Homework Five Solution CSE 355

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1 Homework Five Solution CSE 355 Due: 17 April 01 Please note that there is more than one way to answer most of these questions. The following only represents a sample solution. Problem 1: Linz and : Show that the language L = {w {a, b, c} : n a (w) + n b (w) = n c (w)} is context-free, but not linear. In order to show that the language is context-free, we will give the machine definition of the corresponding npda and by Theorem 7., we know that there exists a CFL for the npda. Define M = ({q 0, q 1 }, Σ, {a, c, z}, δ, q 0, z, {q 1 }) with δ(q 0, a, z) = {(q 0, az)}, δ(q 0, a, a) = {(q 0, aa)}, δ(q 0, a, c) = {(q 0, λ)}, δ(q 0, b, z) = {(q 0, az)}, δ(q 0, b, a) = {(q 0, aa)}, δ(q 0, b, c) = {(q 0, λ)}, δ(q 0, c, z) = {(q 0, cz)}, δ(q 0, c, a) = {(q 0, λ)}, δ(q 0, c, c) = {(q 0, cc)}, δ(q 0, λ, z) = {(q 1, λ)}. Then M accepts L since it keeps count of how many as and bs are read with the stack symbol a and how many cs with the symbol c. For each a and b seen it will cancel a c if that is on the top of the stack or add an a. Similarly, for each c seen, it will add a c to the stack or cancel an a if it is on top of the stack. Finally, we only accept if the number of cs is the same as the number of as and bs combined, that is they all cancel each other on the stack. In order to show that the language is not linear, we will use the pumping lemma from Theorem 8.. Assume that the language is linear and apply the Theorem 8. to the string w = a m c m b m. Inequality (8.5) uvyz m shows that in this case the strings u, v must both consist entirely of a s, that is v = a k and y, z must both consist entirely of b s, that is y = b l. If we pump this string once, we get a m+k c m b m+l, with either k 1 or l 1 due to inequality (8.6), a result that is not in L. This contradiction of Theorem 8. proves that the language is not linear. 8..7: Show that the family of context-free languages is not closed under difference in general, but is closed under regular difference, that is, if L 1 is context-free and L is regular, then L 1 L is context-free. To show that context free languages are not closed under difference in general, we will give two context free languages whose difference is no longer context free. Let L 1 = {a n b n c m : m, n 0} and L = {a m b n c n : m, n 0}. L 1 and L are both context free languages (in fact determinstic context free languages), but L 1 L = L 1 L = {a n b n c m : m, n 0} {a m b n c n : m, n 0} = {a n b n c m : 1

2 m, n 0} {a m b n c n : m, n 0} = {a n b n c n : n 0} which is not context free by Example 8.1 in Linz. Thus, the family of context free languages is not closed under difference. Next, we will show that the class of context free languages is closed under regular difference. Let L 1 be any context free langauge and L be any regular language. We know from Theorm 4.1 that regular languages are closed under complement, therefore L is a regular language. From Theorem 8.5 of Linz we also know that context free languages are closed under regular intersection. Thus, L = L 1 L = L 1 L is a result of a regular intersection and is therefore a context free language. Problem : Linz and : Design Turing machines to compute the following functions for x and y positive integers represented in unary. For each of these problems let u(x) = 1 x where 1 x is 1 written x times. u(0) = λ and will be represented by the tape having only blanks on it. We will assume that (a) f(x) = 3x. The input to our tape will be u(x) and the output will be u(3x). Our strategy follows Example 9.10 of Linz. We will first mark all the 1s and then for every marked 1 we see we will write two 1s at the end of the string, thus tripling the string. Q = {q 0, q 1, q, q 3, q f }, where q 0 is the start state and q f the only final state, Σ = {1}, Γ = {1, 1, } where is the blank tape symbol, and δ(q 0, 1) = (q 0, 1, R), δ(q 0, ) = (q 1,, L), these two transitions mark all the 1s on the tape. δ(q 1, 1) = (q 1, 1, L), go left until a marked 1 is seen, δ(q 1, 1) = (q, 1, R), unmark it δ(q, 1) = (q, 1, R), go to the end of the tape δ(q, ) = (q 3, 1, R), δ(q 3, ) = (q 1, 1, L), write two 1s and repeat looking for the next marked 1. δ(q 1, ) = (q f,, R), if no marked 1 is found, we are done and halt. (b) f(x, y) = x y if x > y; and f(x, y) = 0 otherwise. The input to our tape will be u(x) u(y) and the output will be u(x y) if x > y and the blank tape otherwise. Our strategy will be to remove a 1 from the end of the string (the y portion) and then remove a corresponding 1 from the front of the string (the x portion). We continue until either we are out of 1s in the y portion, in which case we have u(x y) on the tape, or we are out of 1s in the x portion in which case x y, so we clear the tape and halt with the blank tape.

