Logic Puzzles Marathon (2 Viewers)

Paradoxica · Feb 6, 2016

I'm surprised there wasn't one already. Rules as per other marathons, this one is for those problems that are more logical than mathematical.

$\noindent The other $n-1$ statements are false$ \\ $The other $n-1$ statements are false$ \\ $The other $n-1$ statements are false$ \\ \vdots \\ \vdots \\$ The other $n-1$ statements are false$ \\ $The other $n-1$ statements are false$ \\ $The other $n-1$ statements are false$ \\ $The other $n-1$ statements are false$ \\ $The other $n-1$ statements are false$\\ \vdots \\ \vdots \\ $The other $n-1$ statements are false$ \\ $The other $n-1$ statements are false$ \\ $The other $n-1$ statements are false$ \\ \\ $Given there are $n$ statements, where $n>2$, how many of these can be true simultaneously?$

RealiseNothing · Feb 6, 2016

Isn't it just one?

Question: The king has ordered 1,000,000 bottles of wine, but has been told that exactly one is poisoned. To figure out which one is poisoned, he is given access to mice.

He can give any bottle he wants to any mouse, and can do it more than once (so a mouse can have more than one bottle, and one bottle may be shared as much as he wants).

Given that he must give the mice the allocated wine simultaneously, was is the least amount of mice needed to determine the poisoned bottle?

Paradoxica · Feb 6, 2016

RealiseNothing said:
Isn't it just one?

Question: The king has ordered 1,000,000 bottles of wine, but has been told that exactly one is poisoned. To figure out which one is poisoned, he is given access to mice.

He can give any bottle he wants to any mouse, and can do it more than once (so a mouse can have more than one bottle, and one bottle may be shared as much as he wants).

Given that he must give the mice the allocated wine simultaneously, was is the least amount of mice needed to determine the poisoned bottle?

$\noindent Yes, there can only be one true statement.$\\$If we give each of the mice a partition of the wine bottles, so that we can eliminate a large number of the wine bottles at once, we can, in principle, determine the poisoned bottle much faster than giving one bottle to one mouse. This is effective for a very large number of wine bottles, in this case, we have 1,000,000 bottles. At each stage, one mouse dies, so we have to replace that mouse. It then follows that we seek to minimize both the number of mice required per stage, and the number of tests we run in order to find the poisoned bottle. If we perform a partition of 10 groups, we require 6 attempts before we find the bottle, and 10 mice at each stage. So 6 mice will die, and therefore, we need 15 mice.$

$\noindent However, if we choose to use a partition of 5 groups, we find that we only kill off 9 mice, and thus, we only need 13 mice. Following this reasoning, 4 groups only kills 10 mice, and requires 13. 3 groups requires 15 mice, and 2 groups require 21 mice. If we choose a partition size any larger than 4 or 5, we will get a smaller number of mice dying, but will require a lot larger number of mice in order to determine the wine bottle successfully. This is the same as minimising the function $y = x+ 6\log_{x}{10}$ on the integers. So by this logic, the least number of mice required is 13 mice.$

$\noindent However, I find myself unable to prove that this is more effective than changing the partition groups at each stage, which may in principle, prove to be more effective than having a fixed number of partition groups.$

RealiseNothing · Feb 6, 2016

Paradoxica said:
$\noindent Yes, there can only be one true statement.$\\$If we give each of the mice a partition of the wine bottles, so that we can eliminate a large number of the wine bottles at once, we can, in principle, determine the poisoned bottle much faster than giving one bottle to one mouse. This is effective for a very large number of wine bottles, in this case, we have 1,000,000 bottles. At each stage, one mouse dies, so we have to replace that mouse. It then follows that we seek to minimize both the number of mice required per stage, and the number of tests we run in order to find the poisoned bottle. If we perform a partition of 10 groups, we require 6 attempts before we find the bottle, and 10 mice at each stage. So 6 mice will die, and therefore, we need 15 mice.$

