Theory of Computation

Transcription

1 Theory of Computation Dr. Sarmad Abbasi Dr. Sarmad Abbasi () Theory of Computation 1 / 38

2 Lecture 21: Overview Big-Oh notation. Little-o notation. Time Complexity Classes Non-deterministic TMs The Class P Dr. Sarmad Abbasi () Theory of Computation 2 / 38

3 Resource Bounded Computations Let us now look at resource bounded computations The most important resources are: 1 Time 2 Space Dr. Sarmad Abbasi () Theory of Computation 3 / 38

4 Resource Bounded Computations Let A = {0 k 1 k : k 0} we saw an algorithm that accepts A in time O(n 2 ). Dr. Sarmad Abbasi () Theory of Computation 4 / 38

5 Big Oh Notation Let f, g : N R. We say that f (n) = O(g(n)) if there are positive integers c and n 0 such that for all n n 0 f (n) cg(n). When f (n) = O(g(n)) we say that g is an asymptotic upper bound on f. Some times we just say that g is an upper bound on f. Dr. Sarmad Abbasi () Theory of Computation 5 / 38

6 Big Oh Notation Let then Furthermore, But it is not true that f (n) = 13n 4 + 7n f (n) = O(n 4 ). f (n) = O(n 5 ). f (n) = O(n 3 ) Dr. Sarmad Abbasi () Theory of Computation 6 / 38

7 Big Oh Notation Suppose that where b is a constant. Note that Hence, we can say Let then f 2 (n) = O(nlogn). f (n) = log b n log b n = log 2 n log 2 b. f (n) = O(logn). f 2 (n) = 5n log n + 200n log log n + 17 Dr. Sarmad Abbasi () Theory of Computation 7 / 38

8 Big Oh Notation We can use O notation in exponents also. Thus 2 O(n) stands for a function that is upper bounded by for some constant c. 2 cn Dr. Sarmad Abbasi () Theory of Computation 8 / 38

9 Big Oh Notation We have to be careful though. For example consider 2 O(log n). Well remember that O(log n) has a constant attached to it. Thus this function is upper bounded by 2 c log n for some constant c. Now, we get Thus 2 O(log n) is polynomial. 2 c log n = (2 log n ) c = n c. Dr. Sarmad Abbasi () Theory of Computation 9 / 38

10 Big Oh Notation Bounds of the form n c are called polynomial bounds. Where as bounds of the form 2 nc are called exponential bounds. Dr. Sarmad Abbasi () Theory of Computation 10 / 38

11 little Oh Notation The Big-Oh notation roughly says that f is upper bounded by g. Now we want to say f grows strictly slower that g. Let f, g : N R +. We say that if f (n) = o(g(n)) f (n) lim n g(n) = 0. Thus the ratio of f and g becomes closer and closer to 0. Dr. Sarmad Abbasi () Theory of Computation 11 / 38

12 little Oh Notation 1 n = o(n) 2 n = o(n log log n) 3 n log log n = o(n log n) 4 n log n = o(n 2 ) 5 n 2 = o(n 3 ) Dr. Sarmad Abbasi () Theory of Computation 12 / 38

13 little Oh Notation Lets look at a new TM that recognizes A = {0 k 1 k : k 0}. 1 Scan the tape and make sure all the 0s are before 1s. 2 Repeat till there are 0s and 1s on the tape. 1 Scan the tape and if the total number of 0s and 1s is odd reject. 2 Scan and cross of every other 0 and do the same with every other 1. 3 If all 0s and 1s are crossed accept. Thus if we have 13 0s in the next stage there would be 6. Dr. Sarmad Abbasi () Theory of Computation 13 / 38

14 little Oh Notation This algorithm shows that there is O(nlogn) time TM that decides A. Since, nlogn = o(n 2 ). This points to a real algorithmic improvement. Dr. Sarmad Abbasi () Theory of Computation 14 / 38

15 Time Complexity Classes Let t : N N be a function. Define the time complexity class TIME(t(n)) to be: TIME(t(n)) = {L L is decided by an O(t(n))-time Turing machine}. Dr. Sarmad Abbasi () Theory of Computation 15 / 38

