Homework Four Solution CSE 355

Save this PDF as:

Size: px
Start display at page:

Transcription

1 Homework Four Solution CSE 355 Due: 29 March 2012 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 : Find a context-free grammar for the set of all regular expressions on the alphabet {a, b}. Let G be the grammar given by the following productions: S a b λ S + S SS S (S) Let R be the set of all regular expressions over {a, b}. Then any element in R must have one of the seven forms listed in the definition of regular expression given in Definition 3.1. These are precisely the only strings that G can generate, with each production representing one of the generation rules for regular expressions. More formally (but not required in your solution), let G = ({S}, {a, b, λ,, +,, (, )}, S, P ), where P is the set of productions given above. Now, we must show that L(G) = R. Let r L(G). We will show that r R by induction on the number of derivations needed to generate r. Base Case: Assume r is generated using 1 derivation. The only productions in P that end in a terminal will generate one of the following strings: r = a, r = b, r = λ or r =. In each case r is a regular expression and hence r R. Induction Hypothesis: Assume that if G generates r in fewer than n derivations then r R. Induction Step: Assume that G generates r in n derivations, with n 2. Since r is not generated in only one production, one of the following productions of G must be used: S S + S. In this case S S + S r 1 + S r 1 + r 2, where r 1 and r 2 are regular expressions from the induction hypothesis (takes fewer than n derivation to generate since one derivation was used to get the sentential form S + S). Therefore, r = r 1 + r 2 and we get that r R. S SS. In this case S SS r 1 S r 1 r 2, where r 1 and r 2 are regular expressions from the induction hypothesis. Therefore, r = r 1 r 2 and we get that r R. 1

2 S S. In this case S S r1, where r 1 is a regular expression from the induction hypothesis. Therefore, r = r1 and we get that r R. S (S). In this case S (S) (r 1 ), where r 1 is a regular expression from the induction hypothesis. Therefore, r = (r 1 ) and we get that r R. In each case we get that r R Therefore, we conclude that if r L(G) then r R. Whence, L(G) R. We still have to show that R L(G). Assume r R. We will show that r L(G) by induction on the length of r. Base Case: Assume r = 1. Then r is a primitive regular expression and must have one of the following forms: r = x, where x {a, b, λ, }. In each case S x will generate r. Thus, r L(G). Induction Hypothesis: Assume that if r < n then S r, that is r L(G). Induction Step: Assume that r = n, with n 2. Since r > 1, r must have one of the following forms: r = r 1 + r 2, where r 1 and r 2 are regular expressions, each with length less than n. By the induction hypothesis S r 1 and S r 2. Therefore, S S + S r 1 + r 2 = r. Thus, r L(G). r = r 1 r 2, where r 1 and r 2 are regular expressions, each with length less than n. By the induction hypothesis S r 1 and S r 2. Therefore, S SS r 1 r 2 = r. Thus, r L(G). r = r1, where r 1 is a regular expressions with length less than n. By the induction hypothesis S r 1. Therefore, S S r1 = r. Thus, r L(G). r = (r 1 ), where r 1 is a regular expressions with length less than n. By the induction hypothesis S r 1. Therefore, S (S) (r 1 ) = r. Thus, r L(G). In each case we get that r L(G) Therefore, we conclude that if r L(G) then r R. Whence, R L(G). Therefore, we have that L(G) = R and the claim is proved : Transform the grammar with productions into Chomsky normal form. S abab, A bab λ, B BAa A λ. Removing λ productions: Here A and B are nullable variables. Then following the second step of construction in Theorem 6.3, we get S abab aba abb ab, A bab ba, B BAa A Ba Aa a. 2

3 Removing unit-productions(by Theorem 6.4): S abab aba abb ab, A bab ba, B BAa bab ba Ba Aa a. There are no useless productions in the grammar. By step 1 of Theorem 6.6, we introduce variables C and D to substitute terminals a and b. S CDAB CDA CDB CD, A DCB DC, B BAC DCB DC BC AC a, C a, D b. By step 2 of Theorem 6.6, we introduce variables to shorten the right sides of the production. Now the grammar is in CNF. Problem 2: Linz and S EF EA EB CD A GB DC, B BH GB DC BC AC a, E CD, F AB, G DC, H AC, C a, D b : An alternative to Definition 7.2 for language acceptance is to require the stack to be empty when the end of the input string is reached. Formally, an npda M is said to accept the language N(M) by empty stack if N(M) = {w Σ : (q 0, w, z) M (p, λ, λ)}, where p is any element in Q. Show that this notion is effectively equivalent to Definition 7.2, in the sense that for any npda M there exists an npda M such that L(M) = N( M), and vice versa. We will give a procedure to convert an npda M = (Q, Σ, Γ, δ, q 0, z, F ) into a new npda M = ( Q, Σ, Γ, δ, q 0, z, F ) such that L(M) = N( M). 1. Add new states q s and q f to Q. That is Q = Q {q s, q f }. 2. Make q s the start state of M. That is q 0 = q s. This is used to add a new bottom element to our stack. 3

4 3. Add a new stack symbol \$ to Γ. That is Γ = Γ {\$}. This will be used to ensure that M does not accept in a nonfinal state just because the stack of M is empty after reading the input. 4. Create δ as follows: Add the transition δ(q s, λ, z) = {(q 0, z\$)}. This places our new stack symbol on the bottom of the stack. Add the transitions δ(p, λ, a) = {(q f, λ)}, for all p F, a Γ. This nondeterministically checks that if we are in a final state, have no input, but have elements on the stack. Add the transition δ(q f, λ, a) = {(q f, λ)}, for all a Γ. This empties the stack in the new state, thus accepting. All other transitions are the same as in δ. That is, δ(q, b, a) = δ(q, b, a), for all q Q, b Σ, and a Γ. 5. This completes the creation of M Now we must show that L(M) = N( M). If a string is not accepted by L(M), then every state that it can reach with no input left to read must be a non-final state. This string will end in the same set of states in M (since the only transition to q f is from a final state). However, the stack must have the symbol \$ on it since the only way to pop it is from an accepting state or q f. Therefore, the string is also not in N( M). Conversely, if a string is in L(M), then there must be some sequence of moves so that it ends in a final state with no input remaining to be read. Then the string would be in the same set of states in M but would also be in q f with an empty stack by following the transition from the final state and then emptying the stack. Therefore, the string is also in N( M). Thus, L(M) = N( M). More formally (again not required), w L(M) iff there is a sequence of moves in M, (q 0, w, z) M (p, λ, ab) with p F, a Γ {λ} and b Γ iff there is a sequence of moves in M, (q s, w, z) M (q 0, w, z\$) M (p, λ, ab\$) M (q f, λ, b\$) M (q f, λ, λ) (since M has the same moves as M except for the new start and from final states) iff w N( M) : Construct an npda corresponding to the grammar S aabb aaa, A abb a, B bbb A. The grammar does not have any λ productions. So we can convert it into Greibach normal form. By applying Theorem 6.1 on last production, S aabb aaa, A abb a, B bbb abb a. 4

