A party of four travelers comes to a rickety bridge at night. The bridge can hold the weight of at most two of the travelers at a time, and it cannot be crossed without using a flashlight. The travelers have one flashlight among them. Each traveler walks at a different speed: The first can cross the bridge in 1 minute, the second in 2 minutes, the third in 5 minutes, and the fourth takes 10 minutes to cross the bridge. If two travelers cross together, they walk at the speed of the slower traveler.
What is the least amount of time in which all the travelers can cross from one side of the bridge to the other?
Because there is only one flashlight, each trip to the far side of the bridge (except the last trip) must be followed by a trip coming back. Each of these trips consists of either one or two travelers crossing the bridge. To get a net movement of travelers to the far side of the bridge, you probably want to have two travelers on each outbound trip and one on each inbound trip. This strategy gives you a total of five trips, three outbound and two inbound. Your task is to assign travelers to the trips so that you minimize the total time for the five trips. For clarity, you can refer to each traveler by the number of minutes it takes him to cross the bridge.
Number 1 can cross the bridge at least twice as fast as any of the other travelers, so you can minimize the time of the return trips by always having 1 bring the flashlight back. This suggests a strategy whereby 1 escorts each of the other travelers across the bridge one by one.
One possible arrangement of trips using this strategy is illustrated in Figure . The order in which 1 escorts the other travelers doesn’t change the total time: The three outbound trips have times of 2, 5, and 10 minutes, and the two inbound trips are 1 minute each, for a total of 19 minutes.
This solution is logical, obvious, and doesn’t take long to discover. In short, it can’t possibly be the best solution to an interview problem. Your interviewer would tell you that you can do better than 19 minutes, but even without that hint you should guess you arrived at the preceding solution too easily.
This puts you in an uncomfortable, but unfortunately not unusual, position. You know your answer is wrong, yet based on the assumptions you made, it’s the only reasonable answer. It’s easy to get frustrated at this point. You may wonder if this is a trick question: Perhaps you’re supposed to throw the flashlight back or have the second pair use a lantern. No such tricks are necessary here. A more-efficient arrangement of trips exists. Because the only solution that seems logical is wrong, you must have made a false assumption.
Consider your assumptions, checking each one to see if it might be false. First among your assumptions was that outbound and inbound trips must alternate. This seems correct - there’s no way to have an outbound trip followed by another outbound trip because the flashlight would be on the wrong side of the bridge.
Next, you assumed that there would be two travelers on each outbound trip and one on each return trip. This seems logical, but it’s harder to prove. Putting two travelers on an inbound trip seems terribly counterproductive; after all, you’re trying to get them to the far side of the bridge. An outbound trip with only one traveler is potentially more worthwhile, but coupled with the requisite return trip all it really accomplishes is exchanging the positions of two travelers. Exchanging two travelers might be useful, but it will probably waste too much time to be worth it. Because this possibility doesn’t look promising, try looking for a false assumption elsewhere and reconsider this one if necessary.
You also assumed that 1 should always bring the flashlight back. What basis do you have for this assumption? It minimizes the time for the return trips, but the goal is to minimize total time, not return trip time. Perhaps the best overall solution does not involve minimized return trip times. The assumption that 1 should always return the flashlight seems hard to support, so it probably merits further examination.
If you’re not going to have 1 make all the return trips, then how will you arrange the trips? You might try a process of elimination. You obviously can’t have 10 make a return trip, because then he’d have at least three trips, which would take 30 minutes. Even without getting the remaining travelers across, this is already worse than your previous solution. Similarly, if 5 makes a return trip then you have two trips that are at least 5 minutes, plus one that takes 10 minutes (when 10 crosses). Just those three trips total 20 minutes, so you won’t find a better solution by having 5 make a return trip.
You might also try analyzing some of the individual trips from your previous solution. Because 1 escorted everyone else, there was a trip with 1 and 10. In a sense, when you send 1 with 10, 1’s speed is wasted on that trip because the crossing still takes 10 minutes. Looking at that from a different perspective, any trip that includes 10 always takes 10 minutes, no matter which other traveler goes along. Therefore, if you’re going to have to spend 10 minutes on a trip, you might as well take advantage of it and get another slow traveler across. This reasoning indicates that 10 should cross with 5, rather than with 1.
Using this strategy, you might begin by sending 10 and 5 across. However, one of them has to bring the flashlight back, which you already know isn’t the right solution. You’ll want to already have someone faster than 5 waiting on the far side. Try starting by sending 1 and 2 across. Then have 1 bring the flash-light back. Now that there’s someone reasonably fast (2) on the far side, you can send 5 and 10 across together. Then 2 returns the flashlight. Finally, 1 and 2 cross the bridge again. This scheme is illustrated in Figure
The times for the respective trips under this strategy are 2, 1, 10, 2, and 2, for a total of 17 minutes. Identifying the false assumption improved your solution by 2 minutes.
This problem is a slightly unusual example of a class of problems involving optimizing the process of moving a group of items a few at a time from one place to another. More commonly, the goal is to minimize the total number of trips, and there are often restrictions on which items can be left together. This particular problem is difficult because it suggests a false assumption (that 1 should escort each of the other travelers) that seems so obvious you may not even realize you’re making an assumption.