CS2 05: A RTIFI CIA L INTE LLI GEN CE Fall 2006
Problem Set 1
Solutions
Question 1.
a. If we allow the salesperson to “teleport” (transport without cost) to any previously visited city, then the solution for
the remaining cities is the minimum spanning tree of the (yet) unvisited cities, the start city, and the current end city.
This can easily be seen: the optimal solution to this teleport version of the problem is to link all of the (unseen plus
start and end) cities together with the subgraph of minimum cost. This is exactly a minimum spanning tree problem.
We do not care about which city is linked to which; in order to reach a new city, we can “teleport” to the closest city
and then take the trip (and associated cost). This relaxation clearly only reduces the total cost. We have added a move,
but the original solution — not teleporting — still exists.
b. The straight-line distance places a lower bound on the remaining costs by using the distance from the start city to
the city at the end of the current path. If the cities are fully connected and the paths obey the triangle inequality, this is
just the cost of the link that turns the path into a circuit. If the cities aren’t fully connected (or if the link costs do not
obey the triangle inequality), then we can extend this to be the minimum cost path between the two nodes.
This cost clearly dominates the MST cost. The MST cost connects all of the remaining nodes in a minimal subgraph
while the straight-line distance connects just two of those nodes with a minimal subgraph (i.e. a path).
Question 2.
a. Any game which is winnable for a player will be won using either evaluation function. For E2, it will select the
win which has the largest “margin” of victory. For E1, it will select some arbitrary win. Either way, if a win exists, it
will be selected. Alpha-beta search won’t change the result any. It always produces the same result as direct search.
b. The running time for alpha-beta search will be longer in the case of E2 . It must find the maximal margin win (not
just the first win it comes across). Therefore, it cannot prune as much.
c. We expect E2 to do better. E1 risks picking a situation that is only marginally better for the current player. It could
be that the evaluation function miscalculated, and that situation is actually a loss for the current player. Such a reversal
in sign is less likely to happen when E2 is very large. While E2 might not be a completely accurate prediction of who
will win, the larger the absolute value of the metric, the more likely it is to have the correct sign.
d. If the game has chance, things are very different. Consider a chance node with three branches. The first branch
results in a token-count difference of +11. The second and third branches each result in a token-count difference of -1.
In this case, if we get to this node, the true probability of winning is 1
3. In general, perhaps not too good. However,
with a token-count metric, we will report a value of 3for this node (the average of 11,−1, and −1). Matters only get
worse if the first branch’s token-count difference is larger.
Therefore, the evaulation function must be carefully considered. In general, it must be equal to the probability of
winning for the MA X player at a given node (or a positive linear transformation thereof) in order for the chance nodes
to evaluate correctly. This is, of course, in general impossible (if we knew the true probability of winning, we wouldn’t
need to search at all). However, it does mean that we must try hard to insure that our metric is not only correlated with
the chance of winning, but that the correlation is linear in fashion: it can’t just be that bigger scores indicate better
positions for MA X, it must also be true that if a score is twice as big, there must be (at least roughly) twice as great a
chance for MA X to win.
This is true regardless of whether the search can reach the leaves of the game tree.