3 Q = {q 0, q 1, q, q 3, q 4, q 5, q f }, where q 0 is the start state and q f the only final state, Σ = {1, }, Γ = {1,, } where is the blank tape symbol, and δ(q 0, 1) = (q 0, 1, R), δ(q 0, ) = (q 0,, R), δ(q 0, ) = (q 1,, L), moved to the end of the input δ(q 1, 1) = (q,, L), we delete the end 1 δ(q, 1) = (q, 1, L), δ(q, ) = (q,, L), δ(q, ) = (q 3,, R), moved to the start of the input δ(q 3, 1) = (q 0,, R), and delete the first one, essentially substracting one of the sybmols in y from x. We then repeat moving to the end of the string. δ(q 1, ) = (q 4,, L), if out of input with y first (x y), clear the minus and δ(q 4, 1) = (q 4, 1, L), move back to the beginning δ(q 4, ) = (q f,, R), and halt δ(q 3, ) = (q 5,, R), if no 1 in the x portion to match to a 1 in the y portion (y > x) δ(q 5, 1) = (q 5,, R), clear the tape δ(q 5, ) = (q f,, R), and halt (c) f(x, y) = x + 3y. The input to our tape will be u(x) + u(y) and the output will be u(x + 3y). Our strategy follows Examples 9.9 and 9.10 of Linz. We will first mark all the 1s before the plus with one dot and then change the plus to a one with two dots and continue marking every one after it with two dots. We will remove the last 1 as extra to the sum. Then for every 1 with two dots seen we will write two 1s and the end of the string and for every 1 with one dot we will write one 1 and the end of the string, yielding the required total. Q = {q 0, q 1, q, q 3, q 4, q f }, where q 0 is the start state and q f the only final state, Σ = {1, +}, Γ = {1, +, 1, 1, } where is the blank tape symbol, and δ(q 0, 1) = (q 0, 1, R), replace every 1 before the + with a single dot δ(q 0, +) = (q 1, 1, R), δ(q 1, 1) = (q 1, 1, R), replace the + and every 1 after it with two dots 3

4 δ(q 1, ) = (q 1,, L), until we reach the end of the input δ(q 1, 1) = (q,, L), remove the last one (its place was taken by the +) δ(q, 1) = (q, 1, L), search for the next marked one δ(q, 1) = (q 3, 1, R), if the 1 has two marks δ(q 3, 1) = (q 3, 1, R), go to the end of the string δ(q 3, ) = (q 4, 1, R), δ(q 4, ) = (q, 1, L), and write two 1s. Repeat scanning the string for the next mark δ(q, 1) = (q 5, 1, R), if the 1 has one mark δ(q 4, 1) = (q 4, 1, R), go to the end of the string, where we would write a single 1 at the blank (as above for q 4 ) δ(q, ) = (q f,, R), If no mark is found, we have transformed the string, and so we halt. (d) f(x) = x, or f(x) = x if x is even and f(x) = x+1 if x is odd. The input to our tape will be u(x) and the output will be u( x ). Our strategy is to mark a 1 at the start of the string and remove a corresponding one at the end of the string. We will repeat marking the next unmarked 1 at the beginning and removing another 1 at the end. We will continue until there is no unmarked 1 left (x was even) and we cut the output in half, or there is no one left to remove at the end (x was odd) and the output is the fraction rounded up. Then we clear all marks and halt. Q = {q 0, q 1, q, q 3, q f }, where q 0 is the start state and q f the only final state, Σ = {1}, Γ = {1, 1, } where is the blank tape symbol, and δ(q 0, 1) = (q 1, 1, R), Mark the first one δ(q 1, 1) = (q 1, 1, R), δ(q 1, ) = (q,, L), move to the end of the string δ(q, 1) = (q 3,, L), remove the last 1 δ(q 3, 1) = (q 3, 1, L), δ(q 3, 1) = (q 0, 1, R), move back to the first unmarked 1 in the string and repeat from the start δ(q 0, ) = (q,, L), If there are no unmarked 1s remaining to mark (x was even) δ(q, 1) = (q, 1, L), clear the marks moving to the start of the string (This also takes care of the case when x was odd since if there are no unmarked 1s and the end of the string to remove after marking a 1, q will start on a marked 1 and we have to remove the marks). δ(q, ) = (q f,, R), and halt 4

