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Many of you have probably heard about Hilbert's Hotel problem. Mr Hilbert owns a hotel with countably infinite amount of one-bed rooms. All the rooms are, of course, taken.

A (finite or infinite) group of k people walks in and wishes for accommodation. However, here comes the tricky part. The current guests are quite tired and Mr Hilbert does not wish to make them move from one room to another. How does he achieve accommodating the new batch of people without moving the already accommodated ones?

This problem, although it's quite known and documented is nowhere to be found here, or anywhere else on the Internet (which is kind of strange).

DDS
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user238108
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  • Sending the k people to the hotel in front of the street...without moving the current guests and being all the rooms taken I don't think the newly arrived guests-to-be can be accommodated. – Timbuc May 05 '15 at 16:46
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    build a new Hotel? – mandata May 05 '15 at 16:49
  • Well, I've been assured that there actually is a legit solution (even though I'm starting to doubt it). – user238108 May 05 '15 at 16:50
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    I read through the wiki page a bit and it seems you don't want to assume no guests can be moved. Otherwise all rooms are full and there's nowhere to put a new guest. – Gregory Grant May 05 '15 at 16:50
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    That's a peculiar use of "of course". – Travis Willse May 05 '15 at 16:52
  • I wonder if you're confusing with the fact that new guests (even infinitely many) can always be placed so that any number of additional guests can be accommodated without moving anybody. That's on the wiki page you pointed to. – Gregory Grant May 05 '15 at 16:53
  • @GregoryGrant is it? in that case I must've misread something (which is super awkward) – user238108 May 05 '15 at 16:55
  • I read this on the page you linked to "Anticipating the possibility of any number of layers of infinite guests, the hotel may wish to assign rooms such that no guest will need to move, no matter how many guests arrive afterward. One solution is to convert each arrival's address into a binary number in which ones are used as separators at the start of each layer, while a number within a given layer (such as a guests' coach number) is represented with that many zeroes. Thus, a guest with the prior address 2-5-1-3-1 (five infinite layers) would go to room 10010000010100010 (decimal 295458)." – Gregory Grant May 05 '15 at 16:59
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    But the difference is, the hotel will never actually be full using that strategy. – Gregory Grant May 05 '15 at 16:59
  • @user238108 "You've must misread.." or else whoever made the claim was wrong. It happens, even if written on books or even in the super serious and "always infallible" webnet, you know? – Timbuc May 05 '15 at 17:06
  • @GregoryGrant Not only that: the "Thus, a guest with the prior... would go..." part makes it clear that guests already in the hotel will have to move, whether we number the rooms with binary numbers or with red bananas dipped in honey. – Timbuc May 05 '15 at 17:07
  • @Timbuc I think you are misreading it, the claim is that no guest will ever have to move. By "go to" I think it means when they first arrive. – Gregory Grant May 05 '15 at 17:10
  • @GregoryGrant Perhaps I'm misunderstanding something , as in the part you quoted (I haven't gotten the link) it is written "a guess with the prior address ...", thus implying the guest was already there and had to be moved...ain't it? – Timbuc May 05 '15 at 17:11
  • I agree it sounds like that, I didn't read it too closely either yet – Gregory Grant May 05 '15 at 17:14

2 Answers2

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My understanding of the Hilbert Hotel is that the hotel has $n$ rooms. Since the number of rooms is infinite a new guest will get the room $n_{+1}$. So if $k$ people arrive the will stay in the rooms $n_{+1},...,n_{+k}$, and there is no need for the other guests to move.

ASAP_7
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The usual specification is that the hotel is already full when the new guests arrive. If that is true, you cannot accommodate them without moving the guests that are already there. If we just have to have an infinite number of guests in the hotel and accommodate a number more without moving the existing ones, just put the existing guests in the odd numbered rooms and put the new guests in the even numbered rooms. The question is not clearly stated.

Ross Millikan
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