16 So for example: 1 TIME(n) is the set of all languages, L, for which there is a linear time TM that accepts L. 2 TIME(n 2 ) is the set of all languages, L, for which there is a quadratic time TM that accepts L. Dr. Sarmad Abbasi () Theory of Computation 16 / 38

17 Recall A = {0 k 1 k : k 0}. Our first algorithm (or TM) showed that A TIME(n 2 ). The second one showed that A TIME(n log n). Notice that TIME(nlogn) TIME(n 2 ). Dr. Sarmad Abbasi () Theory of Computation 17 / 38

18 What will happen if we allow ourselves to have a 2-tape TM. Can we have a faster TM. Yes, if we have two tapes we can do the following: Dr. Sarmad Abbasi () Theory of Computation 18 / 38

19 1 Scan across to find out if all 0s are before 1s. 2 Copy all the 0s to the second tape. 3 Scan on the first tape in forward direction and the second tape in reverse direction crossing off 0s and 1s. 4 If all 0s have been crossed off and all ones have been crossed off accept else reject. In fact, it one can design a 2-tape TM that accepts A while scanning the input only once! Dr. Sarmad Abbasi () Theory of Computation 19 / 38

20 This leads to an important questions? Question Can we design a 1-tape TM that accepts A in o(n log n) time. Preferably linear time? The answer is NO! Dr. Sarmad Abbasi () Theory of Computation 20 / 38

21 In fact there is a language L such that 1 L is accepted by a 2-tape TM in time O(n). 2 L cannot be accepted by any 1-TM TM in time o(n 2 ). Dr. Sarmad Abbasi () Theory of Computation 21 / 38

22 Thus the definition of TIME(t(n)) changes if we talk about 1-tape TMs or 2-tape TM. Fortunately, it does not change by too much as the following theorem shows. Theorem Let t(n) be a function where t(n) n. Then every t(n)-time k-tape TM has an equivalent O(t 2 (n))-time single tape TM. Dr. Sarmad Abbasi () Theory of Computation 22 / 38

23 Notice the difference from computability theory. Previously we showed: Theorem Every k-tape TM has an equivalent 1-tape TM. (We ignored time). Now, we are doing more detailed analysis. Dr. Sarmad Abbasi () Theory of Computation 23 / 38

24 The proof of the theorem is just a more careful study of the previous proof. Remember given a k-tape TM, M, we made a 1-tape TM, N. N works as follows: 1 On input x convert the input to #q0#x# # the start configuration of M. This configuration says that x is on the first tape. The rest of the tapes are empty and the machine is in q 0. 2 In each pass over the tape change the current configuration to the next one. 3 If an accepting configuration is reached accept 4 If a rejecting configuration is reached reject. Now, we just have to estimate how much time does N require. Dr. Sarmad Abbasi () Theory of Computation 24 / 38

25 On any input x of length n, we make the following claims. 1 M uses at most t(n) cells of its k-tapes. 2 Thus any configuration has length at most kt(n) = O(t(n)). 3 Thus each pass of N requires at most O(t(n)) steps. 4 N makes at most t(n) passes on its tape (after that M must halt). This shows that N runs in time O(t(n) t(n)) = O(t 2 (n)). Dr. Sarmad Abbasi () Theory of Computation 25 / 38

26 Where did we use the fact that t(n) n. In the first step the machine converts x to the initial configuration. This takes time O(n). Thus the total time is actually O(n) + O(t 2 (n)) = O(t 2 (n)). This is a reasonable assumption as M requires time O(n) to read its input. Which is true for most problems. The answer usually depends on the entire input. Dr. Sarmad Abbasi () Theory of Computation 26 / 38

27 You have studied many non-deterministic models such as: 1 NFA: Non-deterministic finite automata. 2 NPDA: Non-deterministic pushdown automata. Now, we will define non-deterministic TMs. We will now define non-deterministic TMs. Dr. Sarmad Abbasi () Theory of Computation 27 / 38

28 A non-deterministic TM is formally given as M = (Σ, Γ,, q 0, q c, q r, δ) where 1 Σ is the input alphabet. 2 Γ is the work alphabet. 3 q 0, q a, q r are start, accept and reject states. 4 δ is now a transition function: For example we may have δ : Q Γ : P(Q Γ {L, R}). δ(q 5, a) = {(q 7, b, R), (q 4, c, L), (q 3, a, R)}. Thus the machine has three possible ways to proceed. Dr. Sarmad Abbasi () Theory of Computation 28 / 38