5 (e) f(x) = x mod 5. The input to our tape will be u(x) and the output will be u(x mod 5). Our strategy is to mark all the 1s as we scan to the right using our states to remember how many 1s we have seen mod 5. Then we will remove the marks from that many 1s and scan back to the left over the remaining marked 1s. Then we will delete all the marked ones and halt with x mod 5 on the tape. Q = {q 0, q 1, q, q 3, q 4, q 5, q f }, where q 0 is the start state and q f the only final state, Σ = {1}, Γ = {1, 1, } where is the blank tape symbol, and δ(q 0, 1) = (q 1, 1, R), Mark a one and move to the state of 1 mod 5 δ(q 1, 1) = (q, 1, R), Mark a one and move to the state of mod 5 δ(q, 1) = (q, 1, R), Mark a one and move to the state of 3 mod 5 δ(q 3, 1) = (q 4, 1, R), Mark a one and move to the state of 4 mod 5 δ(q 4, 1) = (q 0, 1, R), Mark a one and move to the state of 0 mod 5, repeating until the end δ(q 4, ) = (q 4,, L), δ(q 3, ) = (q 3,, L), δ(q, ) = (q,, L), δ(q 1, ) = (q 1,, L), δ(q 0, ) = (q 0,, L), whichever state we ended in we remember δ(q 4, 1) = (q 3, 1, L), δ(q 3, 1) = (q, 1, L), δ(q, 1) = (q 1, 1, L), δ(q 1, 1) = (q 0, 1, L), and unmark that many 1s (this will be our answer) δ(q 0, 1) = (q 0, 1, L), We then scan back to the beginning δ(q 0, ) = (q 5,, R), δ(q 5, 1) = (q 5,, R), and remove the marked 1s δ(q 5, 1) = (q 5, 1, L), until we find an unmarked one δ(q 5, ) = (q f,, R), and then halt. (f) f(x) = x, or f(x) = x if x is even and f(x) = x 1 if x is odd. The input to our tape will be u(x) and the output will be u( x ). Our strategy is similar to part (e) above, but now if we don t find an unmarked 1 at the end of the tape to match with a marked one at the front of the tape, we remove the last marked one before clearing the marks and halting. 5

6 Q = {q 0, q 1, q, q 3, q 4, q f }, where q 0 is the start state and q f the only final state, Σ = {1}, Γ = {1, 1, } where is the blank tape symbol, and δ(q 0, 1) = (q 1, 1, R), Mark the first one δ(q 1, 1) = (q 1, 1, R), δ(q 1, ) = (q,, L), move to the end of the string δ(q, 1) = (q 3,, L), remove the last 1 δ(q 3, 1) = (q 3, 1, L), δ(q 3, 1) = (q 0, 1, R), move back to the first unmarked 1 in the string and repeat from the start δ(q 0, ) = (q 4,, L), If there are no unmarked 1s remaining to mark (x was even) δ(q 4, 1) = (q 4, 1, L), clear the marks moving to the start of the string δ(q 4, ) = (q f,, R), and halt δ(q, 1) = (q 4,, L), If no unmarked 1s to match at the end of the string (x was odd), remove the last marked 1 and go to the state to clear the marks 9..7: Sketch the construction of a Turing machine that can perform the addition and multiplication of positive integers x and y given in the usual decimal notation. For decimal addition f(x, y) = x + y. We will start with x + y on the tape and end with the sum on the tape. The idea is to decrement y and increment x, until y becomes 0. First we decrement y with the following TM: Move to the rightmost digit in y. Repeat the following steps until done. If the current symbol under the tape head is +, then clear y and go to the start of x and stop (the addition is complete). If the current symbol is any digit is anything other than a 0, decrement the current digit under tape head by 1 (replace the digit with the next lower digit) and proceed to increment x by one. The decrementing is done. If the current symbol under tape is 0, replace it with 9 and move the tape head to the left. Continue from Repeat. Then we increment x with the following TM: Move to the rightmost digit in x. Repeat the following steps until done. If the symbol under the tape head is a blank, write a 1. 6

