More Solutions

For those who are still around, it was recently pointed out to me that I'd forgotten to give the solutions to the last couple of unsolved problems from my Skeptic's Circle. Since it's been a while, I'll repeat the problems here to remind you before solving them:

Super-Scammer Secrets

Lord Runolfr recently got a couple of e-mails from scammers, raising my suspicions that something big was afoot behind the scenes. I called up a few contacts, did some research, hired a few James Bond-alikes, and here's what I've figured out:

There's some evil mastermind behind the whole plan, and he's about as supervillanous as they come. This of course means that he wants to capture one of my James Bond-alikes and subject him to an intricately detailed explanation of his evil plan before killing him in a creative way. Apparently, the way he's decided to go about this revelation is through an overhead transparencies with the key points of his evil plan (you'd think he'd have better technology than my high school, but he's out to make money, so he saves it where he can).

Now of course, he's wary of this transparency falling into the wrong hands, so he's come up with a plan to keep things safe. He's figured out some method to spread the information across multiple sheets, arranging it so that if we get a hold of any two, we'll still have no idea what he's planning. So, what we need to figure out now is some possible ways he might have done this, so if we manage to get our hands on more than two, we'll know how to read them (if it's not immediately obvious). What are some possible things he could do? Remember that he's out to save money, so splitting it up to have a single word on each slide or something huge like that doesn't seem too likely.

Solution

While it's possible to split up the information in a simple way over many slides, the trick to this problem is finding a way to divide it up so that absolutely zero information is transmitted in a single slide, or even in two slides. Splitting it up into words or letters (or fragments of a letter) never accomplishes this, as the remaining pieces still give some information. To illustrate a way of transmitting zero information in a simpler case, where any one slide can be lost, I gave the following sample solution:

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 gray text on a mixed black and white background.

Now the trick is moving up to a case where any two slides can be lost, and together they'll give zero information. Part of what makes this problem tricky is that everyone tries to find a solution that will use the absolute minimum number of slides at first (in this case, 3). However, such a solution doesn't exist (at least that I've seen, and I tend to suspect it doesn't at all). There is, however, a solution that uses 4 slides.

Here's how to construct the 4-slide solution: As before, pixelate the message. This time, however, instead of filling the pixels with either gray or white, we'll be filling them with one of three colors of ink. This ink will be designed to absorb one third of the visible light spectrum and let the rest pass through. For instance, we could have ink that absorbs red wavelengths, allowing blue and green light through (which appears as cyan I believe), plus ink to absorb blue (appears as yellow) and ink to absorb green (appears as magenta). If we stack the three different colors on top of each other, no light can get through, and we'll have a black spot on the image. If only two colors are in the stack, it will appear as the remaining color. If just one color, we'll get a mix of the other two.

Now, for each pixel of the image, first determine whether it's part of the message or not. If it is, we want it to appear black. So, through the four slides, we'll arrange it so all three colors show up somewhere, plus one of them appearing twice. We'll randomly choose between all possible permutations that do this. Now, for pixels that aren't part of the message, we want them to not appear black, so we randomly choose one of the permutations that uses only one or two colors. The net result is the message appearing in black on a multicolored (or gray, if the pixels are small enough) background.

To see that this works, image that two slides are stolen, and look at a single pixel on both of them. There are two possibilities here: 1) both slides have the same color for that pixel and 2) the slides have a different color in that pixel. In case 1, it's possible for the remaining two slides to have the other two colors, and it's also possible they might repeat this color. For case 2, it's possible the remaining color will be on one of the other slides, but it's also possible it won't be. In the end, we can't infer anything about whether or not this pixel is part of the message.

Now, why won't this work for only three sheets? Well, let's go back to case 1 if two slides are recovered. If both slides have the same color, there's only one slide left to block more light. This third slide couldn't have both colors, so this pixel cannot be part of the message. We can't infer anything from the cases where the two slides have different colors in a pixel, but we can still gain some information by picking out some pixels we know can't be part of the message (we'll catch 1/3 of them on average, and if the pixels are small enough, we might be able to glean some of the actual message).

What Wifi Woo?

Sandy Shwarc reports that there's been some scare over the ill effects of all the elctromagnetic radiation going through the air, but I'm not buying it. Personally, I think this is all just an excuse to avoid having to work. Confused? Let me explain.

