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eHalcyon

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Re: Logic problems
« Reply #175 on: January 17, 2013, 06:56:20 pm »
0

I think it would be a lot faster to have the maiden who goes into the tower Day 1 to set the switch in the proper off position.

A witch captures 50 maidens. She puts them all in a dungeon and tells them: "Tomorrow morning I will separate you into 50 separate cells, so you will not be able to communicate or see each other in any way after tonight. Then every once in a while, a few times a day or so, I will randomly choose one of you and take you to my tower.

The information isn't exact enough that you can do that.  I don't think it is guaranteed that any of them will be taken the first day.  Depends on how you read the problem, I guess.

That's the problem: no maid can be sure that she's the first to be taken to the switch room.
Doesn't matter As long as there's at least one a day, you can declare Day 1 as off switch day.

Right.  I'd actually thought about that but I made different assumptions based on the initial presentation of the puzzle.  But I can see how you might arrive at different assumptions.  For your solution to work, the following assumptions must be made:

At least 1 maid is taken each day.
The maids are aware of what day it is.
More specifically, they need to know when it is day 1, and at least one maid must be taken on day 1 in order to do the initial setting.
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jotheonah

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Re: Logic problems
« Reply #176 on: January 17, 2013, 10:08:02 pm »
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Somehow this thread moved from "fun logic puzzles" to "math people speaking another language."

But I remember learning about the Axiom of Choice in one of my philosophy classes. I just can't for the life of me remember what it is. Anyone want to venture a layman's explanation?
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theory

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Re: Logic problems
« Reply #177 on: January 17, 2013, 11:14:53 pm »
+5

It's just another term for mathematicians' most preferred axiom at any given time.  (Since given any number of axioms you can always come up with a most preferred axiom.)
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DStu

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Re: Logic problems
« Reply #178 on: January 18, 2013, 02:05:18 am »
+2

But I remember learning about the Axiom of Choice in one of my philosophy classes. I just can't for the life of me remember what it is. Anyone want to venture a layman's explanation?

The Axiom of Choice tells you that when you have a set of sets, that there exists another set which contains exactly one element of each of these sets.  So say you have {{0,1,2},{4,2,1},{5,3,1}}, there exists a set which has exactly one elementof {0,1,2}, one of {4,2,1} and one of {5,3,1}.

With these example this seems quite obvious, and indeed it is.  But if you have more and larger sets, like infintely many infinite large sets, then this is not so clear anymore.  Or better, it maybe looks like that should be true, but it doesn't follow from the other axioms one uses so far (Otherwise one wouldn't need another axiom).

The famousness of AC is that it is somehow on the edge of reasonablefullness.  That is, while if you assume it is false, you get some unreasonable results (most known maybe that there exists Vectorspaces without a basis), it itself implies some unreasonable results (most known maybe the Banach Tarksi Paradox , telling you that you can decompose a ball into subsets, than move and rotate these and end up with two balls).

Therefore, the AC is the most controversial of the usually used axioms, which means that everybody outside of maybe logic and set theory just uses it, when they need (Which is not that often at least explicitely, but as said, it comes in disguise sometimes like with the basis of Vectorspaces)  But I think in these fields people care more, and also historically it was a bit disputed.
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ghostofmars

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Re: Logic problems
« Reply #179 on: January 18, 2013, 03:58:34 am »
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The second problem has a more cheerful setting. A princess has reached a marriageable age and no less then 100 suitors have shown up to the kingdom. The princess is now faced with the task of finding the best candidate for marriage. One by one the 100 suitors will present themselves to the princess and ask for her hand. The princess will have to give her answer right away.  If the princess rejects a suitor, he will be heart broken and will leave the kingdom immediately.

Fortunately our princess is gifted with a magical ability to judge the suitability of a suitor the second she lays eyes on him. We will assume that the suitability of a suitor can be judged on a linear scale. What strategy should the princess follow to maximize the chance of marrying the best suitor?
If my calculations are correct, skip the first 37 and then take the next one, which exceeds all prior suitors. I get 37.1% chance to find the best suitor in this fashion.
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Watno

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Re: Logic problems
« Reply #180 on: January 18, 2013, 05:56:11 am »
+1

The Axiom of Choice tells you that when you have a set of sets, that there exists another set which contains exactly one element of each of these sets.  So say you have {{0,1,2},{4,2,1},{5,3,1}}, there exists a set which has exactly one elementof {0,1,2}, one of {4,2,1} and one of {5,3,1}.
It doesn't quite work that way. Suppose you have {1}, {2} and {1,2}. Then there is no set that contains exactly one element of each set. You neeed to take an indexed family of sets, and then you get an indexed family of elements.