7 If the symbol under the tape head is any digit other than 9, increment the current digit under tape head by 1 (replace the digit with the next higher digit) and proceed to decrement y. We have finished the increment. If the current digit under tape head is 9, replace it with 0 and move the tape head to the left. Continue from Repeat. For decimal multiplication f(x, y) = xy, we start with x y on the tape and stop with the product. The idea is to decrement y until it becomes 0 and for each decrement of y, add x to a running sum. Decrementing y can be done using the decrement module mentioned above. To keep adding x to x we first copy x to another part of the tape and keep a running sum on that part. We then use the adding module above on the copy of x and the running sum. After the sum is complete we decrement y again and repeat. Problem 3: Linz : Let L be a deterministic context-free language and define a new language L 1 = {w : aw L, a Σ}. Is it necessarily true that L 1 is a deterministic contextfree language? No! We will use a variation of Example 7.11 to give a counter-example. Let L = {ca n b n : n 0} {da n b n : n 0}, then L is a deterministic context-free language, if we start with a c we follow a branch where we accept or if we read an a we push one extra a on the stack and then pop one a for each b read, accepting only if the stack is empty. If we start with a d, then we follow another branch where we accept or if we read an a we push two extra as on the stack and then for each b read we pop one a, accepting only if the stack is empty. In every configuration we can only move to one other configuration, meeting the criteria for determinism. However, then L 1 = {a n b n : n 0} {a n b n : n 0} (removing the first symbol from every string in L), which is the language in Example 7.11 which was shown in the example to be nondeterministic. Problem 4: Linz : Provide a high-level description for Turing machines that accept the following languages on {a, b}. For each problem, define a set of appropriate macroinstructions that you feel are reasonably easy to implement. Then use them for the solution. (a) L = {ww R }. We have to match the first input symbol with the last input symbol on the tape, then proceed to match the second input symbol and second last input symbol, and so on until we reach the middle of the string. On input w {a, b} : 1. Repeat the following steps until the input string contains no more a s or b s. Move to the leftmost a or b on the tape. At this stage, if it cannot find a a or b, go to. 7

8 If the symbol on the tape head is an a: Replace it with another symbol 0. Then, scan to a 0 or 1 or blank. If the symbol to the left of it is an a then change it to a 0 and continue from step 1. If the symbol before it is not an a, computation of the TM halts there in a non-accepting state. If the symbol on the tape head is a b: Replace it with another symbol 1. Then, scan to a 0 or 1 or blank. If the symbol to the left of it is b then change it to a 1 and continue from step 1. If the symbol before it is not an b, computation of the TM halts there in a non-accepting state.. Replace all 0s with as and 1s with bs and halt in an accepting state. (b) L = {w 1 w : w 1 w : w 1 = w }. We have to match the length of the first half of the string with the second half of the string and ensure that the first-half substring is not the same as the second-half substring. On input w {a, b} : 1. Repeat the following steps until the input string contains no more a s or b s (This part checks if the string has even length). Move to the leftmost a or b on the tape. At this stage, if it cannot find an a or b, move to the left end of the tape (until it finds a blank symbol) and go to step. If the current symbol is an a, replace it with another symbol 0. If the current symbol is a b, replace it with another symbol 1. Then scan to the first 0, 1, or blank and move left. If the currenty symbol is an a replace it with a 0 and move left. If it is a b replace it with a 1 and move left. If an a or a b is not found at this stage, reject the string (it doesn t have an even length). If the current symbol is a 0, replace it with a #0. If it is a 1, replace it with a #1. (This marks the last symbol of w 1 in the string). Repeat from step 1.. Repeat the following steps until there are no more 0s or 1s on the tape (This part checks if the two halves are equal): Move to the leftmost 0 or 1 on the tape. If a 0 or 1 cannot be found reject and halt. (The strings were identical). If the next symbol is a 0: Replace it with another symbol a. Then, move to the first 0 or 1 after the #. If that symbol is a 1, accept (Different input at this position, so not equal). If it is a 0, replace it with an a. Repeat from. If the next symbol in the input is a 1: 8