5 By applying construction in Theorem 7.1, Define M = ({q 0, q 1, q f }, Σ, {S, A, B, z}, δ, q 0, z, {q f }) with δ(q 0, λ, z) = {(q 1, Sz)}, δ(q 1, a, S) = {(q 1, ABB), (q 1, AA)}, δ(q 1, a, A) = {(q 1, BB), (q 1, λ)}, δ(q 1, b, B) = {(q 1, BB)}, δ(q 1, a, B) = {(q 1, BB), (q 1, λ)}, δ(q 1, λ, z) = {(q f, z)}. Problem 3: Linz and : Is it possible for a regular grammar to be ambiguous? Yes. Consider the grammar S A λ A λ. Then there are two leftmost derivations that generate the string λ. The first is S λ and the second is S A λ. Thus, this regular grammar is ambiguous. Note as problem states that regular languages are not inherently ambiguous. This is because one can make a DFA for the language and convert that DFA to a grammar. Since each string has only one path in a DFA the resulting grammar would also have only one derivation following the same path in the DFA : Eliminate useless productions from S a aa B C, A ab λ, B Aa, C ccd, D ddd. Following the algorithm given in the proof of Theorem 6.2, we will first remove productions that cannot generate any strings. Let V 1 =. Since S a, A λ and D ddd are productions we add S, A and D to V 1. Then, since B Aa is a production we add B to V 1. There are no more variables that can produce a string following the algorithm. Removing these useless productions yields the grammar S a aa B, A ab λ, B Aa, D ddd. Next we remove the unreachable variables. We will start with V 2 = {S} since S is the start variabe, we always reach it. We can reach the variables A and B from S so they are added to V 2. A and B can only reach each other so our algorithm terminates. Since D is not in S 2 it cannot be 5

6 reached by any derivation of this grammar. Therefore we remove all productions with D, yielding the final grammar free of useless productions and variables Problem 4: Linz S a aa B, A ab λ, B Aa, Construct npda s that acccept the following languages on Σ = {a, b, c}. (a) L = {a n b 2n : n 0}. Define M = ({q 0, q 1, q 2, q 3 }, Σ, {a, z}, δ, q 0, z, {q 0, q 3 }) with δ(q 0, a, z) = {(q 1, aaz)}, δ(q 1, a, a) = {(q 1, aaa)}, δ(q 1, b, a) = {(q 2, λ)}, δ(q 2, b, a) = {(q 2, λ)}, δ(q 2, λ, z) = {(q 3, λ)}. Then M accepts L since it accepts the empty string and for each a seen it pushes two additional as onto the stack (3 total). Then it pops one a for each b seen guaranteeing there are twice as many bs as as. Finally it accepts if the input is done and there are no more as on the stack. (b) L = {wcw R : w {a, b}}. Define M = ({q 0, q 1, q 2 }, Σ, {a, b, z}, δ, q 0, z, {q 2 }) with δ(q 0, a, a) = {(q 0, aa)}, δ(q 0, b, a) = {(q 0, ba)}, δ(q 0, a, b) = {(q 0, ab)}, δ(q 0, b, b) = {(q 0, bb)}, δ(q 0, a, z) = {(q 0, az)}, δ(q 0, b, z) = {(q 0, bz)}, δ(q 0, c, z) = {(q 1, z)}, δ(q 0, c, a) = {(q 1, a)}, δ(q 0, c, b) = {(q 1, b)}, δ(q 1, a, a) = {(q 1, λ)}, δ(q 1, b, b) = {(q 1, λ)}, δ(q 1, λ, z) = {(q 2, z)}. Then M accepts L. It accepts c. For each a or b seen (if any) it pushes an a or b respectively onto the stack and stays in state q 0. When it encounters a c, it does not push anything on to the stack, but gets to the next state, q 1. While being in state q 1, if it encounters an a or b, it matches it with the a or b respectively on the stack and pops one a for each a or one b for each b seen guaranteeing that it matches w R against the contents of the stack and the matching starts after it encounters a c. Finally it accepts if the input is done and there are no more as or b on the stack. (c) L = {a n b m c n+m : n 0, m 0}. Define M = ({q 0, q 1, q 2, q 3, q 4 }, Σ, {a, z}, δ, q 0, z, {q 0, q 4 }) with δ(q 0, λ, z) = {(q 1, z)}, δ(q 1, a, z) = {(q 1, az)}, δ(q 1, a, a) = {(q 1, aa)}, δ(q 1, λ, a) = {(q 2, a)}, δ(q 1, λ, z) = {(q 2, z)}, δ(q 2, b, z) = {(q 2, az)}, δ(q 2, b, a) = {(q 2, aa)}, δ(q 2, c, a) = {(q 3, λ)}, δ(q 3, c, a) = {(q 3, λ)}, δ(q 3, λ, z) = {(q 4, λ)}. Then M accepts L since it accepts the empty string and for each a seen (if any) it pushes an as onto the stack. Then it pushes an a onto the stack for each b seen (if any). Thus after reading all as and bs in the input the stack has n a (w) + n b (w) as on it. It then pops one a for each c seen guaranteeing that n a (w) + n b (w) = n c (w) and they are in the right order. Finally it accepts if the input is done and there are no more as on the stack. 6

