6.840 Language Membership

Transcription

1 6.840 Language Membership Michael Bernstein 1 Undecidable INP Practice final Use for A T M. Build a machine that asks EQ REX and then runs M on w. Query INP. If it s in P, accept. Note that the language is Σ if M accepts w and EQ REX if it doesn t. 2 P PATH 260 Breadth-first search with marking algorithm RELPRIME 261 GCD algorithm PRIME Indian guy theorem All CFLs 262 Dynamic programming. Build a table starting from the terminals and building up longer and longer strings using shorter results. For longer ones, try every possible substring to test for a rule A BC. MODEXP HW 4: 7.12 Repeated squaring algorithm. HAPPY-CAT HW 5: 8.15 Use dynamic programming to remember already-explored paths. ALL DF A 7.10 EQ DF A is (I think) in P, so we can test EQ DF A on the input and Σ. UNARY SUBSET-SUM E REG 9.18 Look to see if is concatenated with anything. 1. Cook-Levin Theorem: SAT P iff P = NP. (pg. 272) 2. Remember that if you re mimicing another machine, it may take extra O(n) time to update the tape heads etc. 3. P contains a language not recognizable by a 2DFA. (HW 6: 9.17) 4. Have you tried: 3 NP (a) If P, using dynamic programming? (b) Repeated squaring for each bit? COMPOSITES 265 Give the factors HAMPATH 265 Give the path CLIQUE 268 Give the k-clique SUBSET-SUM 269 Give the set of numbers which sum to t ISOMORPHIC 7.11 Give the isomorphism. See BPP analogue. Not known to be complete. 1. P NP conp.

2 4 NP-Complete SAT 276 Test a tableau for acceptance by constructing φ cell φ start φ move φ accept. φ cell ensures that exactly one light is on in each cell; that is, each cell represents exactly one variable, state marker, or edge marker. φ start makes sure that the first row is a legal start configuration given the input, and φ accept makes sure the accept state is somewhere in the final line. φ move ensures each 2x3 window is legal. O(n 2k ) 3SAT Recitation CLIQUE 274 Create a graph where each clause is represented by three nodes, one for each literal. Connect all edges except those in the same triple and those with disagreeing labels on the same literal. VERTEX-COVER 284 Def n: subset of k nodes where every edge touches one of the nodes. Variable gadget: nodes for x and x, connected by an edge. Clause gadget: cluster of nodes for each literal, connecting each node to its mother node in the clause gadget. Select a true literal in each clause and do not include in the vertex cover; add the other two. Also add each true literal. HAMPATH 286 Diamond reduction. Variable is a diamond-shaped gadget that can be traversed in either direction and includes a link to a clause node if you re traveling in the right direction. UHAMPATH 291 Reduce from HAMPATH. Replace each node of the graph with u in, u mid and u out. s and t are replaced with only s in and s out. Connect in to mid and mid to out. Connect an out to another in if there s an edge in the original graph. Including mid ensures that we hit the node only once (we can t just pass by the out node later). SUBSET-SUM 292 The target is Represent each variable x i by two numbers y i and z i, which both have a 1 in the ith place and zeros elsewhere for the left half, and 1 for every clause they appear in on the right. The bottom half has a pair of 1 s for each clause so that we can bring the number of true literals in each clause up to 3. PUZZLE HW 4: 7.26 U = { M, x, # t NTM M accepts HW 4: 7.32 input x within t steps on at least one branch}. -SAT HW 5: 7.24 Replace each (y 1 y 2 y 3 ) with (y 1 y 2 z i ) (z i y 3 z), where z is constant across all clauses. MAX-CUT HW 5: 7.25 Reduce from SAT. Make 3c nodes for each x and x; connect each x to each x and connect clauses. EQ BP HW 6: A language A is conp-complete iff A is NP-complete. 2. Have you tried: (a) If you can assume P=NP, creating your own NP language and then using it bit-by-bit to solve a difficult problem in polynomial time? 3. If A is NP complete and A P A, then NP conp 4. A language A is conp-complete iff A is NP-complete.