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jotheonah

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Re: Logic problems
« Reply #181 on: January 18, 2013, 08:58:33 am »
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That was enough to jog my memory, thanks.
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Re: Logic problems
« Reply #182 on: January 18, 2013, 10:41:51 am »
0

Hence the famous joke:

"The Axiom of Choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma?"

The Axiom of Choice is intuitive, but implies the well-ordering theorem which is not, and implies Zorn's lemma, which is too advanced for Earthlings to have an intuition about.
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jotheonah

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Re: Logic problems
« Reply #183 on: January 18, 2013, 10:54:14 am »
+3

Sometimes you have to recognize that there's a level of geek you will never achieve. But I do feel like you guys make me smarter, too. I was really proud of myself when I figured out the blue eyes thing well enough to explain it to a friend.

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theory

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Re: Logic problems
« Reply #184 on: January 18, 2013, 12:25:30 pm »
+2

Here's a logic problem:


How many times is 0 a correct answer to this puzzle?
How many times is 1 a correct answer to this puzzle?
How many times is 2 a correct answer to this puzzle?
How many times is 3 a correct answer to this puzzle?
How many times is 4 a correct answer to this puzzle?
How many times is 5 a correct answer to this puzzle?
How many times is 6 a correct answer to this puzzle?
How many times is 7 a correct answer to this puzzle?
How many times is 8 a correct answer to this puzzle?

(The answer is unique.)
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Galzria

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Re: Logic problems
« Reply #185 on: January 18, 2013, 12:43:49 pm »
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Here's a logic problem:


How many times is 0 a correct answer to this puzzle?
How many times is 1 a correct answer to this puzzle?
How many times is 2 a correct answer to this puzzle?
How many times is 3 a correct answer to this puzzle?
How many times is 4 a correct answer to this puzzle?
How many times is 5 a correct answer to this puzzle?
How many times is 6 a correct answer to this puzzle?
How many times is 7 a correct answer to this puzzle?
How many times is 8 a correct answer to this puzzle?

(The answer is unique.)

Gut instinct says One, since the statement "1 is the answer to this puzzle" is true exactly "1" time. If the statement "4 is the correct answer to this puzzle" were to be true, then it would still only be true exactly "1" time.

Perhaps that's poorly explained, but it's what I instinctively see.
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Quote from: Voltgloss
Derphammering is when quickhammers go derp.

Faust has also been incredibly stubborn this game. In other news, it's hot in the summer, and water falls from the sky when it rains.


Mafia Record:
TOWN Wins: M3, M5, M6, M11, M17, M28, M32, M105, M108, M114, M118, M120, M122, DM1, DoM1, OZ2, RM45, RM47, RM48, RM49, RM55
TOWN Losses: M4, M7, M8, M9, M13, M14, M18, M31, M110, M111, M113, M117, M125, RM3, RM4, RM54
SCUM Wins: M2, M19, M23, M100, DM3, RM1, RM2, RM48, RM50
SCUM Losses: M15 (SK), M102 (Tr), OZ1, RM55

Total Wins: 30
Total Losses: 20

jotheonah

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Re: Logic problems
« Reply #186 on: January 18, 2013, 12:45:33 pm »
0

How many times is 0 a correct answer to this puzzle?

Isn't this a paradox (not unlike "This sentence is a lie")?

Assume that 0 is never a correct answer. Then the correct answer would be 0.
Assume that 0 is sometimes a correct answer. Then the correct answer would never be 0.
So the answer can neither be 0, nor can it be anything else.
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theory

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Re: Logic problems
« Reply #187 on: January 18, 2013, 12:49:47 pm »
+1

The answer isn't a single digit, it's a series of digits.  The goal is to come up with the 9 answers that are internally consistent.
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jotheonah

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Re: Logic problems
« Reply #188 on: January 18, 2013, 12:53:05 pm »
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But the very first one seems to be inherently unanswerable.
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Rabid

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Re: Logic problems
« Reply #189 on: January 18, 2013, 12:54:27 pm »
0

How many times is 0 a correct answer to this puzzle? 5
How many times is 1 a correct answer to this puzzle? 2
How many times is 2 a correct answer to this puzzle? 1
How many times is 3 a correct answer to this puzzle? 0
How many times is 4 a correct answer to this puzzle? 0
How many times is 5 a correct answer to this puzzle? 1
How many times is 6 a correct answer to this puzzle? 0
How many times is 7 a correct answer to this puzzle? 0
How many times is 8 a correct answer to this puzzle? 0
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Jack Rudd

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Re: Logic problems
« Reply #190 on: January 18, 2013, 12:55:48 pm »
0

Here's a logic problem:


How many times is 0 a correct answer to this puzzle?
How many times is 1 a correct answer to this puzzle?
How many times is 2 a correct answer to this puzzle?
How many times is 3 a correct answer to this puzzle?
How many times is 4 a correct answer to this puzzle?
How many times is 5 a correct answer to this puzzle?
How many times is 6 a correct answer to this puzzle?
How many times is 7 a correct answer to this puzzle?
How many times is 8 a correct answer to this puzzle?