Take the new Ultra-Mega-Awesome Wifi tower built the other day. It had 1000 power cords going from the bottom to the top, and not one of them was initially hooked up to anything. Worse, they're all tangled in the middle so it's impossible to figure out which bottoms of wires correspond to which tops.

Now, some poor shmuck has to go and sort them all out, and the only tools he's given are a battery and a lightbulb. They somehow expect him to to hook up the battery at the bottom to a couple of wires, then go to the top and see which wires he can connect the lightbulb to to sort them out. Maybe he could pull a few tricks like tying some wires together at the bottom or top to make long wires, but it's still going to take him quite some time.

With that job ahead of him, you can see why he'd want to believe it shouldn't be done. Maybe if we could help him out and figure out the most efficient way to solve this problem, he'll be a bit more likely to accept Wifi. The tower's pretty tall and the only way up is by stairs, so he'd probably appreciate most if we could help minimize the number of trips he has to take, regardless of how much work he has to do at the top or bottom. How can we do this, and what is the minimum number of ascents and descents required?

Solution

The solution to this is pretty complicated, but it helps to look at a simpler case. Let's say we only have 3 wires. In this case, here's how you do it with just one ascent and descent: Start at the bottom. Tie two of the wires together. Mark both of these “2-_”. Mark the loose wire “1-_” (a group of 2 and a group of 1). Now ascend to the top. Hook the battery and lightbulb together, and connect one wire to one end of this assembly. Test each other wire on the other end, and count how many wires will allow the lightbulb to light up. If one wire will cause it to light up, then mark the test wire with “2-_”. If no wires will light it up, mark it with “1-_”. Repeat for all the wires.

Now, at the top you'll have two groups of wires, one of just one wire, and one of two wires. You know that these correspond to the groups you made at the bottom (and the one-wire group has the single wire properly identified). To sort between the wires in the 2-wire group, we need another step. Now, take the wire in the 1-wire group and tie it to one of the wires in the 2-wire group. Mark both of these “X-2” (where X is whatever mark was in the first digit). Mark the other wire “X-1” and leave it unconnected to anything.

Descend to the ground, and repeat what you did when you first got to the top, except this time, fill in the second digit (if it connects to zero, mark 1, if it connects to 1, mark 2). Now, at the top and bottom you'll have wires marked “1-1”, “2-1”, and “2-2”. These IDs match them all up to each other. It says nothing in your job description about untying the wires at either end, so you're done, with one ascent and one descent.

This solution type can be extended simply to any number which is a perfect triangle (1,3,6,10,15...). For instance, with 6 wires, you'd tie up one group of 3 at the bottom, one group of 2, and one group of 1. You can then identify these groups at the top. Then, you can tie up a group of 3 taking one from each bottom group, a group of 2 taking one from the 2 and 3 bottom groups, and a group of 1 from the 3 bottom group. Go back to the bottom to sort out these groups and you've identified them all.

The problem gets trickier, however, when the number you have isn't a perfect triangle. Since 1000 isn't one of these, we'll have to face this. But, it is possible to extend this solution to most non-triangular numbers. The trick is dump the extra wires into the group where the wires aren't connected to any others. Let's look at the 8-wire case to see this.

At the bottom, set up three groups. The first group (1-_) has three wires not connected to each other, and not connected to any others. The second group (2-_) has two wires connected together. The third group (3-_) has three wires all connected together. Go up to the top and identify all of these groups. Now, to set up the top groups. Set up one group which takes one wire from each of the bottom groups, and tie all of these together (X-3). Set up a second group which takes one wire from each of the bottom groups, and leave all of these tied to nothing (X-1). You'll be left with one wire from the 1-bottom group and one wire from the 3-bottom group. Tie these together in the final group (X-2). Go back to the bottom and identify these groups. You'll then have eight wires, all uniquely identified.

However, it turns out there's a problem for some numbers of wires, such as 5 and 9. If you try to do it this way, you'll have too many wires in the group where they aren't connected to anything. I won't go into all the details here, but your best solution with this method is to have the extra wires in the unconnected group on both trips. You'll end up with two wires marked 1-1. Then, connect one of those to another wire when you're at the bottom (1-1a) and leave the other unconnected (1-1b). Go back to the top, and figure out which of the wires is 1-1a by testing its circuit with the one you tied the bottom end to, and the other is then 1-1b. You'll then have figured them all out with just one extra ascent (you also have to descent to go home, but that's just a technicality).