3 5 PSPACE SAT 304 Try all configurations. ALL NF A 305 Def of ALL NF A : accept if the NFA rejects on some string. Nondeterministically guess the rejected string and use linear space to simulate the NFA. Use Savitch s Theorem to get it in deterministic space, and then we can take the complement. TBQF 312 Strip off quantifiers one by one and recursively evaluate with the quantifier s value set to 0 and then 1. Reject or accept based on the or rule. GENERALIZED- GEOGRAPHY 317 Recursive algorithm like with TBQF. Remove the current node and all outgoing arrows, and call recursively on all nodes it originally pointed at. GO-MOKU HW 5: 8.10 Algorithm involves trying to place a marker in every available position and recursively calling as the other player. If player 1 has any available move, accept; if player 2 has no available moves, reject. EQ REX HW 5: 8.8 Decide EQ REX, since PSPACE = copspace. Translate reg. ex. to NFAs, in polynomial space. If p is the max. number of states in the NFA, run the NFAs on all strings of length p 2. If they never disagree, they are equivalent. LADDER Lecture, 8.9 Word ladder problem. Try recursing and replacing one letter at a time, keeping a counter to keep track of how deep you are into the recursion so you can exit if you know you re looping. MIN- FORMULA 6 PSPACE-complete 8.16 Generate all shorter formulas one at a time and test for equivalence. Also, MIN-FORMULA is in conp, and conp PSPACE. TBQF 310 Mirror s Savitch s theorem. Define φ c1,c 2,t = 1 iff M can go from c 1 to c 2 in at most t steps. Test φ cstart,c accept,2 where M has no more than dnk 2 dnk possible configurations. Shrink the formula to be polynomial size by making φ c1,c 2,t = m 1 (a, b {(c 1, m 1 ), (m 1, c 2 )}) [φ a,b, t 2 A LBA HW 5: 8.13 Book s algorithm for TQBF is decidable using an LBA. FORMULA- GAME 315 This is equivalent to TBQF: a formula is true exactly when player E has a winning strategy. GENERALIZED- GEOGRAPHY 316 Reduce from FORMULA-GAME. Given a formula, create a graph where a series of diamond-shaped structures allow the players to alternate in picking truth values for x 1, x n. Then swing around and the A player chooses a clause which he thinks isn t satisfied, E picks a literal which might satisfy it. If that literal was chosen, the node is already passed through and E wins. If not, the next node at the bottom of the diamond has been passed through and A wins. 1. Note that in PSPACE since the nondeterminism is the same as determinism you can also solve the complement language and take the not of it. 2. Have you tried: (a) Savitch s Theorem: Where f(n) log n, NSP ACE(f(n)) SP ACE(f 2 (n)) (pg. 306) (b) Space, time hierarchy theorems.

