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Beef Row
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zex20913 wrote:
Before I get started, does this mean "immediate left" and "immediate right"? That is, no ravens between others if they say "to my left"?

It does, because otherwise the fact that each raven only requires one other raven to be in a particular place, combined with the freedom to face either way means you have no information at all except that either Prahgz or Crangs is west of Quraksch, and the others may be placed arbitrarily.

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06-10-2008 at 11:34 PM
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TripleM
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Logic bash..

So we can ignore left + right and just consider adjacency, and thus we know one of each of these pairs are adjacent:

bd or cd
pq or cq
bk or ck
cp or dp
cg or ck
dg or cg
bp or bg

Case 1: If ck:

Case 1a: if cd, then we have dck => bdck => impossible
Case 1b: if cq, then we have qck => qck,bd => qck,bdp => qck,gbdp => impossible
Case 1c: otherwise, we have ck,bd,pq => either kcpq,bd=>kcpq,bdg => impossible, or ck,bdpq => gck,bdpq => kcgbdpq.

Case 2: Otherwise, must have bk,gc. b is next to p or g, therefore not d, so c is next to d. So we have bk,gcd => bk,gcdp => kb,gcdpq => kbgcdpq.

So, they both work; the former had both of Crang's statements true, which is still logically valid but I guess the second one was intended.


I heard the following cited from a book called Puzzles in Math and Logic. I don't know the answer, so the aim is to convince me your answer is correct :)

If we assume all planet's orbits are perfect circles centered around the sun (and we assume they all move at different speeds so that you could pick a spot for each planet in its orbit and at some point they will all be at the positions you choose), what planet is closest to Pluto most of the time?

[Last edited by TripleM at 07-17-2008 05:13 AM]
07-17-2008 at 05:13 AM
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halyavin
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Because Pluto is not a planet now, are we assume that Pluto's orbit is circular? In reality it is very far from circular and Pluto has a 3:2 resonance with Neptune (what contradicts to the second assumption if we consider Pluto to be a planet).
07-17-2008 at 09:17 AM
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TripleM
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Heh. Well, no, that wasn't the point I don't think. We can assume Pluto doesn't even move, really.
07-17-2008 at 10:47 AM
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See attached.

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07-17-2008 at 01:32 PM
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TripleM
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icon Re: Puzzle "tag" (0)  
I'm not quite sure I follow your logic. While the average vector from Pluto to a planet = the vector from Pluto to the sun, this doesn't mean there isn't a planet which is closer more often.

Consider a case where we have two planets (other than Pluto) - one which is arbitarily close to the sun, and one which is arbitrarily close to Pluto. Then the planet close to the sun is closer to Pluto for 2/3 of the time - see image. (Apologies for my bad hand drawing ;))



[Last edited by TripleM at 07-17-2008 11:48 PM]
07-17-2008 at 11:46 PM
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icon Re: Puzzle "tag" (0)  
The answer is the innermost planet. Math forthcoming as soon as I look up the integral of sqrt(a^2 + 1 - 2acos(theta)) dtheta. (Letting the radius of pluto's orbit be 1, the average distance from a planet with orbital radius r to pluto is 1/pi * integral from 0 to pi of sqrt(r^2 + 1 - 2rcos(theta)) dtheta). If you play around with a plotting program, and vary the values of r, it's a simple matter to see that increasing r increases the value of this integral. I've attached a sketch in Geometer's Sketch Pad that illustrates the idea. The point on the x-axis represents r. Slide it left and right (from 0 to 1) and see how the area under the curve changes. This should convince you of the truth of my assertion, but I'll work on a harder algebraic proof.

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[Last edited by stigant at 07-18-2008 02:33 AM]
07-18-2008 at 02:32 AM
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TripleM
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I agree the average distance is smallest for Mercury, but again that doesn't necessarily mean it is the closest to Pluto for the most time. With a similar example as above, if we have one planet close to the sun and a hundred very close to Pluto, it will only be the closest (2/3)^100 of the time - which means one of the others will be closer more often.

(Perhaps the problem is too hard to have an answer, but the official wording definitely implied closest most of the time.)

[Last edited by TripleM at 07-18-2008 05:56 AM]
07-18-2008 at 05:49 AM
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stigant
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oops, you're right, I answered the wrong question. But lets think about your situation: 1 planet near the sun, 100 planets with almost the same orbital radius of pluto. By your reasoning, the innermost planet has a (2/3)^100 chance of being the closest planet at any given time. However, what is the probability of, say, the second closest planet to the sun (ie the one of the 100 which is closest to the sun) being the closest planet to pluto? Well, it must be closer than all the other planets. It has approximately a 1/3 chance of being closer than "Mercury". Now, lets compare it with any one of the other 99 planets. It should have about a 1/2 chance of being closer than one of the other planets which are, essentially, indistinguishable from planet 2. So shouldn't it's probability of being closest be 1/3 * (1/2) ^ 99? This is less than "Mercury's" probability above. However, (2/3)^100 + 100*1/3*(1/2)^99 <<< 1. So I/we have made a mistake somewhere.

