eytanz
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True. But at least for a low-ranking pirate in Stigant's puzzle, there's a chance he'll leave with both the money and his life.

In your puzzle, the best a low-ranking pirate can hope for, no matter what the sum is, is that he should survive. Even if he's one of lucky "stable" ranks, he can't get the money himself.

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stigant
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Yep, that does it... I'll give the puzzle torch to eytanz.

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stigant
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Well, in fairness, my pirates could divide the treasure. If Hix's pirates had 500 coins to divide, you get a slightly different solution (give 1 coin to every other pirate - there are no chances involved). If my pirates had only one treasure to pass out, they would be in a similar, um, boat to Hix's.

After I posted, I had another thought - what if each pirate gets a different number of votes? (similar to the electoral votes in the US Presidential election). Give the captain the same number of votes as California (55), the first mate gets Texas (34), the second mate gets NY (32) and so on

Here's a list of states vs electoral votes:
Electoral Votes by State

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eytanz
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stigant wrote:
Well, in fairness, my pirates could divide the treasure. If Hix's pirates had 500 coins to divide, you get a slightly different solution (give 1 coin to every other pirate - there are no chances involved). If my pirates had only one treasure to pass out, they would be in a similar, um, boat to Hix's.

Well, sure. The fact that the treasure could be divided was really the big difference between the two puzzles.

Anyway, offering yet another variation on the theme -

Because the first code suggested (Hix's) ended up causing a shortage of low-ranking pirates, and the second code suggested ended up with the low-ranking pirates all becoming rich and retiring, the pirates decided on a new scheme. This time, only the people who don't get any money nominated to them get any votes. Winning requires a clear majority, but in case there is a tie (or if no-one has any votes) the highest ranking person to get money gets the tie breaker.

Everything else remains the same.

One day, 200 pirates got a haul of only 100 coins. What happens?

(yeah, I'm being lazy, but I have a paper to finish before the end of the day)

[Edited by eytanz at Local Time:12-31-2004 at 09:50 PM]

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Hix
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Nice twist on the voting rules. Before the numbers were edited (used to be 200 each of pirates and gold) there was a surprising amount of eqality that resulted from this rule!

I'll solve this one after 24 hours, if necessary.

stigant
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Hmmm, considerably more intricate (and I don't get equality)- you have to consider each possible number of voters for each number of pirates.
Or maybe not... for small numbers of pirates, you can give everybody 1p, and buy the 1st pirate's vote with 1 more than he'll get next round, then keep the rest. Once it becomes to expensive to buy 1's vote and give everybodyelse 1p, just give 1p to everybody except yourself (you'll be the only vote).

Ok

1 pirate - take all
2 - has to give everything to the first pirate or walk the plank. (1st pirate doesn't get to vote the 2nd pirate off the island, err, boat)
3 - Has to divide the 100 between 1 and 2 - each expect to get 50
4 - 50p to 1, 1p to 2 and 3 (to eliminate their votes) and the rest (48p) to himself. Noone has a vote, so 1 breaks the tie and since an assured 50 is better than a maybe 50, he votes for the plan.
5 - 51p to 1, 1p to 2, 3, and 4, rest (46p) to himself
6 - 52p to 1, 1p to 2-5, 44 to 6
7 - 53p to 1, 1p to 2-6, 42 to 7
8 - 54, 1, ... , 40
9 - 55, 1, ... , 38
10 - 56, 1, ..., 36
11 - 57, 1, ..., 34
12 - 58, 1, ..., 32 (heh, this plan is even weirder than the other two - the more mates, the more the captain makes, and the newest guy on the boat makes off second best)
...
27 - 73, 1, ..., 2
28 - can't make any money this way (74, 1, ...., 0) so he can divide the money anyway he like as long as he doesn't get any. so 1-27 have at least 1p each and an expected 3.71 each.
29 - 1+p to each except himself
30 - 1+p to each except himself
...
101 - 1p to each except himself

Ok, now things change
102 - wet (I think - too many votes against even if he tries to buy off the 1st pirate)
103 - 102 and 103 will vote yay, the other 101 split the 100 coins with 1 nay vote (whoever gets 0 coins)
ugh, have to go to the pet store... back in a bit

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Epigone
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So 101 will vote nay to any 0p before his turn.
102 will vote nay to any 0p before 103's turn.
103 will vote nay to any 0p before 103's turn.