7 (d) L = {a n b m+n c m : n 0, m 1}. Define M = ({q 0, q 1, q 2, q 3, q 4 }, Σ, {a, b, z}, δ, q 0, z, {q 4 }) with δ(q 0, λ, z) = {(q 1, z)}, δ(q 0, a, z) = {(q 0, az)}, δ(q 0, a, a) = {(q 0, aa)}, δ(q 0, b, a) = {(q 1, λ)}, δ(q 0, b, z) = {(q 1, bz)}, δ(q 1, b, a) = {(q 1, λ)}, δ(q 1, b, z) = {(q 2, bz)}, δ(q 2, b, b) = {(q 2, bb)}, δ(q 2, c, b) = {(q 3, λ)}, δ(q 3, c, b) = {(q 3, λ)}, δ(q 3, λ, z) = {(q 4, λ)}. Then M accepts L. It does not accept the empty string. If it initially encounters a a (if any), it pushes an a onto the stack. Then it pushes an a onto the stack for each a seen (if any). Then it pops an a for every b seen, if any. Once the stack is empty and it sees a b or if it initially encountered a b, it pushes a b on to the stack. It then pops one b for each c seen guaranteeing that n a (w) + n c (w) = n b (w) and they are in the right order and n 0, m 1. Finally it accepts if the input is done and there are no more bs on the stack. (e) L = {a 3 b n c n : n 0}. Define M = ({q 0, q 1, q 2, q 3, q 4, q 5, q 6 }, Σ, {b, z}, δ, q 0, z, {q 3, q 6 }) with δ(q 0, a, z) = {(q 1, z)}, δ(q 1, a, z) = {(q 2, z)}, δ(q 2, a, z) = {(q 3, z)}, δ(q 3, b, z) = {(q 4, bz)}, δ(q 4, b, b) = {(q 4, bb)}, δ(q 4, c, b) = {(q 5, λ)}, δ(q 5, c, b) = {(q 5, λ)}, δ(q 5, λ, z) = {(q 6, λ)}. Then M accepts L since the first three states ensure that the string starts with three as. Then if nothing else follows the string is accepted. For every b seen it pushes a b onto the stack. Then it pops one b for each c seen guaranteeing that the number of bs and cs match and they are in the right order. Finally it accepts if the input is done and there are no more bs on the stack. (f) L = {a n b m : n m 3n}. Define M = ({q 0, q 1, q 2 }, Σ, {a, b, z}, δ, q 0, z, {q 2 }) with δ(q 0, a, z) = {(q 0, az)}, δ(q 0, a, z) = {(q 0, aaz)}, δ(q 0, a, z) = {(q 0, aaaz)}, δ(q 0, a, a) = {(q 0, aa)}, δ(q 0, a, a) = {(q 0, aaa)}, δ(q 0, a, a) = {(q 0, aaaa)}, δ(q 0, b, a) = {(q 1, λ)}, δ(q 1, b, a) = {(q 1, λ)}, δ(q 1, λ, z) = {(q 2, z)}. Then M accepts L. It accepts empty string. The state q 0 keeps track of the n as encountered in the first part of the string and provides a chance for state q 1 to match the bs to anywhere between n and 3n. It proceeds to state q 2 from state q 1 if the number of bs encountered is anywhere between n and 3n. (g) L = {w : n a (w) = n b (w) + 1}. Define M = ({q 0, q 1, q 2 }, Σ, {a, b, z}, δ, q 0, z, {q 2 }) with δ(q 0, λ, z) = {(q 1, az)}, δ(q 1, a, z) = {(q 1, az)}, δ(q 1, a, a) = {(q 1, aa)}, δ(q 1, a, b) = {(q 1, λ)}, δ(q 1, b, z) = {(q 1, bz)}, δ(q 1, b, a) = {(q 1, λ)}, δ(q 1, b, b) = {(q 1, bb)}, δ(q 1, λ, z) = {(q 2, λ)}, δ(q 1, c, z) = {(q 1, c)}, δ(q 1, c, a) = {(q a, a)}, δ(q 1, c, b) = {(q 1, b)}. Then M accepts L since it starts with one extra a on the stack and then uses the machine given in Example 7.4 to tell if the number of as and bs are the same. If they are and we started with one more a on the stack then the condition of L is met. Additionally, if it sees a c, it just reads it and does not modify the stack. 7

8 (h) L = {w : n a (w) = 2n b (w)}. Define M = ({q 0, q 1, q 2 }, Σ, {a, b, z}, δ, q 0, z, {q 2 }) with δ(q 0, λ, z) = {(q 2, z)}, δ(q 0, c, z) = {(q 0, z)}, δ(q 0, c, a) = {(q 0, a)}, δ(q 0, c, b) = {(q 0, b)}, δ(q 0, a, z) = {(q 0, az)}, δ(q 0, b, z) = {(q 0, bbz)}, δ(q 0, a, a) = {(q 0, aa}, δ(q 0, b, b) = {(q 0, bbb)}, δ(q 0, a, b) = {(q 0, λ)}, δ(q 0, b, a) = {(q 0, b)}, δ(q 0, b, a) = {(q 1, λ)}, δ(q 1, λ, a) = {(q 0, λ)}. Then M accepts L. It accepts empty string. If it sees a c, it just reads it and does not modify the stack. If it sees a a while the stack is empty or the top of stack has a, it pushes a a on to the stack. If it sees a b while the stack is empty or the top of stack has b, it pushes two bs to the stack. If it sees a a while top of stack has b, it pops b off the stack. If it encounters a b while top of stack has a, it could take two paths dependent on whether it is the last b it is encountering or not. If it is not the last b, it replaces the stack with a b. If it is the last b, it goes to state q 1 and pops off a from the stack. It then pops off another a from the stack (if any) and returns to state q 0. Pushing two bs while encountering each b and one a for each input a, and popping off symbols in the aforementioned manner ensures M accepts strings with n a (w) = 2n b (w). (i) L = {w : n a (w) + n b (w) = n c (w)}. 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. (j) L = {w : 2n a (w) n b (w) 3n a (w)}. Define M = ({q 0, q 1, q 2, q 3 }, Σ, {a, b, z}, δ, q 0, z, {q 3 }) with δ(q 0, a, z) = {(q 0, aaz)}, δ(q 0, a, z) = {(q 0, aaaz)}, δ(q 0, a, a) = {(q 0, aaa}, δ(q 0, a, a) = {(q 0, aaaa)}, δ(q 0, b, z) = {(q 0, bz)}, δ(q 0, b, b) = {(q 0, bb)}, δ(q 0, b, a) = {(q 0, λ)}, δ(q 0, a, b) = {(q 1, λ)}, δ(q 1, λ, b) = {(q 0, λ)}, δ(q 1, λ, z) = {(q 0, az)}, δ(q 1, λ, z) = {(q 0, aaz)}, δ(q 1, λ, b) = {(q 2, λ)}, δ(q 2, λ, b) = {(q 0, λ)}, δ(q 2, λ, z) = {(q 0, az)}, δ(q 0, c, z) = {(q 0, c)}, δ(q 0, c, a) = {(q 0, a)}, δ(q 0, c, b) = {(q 0, b)}. Then M accepts L. It accepts empty string. If it sees a c, it just reads it and does not modify the stack. The state q 0 keeps track of as and bs, and provides a chance for states q 0, q 1 and q 2 to match the as when b is encountered with a on the top of the stack by popping the stack, while tracking the condition 2n a (w) n b (w) 3n a (w). q 1 and q 2 also deal with the cases when the bottom of the stack is reached without satisfying the condition but there are still input alphabets to be parsed. Finally, we only accept if the condition 2n a (w) n b (w) 3n a (w) is satisfied. (k) L = {w : n a (w) < n b (w)}. 8