(The answer is unique.)

At least one of {4, 5, 6, 7, 8} has a nonzero answer. Otherwise there would be at least five times that 0 was a correct answer, leading to a contradiction. However, no two of them can each have a nonzero answer. If they did, the pair that did would have to be 4 and 5 (as any other pair would immediately sum to more than 9), but then, even with 5 0s and 4 1s, there couldn't be four numbers that appeared once.

Thus exactly one of {4, 5, 6, 7, 8} has a nonzero answer, and that number is the number of times 0 appears. By the same logic as before, it has to be exactly 1 (because otherwise we end up with, at a minimum, two 4s and a 2 adding up to 10). Therefore there exists at least one 1 - and exactly one 1 would lead to a contradiction in the 1s themselves, so there must be at least two 1s.

This gives us: A 0s, where A is at least 4.
B 1s, where B is at least 2
At least 1 B
Exactly 1 A

These add up to 8, so we have scope for increasing A or B beyond the minimum. Increasing B to 3 would cause an extra 1 to come in as well, so there is no room for it, but we can increase A to 5 with no problem. Thus:

How many times is 0 a correct answer to this puzzle? 5
How many times is 1 a correct answer to this puzzle? 2
How many times is 2 a correct answer to this puzzle? 1
How many times is 3 a correct answer to this puzzle? 0
How many times is 4 a correct answer to this puzzle? 0
How many times is 5 a correct answer to this puzzle? 1
How many times is 6 a correct answer to this puzzle? 0
How many times is 7 a correct answer to this puzzle? 0
How many times is 8 a correct answer to this puzzle? 0

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Centuries later, archaeologists discover the remains of your ancient civilization.

Evidence of thriving towns, Pottery, roads, and a centralized government amaze the startled scientists.

Finally, they come upon a stone tablet, which contains but one mysterious phrase!

'ISOTROPIC WILL RETURN!'

soulnet

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Re: Logic problems
« Reply #191 on: January 18, 2013, 12:57:04 pm »
0

I'm going to guess 521001000, although if I read literally, it does not seem correct. Maybe "How many times is 0 in a correct answer to this puzzle?" is clearer? Or I misinterpreted this?

BTW, search for self describing sequence or Golomb's sequence or similar for LOTS of problems on this sort of thing.

Ninja'd.
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Galzria

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Re: Logic problems
« Reply #192 on: January 18, 2013, 12:59:15 pm »
0

Here's a logic problem:


How many times is 0 a correct answer to this puzzle? 5
How many times is 1 a correct answer to this puzzle? 2
How many times is 2 a correct answer to this puzzle? 1
How many times is 3 a correct answer to this puzzle? 0
How many times is 4 a correct answer to this puzzle? 0
How many times is 5 a correct answer to this puzzle? 1
How many times is 6 a correct answer to this puzzle? 0
How many times is 7 a correct answer to this puzzle? 0
How many times is 8 a correct answer to this puzzle? 0

(The answer is unique.)

521001000?

Starting from "000000000" and going forward - although maybe I mixed up a step along the way and didn't go quite far enough? But I think I've got the general spirit of it down, if not the correct answer.
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Quote from: Voltgloss
Derphammering is when quickhammers go derp.

Faust has also been incredibly stubborn this game. In other news, it's hot in the summer, and water falls from the sky when it rains.


Mafia Record:
TOWN Wins: M3, M5, M6, M11, M17, M28, M32, M105, M108, M114, M118, M120, M122, DM1, DoM1, OZ2, RM45, RM47, RM48, RM49, RM55
TOWN Losses: M4, M7, M8, M9, M13, M14, M18, M31, M110, M111, M113, M117, M125, RM3, RM4, RM54
SCUM Wins: M2, M19, M23, M100, DM3, RM1, RM2, RM48, RM50
SCUM Losses: M15 (SK), M102 (Tr), OZ1, RM55

Total Wins: 30
Total Losses: 20

Galzria

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Re: Logic problems
« Reply #193 on: January 18, 2013, 01:00:31 pm »
0

Seems I was ninja'd by 3 people in the 5 minutes I wrote that up. :P

But if all 4 of us came to the same conclusion, I've got to believe it's correct.
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Quote from: Voltgloss
Derphammering is when quickhammers go derp.