104 is splashed if he has to nominate.
105 gets 104's vote, but that's only 2/5.
106 gets 104's vote and 105's vote for 3/6. The highest ranking pirate, only expecting (100/101)p, will vote yay.

Now 104-106 vote nay for 0p before 106 nominates.

107 splashes, and 108, 109, 110, and 111.

112 will get 6/12 votes, and the highest ranking pirate will again vote yay, getting 1p rather than an expected (100/103)p.

Similarly, 124 will get 12/24 votes and live, and 100+3*2^k will live. So pirates 197-200 will splash, pirates 149-196 will get 0p, and 100 of pirates 1-148 will get 1p.

--
Please pass the abstraction meme.

stigant
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Its not my place to pass the torch (since it was Eytanz's puzzle) but I'm pretty sure you got the answer correct... We've kinda beat this puzzle to death.

Feel free to post another puzzle. If Eytanz gets huffy about it, we'll smite him on the head for you.

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eytanz
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I agree - sorry about that, I saw you posted but I didn't have time to look at the details and then I forgot. You're up.

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Epigone
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Picture the Tower of Pisa (which leans). If it were leaning somewhat more than it is now, it would fall over. Solids can have stable faces on which they can rest without falling and unstable faces on which they cannot rest without falling. Construct a 3-dimensional solid which has no stable face or convince me it can't be done.

[Edited by Epigone at Local Time:01-07-2005 at 10:33 PM]

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Fafnir
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A sphere. The reasoning behind this is that a sphere's single face can be viewed as an infinite number of unstable faces, and that the reason that it can remain motionless without falling to another face is friction - it is effectively being supported by the microscopic imperfections in the surface it is placed on.

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stigant
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Well, if you restrict the domain to convex polyhedra, it wouldn't be possible to find one since the center of mass must lie inside the polyhedron. There must be at least one face on which it can balance. Imagine rotating the solid around the center of mass until one face is parallel with the ground. That face must be stable.

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AlefBet
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Well, working off the nonconvex observation, in the two dimensional area, you could consider a star to have no stable faces since it can't rest on anything but a pair of vertices. A 3 D analog to a star could be constructed, a center with four spires jutting out all equidistant from each other so that the only way the solid could rest stablely would be on three of the vertices of the spires. Is this what you're looking for?

Another example might be a polygona approximation to an hourglass shape with a point at both ends so it can't rest on its top or bottom, but only on its side. But its side is not convex so it ends up resting on edges instead of faces.

If you constrain yourself to convex polyhedra, I would say that they all have at least one stable face. (Probably at least two, I would think.)

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Epigone
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Addendum: a face of size zero is still a face (a.k.a point).

Sphere: Has an infinite number of stable faces.

Convex: True. A convex polyhedra must have a stable face.

Star: Could be balanced on one point.

Hourglass: This looks like it works. Not what I was getting at, but it certainly fits my question. Just make sure it's not symmetric so you can't balance it like the star. Your turn, AlefBet.

[Edited by Epigone at Local Time:01-07-2005 at 10:32 PM]

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AlefBet
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Okay, here you go.

Your TV has one set of RCA inputs that you can connect any RCA device to. However, you have 160 (one hundred sixty) devices you'd like to connect to your TV. You can get four-way RCA switches in order to connect more devices to your TV, and they can be connected and chained in any tree configuration you like.

How many four-way RCA switches do you need to connect all 160 devices to your TV's one RCA input set, and what configuration allows you to do it?

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Hix
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It seems to be...

Every four-way switch you use adds three sets of inputs, regardless of what tree configuration they are in. So you need at least (160-1)/3 = 53 switches to go from 1 set of inputs to 160.

AlefBet
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Yup. Go for it, Hix.

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Tim
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I just thought of something:

Four people (Just call them A,B,C and D) are standing in a large open grid. Everyone occupies a square in the grid looking horizontally or vertically (N,E,W,S). They are also holding a large sword in front of him/her, which also has a size of one grid (like DROD, but only N,E,W,S).

There are three kinds of commands you can give:
1) "A, walk 1 square forward. B, walk 1 square forward. C, walk 1 square forward."
2) "A, walk 2 squares forward. D, walk 2 squares forward."
3) "C, walk 1 square backward if there's no one directly behind you."

