vertex-Cover (VC) to Independent-Set (IS) first. In both VC and IS we are given a graph G and positive integer k, but in VC we are concerned with finding a small vertex cover (with at most k vertices), while IS seeks to find a large independent set (with at least k vertices). If we have a vertex cover for graph G, meaning a set of vertices that cover all edges (include at least one vertex of each edge), then the remaining vertices form an independent set, meaning there are no edges between any two vertices in the independent set. If there were an edge between any pair of the remaining vertices, then this edge would not be covered by the vertex cover. Conversely, if we have an independent set, any edges have at least one vertex in the remaining vertices, so the remaining vertices form a vertex cover. So, the vertex-cover determines an independent set, and vice versa. So, if we have a vertex cover of (at most) k, we have an independent set of (at least) k = V - k, and vice versa. IS is clearly also in NP (given an alleged independent set, we can check there are no edges between any pair of vertices in the set in polynomial time), so it is NP-complete.

7 We can show that the problem of finding the k = V - k independent set of instance <G, k > can be solved by finding the solution to a corresponding LCS instance <B, k >. Given an instance <G, k > of IS with G=(V, E), we construct an instance <R, k > of LCS as follows: The finite set of strings R contains E +1 strings over the alphabet set =V. For each edge, v, v (assuming i< j), the corresponding string is given by listing i j all vertices in order twice, with v i missing from the first list of vertices, and v j missing from the second: vv 1 2Lvi 1vi+ 1LvjLvvv n 1 2LviLvj 1vj+ 1L vn. Besides these E strings, we also have a string t given by. vv Lv 1 2 n Assume that we find a common subsequence s of length at least k (a yes instance). The common subsequence cannot contain both vertices of an edge in G, and all of them are distinctive. This is because the string t in B implies that the common sequence is a vertex sequence with increasing order subscripts. In the string corresponding to any edge v, v (we always let i j), always appears after v j i j < i, and therefore they cannot both appear in the common sequence s. So, if s is of length at least k, it must be composed of distinct vertices, no pair of which are connected by an edge of G. So, the vertices indicated by s compose an independent set of G, and thus we also find the vertex-cover set accordingly. Conversely, if there is an independent set of G of size at least k, then these vertices (in order) are a subsequence of every string in B. So, an instance of the IS problem is a yes instance if and only if it maps to a yes instance of LCS. b. Set-Cover: Given a collection of n sets, are there k of these sets whose union is the same as the union of all n sets? (Hint: transform from Vertex Cover.) 13. The following are flawed proofs purporting to settle the issue of P vs NP. Point out the flaw in each proof. a. We prove P=NP nonconstructively by showing that, for each polynomial-time NTM, there must exist an equivalent polynomial-time (deterministic) TM. By definition, the NTM applies a choice function at each step in order to determine which of several possible next moves it will make. If there is a suitable move that will lead to accepting the string, it will choose; otherwise the choice is irrelevant. In the latter case, the TM can simulate the NTM by using any arbitrary move from the NTM s choice of moves. In the former case, although we do not know which is the correct next move, this move does exist; thus there exists a deterministic machine that correctly simulates the NTM at this step. By merging these steps, we get a deterministic machine that correctly v

8 simulates the NTM (although we don t know don t know what it is). Therefore, P=NP. The flaw is that the argument only applies for a specific yes instance: the NDTM, arriving at some decision point, will presumably take different branches depending on the instance, so that we would have, not one, but many DTMs corresponding to that one NDTM. So, the logic is correct for a single instance but we knew that a single instance is easy to solve anyway; and the logic completely fails for an infinite collection of instances. b. We prove P NP by contradiction. Assume P=NP. Then SAT is in P, and thus, for some k, SAT is in TIME(n k ). But every problem in NP reduces to SAT in polynomial time, so that every problem in NP is also in TIME(n k ). Therefore, NP is a subset of TIME(n k ) and hence so is P, since P NP. But the hierarchy theorem for time (which we didn t cover in class, but you should assume is true) tells us that there exists a problem in TIME(n k+1 ) (and thus also in P) that is not in TIME(n k ), a contradiction. Therefore, P NP. The flaw here is in the assertion that, if Satisfiability can be solved in O(n k ) time, so can any problem in NP, simply because any problem in NP reduces to Satisfiability in polynomial time. The time spent in the polynomial-time reduction could prove a problem (if it is bigger than n k ), but a more fundamental flaw is present: the polynomial-time reduction need not preserve the size of the input! An instance of size n of some problem in NP will be transformed in O(n m ) time (for some value of m) into an instance of Satisfiability of size f(n) where the best we can say is that f(n) is O(n m ). Then the assumed solution to Satisfiability solves the transformed instance in time O(f(n) k ), but that may be as large as n m+k and thus not O(n k ).