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OK, so before the bizarre misunderstandings in my previous post, I had been going to repost question which I thought was an interesting logic puzzle in its own right.

You have five bags of holding. One contains a fabulous treasure. Two contain liches who can't escape until you open the bag. Two contain nothing.

You have a spell which tells you something about the result of a course of action you propose. (This description is slightly altered from the functionality of the original spell to make the puzzle work, feel free to ask for clarification as needed.)

"Weal" for good result (eg. treasure, no liches)
"Woe" for bad result (eg. 1+ lich, no treasure)
"Weal and Woe" for a good and bad result (eg. treasure and also lich)
"Nothing" for a result of no particular good or bad (eg. open no bags or only open empty bags)


What's the minimum number of castings of the spell needed? (I think 3 is easy and 1 is impossible, so basically, can you do 2?)


The course of action has to be 30 minutes or less.

We don't have specifics on how you define the course of action, ask if it needs to be more explicit.

Assume you can include other results in the plan if they help, eg. "if this bad contains nothing, I stab myself in the leg", without necessarily needing to follow through. (This is slightly more generous than the original spell.)

Assume you don't include the castings of further divination spells within the scope of the course of action considered by casting the first spell.

Follow-ups (may be unnecessary depending on the best solution to the original)

If you only have one casting, what's the greatest chance you can give yourself of finding the treasure whilst finding no liches.

The original restrictions of the spell say that if you cast it four times in a day (ignored for the basic puzzle), the second, third and fourth times have a 25%, 50% and 75% chance of giving a random answer. What's the highest chance you can give yourself of finding the treasure and no liches in up to four castings with those failure chances.

Previously we assumed you couldn't create a paradox. If you *can*, and causing a paradox causes the spell to fail to give an answer in a way distinct from "nothing", can you reduce the number of castings?

If you *can* ask about a course of action including further divination spells, does that help?

Does the answer generalise to a larger number of bags (assuming 1 treasure, N liches and N nothing)

ETA: Fix formatting.
jack: (Default)
A friend on facebook linked to a variant of the 100 hats puzzle I talked about at: http://cartesiandaemon.livejournal.com/232415.html?nc=21

Suppose there are 100 scientists (or logicians, or philosophers...) and 101 hats. As in the other puzzle, the scientists are told the rules, allowed to confer, and then not allowed to speak. They're lined up all facing the same way, so the one at the back can see everyone else's hats, and the one at the front can see no-one's hats. And no-one can see their own hat.

The hats are numbered "1" to "101". The hats are placed randomly on the heads of the scientists, with one unknown hat left over. In any order the scientists can guess what hat they're wearing. But can't guess a number that was previously guessed.

What's the maximum number of scientists you can get to guess right?

If you have that, can you do it if there are 102 hats and two are left over? I genuinely don't know the answer to that one.

Hints in the comments.
jack: (Default)
Ptc joked that authors kill off an average of three major characters -- most kill none, but one is George Martin :)

Which got me thinking, which author has really killed off the most characters? I want to say "named" rather than "major" so it's more objective. In which case you maybe want two categories: authors who kill off a whole world at the end of a book; and authors who kill off the most named characters -- but still have plot going on afterwards.

Who are the obvious candidates?

ETA: With thanks to Mark for the link and to seekingferrett, the current frontrunner is the Illiad with 254 :)
ETA: Apparently GoT beats it out with 293!
jack: (Default)
A friend asked on twitter, if you could place two connected portals anywhere on earth (but not in space), what would you do with them. Personally, linking Cambridge and Liv's campus flat would be nicest! Geopolitically, probably linking two disparate regions might be most useful.

But of course, the question turned to free energy. Suppose the portal is 1mx2m, laid horizontally, one at ground level and one 100km up at the edge of space, and you diverted sufficiently much water into the lower one to get an endless waterfall. How much energy would you get out?

I'm not sure I have these equations right any more, but under those assumptions (and that the density of water and the gravitational constant for the whole height are rounded to the nearest power of ten and that the square root of 2 is 1.5), and assuming that extracting the energy from the falling water slows it to rest again, I tried to calculate the speed and the energy. I think freefall from that height (assuming you can somehow construct an airtight tube) lands at 1500m/s. And that translates to 3x10^12 Watts = 3TW. Can anyone confirm if I have that right?

