Circuit Complexity / Counting Problems

Transcription

1 Lecture 5 Circuit Complexity / Counting Problems April 2, 24 Lecturer: Paul eame Notes: William Pentney 5. Circuit Complexity and Uniorm Complexity We will conclude our look at the basic relationship between circuit complexity classes and uniorm complexity classes. First, we will prove that circuits can eiciently solve problems in uniorm complexity classes. Theorem 5. (Pippenger, Fischer). I T (n) n, then TIME(T (n)) SIZE(kT (n) log 2 T (n)). First, consider the ollowing variant o the traditional k-tape TM: Deinition 5.. A multitape TM is oblivious i the motion o each head depends only upon the length o its input. Note that an oblivious TM s motion has nothing to do with the input itsel; given an input o size n, the machine s head(s) will perorm the same series o motions regardless o the original contents o the tape. We will prove two lemmas rom which Theorem 5. ollows immediately. The irst is a more careul version o the original result o Hennie and Stearns showing that k-tape TMs can be eiciently simulated by 2-tape TMs. The key extension is that the resulting TM can be made oblivious. Lemma 5.2 (Hennie-Stearns, Pippenger-Fischer). For T (n) n, i A TIME(T (n)) then A is recognized by a 2-tape deterministic oblivious TM in time O(T (n) log T (n)). The second lemma shows that oblivious TMs can be eiciently simulated by circuits. Lemma 5.3. [Pippenger-Fischer] I A {, } is recognized by a 2-tape deterministic oblivious TM in time O(T (n)), then A k SIZE(kT (n)) (i.e. size linear in T (n). First we will prove Lemma 5.3 whose proo motivates the notion o oblivious TMs. Proo o Lemma 5.3. Consider the tableau generated in the standard proo o the Cook-Levin theorem or an input o size n or a Turing machine running in O(T (n). The tableau yields a circuit o size O(T 2 (n)), with windows o size 2 3. However, on any given input the TM head will only be in one cell at a given time step; i.e., one cell per row in the tableau. With a normal TM, though, it is not possible to anticipate which cell that will be. 22

2 LECTURE 5. CIRCUIT COMPLEXITY / COUNTING PROLEMS 23 = position o TM head. Figure 5.: Representing an oblivious TM using constant-size windows. Since our TM M is oblivious, we know the path that the head shall take, and where it will be at each timestep, simply by knowing the length o the input. We know i and when a given cell on the tape will be revisited, and can ignore it at other times. Thus we can use windows containing simply the contents o a cell rom the last time it was active and the contents o the previous cell visited to see what the head moves, to represent each step taken by that head, and all other cells in the tableau will be irrelevant to that step in the computation. This is represented in Figure 5.. The subsequent coniguration resulting rom any coniguration can thus be computed by a circuit o constant size. Since the tableau is o height T (n), we can express the tableau in size O(T (n)). Now, we prove Lemma 5.2, that i A TIME(T (n)), then A is recognized by a 2-tape deterministic

3 LECTURE 5. CIRCUIT COMPLEXITY / COUNTING PROLEMS 24 g e d c b a h i j k l m Figure 5.2: 4-track oblivious simulation o a TM tape e d c b a g h i j k l m Figure 5.3: The simulation in Figure 5.2 when the head moves let. lock is bunched and stretched. and oblivious TM in time O(T (n) log T (n)). Proo o Lemma 5.2. We will build a 2-tape TM M which will consist o two tapes: one holding all the contents o M s tapes, and one work tape. Traditionally, we think o a TM as having a tape which the head moves back on orth on. For this proo, it may help to instead think o the TM head as being in a ixed position and each o the tapes as moving instead, similar to a typewriter carriage (i anyone remembers this ancient beasts!). O course a tape would be an ininitely long typewriter carriage which would be too much to move in a single step. Instead, we think o moving the tapes back and orth, but we may not do so smoothly; sometimes we will be bunching up a tape, such that some cells may be eectively layered on top o others and other parts will be stretched, parts that are not bunched or stretched are smooth. It will be convenient to concentrate on simulating one tape o M at a time. In reality we will allocate several tracks on tape o M to each o the k-tapes. It is simplest also to imagine that each o M s tapes is 2-way ininite, although M will have only -way ininite tapes. More precisely, imagine the scenario modeled in Figure 5.2. For each tape o M, we have our tracks two or those cells to the let o the current cell, and two or those to the right o the current cell. The outer tracks will contain the contents o the tape when that portion o the tape is smooth; the inner tracks will contain portions o the tape that are bunched up. The current cell is always on the let end o the set o tracks. The tracks wrap around as the head o M moves. We will divide the tracks into blocks, where each block will consist o a portion o the two upper or lower tracks. lock contains the current cell (at the let end o the tape); block is the lower hal o the the cell immediately to the right o block and has capacity or up to 2 cells o M; block 2 is the lower hal o the next 2 cells to the right o that with capacity or up to 4 cells o M, etc. In general, block i has contains the lower hal o the next 2 i cells starting at cell 2 i and has capacity or 2 i cells o M. lock i or i contains the corresponding upper hal o cells o M whose lower hal is the block i. In general, or block i and j, i i < j then the contents o block i will be to the let o those or block

