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The problem I need to solve is written as the following:

Four people want to cross a bridge on a very dark night. They all begin on the same side and want to do the crossing as quickly as possible. A maximum of two persons can cross the bridge at one time. Any group that crosses, either 1 or 2 people, must have a flashlight with them. There is only one flashlight. The flashlight must be walked back and forth, it cannot be thrown. Each person walks at a different speed. A pair must walk together at the rate of the slower person’s pace, no one can be carried.

Here is the list of persons and their speed:

  • $A$ ~ $1\text{ min}$
  • $B$ ~ $2\text{ min}$
  • $C$ ~ $5\text{ min}$
  • $D$ ~ $10\text{ min}$

What is the least amount of time needed to get all four people across the bridge?


Here's how I solved it:

Since each person inevitably has to cross the bridge, the only fact we can manipulate is that someone has to come back each time to get the next person.

$A$ and $D$ cross first, then $A$ goes back. $A$ and $B$ cross second, then $A$ goes back. $A$ and $C$ cross third. In total, that gives me $((10+1) + (2+1) + 5)\text{ min}=(11+3+5)\text{ min}=19\text{ min}$.

The T.A. in class, though, said $19$ was an incorrect answer.

Thoughts?

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    $\begingroup$ This would be better on puzzling.stackexchange, a similar question with different constants is puzzling.stackexchange.com/questions/287/… The point is that you want the two slow ones to cross together, so the second slowest doesn't count. $\endgroup$ Sep 16, 2014 at 2:55

3 Answers 3

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Your solution is this:

$$\begin{align} ABCD&|&0\\ BC&|AD&10\\ ABC&|D&11\\ B&|ACD&16\\ AB&|CD&17\\ &|ABCD&19 \end{align} $$

whereas you can do this:

$$\begin{align} ABCD&|&0\\ CD&|AB&2\\ ACD&|B&3\\ A&|BCD&13\\ AB&|CD&15\\ &|ABCD&17 \end{align}$$

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AB over for 2, A back for 1, CD over for 10, B back for 2, AB over for 2, total 17.

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TedEx solved this!

Not my solution, but this illustrates it.

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