9 Replace it with another symbol b. Then, move to the first 0 or 1 after the #. If that symbol is a 0, accept (Different input at this position, so not equal). If it is a 1, replace it with a b. Repeat from. (c) The complement of the language in part (a). The solution to part (a) has explicitly stated the transitions to accepting and non-accepting states. If the TM in part (a) accepts, reject. If the TM in part (a) rejects, accept. Such a TM will essentially represent the complement of the language in part (a). (d) L = {a n b m : m = n, n 1}. The idea here is to remove n bs for each a seen. This is done with the following TM: On input w {a, b} : 1. Make sure the string is in L(a b ). If it is continue. Otherwise, halt and reject. (This can be done by scanning to the left over the as until a b or blank. If any as are seen after the first b reject.). Mark the first unmarked a. If there are no unmarked as go to Scan to the end of the string and remove n bs. (This can be accomplished by zigzagging between the as and bs giving the as a second mark and deleting bs). If there are not enough bs to remove, then halt and reject (fewer than n bs). 4. Scan to the start of the string and repeat from. 5. If the first symbol after the marked as is a b (more than n bs), halt and reject. If the first symbol is a blank, halt and accept. (e) L = {a n : n is a prime number}. The idea here is to check if there is an 1 < i < n such that n mod i = 0. If there is then n is not prime. We implement this as follows: On input w {a, b} : 1. Make sure the string is in L(a + ). If it is continue. Otherwise, halt and reject. (This can be done by scanning to the left over any as. If a b is seen before the end blank, reject).. If there is exactly one a on the tape, halt and reject (1 is not prime). If there are exactly two as on the tape, halt and accept ( is a prime number). Otherwise, continue. (Those were the base cases). 3. For every a on the tape write an x on another part of the tape. 4. Delete the first x (we only want to check divisibility up to n 1) and replace the next x with a y. 5. Find the next x on the tape and replace it with a y. If there are no more xs halt and accept (n was not divisible by any number smaller than itself and so is prime). 9

10 6. Then for the number of ys on the tape, call it i, check if n mod i = 0. If it is halt and reject. If it is not continue from 5. (This can be done using a modified version of our answer to Problem (e) above, where we zigzag between ys and as placing new marks on them. After we finish the ys we remove the marks on them and start the matching over from the first y. We continue to do so until we run out of as to match (this does the mod ). If we reach the end of the as and ys at the same time, then n is divisible by i and hence not prime; whence, we reject). Problem 5: Linz Write a simple program for a nondeterministic Turing machine that accepts the language L = {xww R y : x, y, w {a, b} +, x y }. How would you solve this problem deterministically? Nondeterminism: We will use the following nondeterministic TM: On input z {a, b} + : 1. Scan to the right and nondeterministically mark any two symbols in the string.. Check if there are at least as many symbols to the left of (and including) the first marked symbol as there are to the right of (and including) the last marked symbol. If there are continue. Otherwise, halt in a nonaccepting state. 3. Check if there is any sybmols between the two marks. If not reject. If there are check if the symbols between the marks are of the form ww R (this can be done by zigzagging from front to the end matching symbols). If it is accept and halt. Otherwise reject. This machine marks two symbols, which breaks the string up into a possible x, y and ww R portions. It does this nondeterministically, meaning that for any way to split the string there is a branch of computation where that split was checked. If at least one of the splits is of the correct form then we accept. Determinism: Here we have to keep updating the marks and check the corresponding split until all possible ways to split are tried. The following TM implements this algorithm: On input z {a, b} + : 1. Mark the first and last symbol in z.. If the inside part of the two marks is not empty and can be expressed as ww R, accept and halt. 3. If the first mark is not on the next to last symbol, continue. Otherwise go to Move the first mark one to the right. 5. Place the last mark on the last symbol. 6. If the middle part of the two marks is not empty and can be expressed as ww R, accept and halt. 10

11 7. If there are more symbols to the left of the first mark than to the right of the second mark and there is an unmarked symbol between the two marks, move the second mark left and repeat from 6. Otherwise repeat from Halt in a non-accepting state (we have tried all possible ways to split the string and none of them produced a string in L). 11