$\noindent However, if we choose to use a partition of 5 groups, we find that we only kill off 9 mice, and thus, we only need 13 mice. Following this reasoning, 4 groups only kills 10 mice, and requires 13. 3 groups requires 15 mice, and 2 groups require 21 mice. If we choose a partition size any larger than 4 or 5, we will get a smaller number of mice dying, but will require a lot larger number of mice in order to determine the wine bottle successfully. This is the same as minimising the function $y = x+ 6\log_{x}{10}$ on the integers. So by this logic, the least number of mice required is 13 mice.$

$\noindent However, I find myself unable to prove that this is more effective than changing the partition groups at each stage, which may in principle, prove to be more effective than having a fixed number of partition groups.$

Note that you can't have stages since the mice have to have the wine all SIMULTANEOUSLY.

InteGrand · Feb 6, 2016

RealiseNothing said:
Note that you can't have stages since the mice have to have the wine all SIMULTANEOUSLY.

Is the answer: 20??

s-f · Feb 6, 2016

InteGrand said:
Is the answer: 20??

Wow, is it really possible to do that with only 20 mice?
Somehow I need ridiculously large number of mice if I can't do a staged operation... namely 500,000.

RealiseNothing · Feb 6, 2016

InteGrand said:
Is the answer: 20??

Yep

InteGrand · Feb 6, 2016

RealiseNothing said:
Yep

Ah ok, good

.Feeling a bit lazy to write out my method now, maybe wil later, unless someone else posts their solution and is similar to mine.

Paradoxica · Feb 6, 2016

InteGrand · Feb 6, 2016

Paradoxica said:
$\noindent The key is to give each bottle a unique combination of mice (assuming we can label each mouse) for every bottle. Using Pascal's triangle, we see that for $n$ mice, we can obtain $2^n$ unique combinations out of all the mice. Note that one of the selections is the empty set, which means that we must exclude one bottle if the number of bottles is a perfect power of two. Since $500\sqrt{2}<1000 < 2^{10}$, we have $500000< 1000000 <2^{20}$. Thus, we need 20 mice to perform the allocation of 1 million bottles of wine.$ \\$Comment: Are these mice time lords or something? How much can they hold in their stomachs? And this poison. How can we dilute it that much and it still be effective? (Insert joke mocking homeopaths here) I'm pretty sure they didn't have castor beans in those days.$

$\noindent Yeah this was similar to my method. The $2^m$ combination of which mouse drinks and does not drink a given bottle (where $m$ is the number of mice) is more easily seen by noting that for a given bottle $i$, we can see that each mouse either drinks it, or does not. Hence $2^m$, as there are $m$ mice and two possibilities for each. In general, if there are $N$ bottles, the minimum number of mice $m^\star $ needed is the smallest $m$ such that $2^m \geq N$ (just generalising). This is so that we are indeed able to assign each of the $N$ bottles a unique combination of mice who drink it. In terms of the ceiling function, this is $m^\star =\lceil \log_2 N\rceil$.$

RealiseNothing · Feb 6, 2016

Paradoxica said:
$\noindent The key is to give each bottle a unique combination of mice (assuming we can label each mouse) for every bottle. Using Pascal's triangle, we see that for $n$ mice, we can obtain $2^n$ unique combinations out of all the mice. Note that one of the selections is the empty set, which means that we must exclude one bottle if the number of bottles is a perfect power of two. Since $500\sqrt{2}<1000 < 2^{10}$, we have $500000< 1000000 <2^{20}$. Thus, we need 20 mice to perform the allocation of 1 million bottles of wine.$ \\$Comment: Are these mice time lords or something? How much can they hold in their stomachs? And this poison. How can we dilute it that much and it still be effective? (Insert joke mocking homeopaths here) I'm pretty sure they didn't have castor beans in those days.$