It turns out that this particular problem comes into play when the number of wires is one less than some triangular number. So, you get problems with 2, 5, 9, 14, and so on. The 2-wire case allows a special solution with only one ascent (attach both wires to the battery at the bottom, marking them with the pole they're attached to. Then go to the top and see which way you have to orient the lightbulb so it'll go on, and match up the poles), but the others will all require 2 ascents and 1 descent with this method to figure it out. Any other number can be done in 1 ascent and 1 descent with this method (3 wires can actually be done in a single ascent with a simple extrapolation from the 2-wire case). Since 1001 isn't a triangular number, our 1000-wire case can be done in just one ascent and one descent using our method.

Side note: There is another, much more complicated method which will allow you to solve this problem in just one ascent and one descent regardless of the number of wires. However, the average time this method takes is proportional to the number of wires to the fourth power; while the method I've given takes time simply proportional to the number of wires squared. For large numbers of wires, the extra time the other method takes at the top or bottom would easily outweigh the time of a single ascent of descent. However, for 5 or 9 wires, it might be worth it. I won't go into it here, though.

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Skeptic's Circle #72 and #73

Bah, I've been remiss about this. I missed out on linking to the 72nd Skeptic's Circle and the 73rd Skeptic's Circle when they came around, so there ya go.

In other news, if you're going to conduct a demonstration claiming that a snake not biting you is evidence of your faith and God's existence, you have to prepared to take the fact that it instead bit and killed you as evidence against this.

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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
Goto SPEEDUP
Goto START
Move 100m forward (label: SPEEDUP)
Goto 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!

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Skeptic's Circle #71

Welcome one, welcome all, to the 71st edition of the Skeptic's Circle. The theme for this week is logic. Logic puzzles to be precise. After all, logic is one of the best razors against irrational thinking, and like any razor it needs to be periodically sharpened. So, for that purpose I've prepared some logic puzzles for you all to work through, each one based on a post submitted.

I've sorted the puzzles by a rough estimate of their difficulty, though the ones each person will find easiest will likely differ. Feel free to discuss the puzzles in the comments, including guesses as to the answers (though if you've heard one before, don't spoil the fun for others). Just be warned that if you go reading the comments, you might run across an answer or two that's already been guessed.

• Easy Puzzles
• Medium Puzzles
• Hard Puzzles

I've also prepared a "Just the links" version if you're short on time or logic, so feel free to take advantage of that.

The next Skeptic's Circle will be hosted at The Quackometer Blog. Check over there for contact information to submit for next week's. So long, and happy puzzle-solving!

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Skeptic's Circle #71: Hard

Following are the hard problems for this Skeptic's Circle. Math isn't as much a requirement as for the medium problems, but you'll have to compensate with a ton of advanced logic.

Great Galileo's Ghost

Cranks have always loved to compare themselves to Galileo. Sick of this, one day Greta Christina decided to set them straight once and for all. To do this, she decided to go on a fictional journey to the afterlife to find Galileo himself so he could explain to them why they're acting like idiots. However, when she reached the afterlife, she ran into a little problem.

It seems that one day, Galileo accidentally touched a noodly appendage he shouldn't have, and ended up with a couple clones. On top of this, the three are cursed to always stay together and that if any yes-or-no question is posed to one of the three, they all must answer it. Any other type of question will be ignored by all three. Each of the clones plus the original has a specific curse on them, but it's unknown who has which curse. One of the curses requires the bearer to tell the truth to any question, another curse requires the bearer to lie in response to any question. The third curse causes the bearer to randomly choose between telling the truth and lying before answering each question.

Each of the Galileos knows whether it's the original or a clone, but without knowing which bears which curse, it would be tricky to figure it out. Is there a questioning procedure that would work here?

