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Hint of final exams jinnjy. Outline Hint of final 2006 (6/28/2007)

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Presentation on theme: "Hint of final exams jinnjy. Outline Hint of final 2006 (6/28/2007)"— Presentation transcript:

1 Hint of final exams jinnjy

2 Outline Hint of final 2006 (6/28/2007)

3 Hint of Problem 1 L 1, L 2, L 3 are RE, there exist M i s.t. M i accepts L i for i = 1,2,3. From condition (i) and (ii), it is obvious that for any of L i If w is in L i, then M i accepts w. If w in not in L i, then there exists j  i, s.t. M j accepts w.

4 Hint of Problem 1 From all above, we can build D i, i=1,2,3, s.t. D i decides L i. D i =“ On input w: For k = 1~infinity Run all M i on w with k steps. If M i accepts, then accept. Else if one of the other TMs accepts, then reject. Else continue.

5 Hint of Problem 2 (a) Alphabet = {1} is decidable. 全部的 domino 都是上面的 1 比下面的多,或是下面的 1 比上面的多,一定無解。 其它情況一定有解。 (b) 原版的 PCP 問題可以 reduce 到 binary 的 PCP 問題。只要用 fix-length encoding 把原本 PCP 問 題的 Alphabet 編成固定長度 0,1 字串,這樣就做 到 reduction 了。

6 Hint of Problem 3 (a) L(T) 基本定義 : L(T)={w | T accepts w} f T (w)= 從 universal TM 的定義和運作: 如果 w 在 L(T) 裡面,則 T accepts w , universal TM 的 simulation 也會做出 T accepts w , 在 L u 裡面。 相對的,如果 w 不在 L(T) 裡面, universal TM 的 simulation 不可能 accept w in L(T) iff in L u. 成功得到 L(T)  m L u.

7 Hint of Problem 3 (b) 見定理 9.8 (Handout 18 頁 ) 注意不要用成定理 9.9 ,定理 9.9 是 L u  m L ne

8 Hint of Problem 3 (a) L(T) 基本定義 : L(T)={w | T accepts w} f T (w)= 從 universal TM 的定義和運作: 如果 w 在 L(T) 裡面,則 T accepts w , universal TM 的 simulation 也會做出 T accepts w , 在 L u 裡面。 相對的,如果 w 不在 L(T) 裡面, universal TM 的 simulation 不可能 accept w in L(T) iff in L u. 成功得到 L(T)  m L u.

9 Hint of Problem 4 A is decidable, there must exists a TM M A s.t. M A decides A. B and B’ are non-empty, there must exists x,y, s.t. x is in B and y is in B’. A is mapping reducible to B if there is a computable function f s.t. for every w. w is in A iff f(w) in B.

10 Hint of Problem 4 We define f as follows: If w is in A, then f(w)=x. Otherwise, f(w)=y.

11 Hint of Problem 5 NP Def 1. NP=the class of languages that have polynomial time verifiers. If A is in NP, there exists another language D s.t. A={x| there exists y s.t. (x,y) is in D} and D is decidable in polynomial time. The style is like problem 4.17 in textbook.

12 Hint of Problem 5 NP Def 2. NP=the class of languages that can be decided by a polynomial time nondeterminstic TM.

13 Hint of Problem 5 NP-Complete: A language L is NP-complete iff 1. L is in NP 2. Every A in NP can be reduced to L in polynomial time.


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