InteGrand said:
$\noindent Yeah this was similar to my method. The $2^m$ combination of which mouse drinks and does not drink a given bottle (where $m$ is the number of mice) is more easily seen by noting that for a given bottle $i$, we can see that each mouse either drinks it, or does not. Hence $2^m$, as there are $m$ mice and two possibilities for each. In general, if there are $N$ bottles, the minimum number of mice $m^\star $ needed is the smallest $m$ such that $2^m \geq N$ (just generalising). This is so that we are indeed able to assign each of the $N$ bottles a unique combination of mice who drink it. In terms of the ceiling function, this is $m^\star =\lceil \log_2 N\rceil$.$

Yep these were exactly my method.

Paradoxica · Feb 6, 2016

RealiseNothing · Feb 7, 2016

Paradoxica said:
$\noindent Alice thinks of two positive integers, secretly tells Bob one, and secretly tells Charlie the other. The following conversation occurs:$ \\ $Alice: ''The product of my numbers was 8. Or maybe it was 16...''$\\$Bob (to Charlie): ''I don't know your number''$\\$Charlie (to Bob): ''I don't know your number''$\\$Bob (to Charlie): ''I don't know your number''$\\$Charlie (to Bob): ''I don't know your number''$\\$Bob (to Charlie): ''I know your number!''$ \\ $Given that Bob and Charlie are perfect logicians with infinitely fast deductive reasoning skills, what are the two numbers?$

Bob has 2, Charlie has 4?

Paradoxica · Feb 7, 2016

RealiseNothing said:
Bob has 2, Charlie has 4?

Correct.

Paradoxica · Feb 13, 2016

Paradoxica · Feb 17, 2016

KingOfActing · Apr 25, 2016

This isn't a logic puzzle per se, but a puzzle nonetheless. I'm currently working on writing a BoS puzzle hunt thing, so this will be a good warm up

$It's 6:05pm, and something very peculiar is happening. From your room, you can hear cries of true misery. Someone is wailing and repeating ``O' woe is me!'' over, and over again, and it's starting to get quite annoying. You decide to follow the sounds, and lo and behold, you find the King of Acting himself, sobbing outside the drama studio.$\\$``What's wrong, your highness?'', you ask, to which he responds by reciting a very strange and contemporary monologue of sorts, throwing the script at you:$

Code:

O' woe is me,
Two terms combined,
Those yews are ours,
Beginning and end, intertwined.

Begging for amnesty,
They flow,
In a sea of decay,
But Never giving up Hope,
In what remains.

$...Aaaaand he's gone back to his wailing. You look at the script he's handed you. It looks like any normal script, except at the bottom of the page, scrawled in messy handwriting:$

Code:

DUVCOPLOZUM

$How incredibly strange... Perhaps you can try and figure out what has upset the King so much?$

*Note, only the text in code tags is relevant to the solving of the puzzle.

Hints: (Read at own risk!)

You heard the monologue before you read it.

The King is quite certainly the opposite of a nihilist.

In the midst of his crying, you're almost certain the King whispered something, but you didn't quite catch what it was he said. It sounded a little like "Subway steel tuition"? Hmm...

KingOfActing · Apr 25, 2016

If no one gets it by tomorrow afternoon-ish ~~I'll post another hint~~ I have posted another hint

Feel free to post partial solutions and I can tell you if you're on the right track

Nailgun · Apr 26, 2016

think i have an idea

Nailgun · Apr 26, 2016

the code is a phrase or something

two terms combined --> its made up of two words
those "u's" are ours --> the u's represent 'ours'
in a 'c' of decay --> not sure what this means, probs relevant
Never and Hope are capitalised, possibly names?
O' woe is me is from Hamlet, but it's meant to be Oh
I think the O refers to the letter O in the code, but I can't decipher how the rest of the line is relevant

edit: more thoughts
de-cay, dk
beginning and end intertwined, one word begins like the other ends?

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