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 TITFORTAT. 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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 Spring '07
 DELCHAMPS
 Algorithms, representative, Formal language, Hillis

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