LRDI Practice Set 24

The Action:

As the game begins, five islands are unconnected by bridges, but contractors are about to remedy that. In a way we are acting as surrogates for the bridge planners: We need to connect the bridges appropriately. Though this is a game situation you probably have not seen before, the Key Issues would seem to be pretty clear, namely:

1) Which islands must, can, or cannot be connected by bridge to which other islands?

2) How many bridges must, can, or cannot an island have?

The Initial Setup:

When a game mentions, or seems to hinge on, a map, you might as well draw one. Since we’re not told anything about the islands’ spatial relationship to each other, and that issue never comes up, it doesn’t matter exactly how you locate them; in this situation, most people tend to try setting the entities in a rough circle. Where to start? A sneak peek at the concrete Rule 7 might suggest that you locate J and M close to O. We can also throw K and L in to complete the circle, and move them elsewhere later on if necessary:

(If you didn’t see this, or if some new information comes along, you can always redraw the arrangement—doing so won’t take long.) Anyway, connecting the letters by straight lines would seem adequate to indicate the bridges.

The Rules:

We’ve already taken care of Rule 7, and so can now fill in the other information around the tentative sketch above:

1) This rule is a loophole-closer—meaning a rule designed to define the setup and limit possibilities that common sense would otherwise leave open. Because of Rule 1, we know that every bridge connects just 2 islands (unlike, say, New York’s Triborough Bridge, which as its name indicates connects three bodies of land), and that no bridges intersect.

2) Another loophole closer. Students have always reported having trouble with this one. But the testmakers are trying to make things clear; maybe the best way to understand it is to try a concrete example or two. Suppose a bridge is built between M and O. Well, then, all the rule is there for is to prohibit any other M/O bridge: One bridge per pair of islands. That’s it! It means that any bridge that connects islands is a single bridge. Again, just a loophole closer meant to forestall any ambiguity in the numbers that govern the game.

3) Not a loophole closer this time, but a concrete piece of information that guides the action: The maximum number of bridges that can be built to any one island is three. So each island will be connected to three islands maximum. Make a note of it.

4) is vague enough that we might profitably skip it for the moment, and move to more concrete rules.

5) “J has exactly two connections.” Well, one of them is established by Rule 7. So we need to link up J to one more island. Which one? No way to tell. Let’s just note it off to the side:

6) “K gets just one bridge.” Fair enough. We don’t know which one it is, so an annotation will do for now:

4) Now back to 4: Of J, K, and L, each one has to link up to M or O or both. A THREE-part rule: Three rules in one. Well, we have a context for that fact now—namely, all the other things we’ve learned—so let’s explore this rule in Step 4 in the context of all the other information:

Key Deductions:

Starting with J in Rule 4: “J must be connected to M or O or both.” Well, that minimum requirement has been met (Rule 7): J is connected to O. So that’s all we can do with J and this rule. J’s other bridge could link J to M, but needn’t. “K is connected to M or O or both.” Well, K only gets one bridge, so Rule 4 tells us it has to be a K/M or K/O bridge. So our sketch can be adjusted to reflect that:

(And you may wish to note that K/J and K/L bridges are therefore prohibited.)

Continuing with Rule 4: “L is connected with M or O or both.” L is otherwise wide open; L could have a minimum of one bridge (L/M or L/O), or two, or the maximum of three. Anything else? We should notice that O, with 2 bridges already, can only have one more maximum; and that M, with 1 bridge already, can get zero, one, or two more.

We should also notice that indeed every island has to have at least one bridge. That wasn’t spelled out as a rule, but proves to be true.

The Final Visualization:

The Questions:

1. (D)

There will be no K/L connection, since K gets only one bridge connection (Rule 6) and that must be to M or O (Rule 4). But nothing stops L from being connected to J, O, and/or M. Any or all of them can link up with L—choice (D).

2. (C)

Since the right answer CAN be true, it follows that the four wrong choices are impossible. This is a variation of the standard “acceptability” question. You can check each rule against the choices, or simply test each choice against the master sketch above. Either way, you should see rather readily that (C) is acceptable (L can have J and M connections with no damage to the rules) and that:

(A) Since J is already connected with O, connecting J with L and M would be a total of three bridges, violating Rule 5.

(B) Since K only gets one connection (Rule 6), (B) is a blatant violation.

(D) Since M is already connected with O, adding (D)’s connections would give M a total of four bridges, violating Rule 3.

(E) Since O is already connected with J, (E) would give O a total of four bridges. Again, a violation of Rule 3.

3. (B)

Creating a new sketch for this question is advisable. In it, repeat the connections we’re sure of (J/O and M/O), and add the new K/O bridge. Two things should now be immediately apparent: (1) That’s it for K, which has the one connection permitted it by Rule 6. (2) That’s also it for O, which has its total of three connecting bridges permitted by Rule 3. L’s the only island not yet hooked up—and all that’s left are J and M. L will have to link up M (Rule 4), and could link up with J as well. Now look through the choices for a possible statement.

(A) Impossible. Not establishing an L/M bridge, with O unavailable to hook up with L, violates Rule 4.

(B) Possible, as noted above. There’s our answer.

(C) Impossible. O’s quota is reached with connections to J, K, and M.

(D) Impossible. L has a maximum of two connections (J and M) here.

(E) Impossible. O, as we’ve seen, has three bridge connections—two from Rule 7, and one from the question stem.

4. (B)

The stem is very concrete. Create a new sketch with the new and old information represented. You see that O has reached its limit (J, L, M = three bridges) and that K is still unconnected by any bridges. That won’t do; and since Rule 4 dictates that K be connected with either O or M, and O is once again maxed out, there’s no way around it—K must link up with M, choice (B).

(A) and (E) are both impossible: Since M must link with K here, M now has three connections—K (see above), L (from the stem), and O (Rule 7). So (E) is dead wrong—M has three, not two connections; and (A) is impossible, since M is now maxed out and therefore cannot connect with J.

(C) is impossible, since O’s limit of three bridges has already been reached through J, L, and M. As described above, this is the fact that forces the KM connection, and hence correct choice (B).

(D) turns out to be impossible as well: If we follow this scenario through to the bitter end, we see that J’s required second connection (Rule 5) must be with L, what with K, O, and M each connected to the max. So J must connect to L, which means that L now has three connections itself (including the two in the stem), not two as (D) would have it.

5. (A)

This stem is more of a critical reading challenge than anything else. We are forbidden, in essence, to connect any island with both M and O. Anything jump out at you at this point? Hopefully, you said to yourself “hey, J is already connected to O, so we can’t have JM.” Following this line of thought, we see that since J can never connect to K—K gets only one bridge and that one has to be with M or O—and J needs two connections to satisfy Rule 5, J must connect to L. Voila, choice (A).

(B) Impossible. O and J are already connected by a bridge; (B) would violate the question stem.

(C) is possible but not necessarily true, since K’s one bridge could be built to M just as readily as to O.

(D), (E) L needs a bridge connection with M or O (Rule 4) and cannot be connected with both (according to the stem). But which bridge is built? We cannot be sure.