29 Time Complexity We can think of a non-deterministic computation as a tree: Dr. Sarmad Abbasi () Theory of Computation 29 / 38

30 We will say that a non-deterministic TM M accepts x. If x is accepted on at least one branch. On the other hand M rejects x if all branches of the tree reach the reject state. Dr. Sarmad Abbasi () Theory of Computation 30 / 38

31 The nondeterministic running time of M is the length of the longest branch. More, formally, The running time of a nondeterministic TM M is the function f : N N such that f (n) is the maximum number of steps M uses on any branch of its computation on any input of length n. Dr. Sarmad Abbasi () Theory of Computation 31 / 38

32 A few comments on the definition: 1 The notion of non-deterministic time does not correspond to any real world machine we can make. 2 It is merely a mathematical definition. 3 We will see that this definition is extremely useful. It allows us to capture many interesting problems. Dr. Sarmad Abbasi () Theory of Computation 32 / 38

33 Let us prove some relation between non-deterministic time and deterministic time. Theorem Let t(n) be a function, where t(n) n. Then every t(n) time nondeterministic single tape TM has an equivalent time deterministic single tape TM. 2 O(t(n)) Thus this theorem says that we can get rid of non-determinism at the cost of exponentiating the time! Dr. Sarmad Abbasi () Theory of Computation 33 / 38

34 Let N be a non-deterministic TM that runs in time t(n). Let us first estimate how big the computation tree of N can be: 1 The tree has depth t(n). 2 Let b be the maximum number of legal choices given by Ns transition function. 3 Then this tree can have b t(n) leaves. 4 The total number of nodes can be at most 1 + b + b b t(n) 2b t(n). Dr. Sarmad Abbasi () Theory of Computation 34 / 38

35 A deterministic TM D simulates the computation tree of N. There are several ways to do this. One is given in the book. Here is a slightly different one. To begin with D is going to be a multi-tape TM. Dr. Sarmad Abbasi () Theory of Computation 35 / 38

36 Consider a string over R = 1, 2,......, b of length t(n). So the string corresponds to a computation of M where 1 Starting from the root go to the first child c 1 of the root. 2 Then the second child of c 1 call it c 2. 3 Then to the third child child of c 2 call it c 3. 4 Then to the first child of c 3 Thus each such string corresponds to a branch in the tree. Dr. Sarmad Abbasi () Theory of Computation 36 / 38

37 Thus D is the following TM 1 On input x of length n. 2 For all strings s R t(n). 3 Simulate N making the choices dictated by s. 4 If N accepts then accept. 5 If all branches of N reject then reject. This method assumes that t(n) can be computed quickly. However, we can also eliminate this requirement. Dr. Sarmad Abbasi () Theory of Computation 37 / 38

38 For each string s the simulation can take O(t(n)) time. There are b t(n) strings to simulate. Thus the total time is O(t(n))b t(n) = 2 O(t(n)). Our last part is to convert this multi-tape TM into a single tape TM. If we use the previous theorem then the total time would be We will come back to NTMs again. (2 O(t(n)) ) 2 = 2 2O(t(n)) = 2 O(t(n)). Dr. Sarmad Abbasi () Theory of Computation 38 / 38

39 We want to know which problems can be solved in practice. Thus we define the class P as follows: P = k TIME(n k ). or more elaborately write is as P = TIME(n) TIME(n 2 ) TIME(n 3 ) TIME(n 4 ). Dr. Sarmad Abbasi () Theory of Computation 39 / 38

40 The class P is important because: 1 P is invariant for all models of computation that are polynomially equivalent to the deterministic single-tape machines. 2 P roughly corresponds to the problems that we can solve realistically on these models. Thus in defining P we can talk about single tape TMs, Java programs, multi-tape TMs. They all give the same definition of P. Dr. Sarmad Abbasi () Theory of Computation 40 / 38

41 The second point is problems in P can be solved in reasonable time. Some people may object to this as they can point out that if a problem can be solved in time O(n 5000 ) then such an algorithm is of no practical value. This objection is valid to a certain point. However calling P to be the threshold of practical and unpractical problems has been extremely useful. Dr. Sarmad Abbasi () Theory of Computation 41 / 38