Question

Logic Puzzle: Four Friends and a Bridge

Objective:
The purpose of this assignment is to apply your knowledge of search algorithms and their applications in solving real-world problems.
Scenario:
Four friends need to cross a bridge at night to get back to their campsite. They have only one flashlight with them, and the bridge can only hold two people at a time. Each of the friends walks at a different speed, and the flashlight must be carried by the faster of the two. The speeds of the friends are:
Friend 1: 1 minute to cross the bridge
Friend 2: 2 minutes to cross the bridge
.Friend 3: 5 minutes to cross the bridge
.Friend 4: 10 minutes to cross the bridge
When two friends cross the bridge together, they must move at the slower friend's pace. What is the minimum time required for all four friends to cross the bridge?

Answer 1

To solve the bridge-crossing puzzle, the optimal strategy involves a sequence of crossings resulting in a total minimum time of 17 minutes, with specific steps to ensure the flashlight is managed efficiently.

Step-by-step explanation:

The logic puzzle presented involves four friends needing to cross a bridge at night with varying speeds and only one flashlight. To solve this problem, we need to find the strategy that results in the minimum time required for all friends to cross.

Let's consider the following step-by-step strategy:

1. Friend 1 and Friend 2 cross with the flashlight. Time taken: 2 minutes.
2. Friend 1 returns with the flashlight. Time taken: 1 minute (Total: 3 minutes).
3. Friend 3 and Friend 4 cross with the flashlight. Time taken: 10 minutes (Total: 13 minutes).
4. Friend 2 returns with the flashlight. Time taken: 2 minutes (Total: 15 minutes).
5. Finally, Friend 1 and Friend 2 cross again. Time taken: 2 minutes (Total: 17 minutes).

The total time taken for all the friends to cross the bridge is 17 minutes, which is the required when applying this strategy. This solution applies search algorithms to explore possible sequences of crossing, assessing the costs of each, and selecting the option with the lowest total time.