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07-18-2008 at 09:09 PM
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TripleM
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All of those halves aren't independent though. Out of the 100 planets near Pluto, each one is equally likely to be the closest of those 100 with probability 1/100, not (1/2)^100. Combine that with the 1-(2/3)^100 chance of being closer than Mercury and they add up to 1.
07-18-2008 at 11:35 PM
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beaver
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TripleM wrote:

If we assume all planet's orbits are perfect circles centered around the sun (and we assume they all move at different speeds so that you could pick a spot for each planet in its orbit and at some point they will all be at the positions you choose), what planet is closest to Pluto most of the time?

The true answer is Pluto, as it always is 0 distance from itself.

A better answer- [EDIT: removed due to lengthy poor unclear logic that lead to a wrong answer]

[Last edited by beaver at 08-15-2008 03:31 AM]
08-13-2008 at 02:57 AM
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TripleM
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I don't quite follow your logic - have a look at the examples I posted above. The case where one planet is virtually at the sun and another with virtually the same radius as Pluto, which I think is the first case you were trying to explain, doesn't give a 50/50 split - but 2/3 and 1/3, as explained above (see my diagram).

Oh, and :hi :)

[Last edited by TripleM at 08-13-2008 03:58 AM]
08-13-2008 at 03:40 AM
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TripleM
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I just heard back from the person who posted this straight from the book in another forum, and he said he'll tell me the author's explanation tomorrow - but the author says Mercury, which I've already shown doesn't work in general, so I guess the solution in the book is wrong like I was starting to suspect.

So, sorry for posting a problem which probably has no easy solution, though hope you had fun playing with it anyway :P Someone else can post a puzzle if they have one.
08-13-2008 at 08:06 AM
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stigant
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Was the original problem supposed to use the actual data for the actual planets orbiting our sun? Because its possible that the real Mercury is closer to the real Pluto more often than any of the other real planets.

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08-14-2008 at 02:52 AM
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beaver
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You're right. I was measuring distance funny- only looking at it in a single direction.
08-15-2008 at 03:29 AM
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Timo006
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beaver wrote:
TripleM wrote:

If we assume all planet's orbits are perfect circles centered around the sun (and we assume they all move at different speeds so that you could pick a spot for each planet in its orbit and at some point they will all be at the positions you choose), what planet is closest to Pluto most of the time?

The true answer is Pluto, as it always is 0 distance from itself.


that answer is actually wrong, because Pluto isn't a planet. O:-

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[Last edited by Timo006 at 08-20-2008 05:21 PM]
08-20-2008 at 05:20 PM
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wonkyth
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It's sad, isn't it? :weep

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08-26-2008 at 09:49 AM
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lopsidation
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So who posts the next puzzle?

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08-30-2008 at 02:18 PM
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wonkyth
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So puzzle the next who posts!

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09-06-2008 at 05:27 AM
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lopsidation
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Okay, sure.
What comes next in this sequence?
I, Z, E, H, S, _,

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09-07-2008 at 04:11 PM
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silver
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I'm going to go with "b" followed by "B" and finally "P" (and series extends to the left with "O" but then stops, in both directions)


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09-07-2008 at 04:53 PM
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lopsidation
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Yep! Your turn.

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09-08-2008 at 09:36 PM
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silver
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Fraid I'm gonna pass - I can't come up with any puzzles right now, and even if I had one, I'm not going through 71 pages of topic to see if it's been done already (which it probably would be, all the puzzles I know are old)


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09-08-2008 at 09:43 PM
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RoboBob3000
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Hmm, what's the reasoning behind this latest puzzle? Even with the answer it's still got me stumped.

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09-08-2008 at 10:45 PM
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lopsidation
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Since silver doesn't have a puzzle, how about the person who posts the reason gets to post the next puzzle?

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09-08-2008 at 10:59 PM
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Maurog
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Umm, silver, I think you forgot an "L" between the "b" and "B"... otherwise seems correct.

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09-08-2008 at 11:35 PM
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silver
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good point, Maurog.
so the whole sequence:

O I Z E H S b L B P
though I personally would have written:
O I Z E h S b L B P

Robobob: the relation is not consistent or perfect, just kinda vague. Dunno if that will help you decipher it.



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[Last edited by silver at 09-09-2008 02:31 AM]
09-09-2008 at 02:30 AM
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TripleM
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Oh.. I finally clicked.
09-09-2008 at 02:42 AM
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lopsidation
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Um, wait, shouldn't it be
Click here to view the secret text


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09-09-2008 at 11:53 PM
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Monkey
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Ohhh...
Click here to view the secret text


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09-10-2008 at 01:14 AM
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