4 7 L {0 k 1 k k 0} 321 Keep two counts of the number of 0 s and the number of 1 s, then check at the end. ADD HW 6: 8.22 Use elementary school algorithm. PAL-ADD HW 6: 8.22 Run the elementary school algorithm for ADD to decide each bit. A DF A 8.7 Keep pointer to the current state and head position, then simulate. Properly nested Section: 8.17 Use a count to make sure you never go negative or end up not at zero. parenthesis Properly nested parenthesis and brackets Section: 8.18 First run the properly nested algorithm. Then use a count for each open brace or paren and make sure when it reaches 0 again it is closed by the same thing that opened it. MULTIPLY 8.20 Def: multiply binary numbers. Use the typical elementary school algorithm. ADD- REVERSE 8.21 Def: add number to its reverse. Use the add algorithm from elementary school, digit-by-digit. ADD- REVERSE Def: add number to its ADD-REVERSE. Figure out digits as in ADD- DIGIT for each palindromic space, then add. ODD-PARITY Practice test Just count. 8 NL PATH 322 Nondeterministically choose up to (# of nodes) edges to follow, starting at s. Accept if we arrive at t. UNIQUE-SAT HW 6: 9.19 BIPARTITE 8.25 We can easily show complement BIP ART IT E is in NL. NL = conl. 9 NL-complete PATH 322 Show NL-Complete. Go through all strings of length c log c to find legal configurations, and output those that are legal. List edges similarly, only recording those which are valid from c 1 to c 2. STRONGLY- CONNECTED HW 6: 8.27 Reduce from PATH. Connect end of path to every node and every node to beginning of path. BOT H NF A 8.28 Reduce from PATH. Take the graph and add ɛ transitions everywhere, and make your other graph two states (start and finish) with an ɛ transition. A NF A 8.29 Same as before, but test for string ɛ. E DF A 8.30 Take the graph and add transitions on any symbol. 1. NL = conl by proving that P AT H NL. 2. If A is NL-complete then overlinea is NL-complete. Since P AT H L A, P AT H L A. P AT H L P AT H since P AT H is in NL, so P AT H L P AT H L A. NL L 2 10 EXPTIME-complete {M, w, i, j, α} Practice final Def: The ith symbol of configurationafter the jth step of computation of M on w. Output {M, w, 1, 2 nk, q accept } 11 EXPSPACE-complete EQ REX 344 To show in EXPSPACE, eliminate the exponentiation, convert to NFAs, and test for NFA equivalence. For completeness, we test whether R 1 = all strings of symbols that may appear in a computation history is equivalent to R 2 = all strings that are not rejecting computation histories = R bad start R bad window R bad reject. These are all strings which don t have the correct start sequence, don t move from configuration to configuration correctly, and don t have a reject configuration. R 1 = R 2 iff M accepts w; if there s a difference, the machine has a rejecting computation history for M on w and thus M doesn t accept w.

5 12 Oracles NONMIN- FORMULA NP SAT UNIQUE-SAT HW Have you tried: 349 NP P SAT, so conp P SAT because P is closed under complement. Since nonequivalence testing is NP, equivalence is conp and thus in P SAT. Nondeterminsistically guess the shorter formula, test for equivalence. (a) If you re dealing with P SAT, try showing it s in NP or conp instead of just using the oracle straightaway. 13 BPP PRIMES 372 Run the Fermat test we know primes pass all Fermat tests. If a number is not pseudoprime (not prime and not a Carmichael number), it fails at least half these tests. If any fail, we reject. EQ ROBP 377 Extend the boolean functions over the integers. If we select a 1,, a m randomly 14 RP COMPOSITES 375 Same as BPP for PRIMES; the error is only on the composite side. 1. Amplification lemma 2. Fermat s little theorem: If p is prime and a is positive, a p 1 1 (mod p) 3. NP IP, BP P IP 4. There are oracles A and B s.t. P A NP A and P B = NP B. 15 IP NONISO 387 Verifier randomly selects G 1 or G 2 and creates an isomorphism; Prover needs to come back with the identity of the original graph. #SAT 392 Extend f() over the nonbooleans. P sends coefficients of, e.g., f 1 (z) as a polynomial. V checks that it matches the previous step and then chooses a random coefficient a 1 to move forward with. Then V must send f 2 (a 1, z), and so on. 1. IP = PSPACE

6 16 Known Relationships L NL = conl P NP P SP ACE = IP EXP SP ACE }{{} BP P P SP ACE (book 10.7) RP NP RP BP P corp conp corp BP P P NP conp 17 Unknown Relationships (8) P = NP? (1) NP = conp? (2) NP BP P? (3) P = NP conp? (4) L = NL( conl)? (5) P = P SP ACE? (6) NP = P SP ACE? (7) 18 Conditional Relationships If P = NP: Every language in P is NP-complete. (7.17) You can find satisfying assignments, factor integers, find the largest clique in polynomial time ( ) If an NL-complete language is proven not NP-complete: P NP (7.19) If every NP-hard laguage is also PSPACE-hard: PSPACE = NP (8.11) every PSPACE language can be reduced to SAT and thus is in NP. If NP = P SAT : NP = conp (9.9) If NEXPTIME EXPTIME: P /neq NP (9.14) If NP BPP: NP = RP