Now, Greta can't go back to the real world with three mixed-up possible Galileos, but fortunately, there's a way to break the curses. If any of them is ever faced with a question they can't answer (for instance, they're bound to tell the truth for this question, but the question has no truthful answer that doesn't result in a paradox), they'll go "Voip!" and vanish (so sayeth the FSM). If the two clones both vanish, the original Galileo will be freed from all the curses on him. However, he could vanish too if he's asked a question he can't answer, and if he does, it's game over. Is there a way to target out the two clones and make them vanish while keeping Galileo safe?

If you can accomplish that, one further challenge: Can you make both clones vanish with a single question, even if you haven't previously figured out which one is the real Galileo?

Super-Scammer Secrets

Lord Runolfr recently got a couple of e-mails from scammers, raising my suspicions that something big was afoot behind the scenes. I called up a few contacts, did some research, hired a few James Bond-alikes, and here's what I've figured out:

There's some evil mastermind behind the whole plan, and he's about as supervillanous as they come. This of course means that he wants to capture one of my James Bond-alikes and subject him to an intricately detailed explanation of his evil plan before killing him in a creative way. Apparently, the way he's decided to go about this revelation is through an overhead transparencies with the key points of his evil plan (you'd think he'd have better technology than my high school, but he's out to make money, so he saves it where he can).

Now of course, he's wary of this transparency falling into the wrong hands, so he's come up with a plan to keep things safe. He's figured out some method to spread the information across multiple sheets, arranging it so that if we get a hold of any two, we'll still have no idea what he's planning. So, what we need to figure out now is some possible ways he might have done this, so if we manage to get our hands on more than two, we'll know how to read them (if it's not immediately obvious). What are some possible things he could do? Remember that he's out to save money, so splitting it up to have a single word on each slide or something huge like that doesn't seem too likely.

Harebrained Hat Help

Orac reports that the University of Maryland's Shock Trauma center has gone to the dark side, citing how Reiki therapy is now being used to "help" patients. Well, I did some digging of my own, and it turns out this isn't the craziest thing they're doing. They've got something even dumber going on, called "Colored Hat Therapy." Here's how it works:

A bunch of patients are seated in a circle, and a hat is placed on each one's head. They don't know the color of their own hat, but they can see everyone else's hat. There are many colors of hats, but each color present shows up on at least two hats. The patients sit in the room for a while, and every couple of minutes an announcement comes on the PA system, asking anyone who's figured out the color of their own hat to announce it and then leave the room. The theory goes that the magical hat energy must have seeped into their brain, and they're now cured.

Of course, most of the time it doesn't work so well, and people guess incorrectly about their hat color, miss their chance to get it and never figure it out, or get screwed up by inferring wrong things from other people's mistakes. However, one day, purely by chance, everyone brought into the room was an expert at logic puzzles, and they all managed to figure out the colors of their hats. How did they do this?

(Aside: When one of these experts left the room, a nurse noticed that the wound he was suffering from hadn't magically healed, and this therapy was quickly abandoned.)

Singular Sword Slashes

Martin Rundkvist recently uncovered a unique 16th century sword, and he was wondering how many lives it might have taken. Well, I did a little Tardis travel archeology of my own, and I managed to come pretty close to getting the answer before I was discovered and had to flee ran out of funding.

It turns out this sword was used in a ceremonial mass execution. 100 condemned prisoners were brought out and given one last chance at life. They were all stood up in a line and had a hat put on their head, colored either red or blue. No one knew the color of their own hat, but they could see the color of every hat in front of them. If they guessed correctly, they were spared, but if they guessed incorrectly, they were killed on the spot. This particular sword was used for the killings, and it was so fast and efficient that none of the prisoners in front would be able to hear a sound from the death, and so would have no idea if the guess was correct (though they would be able to hear the guess).

Unfortunately, I didn't see the actual procedure, but I did learn two things that might give us a clue as to what happened: First, the prisoners were given a chance to discuss a plan amongst themselves before the ceremony and decide on it. However, the executioner was listening in and would likely set up the hats to thwart the plan as best as possible. The second thing I learned is that these prisoners were all condemned to death for being part of a "dangerous cult" - which was actually more along the lines of Mensa. So, these are pretty smart people, and we could trust them to come up with a pretty good plan.

Putting this all together, what probably happened during the ceremony, and how many lives did the sword take?