9 Define M = ({q 0, q 1 }, Σ, {a, b, 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, b) = {(q 0, λ)}, δ(q 0, b, z) = {(q 0, bz)}, δ(q 0, b, a) = {(q 0, λ)}, δ(q 0, b, b) = {(q 0, bb)}, δ(q 0, λ, b) = {(q 1, λ)}, δ(q 0, c, z) = {(q 0, c)}, δ(q 0, c, a) = {(q 0, a)}, δ(q 0, c, b) = {(q 0, b)}. Then M accepts L since it follows Example 7.4 to tell if the number of as and bs are the same, but now it only accepts if there is a b on top of the stack. The symbol on top of the stack is always the symbol we have read more of thus far while processing the string. Therefore, if the input is done and there is a b at the top of the stack then there are more bs than as. Additionally, if it sees a c, it just reads it and does not modify the stack. Problem 5: Linz Find a context-free grammar that generates the language accepted by the npda M = ({q 0, q 1 }, {a, b}, {A, z}, δ, q 0, z, {q 1 }), with transitions δ(q 0, a, z) = {(q 0, Az)}, δ(q 0, b, A) = {(q 0, AA)}, δ(q 0, a, A) = {(q 1, λ)}. This satisfies condition 2 but not 1 (conditions in section 7.2 Context F ree Grammars f or P ushdown Automata). In order to satisfy condition 1, we add two more rules to the CFG and make q 2 the only accepting state (q 1 now empties the stack). δ(q 0, a, z) = {(q 0, Az)}, δ(q 0, b, A) = {(q 0, AA)}, δ(q 0, a, A) = {(q 1, λ)} δ(q 1, λ, A) = {(q 1, λ)} δ(q 1, λ, z) = {(q 2, λ)}. The last three transitions are of the form (7.5). So they yield the corresponding productions (q 0 Aq 1 ) a, (q 1 Aq 1 ) λ, (q 1 zq 2 ) λ. From the first two transitions we get the set of productions (q 0 zq 0 ) a(q 0 Aq 0 )(q 0 zq 0 ) a(q 0 Aq 1 )(q 1 zq 0 ) a(q 0 Aq 2 )(q 2 zq 0 ), (q 0 zq 1 ) a(q 0 Aq 0 )(q 0 zq 1 ) a(q 0 Aq 1 )(q 1 zq 1 ) a(q 0 Aq 2 )(q 2 zq 1 ), (q 0 zq 2 ) a(q 0 Aq 0 )(q 0 zq 2 ) a(q 0 Aq 1 )(q 1 zq 2 ) a(q 0 Aq 2 )(q 2 zq 2 ), (q 0 Aq 0 ) b(q 0 Aa 0 )(q 0 Aq 0 ) b(q 0 Aq 1 )(q 1 Aq 0 ) b(q 0 Aq 2 )(q 2 Aq 0 ), (q 0 Aq 1 ) b(q 0 Aa 0 )(q 0 Aq 1 ) b(q 0 Aq 1 )(q 1 Aq 1 ) b(q 0 Aq 2 )(q 2 Aq 1 ), (q 0 Aq 2 ) b(q 0 Aa 0 )(q 0 Aq 2 ) b(q 0 Aq 1 )(q 1 Aq 2 ) b(q 0 Aq 2 )(q 2 Aq 2 ). (q 1 zq 0 ), (q 1 Aq 0 ), (q 1 zq 1 ), (q 1 Aq 2 ), (q 2 zq 0 ), (q 2 zq 1 ), (q 2 zq 2 ), (q 2 Aq 0 ), (q 2 Aq 1 ) and (q 2 Aq 2 ) do not occur at the left side of any production, so they are useless and can be eliminated. Also noting that the production (q 0 Aq 0 ) b(q 0 Aq 0 )(q 0 Aq 0 ) has no way of terminating and is useless we can 9

10 eliminate it. Then there is no production with (q 0 Aq 0 ) on the left hand side so we eliminate any production containing it on the right hand side. We are then left with the folloing shorter grammar (q 0 Aq 1 ) a (q 1 Aq 1 ) λ, (q 1 zq 2 ) λ, (q 0 zq 2 ) a(q 0 Aq 1 )(q 1 zq 2 ), (q 0 Aq 1 ) b(q 0 Aq 1 )(q 1 Aq 1 ). with (q 0 zq 2 ) as the starting variable. If we did one more step and also eliminated λ-productions and rearranged the productions we are left with the following much shorter grammar (q 0 zq 2 ) a(q 0 Aq 1 ), (q 0 Aq 1 ) b(q 0 Aq 1 ) a. 10

INSTITUTE OF AERONAUTICAL ENGINEERING

INSTITUTE OF AERONAUTICAL ENGINEERING DUNDIGAL 500 043, HYDERABAD COMPUTER SCIENCE AND ENGINEERING TUTORIAL QUESTION BANK Name : FORMAL LANGUAGES AND AUTOMATA THEORY Code : A40509 Class : II B. Tech II

Pushdown Automata. place the input head on the leftmost input symbol. while symbol read = b and pile contains discs advance head remove disc from pile

Pushdown Automata In the last section we found that restricting the computational power of computing devices produced solvable decision problems for the class of sets accepted by finite automata. But along

Formal Languages and Automata Theory - Regular Expressions and Finite Automata -

Formal Languages and Automata Theory - Regular Expressions and Finite Automata - Samarjit Chakraborty Computer Engineering and Networks Laboratory Swiss Federal Institute of Technology (ETH) Zürich March

Deterministic Push-Down Store Automata

Deterministic Push-Down tore Automata 1 Context-Free Languages A B C D... A w 0 w 1 D...E The languagea n b n : a b ab 2 Finite-tate Automata 3 Pushdown Automata Pushdown automata (pda s) is an fsa with

Pushdown automata. Informatics 2A: Lecture 9. Alex Simpson. 3 October, 2014. School of Informatics University of Edinburgh als@inf.ed.ac.