It's hard to tell from wikipedia, but it looks like that world energy consumption was 15 TW the last time someone worked it out and updated that page. So this device would make a worthwhile dent in it, but not obsolete everything else.

Of course, that assumes there are sensible engineering solutions to "build an airtight tube to the edge of space" and "slow water from mach 5 safely without wasting any energy" which there probably aren't.

You could dig the tunnel _down_ instead, but you'd have to actually dig it, although you could put one portal down there and dig up through the other one, until the rock fell freely. And you'd have to be careful not to go down too far because if your experiment starts spewing pressurised magma you've invented your own personal volcano, and you have to hope that it solidifies and buries it well enough to withstand ten atmospheres of pressure.

Of course, if you did the above experiment with rock instead of water, the energy involved would be 10 times greater, although presumably the engineering challenge would still be the limiting factor.

However, leaving the practical impossibilities aside, something bothered me about the physics. It seems like, if you don't slow the falling substance completely but let it go through the portal already at high speed, you get correspondingly more energy out. I guess because the limiting factor for energy is that each atom going through a portal to 100km up creates a certain amount of potential energy, so you get the more energy the more you cause that to happen. It seems dodgy that the energy production could just keep on growing in that case, but I guess the assumptions violated physics, so there's no reason not to expect that to violate a wide number of other principles. Have I actually got that right?
jack: (Default)
Also: The character of swampy is really cute. He doesn't really do much other than have a shower, but he's very expressive.

Also: It's really really good to see modern games hitting a common denominator higher than farmville but lower than AAA FPS games. There's good and bad about games having a $0.65 price point (you need lots of development values to reach a wide audience, as well as to reach an obsessed one), but in general it's nice to see something that can be enjoyed by so many people it's basically free, which gaming previously seemed to have given up on.

Although: I'm always scared by "buy this upgrade, buy that upgrade". I'm happy to buy several chunks when I can see what they are, but I'm extremely nervous of falling into the clutches of someone like zynga, and getting fleeced with "oh, just pay this then you can play the game as intended, no wait, ok, just pay THIS, ok, well, pay that a few more times..."
jack: (Default)
Puzzle the first, I assume you have heard this one, but I'll repeat it in case you haven't.

Suppose a hundred people are buried in the sand one in front of another so each can see all the people in front but none of the people behind. An Evil Dictator (TM) places a red or blue hat on each of them. He goes along the line from the back asking each person what colour their hat is, and at the end kills everyone who gets it wrong.

They're allowed to discuss strategy beforehand, but the dictator listens and doesn't allow them to come with any tone-of-voice signals or the like, information can only be conveyed by saying "red" or "blue" when asked.

What strategy is sure to save the most people? If the hats are colored randomly What strategy has the best Expectaion of number saved? For instance the first person is always at risk because NO-ONE knows anything about his hat at all, so no solution can do better than 99 guaranteed saved or 99.5 expected saved.

Solution )

Puzzle the second. What if the line was (semi) infinite?

Solution )
jack: (Default)
Puzzle: I have two identical cylindrical lemonade bottles I punched small holes into the sides of, just above the bottom1. I find that when I fill them each with water the first empties in 225 seconds and the second more quickly in 75 seconds. Without any other form of measurement, how do I use them to time 25 seconds?

Q. That's just one of those hourglass/numerology puzzles. If one hourglass empties in time p minutes, and the second in time q, then calculate p-1 mod q, and you know pp-1=1 (mod q) so pp-1=1+nq. Run glass Q n times and glass P p-1 times; when the first has finished you have exactly one minute before the second does.
A. If you think that's easy, try it.

Q. It's a trick, isn't it.
A. Probably.

Q. Tilt one of the bottles...
A. If you like. They *are* cylindrical, so you *can* halve the amount of water by tilting them until the water only just covers the base, if you think it'll help.

Q. I could use a...
A. No other equipment is necessary. You can use additional vessels of unknown size if you want to.

Q. How quickly did you solve it?
A. Actually, I made the puzzle up myself last night.

Q. Are you sure your solution works?
A. I may be mistaken. In which case I apologise, and promise to feel very embarassed.

Q. Do I need to use much maths for this?
A. Some. Nothing not taught at A-level IIRC.

Q. Can I find a different solution?
A. I don't think so, but try; it might be better.

[-1] That's not a footnote, that means 'inverse of p'
[1] Really.