Faust has also been incredibly stubborn this game. In other news, it's hot in the summer, and water falls from the sky when it rains.


Mafia Record:
TOWN Wins: M3, M5, M6, M11, M17, M28, M32, M105, M108, M114, M118, M120, M122, DM1, DoM1, OZ2, RM45, RM47, RM48, RM49, RM55
TOWN Losses: M4, M7, M8, M9, M13, M14, M18, M31, M110, M111, M113, M117, M125, RM3, RM4, RM54
SCUM Wins: M2, M19, M23, M100, DM3, RM1, RM2, RM48, RM50
SCUM Losses: M15 (SK), M102 (Tr), OZ1, RM55

Total Wins: 30
Total Losses: 20

jotheonah

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Re: Logic problems
« Reply #194 on: January 18, 2013, 01:00:47 pm »
0

Oh, now I get how it works it's a fun puzzle.
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jotheonah

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Re: Logic problems
« Reply #195 on: January 18, 2013, 01:06:48 pm »
0

I just found this one on wikipedia:
http://en.wikipedia.org/wiki/Zebra_Puzzle

Have not worked it out yet, looks fun to solve
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Jimmmmm

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Re: Logic problems
« Reply #196 on: January 19, 2013, 08:01:49 am »
0

I just found this one on wikipedia:
http://en.wikipedia.org/wiki/Zebra_Puzzle

Have not worked it out yet, looks fun to solve

I started this this morning and finally have an answer (I did other stuff too!).

My solution, before checking:

The Norwegian drinks water, and the Japanese owns the zebra.

Checking... Hooray!
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DStu

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Re: Logic problems
« Reply #197 on: January 19, 2013, 12:48:43 pm »
+1

@infinite hats:

Again we have here that everybody guesses wrong in 1/2 of the cases.  This also implies that on average, there are infinitely many people guessing wrong.
Now it's still possible that this is the case when just finitely many people guessing wrong in every case. For example, you can have person 1 guessing wrong in always, person 2 and 3 guessing wrong in 1/2 the cases, 4-7 in 1/4 and so on, so that you have like 2ˆn person guessing wrong in 1/2ˆn of the cases.  This would always be finite,  but with infinite mean.

But I don't the how to do this, we had like 1/2ˆn in the riddle with the 4 hats, but there everybody sees everyone else hats, and they didn't had to guess...

I don't see any method how to guarantee that the guesses are coupled in this sense here, so that one person guessing wrong/right implies that one other person is also wrong/right.
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Re: Logic problems
« Reply #198 on: January 19, 2013, 01:15:15 pm »
0

Again we have here that everybody guesses wrong in 1/2 of the cases.  This also implies that on average, there are infinitely many people guessing wrong.

This is a nice comment that I haven't seen before.
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Re: Logic problems
« Reply #199 on: January 20, 2013, 05:04:59 am »
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The second problem has a more cheerful setting. A princess has reached a marriageable age and no less then 100 suitors have shown up to the kingdom. The princess is now faced with the task of finding the best candidate for marriage. One by one the 100 suitors will present themselves to the princess and ask for her hand. The princess will have to give her answer right away.  If the princess rejects a suitor, he will be heart broken and will leave the kingdom immediately.

Fortunately our princess is gifted with a magical ability to judge the suitability of a suitor the second she lays eyes on him. We will assume that the suitability of a suitor can be judged on a linear scale. What strategy should the princess follow to maximize the chance of marrying the best suitor?
If my calculations are correct, skip the first 37 and then take the next one, which exceeds all prior suitors. I get 37.1% chance to find the best suitor in this fashion.

That is indeed correct. In general the princess should reject the first 1/e part of the suitors, where e=2.71..., Euler's number. Here's the proof:

Assume there are n suitors and that the princess will reject the first k of them and then pick the
first suitor that is better than all the previous ones. The chance she'll find the best is
P(princess finds best suitor) = \sum_i=1 to n P(princess chooses suitor i| suitor i is best) x P(suitor i is best)
= \sum_i=1 to k 0 x 1/n+\sum_i=k+1 to n P(best suitor before i was one of the first k) x 1/n
= \sum_i=k+1 to n k/(i-1) x 1/n = k/n x \sum_i=k+1 to n 1/(i-1) = k/n x (1/k + 1/k+1 ... +1/(n-1))

We can approximate this sum by log(n-1)-log(k-1) = log( (n-1)/(k-1) ) ~=~ log(n/k) = - log(k/n).
So the princes has -k/n x log(k/n) chance of finding the best suitor. The function f(x) = -x log(x) has a maximum at x = 1/e and this maximum is also 1/e.

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