Pushdown automata Informatics 2A: Lecture 9 Alex Simpson School of Informatics University of Edinburgh als@inf.ed.ac.uk 3 October, 2014 1 / 17 Recap of lecture 8 Context-free languages are defined by context-free

Automata and Computability. Solutions to Exercises

Automata and Computability Solutions to Exercises Fall 25 Alexis Maciel Department of Computer Science Clarkson University Copyright c 25 Alexis Maciel ii Contents Preface vii Introduction 2 Finite Automata

C H A P T E R Regular Expressions regular expression

7 CHAPTER Regular Expressions Most programmers and other power-users of computer systems have used tools that match text patterns. You may have used a Web search engine with a pattern like travel cancun

Finite Automata and Formal Languages

Finite Automata and Formal Languages TMV026/DIT321 LP4 2011 Lecture 14 May 19th 2011 Overview of today s lecture: Turing Machines Push-down Automata Overview of the Course Undecidable and Intractable Problems

ÖVNINGSUPPGIFTER I SAMMANHANGSFRIA SPRÅK. 15 april 2003. Master Edition

ÖVNINGSUPPGIFTER I SAMMANHANGSFRIA SPRÅK 5 april 23 Master Edition CONTEXT FREE LANGUAGES & PUSH-DOWN AUTOMATA CONTEXT-FREE GRAMMARS, CFG Problems Sudkamp Problem. (3.2.) Which language generates the grammar

CS103B Handout 17 Winter 2007 February 26, 2007 Languages and Regular Expressions

CS103B Handout 17 Winter 2007 February 26, 2007 Languages and Regular Expressions Theory of Formal Languages In the English language, we distinguish between three different identities: letter, word, sentence.

CS154. Turing Machines. Turing Machine. Turing Machines versus DFAs FINITE STATE CONTROL AI N P U T INFINITE TAPE. read write move.

CS54 Turing Machines Turing Machine q 0 AI N P U T IN TAPE read write move read write move Language = {0} q This Turing machine recognizes the language {0} Turing Machines versus DFAs TM can both write

The Halting Problem is Undecidable

185 Corollary G = { M, w w L(M) } is not Turing-recognizable. Proof. = ERR, where ERR is the easy to decide language: ERR = { x { 0, 1 }* x does not have a prefix that is a valid code for a Turing machine

ให p, q, r และ s เป นพห นามใดๆ จะได ว า

เศษส วนของพห นาม ให A และ B เป นพห นาม พห นามใดๆ โดยท B 0 เร ยก B A ว า เศษส วนของพห นาม การดาเน นการของเศษส วนของพห นาม ให p, q, r และ s เป นพห นามใดๆ จะได ว า Q P R Q P Q R Q P R Q P Q R R Q P S P Q

Fundamentele Informatica II

Fundamentele Informatica II Answer to selected exercises 1 John C Martin: Introduction to Languages and the Theory of Computation M.M. Bonsangue (and J. Kleijn) Fall 2011 Let L be a language. It is clear

6.045: Automata, Computability, and Complexity Or, Great Ideas in Theoretical Computer Science Spring, 2010. Class 4 Nancy Lynch

6.045: Automata, Computability, and Complexity Or, Great Ideas in Theoretical Computer Science Spring, 2010 Class 4 Nancy Lynch Today Two more models of computation: Nondeterministic Finite Automata (NFAs)

Regular Expressions. Languages. Recall. A language is a set of strings made up of symbols from a given alphabet. Computer Science Theory 2

Regular Expressions Languages Recall. A language is a set of strings made up of symbols from a given alphabet. Computer Science Theory 2 1 String Recognition Machine Given a string and a definition of

Theory of Computation

Theory of Computation For Computer Science & Information Technology By www.thegateacademy.com Syllabus Syllabus for Theory of Computation Regular Expressions and Finite Automata, Context-Free Grammar s

Regular Languages and Finite State Machines

Regular Languages and Finite State Machines Plan for the Day: Mathematical preliminaries - some review One application formal definition of finite automata Examples 1 Sets A set is an unordered collection

Honors Class (Foundations of) Informatics. Tom Verhoeff. Department of Mathematics & Computer Science Software Engineering & Technology

Honors Class (Foundations of) Informatics Tom Verhoeff Department of Mathematics & Computer Science Software Engineering & Technology www.win.tue.nl/~wstomv/edu/hci c 2011, T. Verhoeff @ TUE.NL 1/20 Information

3515ICT Theory of Computation Turing Machines

Griffith University 3515ICT Theory of Computation Turing Machines (Based loosely on slides by Harald Søndergaard of The University of Melbourne) 9-0 Overview Turing machines: a general model of computation

Course Manual Automata & Complexity 2015

Course Manual Automata & Complexity 2015 Course code: Course homepage: Coordinator: Teachers lectures: Teacher exercise classes: Credits: X_401049 http://www.cs.vu.nl/~tcs/ac prof. dr. W.J. Fokkink home:

Computing Functions with Turing Machines

CS 30 - Lecture 20 Combining Turing Machines and Turing s Thesis Fall 2008 Review Languages and Grammars Alphabets, strings, languages Regular Languages Deterministic Finite and Nondeterministic Automata

Chapter 7 Uncomputability

Chapter 7 Uncomputability 190 7.1 Introduction Undecidability of concrete problems. First undecidable problem obtained by diagonalisation. Other undecidable problems obtained by means of the reduction

Pushdown Automata. International PhD School in Formal Languages and Applications Rovira i Virgili University Tarragona, Spain

Pushdown Automata transparencies made for a course at the International PhD School in Formal Languages and Applications Rovira i Virgili University Tarragona, Spain Hendrik Jan Hoogeboom, Leiden http://www.liacs.nl/

Motivation. Automata = abstract computing devices. Turing studied Turing Machines (= computers) before there were any real computers

Motivation Automata = abstract computing devices Turing studied Turing Machines (= computers) before there were any real computers We will also look at simpler devices than Turing machines (Finite State

Scanner. tokens scanner parser IR. source code. errors

Scanner source code tokens scanner parser IR errors maps characters into tokens the basic unit of syntax x = x + y; becomes = + ; character string value for a token is a lexeme

Introduction to Automata Theory. Reading: Chapter 1

Introduction to Automata Theory Reading: Chapter 1 1 What is Automata Theory? Study of abstract computing devices, or machines Automaton = an abstract computing device Note: A device need not even be a

a 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2.

