noma
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Well, it looks like I made a mistake. Let me just apologize for my derailment of this thread and ask that you continue on back here as though nothing ever happened. (Nothing to see here officer.)

Maurog, since you solved the last puzzle in the Math Puzzle thread, would you post your next one here? Thanks.

(Once this is back on track, I'll get rid of the other puzzle threads to avoid confusion.)

wonkyth
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theres nothing wrong with a trick question, except if its impossible.

unless it's in the rules (which it's not), I think trick questions should be allowed.


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RoboBob3000
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The thing about trick questions, as it turns out, is that they're not actually any fun.

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stigant
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I'm angered and confused. Why are we all back in this thread?

Who cares? Here's a puzzle my brother-in-law-in-law gave me: (Solve it or go to another thread)

There's this king, see. And he has a huge wine collection of 1000 bottles. And he's going to throw this huge celebration in a month for his birthday. But one of his rivals sends a spy to poison the wine collection. Luckily, the spy is caught after poisoning only one bottle. But he takes an iocane capsule and dies before anybody can, uh, sternly question him as to which bottle is poisoned. So now the king is in a delima. He knows that the poison (which is not iocane) will kill anybody who drinks even a drop of the poisoned wine, but it will take exactly 3 weeks to kill whoever drinks it. These parameters are unchanged by how much the victim drinks. Furthermore, he has a foolproof way of resealing the wine bottles so that the wine won't go bad (other than possibly having poison in it) before the party. The king doesn't want to poison any of his friends, so he rounds up 10 servants in red shirts (ie they're expendable). How can he determine which bottle is poisoned before his party?

by the way, there are (at least) 2 (qualitatively different) ways to solve this problem: A hard way and an easy way. Either solution will of course be accepted.

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[Last edited by stigant at 04-22-2008 04:50 AM]

calamarain
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Well, there's always the trivial and brute-force solution.

Pour out a sip of all 1000 bottles into 1000 separate glasses.

Separate those into 10 sets of 100. Allocate one set to each servant.

Make each servant drink each glass (from the set of 100) at semi-regular intervals over a 1 week period. Note down which servant dies and at what exact time over the next 3 weeks.

Thus, you'll figure out which bottle had the poison in.

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[Last edited by calamarain at 04-22-2008 05:19 AM]

Rabscuttle
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I'm guessing the simple way is just using binary - number the bottles, assign each servant a power of 2, and get each servant to take a sip from the bottles that have a 1 in their position.

servant 1: 1, 3, 5, 7, 9...
servant 2: 2, 3, 6, 7, 10, 11...
servant 3: 4, 5, 6, 7, 12, 13...

Then after 3 weeks you can look at who is dead and easily work out which bottle is poisoned.

--

But there's an extra week to play with, so you can be more efficient - eg use 8 servants, have them test 255 bottles on 4 separate days (not 256 because of everyone living)

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ooh, factorials are better, then you only need 7 servants (and 7 days)
Divide up the wine into 7 lots, divide each of those up into 6, each of those into 5 etc etc.
On day 1, the servants each drink 1/7 of the wine.
On day 2, the servants each drink 1/6 of each of the 1/7 that they didn't drink from the previous day
On day 3, they drink from 1/5 of the 1/6s of 1/7s they didn't drink from earlier.

hrmm.. example?

4 servants, 24 bottles
Day 1:
s1: 1, 2, 3, 4, 5, 6
s2: 7, 8, 9, 10, 11, 12
s3: 13, 14, 15, 16, 17, 18
s4: 19, 20, 21, 22, 23, 24

Day 2:
s1: 7, 8, 13, 14, 19, 20
s2: 1, 2, 15, 16, 21, 22
s3: 3, 4, 9, 10, 23, 24
s4: 5, 6, 11, 12, 17, 18

Day 3:
s1: 9, 11, 15, 17, 21, 23
s2: 3, 5, 13, 18, 19, 24
s3: 1, 6, 7, 12, 20, 22
s4: 2, 4, 8, 10, 14, 16

Day 4:
um.. not needed, unless you want to just kill someone off.

So then if the order of deaths is 3, 1, 2, (4)
then the poisoned wine is, um, bottle 13

7! > 1000 so 7 servants are sufficient for 1000 bottles.

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And now I am wondering if there isn't anything you could do with multiple people dying on one day
(but now I have to fly)

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[Last edited by Rabscuttle at 04-22-2008 05:35 AM]

RoboBob3000
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I dunno Rabs, if I were you, I'd just look for the bottle that was uncorked.

