I've never used spotlight, and beagle only very little, but I've had extremely good results with google desktop. My primary use is searching pdf-files (whenever I want to save a web page I print it with pdfcreator). I can't really recognise the problems you're describing.
"and he can call and recall each prisoner as many times as he wants"
as meaning that he can choose to only call on one of the prisoners a finite number of times, no matter how long we continue, but
"But eventually we may be certain that each prisoner will be called in ten times, or twenty times, or any number you choose. "
clearly states that this isn't what the poster means.
If that still doesn't convince you, think of this example:
Let k>n.
Let us assume that he only calls on each prisoner once.
Because k>n, he can completely destroy all the information the prisoners may try to send each other with the chalice, and there is obviously no way for the n'th prisoner to be called, to know that he is the last. Therefore the riddle is unsolvable in this situation, but the poster clearly states that there IS a solution!
No, it doesn't work in all situations:
Lets say the king doesn't interfere at all, eventually all of the non-watchers will have flipped the cup exactly once, so the watcher will have flipped the cup n-1 times. But noone will ever flip it again and n-1 < n + k + 1, so they are stuck forever.
I have a solution, but the prisoners have to know the number k:
They choose one of them as a leader. When he finds a chalice "wrong side up" he flips it over. If it has the right side up, he doesn't touch it. He keeps track of the number of times, he has flipped it.
The others flip it only when it has the right side up. If it has the wrong side up they don't touch it. They also keep track of the the number of times they flip it. They each only flip it (2k+1) times.
When the leader is about to flip it back for the ((n-1)*(2k+1) - k) time, he answers yes to the question. Proof that it will always work:
If only (n-2) of the non-leaders have been in the room, the max times the leader can have flipped it is: (n-2)*(2k+1) + k = (n-1)*(2k+1) - k - 1, so he cannot possibly answer yes too early.
Every time one of the non-leaders flip it, it has to be reset. The king will reset it no more than k times, the leader will reset it no more than ((n-1)*(2k+1) - k -1) times, combined they will max reset it (n-1)*2(k+1) - 1 times. But the non-leaders will flip it (n-2)*(2k+1) times, so we can't get "stuck" either.
The CXV would be attached to its booster and carried aloft under the belly of a large carrier aircraft to an altitude of 7,600 meters for release and launch - an approach t/Space thinks has significant safety advantages over a pad launch in the event of a booster failure.
See, that's the part I don't understand. Mac OS X Server makes Linux look like chopped liver in every single respect. Why would you ever want to take a giant step backward?
There's a lot more incentive to upgrade away from the linux desktop, than the linux server.
If your server is already doing what you want it too, why upgrade it?
You clearly know a lot more than just the market-speak about Quartz. What direction do you think the open-source display systems should move in? How would you design it?
Transplanting the brain wouldn't make you immortal no matter what, the brain would still be deteriorating.
But what makes you think the brain isn't the whole person?
Slashdot Mirror
Yes, it's in File->Export->3D Studio (.3ds).
(It's actually the first choice in the Export menu).
At least version 2.41 also had it, but I'm not sure when it became a part of blender.
How long do your searches usually take, say if you let it search your entire home folder? Are we talking ms, s, minutes?
I've never used spotlight, and beagle only very little, but I've had extremely good results with google desktop. My primary use is searching pdf-files (whenever I want to save a web page I print it with pdfcreator). I can't really recognise the problems you're describing.
You're not using indexing? That is kinda the selling point of Spotlight, Beagle, Google Desktop, etc.
(or rather, the selling point is that the speed of searching is so fast that it's actually useful as a way of navigating your files)
It means that Zonk wrote the headline.
If that still doesn't convince you, think of this example: Let k>n.
Let us assume that he only calls on each prisoner once.
Because k>n, he can completely destroy all the information the prisoners may try to send each other with the chalice, and there is obviously no way for the n'th prisoner to be called, to know that he is the last. Therefore the riddle is unsolvable in this situation, but the poster clearly states that there IS a solution!
No, you're wrong. The part of the riddle you quote, actually means that he has to keep calling them. Re-read the last sentence.
No, the king could simply decide to never flip the chalise, and they would never finish.
8 02205
I think I have solved it:
http://slashdot.org/comments.pl?sid=165444&cid=13
No, it doesn't work in all situations:
8 02205/ and 8 02235.
Lets say the king doesn't interfere at all, eventually all of the non-watchers will have flipped the cup exactly once, so the watcher will have flipped the cup n-1 times. But noone will ever flip it again and n-1 < n + k + 1, so they are stuck forever.
I think I got it right:
http://slashdot.org/comments.pl?sid=165444&cid=13
http://slashdot.org/comments.pl?sid=165444&cid=13
The above solution assumes that the chalice starts out being "right side up", if we don't want that asumption, just pick k:=k+1 instead.
I have a solution, but the prisoners have to know the number k:
They choose one of them as a leader.
When he finds a chalice "wrong side up" he flips it over. If it has the right side up, he doesn't touch it. He keeps track of the number of times, he has flipped it.
The others flip it only when it has the right side up. If it has the wrong side up they don't touch it. They also keep track of the the number of times they flip it. They each only flip it (2k+1) times.
When the leader is about to flip it back for the ((n-1)*(2k+1) - k) time, he answers yes to the question.
Proof that it will always work:
If only (n-2) of the non-leaders have been in the room, the max times the leader can have flipped it is:
(n-2)*(2k+1) + k = (n-1)*(2k+1) - k - 1, so he cannot possibly answer yes too early.
Every time one of the non-leaders flip it, it has to be reset. The king will reset it no more than k times, the leader will reset it no more than ((n-1)*(2k+1) - k -1) times, combined they will max reset it (n-1)*2(k+1) - 1 times. But the non-leaders will flip it (n-2)*(2k+1) times, so we can't get "stuck" either.
It probably has more to do with the fact that Katamari Damacy isn't out in europe yet. It probably isn't even eligible.
I was commenting on why the switch to OS X was a little slower on the server side, than on the desktop, not whether or not it's appropriate.
There's a lot more incentive to upgrade away from the linux desktop, than the linux server.
If your server is already doing what you want it too, why upgrade it?
The first sun ceo, I think.
You clearly know a lot more than just the market-speak about Quartz.
What direction do you think the open-source display systems should move in?
How would you design it?
Lots of people die from strokes, aneurisms and tumors in the brain.
Transplanting the brain wouldn't make you immortal no matter what, the brain would still be deteriorating. But what makes you think the brain isn't the whole person?
Please point out where and when these comments where made.