Chapter 1 LINEAR EQUATIONS 1.1 Introduction to linear equations A linear equation in n unknowns x 1, x,, x n is an equation of the form a 1 x 1 + a x + + a n x n = b, where a 1, a,..., a n, b are given

2110711 THEORY of COMPUTATION

2110711 THEORY of COMPUTATION ATHASIT SURARERKS ELITE Athasit Surarerks ELITE Engineering Laboratory in Theoretical Enumerable System Computer Engineering, Faculty of Engineering Chulalongkorn University

Formal language and Automata Theory Course File

Formal language and Automata Theory Course File D.HIMAGIRI, Asst.Professor, CSE Department, JBIET. 2013-14 COURSE PLAN FACULTY DETAILS: Name of the Faculty:: Designation: Department:: D. HIMAGIRI Asst.Professor

SRM UNIVERSITY FACULTY OF ENGINEERING & TECHNOLOGY SCHOOL OF COMPUTING DEPARTMENT OF SOFTWARE ENGINEERING COURSE PLAN

Course Code : CS0355 SRM UNIVERSITY FACULTY OF ENGINEERING & TECHNOLOGY SCHOOL OF COMPUTING DEPARTMENT OF SOFTWARE ENGINEERING COURSE PLAN Course Title : THEORY OF COMPUTATION Semester : VI Course : June

Overview of E0222: Automata and Computability

Overview of E0222: Automata and Computability Deepak D Souza Department of Computer Science and Automation Indian Institute of Science, Bangalore. August 3, 2011 What this course is about What we study

(IALC, Chapters 8 and 9) Introduction to Turing s life, Turing machines, universal machines, unsolvable problems.

3130CIT: Theory of Computation Turing machines and undecidability (IALC, Chapters 8 and 9) Introduction to Turing s life, Turing machines, universal machines, unsolvable problems. An undecidable problem

Philadelphia University Faculty of Information Technology Department of Computer Science First Semester, 2007/2008.

Philadelphia University Faculty of Information Technology Department of Computer Science First Semester, 2007/2008 Course Syllabus Course Title: Theory of Computation Course Level: 3 Lecture Time: Course

The Inverse of a Square Matrix

These notes closely follow the presentation of the material given in David C Lay s textbook Linear Algebra and its Applications (3rd edition) These notes are intended primarily for in-class presentation

CSC4510 AUTOMATA 2.1 Finite Automata: Examples and D efinitions Definitions

CSC45 AUTOMATA 2. Finite Automata: Examples and Definitions Finite Automata: Examples and Definitions A finite automaton is a simple type of computer. Itsoutputislimitedto yes to or no. It has very primitive

MODULAR ARITHMETIC. a smallest member. It is equivalent to the Principle of Mathematical Induction.

MODULAR ARITHMETIC 1 Working With Integers The usual arithmetic operations of addition, subtraction and multiplication can be performed on integers, and the result is always another integer Division, on

OHJ-2306 Fall 2011 Introduction to Theoretical Computer Science

1 OHJ-2306 Fall 2011 Introduction to Theoretical Computer Science 2 3 Organization & timetable Lectures: prof. Tapio Elomaa, M.Sc. Timo Aho Tue and Thu 14 16 TB220 Aug. 30 Dec. 8, 2011 Exceptions: Thu

COMPRESSION SPRINGS: STANDARD SERIES (INCH)

: STANDARD SERIES (INCH) LC 014A 01 0.250 6.35 11.25 0.200 0.088 2.24 F F M LC 014A 02 0.313 7.94 8.90 0.159 0.105 2.67 F F M LC 014A 03 0.375 9.52 7.10 0.126 0.122 3.10 F F M LC 014A 04 0.438 11.11 6.00

Model 2.4 Faculty member + student

Model 2.4 Faculty member + student Course syllabus for Formal languages and Automata Theory. Faculty member information: Name of faculty member responsible for the course Office Hours Office Number Email

26 Integers: Multiplication, Division, and Order

26 Integers: Multiplication, Division, and Order Integer multiplication and division are extensions of whole number multiplication and division. In multiplying and dividing integers, the one new issue

Regular Expressions and Automata using Haskell

Regular Expressions and Automata using Haskell Simon Thompson Computing Laboratory University of Kent at Canterbury January 2000 Contents 1 Introduction 2 2 Regular Expressions 2 3 Matching regular expressions

Compiler I: Syntax Analysis Human Thought

Course map Compiler I: Syntax Analysis Human Thought Abstract design Chapters 9, 12 H.L. Language & Operating Sys. Compiler Chapters 10-11 Virtual Machine Software hierarchy Translator Chapters 7-8 Assembly

Turing Machines: An Introduction

CIT 596 Theory of Computation 1 We have seen several abstract models of computing devices: Deterministic Finite Automata, Nondeterministic Finite Automata, Nondeterministic Finite Automata with ɛ-transitions,

Markov Algorithm. CHEN Yuanmi December 18, 2007

Markov Algorithm CHEN Yuanmi December 18, 2007 1 Abstract Markov Algorithm can be understood as a priority string rewriting system. In this short paper we give the definition of Markov algorithm and also

Lempel-Ziv Coding Adaptive Dictionary Compression Algorithm

Lempel-Ziv Coding Adaptive Dictionary Compression Algorithm 1. LZ77:Sliding Window Lempel-Ziv Algorithm [gzip, pkzip] Encode a string by finding the longest match anywhere within a window of past symbols

Deterministic Finite Automata

1 Deterministic Finite Automata Definition: A deterministic finite automaton (DFA) consists of 1. a finite set of states (often denoted Q) 2. a finite set Σ of symbols (alphabet) 3. a transition function

Casey s Theorem and its Applications

Casey s Theorem and its Applications Luis González Maracaibo. Venezuela July 011 Abstract. We present a proof of the generalized Ptolemy s theorem, also known as Casey s theorem and its applications in

Learning Analysis by Reduction from Positive Data

Learning Analysis by Reduction from Positive Data František Mráz 1, Friedrich Otto 1, and Martin Plátek 2 1 Fachbereich Mathematik/Informatik, Universität Kassel 34109 Kassel, Germany {mraz,otto}@theory.informatik.uni-kassel.de

