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lat:homework [2009/11/27 22:52]
barbara.jobstmann
lat:homework [2009/12/06 12:32] (current)
radu.iosif
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 ===== 02.10.2009 ===== ===== 02.10.2009 =====
- Consider the automata construction from proof of the Myhill-Nerode theorem for a language L. Prove + 
 +Consider the automata construction from proof of the Myhill-Nerode theorem for a language L. Prove 
   * (a) the constructed automaton A is the minimal automaton (in the number of states) that recognizes L;   * (a) the constructed automaton A is the minimal automaton (in the number of states) that recognizes L;
   * (b) any minimal automaton recognizing L is isomorphic to A.   * (b) any minimal automaton recognizing L is isomorphic to A.
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 ===== 06.11.2009 ===== ===== 06.11.2009 =====
 +
      - Given a reachability game (G,F), give an algorithm that computes the 0-Attractor(F) set in time O(|E|).      - Given a reachability game (G,F), give an algorithm that computes the 0-Attractor(F) set in time O(|E|).
      - Consider the game graph shown in below and the following winning conditions: a) Occ(ρ) ∩ {1} ≠ ∅ and b) Occ(ρ) ⊆ {1,​2,​3,​4,​5,​6} and c) Inf(ρ) ∩ {4,5} ≠ ∅. Compute the winning regions and corresponding winning strategies showing the intermediate steps (i.e., the Attractor and Recurrence sets) of the computation.\\ {{hw1.gif|Figure 1}}      - Consider the game graph shown in below and the following winning conditions: a) Occ(ρ) ∩ {1} ≠ ∅ and b) Occ(ρ) ⊆ {1,​2,​3,​4,​5,​6} and c) Inf(ρ) ∩ {4,5} ≠ ∅. Compute the winning regions and corresponding winning strategies showing the intermediate steps (i.e., the Attractor and Recurrence sets) of the computation.\\ {{hw1.gif|Figure 1}}
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 ===== 13.11.2009 ===== ===== 13.11.2009 =====
 +
      - Consider the game graph shown below. ​ Let the winning condition for Player 0 be Occ(ρ)={1,​2,​3,​4,​5,​6,​7}      - Consider the game graph shown below. ​ Let the winning condition for Player 0 be Occ(ρ)={1,​2,​3,​4,​5,​6,​7}
         - Find the winning region for Player 0 and describe a winning strategy         - Find the winning region for Player 0 and describe a winning strategy
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 ===== 27.11.2009 ===== ===== 27.11.2009 =====
 +
      - Consider the game graph below and the Muller condition F={{2,​4,​5,​7},​{1,​2,​3,​4,​5,​6,​7}}. Find an automaton winning strategy for Player 0 in the Muller game with as few states as possible and show that any other automaton strategy with less states is not winning for Player 0.\\ {{hw4.gif}}      - Consider the game graph below and the Muller condition F={{2,​4,​5,​7},​{1,​2,​3,​4,​5,​6,​7}}. Find an automaton winning strategy for Player 0 in the Muller game with as few states as possible and show that any other automaton strategy with less states is not winning for Player 0.\\ {{hw4.gif}}
-     - Consider the game shown on page 6 of {{:​slide10.pdf}}. We propose the following "​latest appearance queue" (LAQ) strategy for Player 0, initializing the queue with s<​sub>​1</​sub>​s<​sub>​2</​sub>​s<​sub>​3</​sub>​s<​sub>​4</​sub>, ​(1) add the current state at the front of the LAQ and delete the last state (2) move to the state t<​sub>​i</​sub>​ whose number i is the number of different states in the current LAQ, e.g., for the sequence of states s<​sub>​1</​sub>,​ s<​sub>​3</​sub>,​ s<​sub>​3</​sub>,​ s<​sub>​4</​sub>,​ s<​sub>​1</​sub>,​... we obtain the following sequence of LAQs: s<​sub>​1</​sub>​s<​sub>​2</​sub>​s<​sub>​3</​sub>​s<​sub>​4</​sub>,​ s<​sub>​1</​sub>​s<​sub>​1</​sub>​s<​sub>​2</​sub>​s<​sub>​3</​sub>,​ s<​sub>​3</​sub>​s<​sub>​1</​sub>​s<​sub>​1</​sub>​s<​sub>​2</​sub>,​ s<​sub>​3</​sub>​s<​sub>​3</​sub>​s<​sub>​1</​sub>​s<​sub>​1</​sub>,​ s<​sub>​4</​sub>​s<​sub>​3</​sub>​s<​sub>​3</​sub>​s<​sub>​1</​sub>,​s<​sub>​1</​sub>​s<​sub>​4</​sub>​s<​sub>​3</​sub>​s<​sub>​3</​sub>,​... ​    ​Decide wheater the new LAQ strategy is a winning strategy for Player 0. Prove this or give a counter-example.+     - Consider the game shown on page 6 of {{:​slide10.pdf}}. We propose the following "​latest appearance queue" (LAQ) strategy for Player 0, initializing the queue with s<​sub>​1</​sub>​s<​sub>​2</​sub>​s<​sub>​3</​sub>​s<​sub>​4</​sub>,​ 
 +        - add the current state at the front of the LAQ and delete the last state 
 +        - move to the state t<​sub>​i</​sub>​ whose number i is the number of different states in the current LAQ,\\ e.g., for the sequence of states s<​sub>​1</​sub>,​ s<​sub>​3</​sub>,​ s<​sub>​3</​sub>,​ s<​sub>​4</​sub>,​ s<​sub>​1</​sub>,​... we obtain the following sequence of LAQs: s<​sub>​1</​sub>​s<​sub>​2</​sub>​s<​sub>​3</​sub>​s<​sub>​4</​sub>,​ s<​sub>​1</​sub>​s<​sub>​1</​sub>​s<​sub>​2</​sub>​s<​sub>​3</​sub>,​ s<​sub>​3</​sub>​s<​sub>​1</​sub>​s<​sub>​1</​sub>​s<​sub>​2</​sub>,​ s<​sub>​3</​sub>​s<​sub>​3</​sub>​s<​sub>​1</​sub>​s<​sub>​1</​sub>,​ s<​sub>​4</​sub>​s<​sub>​3</​sub>​s<​sub>​3</​sub>​s<​sub>​1</​sub>,​s<​sub>​1</​sub>​s<​sub>​4</​sub>​s<​sub>​3</​sub>​s<​sub>​3</​sub>,​...\\ Decide wheater the new LAQ strategy is a winning strategy for Player 0. Prove this or give a counter-example.
      - Let A=(S,​s<​sub>​0</​sub>,​T,​p) be a parity automaton with p:S -> {1,..,k}. Construct a Streett automaton B=(S,​s<​sub>​0</​sub>,​T,​F) equivalent to A that has the same transition structure as A. (i.e., find a suitable set of Streett pairs (F_i,E_i))      - Let A=(S,​s<​sub>​0</​sub>,​T,​p) be a parity automaton with p:S -> {1,..,k}. Construct a Streett automaton B=(S,​s<​sub>​0</​sub>,​T,​F) equivalent to A that has the same transition structure as A. (i.e., find a suitable set of Streett pairs (F_i,E_i))
  
 +===== 6.12.2009 =====
  
 + - Exam-like problems on automata and Presburger Arithmetic {{homework.pdf}}