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lat:homework [2009/11/27 22:51] barbara.jobstmann |
lat:homework [2009/12/06 12:30] 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 ===== | ||
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- 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. | + | - 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}} |
- Given a game graph G=(S,S<sub>0</sub>,T) and a set F ⊆ S. Let W<sub>0</sub> and W_1 be the winning regions of Player 0 and Player 1, respectively, in the Buchi game (G,F). Prove or disprove: | - Given a game graph G=(S,S<sub>0</sub>,T) and a set F ⊆ S. Let W<sub>0</sub> and W_1 be the winning regions of Player 0 and Player 1, respectively, in the Buchi game (G,F). Prove or disprove: | ||
- The winning set of Player 0 in the safety game for (G,W<sub>0</sub>) is W<sub>0</sub>, | - The winning set of Player 0 in the safety game for (G,W<sub>0</sub>) is W<sub>0</sub>, | ||
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- the winning set of Player 1 in the guaranty game for (G,W_1) is W_1, and | - the winning set of Player 1 in the guaranty game for (G,W_1) is W_1, and | ||
- if f_1 is a winning strategy for Player 1 in the guaranty game for (G,W<sub>0</sub>), then f_1 is a winning strategy in the Buchi game (G,F). | - if f_1 is a winning strategy for Player 1 in the guaranty game for (G,W<sub>0</sub>), then f_1 is a winning strategy in the Buchi game (G,F). | ||
- | |||
- | {{hw1.gif|Figure 1}} | ||
===== 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 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 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>, | ||
+ | - 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.ps}} | ||