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1. 1 Game Playing
(a) UsetheMinimaxalgorithmtocomputetheminimaxvalueat
each node for the game tree below.
4
-11
1
1
14 1
-15
8 -15
-11
-11
-5 -11
-12
-12 5
4
5
5
5 10
-6
14 -6
4
4
9 4
-13
-13 3
min
max
min
max
(b) UseAlpha-BetaPruningtocomputetheminimaxvalueateach
node for the game tree below, assuming children are visited left to right.
Show the alpha and beta values at each node. Show which branches are
pruned.
1
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2. −∞, ∞
−∞, ∞
−∞, ∞
−∞, ∞
8 -14
−14, ∞
-5 -13
−∞, −13
−∞, −13
15 15
pruned
6 4
−13, ∞
−13, ∞
−13, ∞
-12 -1
−12, ∞
2 15
−13, 2
−13, 2
11 0
0, 2
10 13
min
max
min
max
Figure 1: Upon Arrival
2, ∞
−∞, −13
−13, ∞
−∞, −14
8 -14
−14, −13
-5 -13
−13, −13
−∞, −13
15 15
pruned
6 4
−13, 2
2, ∞
−13, −12
-12 -1
−12, 2
2 15
2, 2
−13, 0
11 0
0, 2
10 13
min
max
min
max
Figure 2: Upon Departure
2

3. 2 A∗
A
B C
D E
F G
8
3
9
2
7
10
3
5
1
1
(a) Inthefigureabove,letAbethestartstateandEbethegoal
state. The weights on the edges reflect the cost to traverse them. Show
the ordered contents of the open and closed queues for each stage of the A∗
algorithm. The heuristic function values are given by the table below.
node A B C D E F G
h(n) 10 3 9 5 0 1 1
The following shows the state of the queues after each round. The queue
objects are in the form XY (z) where X is the node identity, Y is the pre-
ceding node in the path under consideration to reach X, and z is the value
f(z) = g(z) + h(z) where h is given above and g is the cost of the path
being considered that leads to X.
Open Closed
{A(10)} {}
{BA(11), DA(14)} {A(10)}
{DA(14), EB(15), CB(20)} {A(10), BA(11)}
{FD(13), EB(15), CB(20)} {A(10), BA(11), DA(14)}
{GF (14), EB(15), CB(20)} {A(10), BA(11), DA(14), FD(13)}
{EG(14), CB(20)} {A(10), BA(11), DA(14), FD(13), GF (14)}
After this round, EG is pulled off the queue and the algorithm ends. Fol-
lowing backpointers will yield the path ADFGE.
3

4. (b) Whatpathisproducedbyrunningthebestfirstgreedyalgo-
rithm?
A → B → E
This can be seen by repeating the above algorithm but using f = h to
compute the z value.
Open Closed
{A(10)} {}
{BA(3), DA(5)} {A(10)}
{EB(0), DA(5), CB(9)} {A(10), BA(3)}
3 Simulated Annealing
f(x) = max{(4−|x|, 2−|x−6|, 2−|x+6|}. It
has three peaks with one forming a unique global maximum. Perform 8 rounds of
simulated annealing using 4 as your start point, letting the temperature decrease
by a factor of 0.9 each round. A point x has four successors: {x + 2, x + 1, x −
1, x − 2}. Show all of your work, including the current point, the successor
chosen, the round temperature, and the probability of changing position given
the successor and the temperature.
Answers will respect the following constraints.
1. 0 ≤ i ≤ 7
2. Xi+1 ∈ {Xi, Yi}
3. |Yi − Xi| ∈ {1, 2}
4. Ti = 2(0.9)i
5. Pi ∈ {1, e−1/Ti
, e−2/Ti
}
6. Pi = min{1, e(f(Yi)−f(Xi))/Ti
}
7. F(Yi) ≥ f(Xi) ⇒ Xi+1 = Yi, Pi = 1
This can also be interpreted using the following table. Note that here we assume
an initial temperature of 2. As this was not specified in the original assignment
wording, it will not be enforced during grading.
4

5. i Xi Yi Ti Pi
0 4 Y0 2.0000 {1}
1 {X0, Y0} Y1 1.8000 {1, 0.3292, 0.5738}
2 {X1, Y1} Y2 1.6200 {1, 0.2910, 0.5394}
3 {X2, Y2} Y3 1.4580 {1, 0.2537, 0.5037}
4 {X3, Y3} Y4 1.3122 {1, 0.2178, 0.4667}
5 {X4, Y4} Y5 1.1810 {1, 0.1839, 0.4288}
6 {X5, Y5} Y6 1.0629 {1, 0.1523, 0.3903}
7 {X6, Y6} Y7 0.9566 {1, 0.1236, 0.3516}
8 {X7, Y7}
My sample run through the algorithm produced the following answer.
i Xi Yi Ti Pi move?
0 4 3 2.0000 1 yes
1 3 5 1.8000 1 yes
2 5 3 1.6200 1 yes
3 3 4 1.4580 0.5037 yes
4 4 5 1.3122 1 yes
5 5 6 1.1810 1 yes
6 6 7 1.0629 0.3903 no
7 6 5 0.9566 0.3516 no
8 6
This particular run converged to the local maximum at 6.
5
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