Ending Erroneous Expectations

The Sexy Secularist writes in about how he was able to persuade his mother against recommending the atrocious woo film What the (Bleep) Do We Know?. You see, personally, I know that the whole Law of Attraction thing is bull. Why? Because I was expecting at least one post sent in would have some tenuous connection to pirates and allow me to bring you this classic puzzle. But nope, no pirates. Well, screw you guys, I'm doing pirates anyways!

A crew of 5 logic pirates comes upon a treasure of 100 gold coins. According to the rules of the logic pirates, to distribute the loot the most senior pirate must first propose a plan, and then they'll all vote on it. If at least half agree with the plan, they'll go with it. If not, the most senior pirate has to walk the plank, and then the next most senior pirate has a chance to propose a plan. This continues until a plan is accepted.

Each pirate is perfectly logical and has the following priorities they will strictly pursue: First, they don't want to die. Second, they want to get as much money as possible. If everything else is equal, they'd rather see more of their seniors walk the plank.

So, what plan can the most senior pirate of this group of 5 propose that will get him the most coins? Figured that out? Now, try the situation with 6 pirates and only one coin; what happens there?

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Skeptic's Circle #71: Medium

Following are the medium-difficulty problems for this Skeptic's Circle. Some math skills may be useful here.

Poor Poisoned Pinheads

You just knew one of these days the Moon Hoaxers would push Buzz Aldrin too far, and they finally have. This one started with a post at Space Files that showed a clip of Buzz explaining that the "UFOs" seen during the Apollo 11 flight had actually been identified. Well, a group of five hoaxers really didn't like this, and so they tried to take Buzz out of the picture.

They prepared a special poison which would cause retrograde amnesia in the person to consume it. It's extremely potent, and you'd have to dilute it to homeopathic portions to make it safe, but it takes a day to kick in. They sneaked into Buzz's wine cellar (Didn't know he had one? Probably because I'm just making it up), and went about spiking his bottles. However, they'd only managed to spike a single bottle before Buzz found them. In the ensuing chaos, Buzz managed to catch them all, tie them up, and get them to spill the beans on their plan, but he lost track of which bottle they'd spiked.

At this point, Buzz is so pissed off that he cares more about figuring out which bottle was spiked than whether he might inflict amnesia on his new prisoners. His cellar has a total of 874 wine bottles in it he has to check. With his five prisoners and one day for the poison to take effect, what's the shortest time it might take him to figure out which is the poisoned bottle, and how can he do it in this time?

What Wifi Woo?

Sandy Shwarc reports that there's been some scare over the ill effects of all the elctromagnetic radiation going through the air, but I'm not buying it. Personally, I think this is all just an excuse to avoid having to work. Confused? Let me explain.

Take the new Ultra-Mega-Awesome Wifi tower built the other day. It had 1000 power cords going from the bottom to the top, and not one of them was initially hooked up to anything. Worse, they're all tangled in the middle so it's impossible to figure out which bottoms of wires correspond to which tops.

Now, some poor shmuck has to go and sort them all out, and the only tools he's given are a battery and a lightbulb. They somehow expect him to to hook up the battery at the bottom to a couple of wires, then go to the top and see which wires he can connect the lightbulb to to sort them out. Maybe he could pull a few tricks like tying some wires together at the bottom or top to make long wires, but it's still going to take him quite some time.

With that job ahead of him, you can see why he'd want to believe it shouldn't be done. Maybe if we could help him out and figure out the most efficient way to solve this problem, he'll be a bit more likely to accept Wifi. The tower's pretty tall and the only way up is by stairs, so he'd probably appreciate most if we could help minimize the number of trips he has to take, regardless of how much work he has to do at the top or bottom. How can we do this, and what is the minimum number of ascents and descents required?

Manic Motor Mythbusting

Paddy K recently demolished some myths about the efficacy of so-called "green" cars, and now that he's done with that, he wants to actually demolish the cars. To do that, he's sent them to - who else? - the Mythbusters team. Now, they can't simply destroy the cars, they have to do it in an interesting way. Here's what they've set up:

Two cars are set up with a robot controller each, and this robot will have some programmed instructions. The two cars will be airlifted and dropped at a random point on a very long line marked out in the desert. Both cars will be initially facing north along the line, but we don't know which one will be in front of the other. Once they hit the ground, their parachutes will detach and remain where they landed. After this, the robots will take control of the cars and start driving according to their programming.