SYSTEMS OF EQUATIONS AND MATRICES WITH THE TI-89. by Joseph Collison

SYSTEMS OF EQUATIONS AND MATRICES WITH THE TI-89 by Joseph Collison Copyright 2000 by Joseph Collison All rights reserved Reproduction or translation of any part of this work beyond that permitted by Sections

CMPSCI 250: Introduction to Computation. Lecture #19: Regular Expressions and Their Languages David Mix Barrington 11 April 2013

CMPSCI 250: Introduction to Computation Lecture #19: Regular Expressions and Their Languages David Mix Barrington 11 April 2013 Regular Expressions and Their Languages Alphabets, Strings and Languages

Online EFFECTIVE AS OF JANUARY 2013

2013 A and C Session Start Dates (A-B Quarter Sequence*) 2013 B and D Session Start Dates (B-A Quarter Sequence*) Quarter 5 2012 1205A&C Begins November 5, 2012 1205A Ends December 9, 2012 Session Break

Push-down Automata and Context-free Grammars

14 Push-down Automata and Context-free Grammars This chapter details the design of push-down automata (PDA) for various languages, the conversion of CFGs to PDAs, and vice versa. In particular, after formally

Estimating Probability Distributions

Estimating Probability Distributions Readings: Manning and Schutze, Section 6.2 Jurafsky & Martin, Section 6.3 One of the central problems we face in using probability models for NLP is obtaining the actual

Using Hands-On Visualizations to Teach Computer Science from Beginning Courses to Advanced Courses

Using Hands-On Visualizations to Teach Computer Science from Beginning Courses to Advanced Courses Susan H. Rodger Department of Computer Science Duke University Durham, NC 27705 rodger@cs.duke.edu Abstract

LAMC Junior Circle March 18, 2012. The Sierpinski triangle and the Hanoi Tower Puzzle.

LAMC Junior Circle March 18, 2012 Olga Radko radko@math.ucla.edu Oleg Gleizer oleg1140@gmail.com The Sierpinski triangle and the Hanoi Tower Puzzle. Copyright: for home use only. This handout is a part

CAs and Turing Machines. The Basis for Universal Computation

CAs and Turing Machines The Basis for Universal Computation What We Mean By Universal When we claim universal computation we mean that the CA is capable of calculating anything that could possibly be calculated*.

Local periods and binary partial words: An algorithm

Local periods and binary partial words: An algorithm F. Blanchet-Sadri and Ajay Chriscoe Department of Mathematical Sciences University of North Carolina P.O. Box 26170 Greensboro, NC 27402 6170, USA E-mail:

ECS 120 Lesson 17 Church s Thesis, The Universal Turing Machine

ECS 120 Lesson 17 Church s Thesis, The Universal Turing Machine Oliver Kreylos Monday, May 7th, 2001 In the last lecture, we looked at the computation of Turing Machines, and also at some variants of Turing

Theory of Computation Class Notes 1

Theory of Computation Class Notes 1 1 based on the books by Sudkamp and by Hopcroft, Motwani and Ullman ii Contents 1 Introduction 1 1.1 Sets.............................................. 1 1.2 Functions

Lecture 1: Time Complexity

Computational Complexity Theory, Fall 2010 August 25 Lecture 1: Time Complexity Lecturer: Peter Bro Miltersen Scribe: Søren Valentin Haagerup 1 Introduction to the course The field of Computational Complexity

Graph Theory Problems and Solutions

raph Theory Problems and Solutions Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles November, 005 Problems. Prove that the sum of the degrees of the vertices of any finite graph is

Reading 13 : Finite State Automata and Regular Expressions

CS/Math 24: Introduction to Discrete Mathematics Fall 25 Reading 3 : Finite State Automata and Regular Expressions Instructors: Beck Hasti, Gautam Prakriya In this reading we study a mathematical model

Automata and Rational Numbers

Automata and Rational Numbers Jeffrey Shallit School of Computer Science University of Waterloo Waterloo, Ontario N2L 3G1 Canada shallit@cs.uwaterloo.ca http://www.cs.uwaterloo.ca/~shallit 1/40 Representations

Compiler Construction

Compiler Construction Regular expressions Scanning Görel Hedin Reviderad 2013 01 23.a 2013 Compiler Construction 2013 F02-1 Compiler overview source code lexical analysis tokens intermediate code generation

CS 301 Course Information

CS 301: Languages and Automata January 9, 2009 CS 301 Course Information Prof. Robert H. Sloan Handout 1 Lecture: Tuesday Thursday, 2:00 3:15, LC A5 Weekly Problem Session: Wednesday, 4:00 4:50 p.m., LC

Arithmetic Coding: Introduction

Data Compression Arithmetic coding Arithmetic Coding: Introduction Allows using fractional parts of bits!! Used in PPM, JPEG/MPEG (as option), Bzip More time costly than Huffman, but integer implementation

Automata and Formal Languages

Automata and Formal Languages Winter 2009-2010 Yacov Hel-Or 1 What this course is all about This course is about mathematical models of computation We ll study different machine models (finite automata,

Math 313 Lecture #10 2.2: The Inverse of a Matrix

Math 1 Lecture #10 2.2: The Inverse of a Matrix Matrix algebra provides tools for creating many useful formulas just like real number algebra does. For example, a real number a is invertible if there is

Matrix Algebra. Some Basic Matrix Laws. Before reading the text or the following notes glance at the following list of basic matrix algebra laws.

Matrix Algebra A. Doerr Before reading the text or the following notes glance at the following list of basic matrix algebra laws. Some Basic Matrix Laws Assume the orders of the matrices are such that

BEGINNING ALGEBRA ACKNOWLEDMENTS

BEGINNING ALGEBRA The Nursing Department of Labouré College requested the Department of Academic Planning and Support Services to help with mathematics preparatory materials for its Bachelor of Science

CS 103X: Discrete Structures Homework Assignment 3 Solutions

CS 103X: Discrete Structures Homework Assignment 3 s Exercise 1 (20 points). On well-ordering and induction: (a) Prove the induction principle from the well-ordering principle. (b) Prove the well-ordering

Theory of Computation Prof. Kamala Krithivasan Department of Computer Science and Engineering Indian Institute of Technology, Madras

Theory of Computation Prof. Kamala Krithivasan Department of Computer Science and Engineering Indian Institute of Technology, Madras Lecture No. # 31 Recursive Sets, Recursively Innumerable Sets, Encoding

