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SolutionsI

# SolutionsI - ECE 496 SOLUTIONS TO HOMEWORK ASSIGNMENT I...

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ECE 496 SOLUTIONS TO HOMEWORK ASSIGNMENT I Spring 2007 The numbers correspond with the numbering of the Thought Problems at the end of Chapter 1 of the book. 2. In parts (a) through (c) we list the outcomes of the games in the standard way assuming that Player 1 is playing TIT-FOR-TAT. The resulting payoff pairs are in parentheses. Part (d) is a bit trickier. (a) CD (0 , 5) DD (1 , 1) DD (1 , 1) DD (1 , 1) DD (1 , 1) DD (1 , 1) DD (1 , 1) DD (1 , 1) DD (1 , 1) DD (1 , 1) Total payoff to Player 1 = 9. Total payoff to Player 2 = 14. (b) CC (3 , 3) CC (3 , 3) CC (3 , 3) CC (3 , 3) CC (3 , 3) CC (3 , 3) CC (3 , 3) CC (3 , 3) CC (3 , 3) CC (3 , 3) Total payoff to Player 1 = 30. Total payoff to Player 2 = 30. (c) CD (0 , 5) DD (1 , 1) DC (5 , 0) CC (5 , 5) CD (0 , 5) DD (1 , 1) DC (5 , 0) CC (5 , 5) CD (0 , 5) DD (1 , 1) Total payoff to Player 1 = 23. Total payoff to Player 2 = 28 (d) You are Player 1 and are playing TFT. The other player is playing “random moves.” That description is a bit vague, but I’ll assume what Melanie Mitchell intended was for the random player to flip a fair coin before each move — i.e., to choose C or D with probability 1/2 in each repetition of the game, with all the choices independent of each other. You play C in the first game. Player 2 moves at random; hence the outcome is either CC or CD with probability 1/2 each. So your expected payoff from that first game is 1 / 2(3 + 0) = 3 / 2, and Player 2’s expected payoff is 1 / 2(3 + 5) = 4. Now, what about subsequent games? Note that Player 2’s move in the first stage determines not only the payoffs for that stage, but also what move you make in the second stage. Hence you will play C with probability 1/2 and play D with probability 1/2 in that stage. If you play C, you have an expected payoff of 3/2 again; if you play D, you have an expected payoff of 1 / 2(5 + 1) = 3. (Note that Player 2’s move in the second stage is independent of yours, since the move you make is determined solely by Player 2’s “coin flip” in the first stage.) Since these eventualities occur with equal probability 1/2, your overall expected payoff in the second stage is 1 / 2(3 / 2 + 3) = 9 / 4. Similarly, Player 2 plays C or D with probability 1/2 in the second stage; he faces a C or a D from you with equal probability independent of his choice here (again because what determines your move is what Player 2 did in the first stage). By the same reasoning as for you, Player 2’s expected payoff in the second stage is also 9/4. You can reason similarly about the rest of the stages and conclude both players have expected payoff 9/4 in every stage after the first. Conclusion: Player 1’s expected payoff is 3 / 2+81 / 4 = 87 / 4 and Player 2’s expected payoff is 4+81 / 4 = 97 / 4. (e) The answers are Player 2’s payoffs in (a) through (d). 3. The easiest way to do this is to refer to the discussion near the bottom of page 22. Think of a sorting network with m comparisons as a list of m pairs of numbers; pairs such as (11 , 11) are not allowed, but a given permitted pair might appear more than once in the sequence. There are ` 16 2 ´ = 120 permitted pairs. Thus there are 120 m sorting networks with m comparisons. Summing over 60 m

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