Obviously, the swords cannot be in the same place as a person, if that happens, the person will die. And the grid is so large that there are no walls either.

Can you set up a configuration, and a set of commands, so that all 4 persons will stay alive and all touch each other?

-- Tim

PS. There's an evil trick in there...

[Edited by Tim at Local Time:02-12-2005 at 12:00 PM]

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stigant
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What do you mean by "touch" - occupy a square horizontally or vertically adjacent to? Can it be diagonally? Does being next to someone's sword count as touching? etc.

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Tim
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stigant wrote:
What do you mean by "touch" - occupy a square horizontally or vertically adjacent to? Can it be diagonally? Does being next to someone's sword count as touching? etc.

1) Horizontally, Vertically or Diagonally adjacent.
2) Bodies only.
3) A square has only room for one body.



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silver
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just for clarification, can they touch each other before the first "turn"? so if you start out with "AB" in one row and "CD" in a second row (and aligned), now you just need A and D to touch each other, and B and C to touch each other to "finish"? (not that I think that would be possible, if they started out that way, but wanted to clarify the touch rule)


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Maurog
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Hmm, let's see:

 A                                         
 ||               |             |        
  C              AC             C          |     
    -D           |-D          -D         -DC     
  B               B            AB         AB       
  |               |            ||         || 

Nobody        After         Another     Now,
touches       command       command     command
anybody.      no. 2         no. 2       no. 3

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Tim
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silver wrote:
just for clarification, can they touch each other before the first "turn"? so if you start out with "AB" in one row and "CD" in a second row (and aligned), now you just need A and D to touch each other, and B and C to touch each other to "finish"? (not that I think that would be possible, if they started out that way, but wanted to clarify the touch rule)

Yes. [Edit: I mean: they should all touch each other at once, like in a square.] So, that's why even though Maurog example is correct, it can be much shorter

Maurog, feel free to post a next puzzle, because it is a valid solution...

(And silver, I will give you a point if you can tell me what the shortest solution is...)

[Edited by Tim at Local Time:02-13-2005 at 02:01 AM]

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Maurog
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Here is your next puzzle:

We send two paratroopers to land on a railroad track which goes east-west, and we need them to meet. Note that we have no idea where exactly on the track each one landed. In addition, the paratroopers are kinda stupid, and understand only 4 commands:

- Move one step east.
- Move one step west.
- Goto [label].
- If there is a parachute here, Goto [label].

Now we need to provide a list of commands, which must be identical for both of the troopers. The plan is - they land somewhere, leave the parachute there, open the commands list and just follow blindly. Each does one command each second. We must be certain that they meet.
for example the commands list

   [1] Move one step west.
   [2] Goto [1].

will make them both just go west indefinetely (they will not meet obv).
Game on!

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Tim
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Wow! I never thought that was possible...

Here's a list of commands that I think should work:


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TripleM
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Wow, that is cool. Nice solution, Tim.. I wonder if theres any clever variants of that.

edit - it looks like that idea can be generalised to any (fixed) number of paratroopers also. Something like:


[Edited by TripleM at Local Time:02-13-2005 at 04:34 AM]

Maurog
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You got it Tim, although you don't need the going west at all.
They spend one second on checking if there is a parachute anyway, so the shortest solution I know is:

[1] Move one step east.
    If there is a parachute here, Goto [2].
    Goto [1].
[2] Move one step east.
    Goto [2].

Game on!

PS: I actually gave you a clue when I said they both have the same list. In the original puzzle I heard, you can give them different lists, and it takes some thinking to realize they both need to be the same anyways.

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Tim
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Here's a new puzzle:

Using the same rules as my previous puzzle (see a few posts above), give me an non-touching starting position, and a set of commands, to give me the following final position:

...
.X.
.X.
.X.
.X.
...

Which X stand for which person and which orientation of the swords they have is up to you to solve...

This one is easy as well...

--Tim

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Maurog
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This is ridiculously easy:

     |           |         Going from no touches at all to
     B           B          the required position in one
   A-            A-         command of the second type.
      -D        -D
     C           C
     |           |

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Tim
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It was meant to be easy But I didn't know it was that easy...

Next.

[Edited by Tim at Local Time:02-13-2005 at 09:42 PM]

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