Formal Grammars and Languages

Formal Grammars and Languages Tao Jiang Department of Computer Science McMaster University Hamilton, Ontario L8S 4K1, Canada Bala Ravikumar Department of Computer Science University of Rhode Island Kingston,

Session 5 Dissections and Proof

Key Terms for This Session Session 5 Dissections and Proof Previously Introduced midline parallelogram quadrilateral rectangle side-angle-side (SAS) congruence square trapezoid vertex New in This Session

Finite Automata and Formal Languages

Finite Automata and Formal Languages TMV026/DIT321 LP4 2011 Ana Bove Lecture 1 March 21st 2011 Course Organisation Overview of the Course Overview of today s lecture: Course Organisation Level: This course

Lecture summary for Theory of Computation

Lecture summary for Theory of Computation Sandeep Sen 1 January 8, 2015 1 Department of Computer Science and Engineering, IIT Delhi, New Delhi 110016, India. E- mail:ssen@cse.iitd.ernet.in Contents 1 The

CPS 140 - Mathematical Foundations of CS Dr. S. Rodger Section: Properties of Context-Free Languages èhandoutè Section 3.5 Which of the following languages are CFL? æ L=fa n b n c j j 0 énçjg æ L=fa n

4 BOOLEAN ALGEBRA AND LOGIC SIMPLIFICATION

4 BOOLEAN ALGEBRA AND LOGIC SIMPLIFICATION BOOLEAN OPERATIONS AND EXPRESSIONS Variable, complement, and literal are terms used in Boolean algebra. A variable is a symbol used to represent a logical quantity.

CS5236 Advanced Automata Theory Frank Stephan Semester I, Academic Year 2012-2013 Advanced Automata Theory is a lecture which will first review the basics of formal languages and automata theory and then

Ambiguity, closure properties, Overview of ambiguity, and Chomsky s standard form

Ambiguity, closure properties, and Chomsky s normal form Overview of ambiguity, and Chomsky s standard form The notion of ambiguity Convention: we assume that in every substitution, the leftmost remaining

Warm-up Theorems about triangles. Geometry. Theorems about triangles. Misha Lavrov. ARML Practice 12/15/2013

ARML Practice 12/15/2013 Problem Solution Warm-up problem Lunes of Hippocrates In the diagram below, the blue triangle is a right triangle with side lengths 3, 4, and 5. What is the total area of the green

25 The Law of Cosines and Its Applications

Arkansas Tech University MATH 103: Trigonometry Dr Marcel B Finan 5 The Law of Cosines and Its Applications The Law of Sines is applicable when either two angles and a side are given or two sides and an

MACM 101 Discrete Mathematics I

MACM 101 Discrete Mathematics I Exercises on Combinatorics, Probability, Languages and Integers. Due: Tuesday, November 2th (at the beginning of the class) Reminder: the work you submit must be your own.

CSE 135: Introduction to Theory of Computation Decidability and Recognizability

CSE 135: Introduction to Theory of Computation Decidability and Recognizability Sungjin Im University of California, Merced 04-28, 30-2014 High-Level Descriptions of Computation Instead of giving a Turing

Lecture 6. Inverse of Matrix

Lecture 6 Inverse of Matrix Recall that any linear system can be written as a matrix equation In one dimension case, ie, A is 1 1, then can be easily solved as A x b Ax b x b A 1 A b A 1 b provided that

Qualifying examination in the software area

Qualifying examination in the software area This is a sample examination involving the following 4 areas: Database systems, compiler design, programming languages, and software engineering. Each area is

Lecture 5: Context Free Grammars

Lecture 5: Context Free Grammars Introduction to Natural Language Processing CS 585 Fall 2007 Andrew McCallum Also includes material from Chris Manning. Today s Main Points In-class hands-on exercise A

Fast nondeterministic recognition of context-free languages using two queues

Fast nondeterministic recognition of context-free languages using two queues Burton Rosenberg University of Miami Abstract We show how to accept a context-free language nondeterministically in O( n log

2) Write in detail the issues in the design of code generator.

COMPUTER SCIENCE AND ENGINEERING VI SEM CSE Principles of Compiler Design Unit-IV Question and answers UNIT IV CODE GENERATION 9 Issues in the design of code generator The target machine Runtime Storage

Section 4.2: The Division Algorithm and Greatest Common Divisors

Section 4.2: The Division Algorithm and Greatest Common Divisors The Division Algorithm The Division Algorithm is merely long division restated as an equation. For example, the division 29 r. 20 32 948

Exercises with solutions (1)

Exercises with solutions (). Investigate the relationship between independence and correlation. (a) Two random variables X and Y are said to be correlated if and only if their covariance C XY is not equal

recursively enumerable languages context-free languages regular languages

CPS 140 - Mathematical Foundations of CS Dr. S. Rodger Section: Recursively Enumerable Languages èhandoutè Deænition: A language L is recursively enumerable if there exists a TM M such that L=LèMè. if

On the generation of elliptic curves with 16 rational torsion points by Pythagorean triples

On the generation of elliptic curves with 16 rational torsion points by Pythagorean triples Brian Hilley Boston College MT695 Honors Seminar March 3, 2006 1 Introduction 1.1 Mazur s Theorem Let C be a

Inf1A: Deterministic Finite State Machines

Lecture 2 InfA: Deterministic Finite State Machines 2. Introduction In this lecture we will focus a little more on deterministic Finite State Machines. Also, we will be mostly concerned with Finite State

Jan 31 Homework Solutions Math 151, Winter 2012. Chapter 3 Problems (pages 102-110)

Jan 31 Homework Solutions Math 151, Winter 01 Chapter 3 Problems (pages 10-110) Problem 61 Genes relating to albinism are denoted by A and a. Only those people who receive the a gene from both parents

Theory of Computation Lecture Notes

Theory of Computation Lecture Notes Abhijat Vichare August 2005 Contents 1 Introduction 2 What is Computation? 3 The λ Calculus 3.1 Conversions: 3.2 The calculus in use 3.3 Few Important Theorems 3.4 Worked

1. LINEAR EQUATIONS. A linear equation in n unknowns x 1, x 2,, x n is an equation of the form

1. LINEAR EQUATIONS A linear equation in n unknowns x 1, x 2,, x n is an equation of the form a 1 x 1 + a 2 x 2 + + a n x n = b, where a 1, a 2,..., a n, b are given real numbers. For example, with x and