Author |
Topic: 100 people, 100 numbers (Read 6627 times) |
|
JiNbOtAk
Uberpuzzler
Hana Hana No Mi
Gender: 
Posts: 1187
|
 |
Re: 100 people, 100 numbers
« Reply #25 on: Sep 7th, 2007, 4:46am » |
 Quote  Modify
|
on Sep 4th, 2007, 9:05pm, Obob wrote:| They have NO CONNECTION. For the love of god, at least read the riddle before you post. |
|
Hee hee, calm down Obob, he's not worth it. 
|
|
 IP Logged |
Quis custodiet ipsos custodes?
|
|
|
Obob
Senior Riddler
Gender:
Posts: 489
|
|
Re: 100 people, 100 numbers
« Reply #26 on: Sep 7th, 2007, 7:00am » |
Quote Modify
|
I know he's not. It's just so annoying though, that he and some other newbies, most of whom have become bored and left, conduct themselves in such a manner. While I've been reading most of the forums for about 4 years now, I have not even attained 150 posts yet. The reason for this? I think before I post, and try to only post when I have something interesting to add to the conversation.
To see the boards watered down with so many terrible posts pains me. It pains me even more to see the hard and putnam sections corrupted by somebody who has absolutely no business posting "solutions" within 2 minutes of "reading" the threads. Having to sift through so much crap because a 13-year-old thinks its fun to make a fool and an ass of himself sucks.
Of course, srn347 won't read this post. But yes, I am talking about him.
|
| « Last Edit: Sep 7th, 2007, 7:02am by Obob » |
IP Logged |
|
|
|
Barukh
Uberpuzzler
Gender:
Posts: 2276
|
|
Re: 100 people, 100 numbers
« Reply #27 on: Sep 7th, 2007, 9:28am » |
Quote Modify
|
on Sep 7th, 2007, 7:00am, Obob wrote:| While I've been reading most of the forums for about 4 years now, I have not even attained 150 posts yet. The reason for this? I think before I post, and try to only post when I have something interesting to add to the conversation. |
|
I would like to tell you, Obob, that I always read your posts with great interest!
|
|
IP Logged |
|
|
|
Aryabhatta
Uberpuzzler
Gender:
Posts: 1321
|
|
Re: 100 people, 100 numbers
« Reply #28 on: Sep 7th, 2007, 3:49pm » |
Quote Modify
|
on Sep 7th, 2007, 9:28am, Barukh wrote:
I would like to tell you, Obob, that I always read your posts with great interest!
|
|
Second that.
|
|
IP Logged |
|
|
|
Obob
Senior Riddler
Gender:
Posts: 489
|
|
Re: 100 people, 100 numbers
« Reply #29 on: Sep 7th, 2007, 11:55pm » |
Quote Modify
|
Thanks, guys. I didn't mean for that to come out so harsh, but you can imagine my frustration with all my posts directed at him being nearly completely ignored.
|
|
IP Logged |
|
|
|
mikedagr8
Uberpuzzler
A rich man is one who is content; not wealthy.
Gender:
Posts: 1105
|
|
Re: 100 people, 100 numbers
« Reply #30 on: Sep 8th, 2007, 3:22am » |
Quote Modify
|
on Sep 7th, 2007, 11:55pm, Obob wrote:| Thanks, guys. I didn't mean for that to come out so harsh, but you can imagine my frustration with all my posts directed at him being nearly completely ignored. |
|
I think totally. Nearly would imply that he hasn't ignored them. This is not the case.
|
|
IP Logged |
"It's not that I'm correct, it's that you're just not correct, and so; I am right." - M.P.E.
|
|
|
Whiskey Tango Foxtrot
Uberpuzzler
Sorry Goose, it's time to buzz a tower.
Gender:
Posts: 1672
|
|
Re: 100 people, 100 numbers
« Reply #31 on: Sep 8th, 2007, 10:22am » |
Quote Modify
|
on Sep 8th, 2007, 3:22am, mikedagr8 wrote:
I think totally. Nearly would imply that he hasn't ignored them. This is not the case.
|
|
This is a little ridiculous.
|
| « Last Edit: Sep 8th, 2007, 10:22am by Whiskey Tango Foxtrot » |
IP Logged |
"I do not feel obliged to believe that the same God who has endowed us with sense, reason, and intellect has intended us to forgo their use." - Galileo Galilei
|
|
|
Ajax
Full Member
Gender:
Posts: 221
|
|
Re: 100 people, 100 numbers
« Reply #32 on: Sep 10th, 2007, 3:07pm » |
Quote Modify
|
A predefined someone will go stand somewhere.
Whatever his number is, that many will move to another place (they will have already discussed who will walk first, second, etc).
By counting them, he will know his number.
