The spiders and the fly can move at any speed they want, up to their maximal speeds. The main spider moves at its full speed around a four edge loop, while the other two move at half speed, closing off the other two edges so that the fly has to stay in the chaser's loop and be run down.

Yep, now that makes sense.

@ghug- I don't quite understand what you are saying, but it is true that a lot of opportunities to slip between the two enforcers are given, so it is definitely true that my way doesn't work. That is probably what you are saying. Ah well.

About the starting position, yeah. I hadn't read the question in days, so if you don't know that it starts off, just make a trivial modification at the beginning and you are set.

@Draug- If the spiders had to move at the same speed, you could just have the "enforcer" spiders go 3/4 of the way, then back 1/2 the way, then forward 3/4 of the way to take twice as long as it usually takes them to get from one edge to the other, for example. Any slower speed can be simulated in similar fashion.

Also interpretted as they could only turn at verteces. But that has been the general interpretation all along here. In fact, I thought there was something at one point about them not being able toi change direction except on the verteces. But I don't mind the half speed thing.

@uclabb, the modification is more than trivial for the light switch problem, as you don't know when people are going in, so you can't do something simple like making the first person turn it off. You have to make the counter flip the switch every time he goes in (thereby making the time estimate about 2/3 of a century, assuming an average of one prisoner per day. What happens when they start dying?).

@ghug, uclabb.

Remember there are 2 switches, not 1, and "not flipping" isn't an option. One and only one switch MUST be flipped each time a prisoner enters.

You are on the right track though by picking a "counter."

YJ, I gave you a full solution on the last phase.

page*

Oh, yes ghug you basically got it I think. How many times will the leader have to flip switch 1 down to be sure everybody has visited once?

@ghug You are right, it is nontrivial since we don't know when the prisoners are called up. For some reason I thought that it was one prisoner a day. That's what I get for trying to write a solution without going back and reading the question.

@YJ- Having two switches is strictly easier than having one switch plus the option of doing nothing, as doing nothing can just become f*ing with the second switch.

I have a new one:

A certain rectangle can be covered completely by 25 circles with radius 2 inches. Can it be covered completely by 100 circles with radius 1 inch? By covered completely I mean that when the circles are put on the rectangle , no portion of the rectangle is still showing. Of course, the circles are allowed to overlap each other.

uclabb:

Sure, if the rectangle is 1 inch square.... sure you're not missing some information here?

in fact, you almost certainly have to be missing some... if the rectangle is .00001 inches wide, and 99.9 inches long it'll be covered completely by only 50 radius 1 circles... the question is trivial as you asked it.

I have no idea what you are saying. The question is not trivial, though.

The question is given *any* rectangle that can be covered by 25 circles with radius 2, can that same rectangle be covered by 100 circles of radius one?

ahh well that's a bit different.

I'm thinking no.

If r of the big circle = 2, then the length of the square inscribed by it is 2sqrt(2), area 8.
If r of the little circle = 1, then the length of the square inscribed by it is sqrt(2), area 2.

therefore with 4x the number of little circles, you have the same area of possible rectangle to cover, but with more versatility on how you want to cover it. (i.e. there is no rectangle you could cover with 1 big circle that you could also not cover with 4 small ones).

I think that is false. How do you cover the square that is exactly inscribed into the big circle with 4 half-radius ones? There will be gaps or enough overlap to not to fit the area.

This may not be the solution to the original puzzle, mind you.

@YJ- That was my first thought when I heard the puzzle as well, but it certainly isn't a proof. For example, consider the hexagon that can be inscribed in 1 circle with radius 2. This hexagon cannot be covered by 4 circles with radius 1. The fact that we are considering rectangles is important.

@uclabb:

OK I see a problem with my thinking now, though I think "inscribed squares" is important.

I'm changing my answer to "yes."

Consider the rectangle covered by the squares inscribed by lining up the 25 big circles in horizontal rows, each offset from its neighbors by sqrt(2) :

Row 1: 4
Row 2: 3
Row 3: 4
Row 4: 3
Row 5 4

What's the area of the biggest rectangle covered? You'd think it's 10sqrt(2) x 6sqrt(2), but in reality it is more than that because in the region where row 1,3,5 sticks out further than row 2&4, the vertical edges from 1,3,5 overlap the horizontal edges from 2,4.

So what? One big circle has "overlap region" that is more constructive to this kind of overlap than 4 small circles does, so the area of the final rectangle generated by big circles will be slightly larger in the horizontal direction.

OK?

when I said "each offset from its neighbors" I mean, of course, a bulk shift in the horizontal direction. There is no overlap of the squares.

I like this one!

I *think* that you are switching "yes" and "no" with the way I am thinking (and think I said it). By yes you mean that there is a rectangle that can be covered by 25 big circles but not 100 small circles?

yes.

I guess I didn't put it all in a conclusion statement :P

So let me try to check to see if I understand what you are saying: You are saying that if you arrange the big circles in a Olympic rings- like style it is more efficient to do this with bigger circles than smaller circles? Is that what you are saying?

yeah, that's pretty much an easier way of saying it.

Ok, but why would Olympic rings always be the most efficient style? It turns out that its not always. For example. you can cover a 3 by 33 rectangle by just placing the 25 circles next to each other with appropriate spacing, but not with an olympic rings style with the spacing you suggested above.

It doesn't have to be always the most efficient style. All I had to do was disprove the hypothesis by providing one counterexample :P