Now, there are a lot of fancy things that could be done in programming them, but Adam has decided to give Jamie a challenge. The first restriction is that both robots must use the same instructions. The second restriction is the the instructions are limited to the following commands:

Move 100m forward
Move 100m backward
Skip next command unless a parachute is nearby
Go to [label] (any line may be labeled for this purpose, and as many labels as necessary may be used. This line means the robot's "mind" will go to the instruction at this label and start working forward from there.)

How can the robots be programmed to guarantee a collision? Try to use as few lines of code as possible.

Paddling Pooch Problem

Well, Bronze Dog has done it again. He went and pointed out that made a completely unfounded accusation that the way many fundies practice is akin to the devil worship atheists and D&D players have been accused of. Now, they're accusing him of being a devil (as a talking dog made out of bronze, he's definitely not normal, but I don't see many mentions of devils like this in the Bible).

In the hot pursuit, eventually Bronze Dog found himself paddling in the middle of a circular lake, with a fundie waiting on shore for him. This fundie can run about four times as fast as Bronze Dog can swim, though fortunately he never learned how to swim and won't enter the water. If Bronze Dog can make it to land, and the fundie isn't waiting right at the shor for him, he should have no problem outrunning him.

The fundie isn't going to listen to anything Bronze Dog might say, so tricking him is out of the question. Bronze Dog is pretty fit and can paddle for quite a while, but he'll still likely run out of energy before the fundie gives up, so he's going to have to do something. Assuming the fundie makes his best effort to catch Bronze Dog, what can Bronze Dog do to guarantee he'll be able to get to shore safely?

Action/Adventure Akusai

Akusai recently "psychically" predicted a phone-call from his long estranged cousin. After hearing that his father had gotten this call, he decided to go out and surprise this cousin with a visit. Well, things didn't go quite as planned on his journey, and, long story short, he found himself at the mercy of a deranged hermit.

This hermit offered to play Akusai in a game, with his freedom at stake. The game is played on a circular table, and the players take turns laying circular disks down at any location on the table. If a player can't lay down a disk anywhere without overlapping a disk that has already been placed, they lose. The hermit gives Akusai the choice of playing first or second. Which should he choose, and what strategy should he use in the game?

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Skeptic's Circle #71: Easy

Following are the easy problems for this Skeptic's Circle.

Personalized Perfume Peril

Bad has been captured by perfume manufacturers he recently angered. They're utterly convinced that their personalized perfume works, so they've arranged an elaborate puzzle to force Bad to admit this as well. They've thrown him into a dark room, in which 100 disks of perfume samples are placed. Each disk has a different scent on each side, taken from "completely different" people. The sides are color-coded red and blue for easy counting later, but it's too dim in the room for Bad's eyes to be able to make out the colors. Bad's task is to sort the disks into two piles, each with the same number of red sides facing up. If he succeeds, he'll be set free, but if he fails, he'll be shot. He'll also be shot if he tries anything "clever" such as balancing disks on their edges or throwing them out the window.

Bad knows that the task is hopeless, so he pleads with his captors to give him a hint, just one little hint. Eventually, one of them takes pity on him and gives him the following hint: In the initial set-up, 48 of the disks are placed with the red side facing up. With this information, is it possible for Bad to come up with a plan to separate the disks into two piles with the same number of red sides facing up? If so, how?

Creative Cake Capers

Christian of Med Journal Watch has been attempting to combat rumors that overweight women should lose weight when they get pregnant. In his quest to inform women that this isn't always going to be the case, he's stumbled upon one particularly tricky customer.

This particular woman has quite a sweet tooth, and figures that since she should expect to lose some weight, she can afford to binge a little. She's got a nice big rectangular cake which she plans to eat, but Christian is able to convince her down to only consuming half of it. However, when they get out the cake they find that her husband has already cut out a rectangular slice from it. The woman wants half of the full cake, but she settles on half of the remainder.

The woman knows that each slice takes away a small amount of the cake and so she won't let him use more than one. She also won't settle for anything less than half the cake, but Christian wants to make sure she gets no more than half. So, the problem is, how can Christian cut the cake perfectly in half with a single cut?