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[Last edited by RoboBob3000 at 04-22-2008 05:50 AM]

calamarain
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RoboBob3000 wrote:
I dunno Rabs, if I were you, I'd just look for the bottle that was uncorked.

Indeed. It says that the King has a way of resealing the bottles. Never mentioned that the poisoner did.

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TripleM
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Calamarains method actually lets you just use one servant; get him to drink a sip from all wine bottles a few seconds apart. I'm guessing '3 weeks' is only valid up to a day, rather than hours/minutes/seconds though, like Rabscuttle did, otherwise its too easy..

calamarain
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TripleM wrote:
Calamarains method actually lets you just use one servant; get him to drink a sip from all wine bottles a few seconds apart. I'm guessing '3 weeks' is only valid up to a day, rather than hours/minutes/seconds though, like Rabscuttle did, otherwise its too easy..

Oh, I suspect so as well. But that wasn't specified in the question. And *someone* has to provide the trivial obvious case. That way it's ruled out, so the actual intended solution can be arrived at

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Rabscuttle
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Rabscuttle wrote:
And now I am wondering if there isn't anything you could do with multiple people dying on one day

...which is actually simple - you do the first method but numbering things in base (days + 1) rather than binary.

So with a week extra you do it base 8, and then you only need 4 redshirts.

The day each servant dies gives you a digit (0 for survival) and that lets you create a base-8 number.

Rabscuttle
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RoboBob3000 wrote:
I dunno Rabs, if I were you, I'd just look for the bottle that was uncorked.

But then you don't get the killing of servants - think of the gambling amongst the members of the court!

stigant
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so to clear up questions:
1. The spy was able to seal the bottle before he died (otherwise the problem is pointless)
2. The granularity on the deaths is 1 day (ie they'll die sometime on the 21st day after ingesting the poison, but no telling when) otherwise the problem is trivial.

And Rabscuttle provided two good solutions.

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Vike91
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This should be "Sticky topic".

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Rabscuttle
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An easy one:

What is the missing number?

3 1 1 2 1 _ 2 1 3 1

Mr. Slice
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1 or 2. I'll assume 2 because the last two ones have no 'buddy' ones.

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bachus
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Then I'll assume 1.

Rabscuttle
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I'll revise the question to "What is the missing number and why?" :P

bachus
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OK, revised answer:

The missing number may be 0 because 31121 is prime and also would be 02131.

Or the missing number may be 1 because then the first 5 digits would have 3 1's, and one of 2 and 3, just like the last 5 digits.

Or the missing number may be 2 because with the first 5 digits we have 3 x 1 - 1 x 2 = 1, so with the last 5 digits we would have 2 x 2 - 1 x 3 = 1.

And probably there are explanations for higher numbers...

TripleM
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Sequence problems can always have lots of bad solutions. For example, finding polynomials which go through all the points. Its usually quite obvious once you've found the intended answer, and I'm sure you haven't

Rabscuttle
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Indeed, those are incorrect.

clue1: The next number in the sequence is 4, the previous number is 2

brian_s
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Can we get a hint for the sequence? Is it purely mathematical based on factors, different bases, or some other property? Is it based on some list of historical events? Does 0 or 5 ever appear?

Jutt
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The OEIS finds three matches, all of them with a 1 as the missing number…
Unfortunately none of them holds under the restriction that 2 precedes and 4 follows.

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Rabscuttle
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I adapted it from a puzzle in a game I was playing recently. You could say I localised it?


0, 1, 2, 3, 1, 1, 2, 1, _, 2, 1, 3, 1, 4, 1, 5

Rabscuttle
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Clue 2: Just giving the next and previous digits may be slightly misleading, because you should group some of them together.

Rabscuttle
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Clue 3:

brian_s wrote:
Is it based on some list of historical events?

Yes, you could say that.

noma
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Rabscuttle wrote:
You could say I localised it?

I've been playing with all kinds of combinations. Just in case I'm on the wrong track... Localised, "historical" events : is the answer something to do with this forum?

Rabscuttle
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No...

I am in Australia. The original puzzle used an American localisation thingy, despite the fact it was an English character in a French town. :/

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[Last edited by Rabscuttle at 05-03-2008 04:58 PM]

bachus
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Perhaps it is time to give us hints no. 5, 6 and 7....

Rabscuttle
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hrm...

look at the original set of 10 digits.

Now consider: 4, 2, 2, 2.