Another very old type of game is the network map. It was tested several times in the early and mid 1990s and then disappeared until it was used in June 2003. Whether it will reappear again is for LSAC to decide, but you can familiarize yourself with it to eliminate any surprises. The network map involves drawing connecting lines between subway stations, islands, etc. Each line represents a subway line, a bridge, etc., that connects two members. Keep close count of the number of lines coming and going from each member; occupancy limits are important in network maps. Apart from that, there is nothing
particularly difficult about the network maps. The following is an example of a network map.
Train tracks connect six cities: Arborville, Benson, Cherryville, Dawson, Eagle, and Frederick. Two cities are connected by one track that runs between the two cities and does not pass through any other cities. Tracks do not intersect. Every city is connected to at least one other city. The following regulations govern the placement of tracks:
Frederick is connected to Arborville and Cherryville, and no other cities. Arborville is connected to exactly one more city than Eagle is.
Dawson is connected to Eagle and to exactly one other city. Cherryville is connected to exactly one city.
This network map is pretty typical. There is no special order you must list the members, just allow enough space to draw the connecting lines. (Figure 1) Although the setup says that tracks do not intersect, if you can simply remember this, there is no need to draw awkward loops and curved lines in order to avoid intersecting two tracks. Draw the lines from F to A and C, and from D to E. Then consider the limits. F cannot be connected to any more cities, nor can C. D can be connected to one other city. B has no specified limit. A has to connect to one more city than E does. How many cities can A possibly connect to? A could connect to B, D, F, and E, for a total of four cities. This would require E to connect to a total of three cities, A, B, and D. If E connects to two cities, A would have to connect to three cities. If E connects to one city, A would have to connect to two cities. This A/E occupancy feature is the important part of this game.
=E+1 =? =1
A B C
D E F
=2 =2
1
1
Questions 7–11
Six cities—X, Y, Z, Q, R, and S—are to be connected by a new highway system. Each city has at least one new highway connecting it to another city. Each new highway connects two cities, and no other cities. No highways intersect. The following regulations determine the placement of the new highways:
X, Y, and Z are each connected to exactly two other cities.
Q, R, and S are each connected to no more than three other cities.
X is connected to Z and to Y.
Y and Z each must connect to at least one of Q, R, and S.
7. Which one of the following is a complete and accurate list of the cities, any one of which could be directly connected to Q? (A) X (B) Y (C) Y, Z, R (D) Y, Z, R, S (E) Y, Z, R, X, S
8. If a highway connects R with Z, and a highway connects Y and S, which of the following could R connect with? (A) X (B) Y (C) X and Y (D) S and Q (E) Y, Q, and S
9. If S has only one highway and it connects to Q, and Q has exactly two highways, one of which connects to R, then what is the minimum number of cities that a traveler leaving S must pass through to reach X? (A) 1
(B) 2 (C) 3 (D) 4 (E) 5
10. Each of the following could be a possible listing of two cities connected by one highway EXCEPT:
(A) X and Z. (B) Q and R. (C) Y and Z. (D) Y and Q. (E) R and S.
11. Suppose that the condition that requires X to be connected to Z and Y is replaced with the condition that X is connected to Q. Which one of the following CANNOT be true?
(A) Y is connected to X and Z.
(B) Q is connected to Y, and Y is connected to Z. (C) R is connected to Z and to no other city. (D) S is connected to Q and no other city.
(E) Q is connected to Y and Z, and Y is connected to Z.
X Y Z Q Q RR SS X Y Z Q R S =2 =3 =1 initial analysis
X
Y
Z
Q
R
S
3
3
3
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X Y Z Q Q R S =2 =2 =2 =2 =2 =2Unusual Games 159 New highways
When the members are physical locations with something physically connecting them, you should automatically think of a network map. The first step is to decide how to draw the map. Allow enough space for the lines and the occupancy limits. (Figure 1) Rule 1 specifies that X, Y, and Z each have exactly two highways. Note this above them. Rule 2 specifies that Q, R, and S have three or fewer highways. Note this below them. Rule 3 is simple to diagram. Rule 4 is more difficult to diagram, so keep it in the back of your mind. The two things you will need to keep track of are the occupancy limits for each city and the requirement that Y and Z connect to at least one of Q, R, and S. Now, what conclusions can be made? You know that X cannot have any more highways. You can conclude that Y and Z cannot be connected by a highway, since doing so would prevent both of them from connecting to at least one of Q, R, and S. You can also conclude that of Q, R, and S, at least one of them will not be connected to Y or Z.
Each = 2
X Y x Z
Q R S
Each = 1, 2, 3 Fig. 1
7. (D) – Answer elimination will not work for this question, but you can still use a timesaving trick. Since Y is featured in four of the five answers, you don’t need to check for Y; you can assume it is true. Checking for X, you see that X has reached its limit of two highways, and so cannot connect to Q. This eliminates two answer choices. Then you must determine that Z, R, S, and Y can potentially connect to Q.
(A) See the analysis. (B) See the analysis. (C) See the analysis. (D) * See the analysis. (E) See the analysis.
8. (D) – After graphing this new information, you see that X, Y and Z have all reached their limit of two highways, so R cannot connect to X or Y.
(A) See the analysis. (B) See the analysis. (C) See the analysis. (D) * See the analysis. (E) See the analysis.
Unusual Games 160
9. (C) – Graphing this new information allows us to reach the maximum number of highways for each city. Then count the number of cities between X and S. The route must be S to Q, to R, to Z, to X, or it must be S to Q, to R, to Y, to X. So there are three cities to pass through.
(A) See the analysis. (B) See the analysis. (C) * See the analysis. (D) See the analysis.
(E) See the analysis. There are not even five cities that could be passed through.
10. (C) – When a question does not add new information, the answer can usually be found in the initial analysis. The initial analysis showed that Y and Z could not be connected.
(A) See the analysis. Also, Rule 3 connects X and Z. (B) See the analysis.
(C) * See the analysis. (D) See the analysis. (E) See the analysis.
11. (A) – Because this question alters the initial rules, you must go back and recreate the diagram from scratch. After doing so, consider the answer choices.
(A) * If Y were connected to X and Z, then Y would not be able to connect to Q, R, or S because Y must connect to exactly two other cities. This would violate Rule 4.
(B) There is no reason Q cannot be connected to Y and Y be connected to Z. (C) R can be connected only to Z.
(D) S can be connected only to Q. Y and Z would not be able to connect to S, but that is fine.
(E) Y and Z can still fulfill their requirements if they are connected to Q. They would both be connected to exactly two cities. Q would be connected to exactly three cities. X would have to connect to one other city, and R and S would also have to connect to at least on other city.
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