And no, making a big horizontal slice through the center of the cake isn't an option, as the cake has icing on the top and thus isn't symmetric in that direction.

Popping Placebo Pills

After a recent article by Mark Hoofnagle on the diagnosis of Chronic Lyme Disease, an enterprising researcher decided to conduct a placebo-controlled study to see if there was any benefit to the use of the antibiotics typically prescribed for treatment of it. During the conduction of this experiment, the grad student assigned to sort out the bottles of pills (some placebos, some real) ran across a problem when the record sheet was smudged and she couldn't identify whether five of the bottles had the placebos or real pills. Counting up the identified bottles, she figures out that one of the bottles should have real pills, while the rest should be placebos.

She does some quick research and figures out that the easiest way to tell apart the pills is by weight. The real pills weigh 1 gram more than the placebos, so a few comparative weighings should be able to sort them out. However, when she gets to the lab to weigh them, she finds that a lab class is going on and there's a huge line-up to use the scale. The pills need to be in before the class is over, so she'll have to put up with it. Asking the instructor for permission to step in, he relents, but allows her only the time to perform one weighing. Is it possible for her to sort out which of the bottles contains the real pills with only one weighing? If so, how?

Perilous Peace Problems

The Factician recently dismantled some biased thinking which led to the bizarre conclusion that peacetime is more dangerous than war, and for his trouble, he received a mysterious package in the mail one day. Opening it, he finds a corked wine bottle with a pill inside of it and a note. The note reads:

So, you think you're a smart guy pointing out accidental deaths, huh? Well, here's the situation: When you opened this box, a specially-prepared poison was released into the air. The pill in the bottle is the antidote for it, but I've got a little challenge for such a smart guy. I want to see if you can get the pill out of the bottle without removing the cork or breaking the bottle. If you do either of those things, I can't guarantee you won't have an "accidental" death of your own.

The Factician suspects they're just bluffing about the whole thing, but he decides to go ahead with it anyways, as he's already come up with the solution. What does he do?

Crazed Canting Christians

Romeo Vitelli tells us a story of some strange convulsing women, which is apparently a miracle. Personally, I'd chalk up curing something like this to be more miraculous, but I guess that just goes to show I don't have faith.

Anyways, it seems that a group of 20 of these women decided that it if their strange behavior led to their death, they'd go straight to heaven. So, they set up a weird ritual suicide type of thing, where the 20 of them get out on a 100-meter long raft in the middle of the ocean, each randomly selecting a direction to face and a starting point from marks laid out every meter (the first a meter from one end, up to one a meter from the other end).

At a cue to start, each woman starts convulsing forward at 0.1 m/s. If she bumps into another woman, both will immediately turn around and start walking in the other direction. They'll keep walking until they inevitably all fall off one end and (hopefully) meet their end. If the woman are miraculously set up in the right configuration, what's the maximum time it might take for all of them to fall off the raft?

Hidden Handbook Hassle

After a perilous journey into the land of the woos, Skeptico managed to escape with the Woo Handbook. However, he's now on the run from woos who want it back, and he needs to pass off the book to a fellow skeptic. He's under close observation, so he won't be able to make personal contact with this other skeptic, but they've arranged a plan to get the hand-off to take place. The plan was to have them both lodge at the same hotel, and during the night, hire one of the employees there to pass it off.

But they ran into a problem with this plan, as it turns out that everyone that works at this particular hotel is a rabid kleptomaniac and would steal anything in their hands before passing it off to another guest. Each room did come equipped with a small portable safe though, and these are equipped with tracking devices to make sure no guests would run off with them. It also fortunately means that the employees wouldn't run off with them, so the trick is to transport the handbook within a safe.

Of course, there's still a catch. The safes are closed through a clasp, which a padlock can be put on to steal it shut. The padlocks can only be unlocked with keys found in the hotel rooms and safely wired down, and each key is unique to each lock. So even if the handbook were passed off in a locked safe, Skeptico's friend would have no way to unlock it. Is there any way to solve this problem without either Skeptico or his friend leaving their room and thus risking being caught by a rabid woo?

