Computability and Computational Complexity Solutions of exercise set 2
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1 Computability and Computational Complexity Solutions of exercise set 2 Exercise 1 Give a different proof for the theorem in lecture 2, slide 26 by representing the k strings not next to each other, but on top of each other. In other words, a single symbol of the simulating machine will encode k symbols of the simulated one. Hint: careful about the cursor positions. Theorem For any k-tape Turing machine M operating within time f(n) we can construct a single-tape TM M operating within time O(f 2 (n)) such that for any input x, M(x)=M (x). Let M = (K, Σ, δ, s) be our initial k-tape TM. We can build an equivalent single-tape TM M =(K, Σ, δ, s) by overlapping the k tapes on top of the first one. M must remember the position of each of the k cursors of M, and simulate the functioning of M. First, let us consider the encoding of the cursor positions. We can mark the position of a cursor by underlining the character pointed by the cursor (e.g. 0 stands for a cell that contains character 0, and is pointed by the cursor). This means we consider the tapes contain characters from Σ Σ, where Σ = {σ σ Σ}. An example of a possible configuration for a 4-tape TM is given in the following table. Here, the cursor of the first tape is on the third cell, the cursor of the second tape is on the second position, the cursor of the third tape is on the fifth cell, and the last cursor is in the initial position. The single-string machine M will have on its tape a string σ 1 σ 2 σ 3 σ 4 σ 5, where σ 1 is the symbol encoding >>> > (the first column), σ 2 is the symbol encoding 011# (second column), etc. > # > 1 1 # # > # > # # # # The alphabet for M will contain characters that encode combinations of k characters from Σ Σ. Considering Σ contains n elements, the new alphabet Σ will have (2 n) k characters. M represents a configuration (q, w 1, u 1,, w k, u k ) of M through a configuration (q, w, u), where w is obtained from all strings w i u i by taking the encoding for the first w k groups of k column-wise characters, and u is obtained from all strings w i u i by taking the encoding characters of the last u k groups of k column-wise characters. Here, w i is obtained from w i by replacing its last character a i with the underlined corresponding symbol a i, and the notation x stands for the number of characters in the word x. For the configuration in the table, w = σ 1 and u = σ 2 σ 3 σ 4 σ 5. Mind the fact that we are interested in the output string from the k-th tape, so when writing a configuration, the cursor of M will have the position of M s k-th tape cursor. Simulating M with M M will start in the initial state s with the encoding for all cursors being on the first symbol(> k ) and all tapes except the first one being empty. A move in M will need two scans in M (left-to-right and back): - The first scan (left-to-right) will determine the positions of the cursors and the symbols they point to (the 1
2 characters in Σ containing underlined symbols), remember them through new states and thus get the transition M is supposed to make - The second scan (right-to-left) will change the symbols pointed to by cursors and move the cursors. For example, if a symbol σ i Σ is an encoding containing an underlined symbol a j on position j, then to simulate replacing a j with b j and moving the cursor to the right on tape i of M, the machine M will: 1) replace σ i with the symbol that encodes b j on position j and all the rest symbols are like in σ i 2) replace the the next symbol σ i+1 with the symbol that encodes the same characters as σ i+1, except on the j-th position the symbol is underlined - The cursor needs to first go to the beginning of the tape, and after the scan go to the position it is supposed to currently be in. On input x, M halts within f( x ) steps and so, none of its strings can become larger than f( x ). The same is true for M. Because of the two scans M does for each step in M plus the cursor move, the run time of one step in M is (4 f( x )), which gives a total time of O(f 2 ( x )). 2
3 Exercise 2 Construct a Turing machine deciding the language L = {w$w w Σ }, where $ Σ. What are the time and space complexities of your machine? We can build a two-tape TM M = (K = {s, q, q }, Σ {$}, δ, s) that works as follows: it will read the input and copy it on the second tape until it reaches $ (or # in case the input string has no $ in it). If it reached $, the cursor on the second tape will move at the beginning of the tape. The machine will continue reading a symbol from both tapes as long as the symbols are equal. If it reaches # on both tapes at the same time, it will accept the string. In all other cases, it will end in the rejecting state. The transitions are listed in the table below: q K σ 1, σ 2 Σ {$} δ(q, σ 1, σ 2 ) s >, > (s, >,, >, ) s X, # (s, X,, X, ) s $, # (q, $,, #, ) s #, # (no, #,, #, ) q $, X (q, $,, X, ) q $, > (q, $,, >, ) q X, X (q, X,, X, ) q #, # (yes, #,, #, ) q X, X (no, X,, X, ) q #, X (no, #,, X, ) Here, by X we denote any symbol of Σ \ {>, #}, and by X any symbol in Σ \ {X}. Let m denote the length of the input. The machine reads the whole input at most once, and has an additional right-to-left move at the beginning of its second tape of length at most m. It follows that M runs in time O(m). The additional space it uses on its second tape is at most m, so the space complexity is O(m) as well. 3
4 Exercise 3 Construct a Turing machine deciding the language L = {ww w {0, 1} }. Example: L and 00111, 0100 L. What are the time and space complexities of your machine? We can build a nondeterministic two-tape TM M = (K = {s, q, q }, Σ = {0, 1, #, >}, δ, s) that works as follows: at each step, it nondeterministically chooses to either copy the input symbol to the second tape and move to the right, or move the cursor of the second tape to the beginning and start comparing the remaining string on the first tape with the string on the second tape. The machine stops when it finished reading the input string or it has found a mismatch while comparing the two tapes. If it reaches # on both tapes at the same time while doing the comparison, it will accept the string. In all other cases, it will end in the no state. The transitions are listed in the table below: q K σ 1, σ 2 Σ {$} δ(q, σ 1, σ 2 ) s >, > (s, >,, >, ) s X, # (s, X,, X, ) s X, # (q, X,, #, ) s #, # (no, #,, #, ) q X, Y (q, X,, Y, ) q X, > (q, X,, >, ) q X, X (q, X,, X, ) q #, # (yes, #,, #, ) q X, X (no, X,, X, ) q #, X (no, #,, X, ) Here, X, Y {0, 1} and X Σ \ {X}. Let m denote the length of the input. The machine reads the whole input at most once, and has an additional right-to-left move at the beginning of its second tape of length at most m. It follows that M runs in time O(m). The additional space it uses on its second tape is at most m, so the space complexity is O(m) as well. 4
5 Exercise 4 How much space do you need to encode as an input to a Turing machine: (a) 1,...,n (the set of natural numbers from 1 to n); We consider binary representation of numbers. Each number is encoded using log(n) +1 bits. Then to encode all numbers from 1 to n we need log(1) log(2) log(n) + 1 bits. We have that log(k) + n log(n) + n = n log(n) + n We can say now that we need O(n log(n) + n) space. Since the term n is significantly smaller than n log(n) for large values of n, we can ignore it and we get that we need O(n log(n)) space to encode the given input. (b) 1 2,2 2,,n 2 (the set of squares of natural numbers from 1 to n). We follow the same reasoning as before: to encode all squares of numbers from 1 to n we need log(1 2 ) log(2 2 ) log(n 2 ) + 1 bits. In other words we need log(k 2 ) + n = 2 2 log(k) + n = log(k) + n 2 n log(n) + n bits. We can conclude we need O(2 n log(n)) = O(n log(n)) space to encode the given input. Mind that log(n) is a shorthand notation for log 2 (n). 5
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