Thursday, October 18, 2007

Skeptic's Circle #71: Solutions, part 1

For those of you still hanging around and trying to solve the problems I posed you in the last Skeptic's Circle, I thought I'd do you the favor of compiling some of the solutions that have been posted. So far, I'm just going to give solutions to the problems that someone has solved in the comments, so anyone who wants to can still work on the as-yet unsolved problems. If those don't get solved in a while, I'll post the solutions for them as well.

If you still want to solve them on your own, don't read on. Also, note that for parsimony, I'm not going to be repeating the problems here. Go back here if you need a refresher.

Personalized Perfume Peril

Flip over 48 disks, and then separate those 48 into one pile, with the other 52 in the other pile.

Creative Cake Capers

As yet unsolved, at least here. This problem has been posted with solutions elsewhere on the internet.

Popping Placebo Pills

Take out one pill from the first jar, two from the second, three from the third, four from the fourth, and five (or zero would work too) from the fifth, and weight them. The weight should be the expected weight of 15 placebo pills + x grams, where x corresponds to the number of pills you took out of the jar which has the real pills.

Perilous Peace Problems

Push the cork into the bottle, somehow destroy the cork while making sure any remnants fall into the bottle, melt a hole in the bottle, or simply ignore the whole problem as it's more likely there is no poison gas and it's instead the pill that's poisoned.

Crazed Canting Christians

As yet unsolved here. One little hint: If you make a certain observation about the problem, it becomes trivial math to find the solution.

Hidden Handbook Hassle

Skeptico puts the book in his safe, and his lock on it. He sends the safe to his friend, who puts his lock on it as well, and then returns it. Skeptico removes his lock and sends his safe back. His friend removes his lock and takes out the handbook.

Weird Water Woo

Tilt the glass to the side until the water just reaches the rim. If the water at the bottom also meets the edge there, it's half full. If it's above the edge, you have more than half; below, less than half. If you accidentally spill the water, you now have less than half.

Screwy Scarfe's Secrets

Referring to the guys by the time it takes them to cross:

1 and 2 cross (2 minutes)
1 returns (1 minute)
4 and 8 cross (8 minutes)
2 returns (2 minutes)
1 and 2 cross (2 minutes)

Total time: 15 minutes. Most people end up with some solution that results in 16 minutes, but it's not optimal.

Poor Poisoned Pinheads

Number all the bottles in binary, from 0000000001 to 1101101010 (which corresponds to 874. On the first day, give each of Buzz's captives a number from 1 to 5. Have each of them take a sip from each bottle that has a 1 in its binary digit corresponding to their number. For each captive that gets amnesia the next day, write a 1 in that digit, and a zero for captives who didn't get amnesia. The second day, do the same thing with the 6th through 10th digits. Once all the digits are written down, you'll have uniquely identified the poisoned bottle. At this point, make them all drink from it and throw them out on the street so they won't be able to tell anyone what you did.

What Wifi Woo?

As yet unsolved here. The best solution given can do it in 4 total trips, but it's possible to do it in only 2.

Manic Motor Mythbusting

Use the following program:

Move 100m forward (label: START)
Move 100m forward
Move 100m backward
Skip next command unless a parachute is nearby
Move 100m forward (label: SPEEDUP)

Paddling Pooch Problem

As yet unsolved here.

Action/Adventure Akusai

Choose to play first, and place your first disk at the center of the table. After this, match each of your opponents moves with a symmetric move across the table from him. He'll run out of moves first.

Great Galileo's Ghost

I'll just quote the solutions from my comments here. Figuring out which is which is trivial, as you just have to ask questions you know the answer to, and keep repeating to sort out who's answering randomly. First, from Miller:

For the Galileo puzzle, you can ask the following compound question to voip out lying clones:

Is the following true: You will answer this with "yes" or (inclusive) you are the real Galileo.

Similarly, the following will voip truth-telling clones.

Is the following true: You will answer this with "no" or you are the real Galileo.

Alternating between these two questions will eventually voip a clone with the random curse.

From Simon, a single-shot question:

"Is it the case that you are a clone and that you will either answer this question truthfully with a 'no' or falsely with a 'yes'?"

My original answer to this last part was the question: "Is the statement, 'You are a clone and this statement is false,' true?"

Super-Scammer Secrets

As yet unsolved completely. There are many ways you could decrease the amount of information found on a single slide, but the true puzzle is to figure out a way that whichever two slides are found, absolutely zero information is passed on.

For a hint, imagine if the puzzle were to instead use two slides, and either alone would carry no information. The following solution would work in this case: Pixelate the presentation, and make sure the pixels are quite large and easily distinguishable. Generate one transparency that's completely random, with each pixel being randomly either transparent or 50% opaque. Then, for the second sheet, for the places where the message is spelled out, choose either transparent or 50% opaque as necessary to make the pixel result in 50% opaque. For places where the message isn't, match the pixel to the one on the other slide. The result will be the message appearing in grey text on a mixed black and white background.