4 LECTURE 5. CIRCUIT COMPLEXITY / COUNTING PROLEMS e d c b a e d c b a j g h i k l m g h i j k l m Figure 5.4: The smoothing o blocks 2 and block 2. j. The outer tracks o block i or i will contain contents o cells o M that are closer to the cell in block than the inner tracks. In our simulation, when we wrap the tracks around, we permit the tape to bunch up within a block; when this happens, the contents o the cell o M that would ordinarily be in the outer track o the next block is now placed in the next ree cell on the inner track o the current block. We can later undo this by moving the bunched-up data in block i into i+ ; this process is called smoothing the block. Similarly, a block i may be stretched, so that it contains ewer than 2 i cells rom the tape; i other data is pushed into it, the vacant area on the track is illed. The tape is smoothed when it is no longer stretched. Figure 5.3 shows the movement o the tracks in our simulation depicted in Figure 5.2 when the head moves let; block is bunched and block stretched. In Figure 5.4, we see the smoothing o blocks 2 and 2 on such a simulation. We maintain the ollowing invariants ater simulating each time step t o the simulation:. Every block i with i > log 2 t + is smooth. 2. i + i = 2 i i.e. i i is bunched, i is stretched an equal amount. 3. Each i consists o consecutive elements o the tape o M. 4. I t is a multiple o 2 i then blocks i, i, i 2,,,, 2, i are smooth. To maintain our invariants, whenever t is 2 i t or some odd t, we smooth out blocks i i by borrowing rom or adding entries to i+ and (i+). We do this by copying the contents o all the cells in,..., i, i+ onto tape 2 and then writing them back into those cells only on the outer tracks, except, possibly, or block i+. The same is done or blocks,..., (i+). We need to argue that this smoothing can be accomplished. Ater smoothing blocks (i+),..., (i+) when t is a multiple o 2 i+, we may simulate M or up 2 i steps without accessing blocks i+ or (i+). On the 2 i -th step the number o entries that will need to appear in blocks,,..., i (or i, i,..., ) is between and 2 i+ since this number can have changed by at most 2 i during this time. Since i+ has remained smooth during this time, there are precisely 2 i occupied cells in i+ and space or 2 i more inputs. Thereore there enough entries in i+ to borrow to ill the up to 2 i spaces in blocks,..., i that need to be illed, or enough space to take the overlow rom those blocks. A similar argument applies to blocks i,...,. Thereore this smoothing can be accomplished. y careully making sure that the inner and outer tracks are scanned whether or not they contain entries, the smoothing can be done with oblivious motion. How much does the smoothing cost when t is a multiple o 2 i but not 2 i+? It will be less than or equal to c2 i steps in tota or some constant c. We will perorm smoothing 2 i o the time, with cost c2 i per smoothing. Thereore, the total time will be i log 2 T (n) + c2i T (n) 2 i which is O(T (n) log T (n).

5 LECTURE 5. CIRCUIT COMPLEXITY / COUNTING PROLEMS 26 We can use the proo o THeoremm 5. to prove the ollowing theorem: Theorem 5.4 (Cook 83). I T (n) n and L NTIME(T (n)) then there exists a reduction that maps input x to a 3-CNF ormula φ x in O(T (n) log T (n)) time and O(log T (n)) space such that φ x has size O(T (n) log T (n)) and x L φ x 3SAT. Proo. φ x is usual 3-CNF based on CIRCUIT-SAT. Check that producing the circuit in the proo o the previous lemmas can be done in O(T (n) log T (n)) time and O(log T (n)) space. 5.2 Function Complexity Classes, Counting Problems, and #P We will now deine some complexity classes o numerical unctions on input strings rather than decision problems. Deinition 5.2. The class FP is the set o all unctions : {, } N that are computable in polynomial time. Among the algorithmic problems representable by such unctions are counting problems pertaining to common decision problems. For example, Deinition 5.3. Deine #3-SAT as the problem o counting the number o satisying assignments to a given 3-SAT ormula φ. Remark. Note that i we can solve #3-SAT in polynomial time, then clearly we can solve any NP-complete problem in polynomial time and thus P = NP. We can deine new complexity classes to represent such counting problems: Deinition 5.4. For any complexity class C, let #C be the set o all unctions { : {, } N or which there exists R C and a polynomial p : N N such that (x) = #{y {, } p( x ) : (x, y) R} = {y {, } p( x ) : (x, y) R}. In particular, or C = P, we obtain Deinition 5.5. Deine the class #P to be the set o all unctions { : {, } N or which there exists R P and a polynomial p : N N such that (x) = #{y {, } p( x ) : (x, y) R} = {y {, } p( x ) : (x, y) R} Function classes and oracles Let us consider the use o oracle TMs in the context o unction classes. For a language A, let FP A be the set o unctions : {, } N that can be solved in polynomial time by an oracle TM M? that has A as an oracle. Similarly, we can deine oracle TM s M? that allow unctions as oracles rather than sets. In this case, rather than receiving the answer rom the oracle by entering one o two states, the machine can receive a binary encoded version o the oracle answer on an oracle answer tape. Thus or unctions : {, } N and a complexity class C or which it makes sense to deine oracle versions, we can deine C, and or a complexity class FC o unctions we can deine C FC = FC C.