In case the number is 100 (only 99 will be left), then no one will move.
|
|
IP Logged |
mmm
|
|
|
RandomSam
Newbie
Gender:
Posts: 20
|
|
Re: 100 people, 100 numbers
« Reply #33 on: Oct 3rd, 2007, 4:38pm » |
Quote Modify
|
on Mar 25th, 2007, 3:19pm, Eigenray wrote:N=2: prisoner 1 assumes the sum is  1, i.e., they're different; and prisoner 2 assumes the sum is 2, i.e., they're the same. |
|
For N=3, the three prisoners A, B and C can guess B+C, A+2C+2 and A + 2B + 1 (mod 3) respectively. I found that result by trial and error using a truth table, but can't yet find a similar solution or N=4. 
EDIT: Woo! Found a general solution for any size! Will explain soon - it's stupidly late at night here.
|
| « Last Edit: Oct 3rd, 2007, 5:49pm by RandomSam » |
IP Logged |
|
|
|
SMQ
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 2084
|
|
Re: 100 people, 100 numbers
« Reply #34 on: Oct 3rd, 2007, 7:32pm » |
Quote Modify
|
on Oct 3rd, 2007, 4:38pm, RandomSam wrote:| EDIT: Woo! Found a general solution for any size! Will explain soon - it's stupidly late at night here. |
|
Does it look anything like this solution posted earlier?  If so, congratulations! That solution is both elegant and non-trivial to come up with.
--SMQ
|
|
IP Logged |
--SMQ
|
|
|
RandomSam
Newbie
Gender:
Posts: 20
|
|
Re: 100 people, 100 numbers
« Reply #35 on: Oct 4th, 2007, 6:05am » |
Quote Modify
|
on Oct 3rd, 2007, 7:32pm, SMQ wrote:| Does it look anything like this solution posted earlier? If so, congratulations! That solution is both elegant and non-trivial to come up with. |
|
Heh.... when reading the thread through, I was sure that solution had been proved wrong! On closer inspection, it turns out our algorithms aren't that different, but Eigenray expresses his much more elegantly than I did. | hidden: | Given the condition that only one prisoner must guess correctly, I realised that the guesses g must be some linear function of the hat numbers x.
g = M x + c (mod 100)
Where M is a matrix which must have zeros along its diagonal, since your guess gi can't depend on your own hat xi.
g-x = (M - I)x + c (mod 100) must have one and only one zero element.
The solution to this that I found is
| / | -1 | 1 | 1 | 1 | ... | 1 | \ | | | | 1 | -1 | -1 | -1 | ... | -1 | | | | (M - I)= | | | 1 | -1 | -1 | -1 | ... | -1 | | | | | | 1 | -1 | -1 | -1 | ... | -1 | | | | | | : | : | : | : | :. | : | | | | \ | 1 | -1 | -1 | -1 | ... | -1 | / |
which basically results in a "leader" (the top row) guessing the sum of everyone else's hats (mod 100), and each other prisoner guessing with the leader's hat minus the sum of everyone else's hat, minus their own uniquely pre-allocated badge (again, mod 100).
Eigenray's solution equates to this matrix:
| / | 1 | 1 | 1 | 1 | ... | 1 | \ | | | | 1 | 1 | 1 | 1 | ... | 1 | | | | (M - I)= - | | | 1 | 1 | 1 | 1 | ... | 1 | | | | | | 1 | 1 | 1 | 1 | ... | 1 | | | | | | : | : | : | : | :. | : | | | | \ | 1 | 1 | 1 | 1 | ... | 1 | / |
Perhaps it is obvious why that is simpler.
|
Whoa... hidden tables are difficult in BBCode... sorry for repeated edits as I try to get it all right!
|
| « Last Edit: Oct 4th, 2007, 10:58am by RandomSam » |
IP Logged |
|
|
|
Hippo
Uberpuzzler
Gender:
Posts: 919
|
|
Re: 100 people, 100 numbers
« Reply #36 on: Oct 6th, 2007, 3:22pm » |
Quote Modify
|
It seems to me both solutions correspond to general concept of latin hypercube solution.
The original solution was numbered by distance from origin, your numbering is more complicated.
|
|
IP Logged |
|
|
|
Hippo
Uberpuzzler
Gender:
Posts: 919
|
|
Re: 100 people, 100 numbers
« Reply #37 on: Jan 20th, 2008, 8:53am » |
Quote Modify
|
on Jul 11th, 2007, 3:31pm, Hippo wrote:
I have tought about the problem a litlle bit ... any function from n^n->n which gives all n values when you fix arbitrary n-1 coordinates can be used instead of sum mod n. k-th prisoner chooses his coordinate such that the function gives value k ...
this guaranties exactly one prisoner guesses correctly. It seems to me every solution can be described this way, otherwise you cannot guarantee "synchronised answers" ... no more, no less than one prisoner guesses correctly.
(Probably Eigenray told the same in previous posts? ... how many such functions ("latin hypercubes") are there?) |
|
Actually the first paragraph is correct, but not only latin hypercubes are the solutions.
(0 0 1 | 1 2 0 | 2 1 2
0 2 1 | 2 0 1 | 1 2 0
2 1 2 | 0 1 2 | 1 0 0) is not latin hypercube and is solution as well.
|
| « Last Edit: Jan 25th, 2008, 1:23pm by Hippo » |
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print