good question and bad question

You get the key point in the problem, but you also miss the key point in my solution.

I explicitly emphasized the following:

in step b the car picks up 4 first and then 5, <-- IMPORTANT!

It is true that at a * state, 5 is falling behind. But after step a and when we get into step b where the car is coming back to meet 4 and pick him up, it must be the case that, 5 has already passed 4.

And this is exactly the reason why from one "* to * iteration" to the next, the distance that 5 is following behind is reducing at a constant rate r, where r must greater or equal than 0.5357, and it cannot possibly reduce any faster than that. This magic number is computed for the case that, when the car comes back to meet 4 and pick him up, 5 is right catching 4 at the same point.

Again

(a) as long as in step b, when the car picks up 4, 5 has already passed 4, the speed of the 2,3 and 4 will be the same as the 4 people case (in fact, it's easy to see 5 can be ignored).

(b) while (a) satisfied, the distance that 5 is falling behind is reducing at a constant rate r, where r >= 0.5357. That means, while 2,3 and 4 are advancing at a constant speed, if in one iteration 5 is falling behind d, then the next iteration he's only falling behind d * r. As long as r is less than 1, it converges and 5 is catch up with 2,3 and 4 the end of the day.

所有跟帖： 

• Same as 康MM's solution -15少- ♂ (0 bytes) () 10/09/2008 postreply 16:10:47

• ok, but -火球魔法- ♂ (363 bytes) () 10/09/2008 postreply 16:23:13

• 关于 康MM's solution -koushi321- ♀ (526 bytes) () 10/14/2008 postreply 15:08:06

• It's still not very clear to me -endofsuburbia- ♂ (213 bytes) () 10/09/2008 postreply 17:59:01

• 回复：It's still not very clear to me -火球魔法- ♂ (46 bytes) () 10/09/2008 postreply 18:38:58

• 回复：It's still not very clear to me -火球魔法- ♂ (162 bytes) () 10/09/2008 postreply 18:40:06