Note that this is possibly the most difficult problem here. Though it's been posed on the internet, I haven't found any solutions posted. I have solved it myself, however, so don't worry.

Harebrained Hat Help

Solution by RodeoBob:

Got the Hats puzzle solved. It does, however, depend on everyone being an expert at logic, and everyone following the same game plan...

The color of the wearer's hat is the same color as the smallest group of colored hats he or she can see, and they must make their guess (and leave the circle!) at the first opportunity allowed.

To make it clearer, let's break the process up into 5-minute rounds. (at the end of each 'round', the announcement comes on asking folks to announce the color of their hat)

In the first round, anyone who can only see one hat of a specific color is wearing that color hat. (we know there must be at least two, right?)

In the second round, anyone who can see only two hats of a given color is wearing that color. (we know that there must be three of each color now, since any color that were only present on two heads should have left last round...)

The puzzle only works if everyone is looking, and if everyone leaves at the right time. If somebody falls asleep, or isn't paying attention, or loses count and misses a round, the whole thing falls apart.

Singular Sword Slashes

From Rick Taylor in the comments:

In the singular sword slashes, none of the prisoners were killed.

They all got together and agreed as follows. Whoever was last in line would call out his own hat based on the parity of red hats he saw before him. If he saw an even number of red hats, he'd call his red; if he saw an odd number red hats, he'd call his blue. That man might die, but the next in line, seeing the hats in front of him and knowing the parity of red hats including his own could deduce his hat color. The man in front of him, now knowing both the color of the hat behind him and the parity of all the hats besides his own could deduce his own, and so on to the front of the line. The executioner, hearing this and seeing he could not avoid sparing all but the last in line, arranged the hats to ensure at least he was killed, even though the 99 others were spared, and that was that.

Only it wasn't. All one hundred silently reasoned that the executioner would have to place an even number of red hats in order to kill the last one in line. And so they abandoned their plan and used that information to save them all from last to first.

The truly delicious part of that last solution is that even if we assume the executioner anticipated they would change their strategy to trick him (no reason to as he isn't part of the mensa cult) and put an odd number of red hats to on them, the last man in line would die, but the 99 remaining would still live, even using the wrong information. So there's no reason for them not to try!

Ending Erroneous Expectations

From Edward in the comments:

We can answer the pirates problem using induction, of sorts.

Consider the situation with 5 pirates. If it ever gets down to two pirates, the senior one can simply award all the money to himself and vote for it. With three pirates, the senior one has to convince one other pirate to vote for his plan. The cheapest way of doing this is to award the junior pirate 1 coin and keep 99. Then with four pirates, the senior one only has to convince one other pirate to join him. If it gets down to three, the middle one can't expect to make anything, so he can be bought with 1 coin. With five pirates, the senior pirate needs two others to join him. He can do this by giving one coin to each of the 3rd and 5th most senior pirates, since they'll get nothing if he dies. He would keep the other 98 coins to himself.

Now consider six pirates and only one coin. As before, with only two pirates, the senior pirate can award all money to himself. With three, the senior pirate needs to award the one coin to the junior pirate. With four, the senior pirate can award the coin to either of the two pirates immediately below him. With five pirates, the senior pirate need to convince two pirates to join him, which is impossible. Therefore the second most senior pirate will die if it gets to him, so he will vote for absolutely any plan the most senior pirate proposes. The most senior pirate can then avoid death by awarding the coin to the most junior pirate.

That's it for now, so go give those unsolved problems another try if you think you're up for it!


Maronan said...

For the sword, executioner, and hat question (or whatever you want to call it), would it be cheating to have each person in line tell the next person what colour his hat is? What would stop you?

I guess you can't save the last person, though.

Infophile said...

If they speak out of turn to do so, then yes, that's cheating, and would earn them a swift execution. However, they could forfeit their guess and instead say the color of the hat in front of them, though this method would only save have of them.

Maronan said...

Oh, they can only speak when they're up for execution themselves? In that case, wouldn't it be the first person in line who gives the initial information?

When it says they can see all the hats in front of them, does that mean before them in line or while standing up awaiting execution and guessing?

Infophile said...

They can only see the hats in front of them in the line; the hats aren't put on until they're lined up. (And if they try to look backwards, they're immediately executed.) Also, the executioner starts from the very back of the line and works his way forward.

Anonymous said...

Wifi Woo - When you say this could be done in 2 trips, do you mean round trips?

Infophile said...

Nope. One trip up, one trip back down.

Quantis said...

For the Galileo puzzle;
What about the question, "Would the real Galileo say you are the real Galileo?"

The clones would be unable to answer the predictive question with a certain truth or lie but the real one would be able to declare yes or no with certainty since his answer to this question is the same as his answer to the theoretical question. But I may be missing a flaw in my logic....