6 LECTURE 5. CIRCUIT COMPLEXITY / COUNTING PROLEMS 27 u v edge in G u v.. gadget in G N = n log n layers 2 Figure 5.5: Gadgets used in reduction o #HAM-CYCLE to #CYCLE. Deinition 5.6. A unction is #P-complete i. #P. 2. For all g #P, g FP. As 3-SAT is NP-complete, #3-SAT is #P-complete: Theorem 5.5. #3-SAT is #P-complete. Proo. Cook s reduction is parsimonious, in that it preserves the number o solutions. More precisely, in circuit orm there is precisely one satisying assignment or the circuit or each NP witness y. Moreover, the conversion o the circuit to 3-SAT enorces precisely one satisying assignment or each o the extension variables associated with each gate. Since the standard reductions are requently parsimonious, and can be used to prove #P-completeness o many counting problems relating to NP-complete problems. In some instances they are not parsimonious but can be made parsimonious. For example we have the ollowing. Theorem 5.6. #HAM-CYCLE is #P-complete. The set o #P-complete problems is not restricted to the counting versions o NP-complete problems, however; interestingly, problems in P can have #P-complete counting problems as well. Consider #CY- CLE, the problem o inding the number o directed simple cycles in a graph G. (The corresponding problem CYCLE is in P). Theorem 5.7. #CYCLE is #P-complete. Proo. We reduce rom #HAM-CYCLE. We will map the input graph G or #HAM-CYCLE to a graph G or #CYCLE. Say G has n vertices. G will have a copy u o each vertex u G, and or each edge (u, v) G the gadget in Figure 5.5 will be added between u and v in G. This gadget consists o N = n log 2 n + layers o pairs o vertices, connected to u and v and connected by 4N edges within. The number o paths rom u to v in G is 2 N > n n. Each simple cycle o length l in G yields 2 Nl simple cycles in G. I G has k Hamiltonian cycles, there will be k2 Nn corresponding simple cycles in G. G has at most n n simple cycles o length n. The total number o simple cycles in G corresponding to these is n n 2 N(n ) < 2 Nn. Thereore we compute #HAM-CYCLE(G) = #CYCLE(G )/2 Nn. The ollowing theorem is let as an exercise: Theorem 5.8. #2-SAT is #P-complete.

7 LECTURE 5. CIRCUIT COMPLEXITY / COUNTING PROLEMS Determinant and Permanent Some interesting problems in matrix algebra are represented in unction complexity classes. Given an n n matrix A, the determinant o A is det(a) = σ S n ( ) sgn(σ) n a i,σ(i), where S n is the set o permutations o n and sgn(σ) is the number o trnaspositions required to produce σ modulo 2. This problem is in FP. The sgn(σ) is apparently a complicating actor in the deinition o det(a), but i we remove it we will see that the problem actually becomes harder. Given an n n matrix A, the permanent o A is equal to perm(a) = σ S n i= i= n a i,σ(i). Let -PERM be the problem o inding the permanent o a - matrix. When we continue, we will prove the ollowing theorem: Theorem 5.9 (Valiant). -PERM is #P-complete. The ollowing is an interesting interpretation o the permanent. We can view the matrix A the adjacency matrix o a weighted bipartite graph on vertices [n] [n] where [n] = {,..., n}. Each σ S n corresponds to a perect matching o this graph. I we view the weight o a matching as the product o the weights o its edges the permanent is the total weight o all matchings in the graph. In particular a - matrix A corresponds to an unweighted bipartite graph G or which A is the adjacency matrix, and Perm(A) represents the total weight o all perect matchings on G. Let #IPARTITE- MATCHING be the problem o counting all such matchings. Thus we obtain the ollowing corollary as well: Corollary 5.. #IPARTITE-MATCHINGS is #P-complete