Weird Water Woo

PalMD recently made a post discussing hydrogen peroxide woo, and, true to the nature of events these two weeks, has been kidnapped by a crazed woo and forced to solve a logic puzzle if he wishes to live. He's locked in an empty room and given a glass that's around half full of water. His task is to determine precisely whether the glass is half full, less than half full, or more than half full. There are a few ways to do this, but some of them are pretty tricky and inaccurate if you don't have a very steady hand. What are some good methods?

Screwy Scarfe's Secrets

The guys at Holford Watch recently exposed Chistopher Scarfe as the fraud he is. Unfortunately, they didn't realize that Scarfe is also an insane supervillain, and they were promptly captured and imprisoned in his mountain fortress. They managed to escape from the fortress (Scarfe forgot to lock the cell door), but on the way out they came across a rickety bridge they'll need to cross.

It's night, and they only have one flashlight among them which anyone crossing the bridge will need. The bridge is unable to support more than two people at any time, so they'll have to make multiple trips to get everyone across, passing off the flashlight as necessary. The guys each incurred various injuries in the escape, so they're all able to move at different rates. One guy is pretty much uninjured and could make it across the bridge in two minute. Another of them is a marathon runner and could easily do it in a single minute. A third stepped on some caltrops on the way out, and it will take him four minutes to cross. The fourth had his leg broken in a fight with a guard dog, and it'll take him eight minutes to cross (if there aren't actually four guys behind this blog, pretend there are). Of course, if two are crossing together, they have to cross at the speed of the slower person.

Scarfe's hot on their tails, so they want to get across the bridge as quickly as possible. How can this be done, and how long will it take them?

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Skeptic's Circle #71: Quick Links Version

Here ya go, all the links for this Skeptic's Circle in one small place, for those of you too intellectually lazy (or time-deprived) to work on a few logic puzzles.

• NeuroLogica Blog: The Framing of Alternative Medicine
  
• The Bronze Blog: Religion is Devil Worship
• Respectful Insolence: Woo infiltrates one of the premiere trauma hospitals in the U.S.
• White Coat Underground: It burns! It burns!
• Holford Watch: Christopher Scarfe - former partner in Institute for Optimum Nutrition - is nasty about cute little cats
• Sexy Secularist: Woo: It's All in the Family (Part II)
  
• Skeptico: The Woo Handbook
• Aardvarchaeology: Doctoring My Spin
• Action Skeptics: The Statistical Outlier
• Paddy K: Greenwash
• The Saga of Runolfr: Scam Email Double Feature
• Providentia: In The Saint's Name
• Junkfood Science: W is for wifi woo
• Space files: UFOs seen during Apollo missions
• denialism blog: A Critical Appraisal of "Chronic Lyme" in the NEJM
• Conspiracy Factory: Peace is Dangerous - wait, WTF?
• Med Journal Watch: Yet pregnant women target of weight loss hysteria
• Greta Christina's Blog: The Galileo Fallacy, and the Gadfly Corollary
• The Bad Idea Blog: Personalized perfume based on your DNA?

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Deadline Update

Well, it looks like I'm not going to have any conflicts with hosting, so I'm going to extend the deadline for Skeptic's Circle submissions to Wednesday at midnight GMT. Even if you miss that, don't worry too much, I'll still slip in a link for ya.

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Skeptic's Circle #70 + Submission Guidelines

The latest edition of the Skeptic's Circle is now up, courtesy of the Factician.

Two weeks from now, the Skeptic's Circle will be circling right back here to Infophilia. No big clues to the theme this time, except to promise to do my best to make you think. You can send your best skepticism-related posts to TheInfophile {at} gmail {that punctuation mark which is like a comma but without the tail} com. If you're unsure what might be appropriate, check out the submission guidelines.

Now, I'll warn you guys in advance that there may be something coming up with me around the time the circle is scheduled (probably not, but it's a possibility), so I might have to put it up a day early or late. So, in case it's early, I'm going to make the tentative deadline for submissions Tuesday at midnight GMT. I'll know by next weekend if there will be a conflict, so I'll be able to confirm a deadline then.

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Skeptic's Circle #69

The 69th edition of the Skeptic's Circle has now been posted at Unscrewing the Inscrutable. I'll have to give Brent props for not going the obvious way with the theme for this one, or even the slightly-less-obvious-but-works-when-you-think-about-it way.

Open thread as usual, but discussion of the "obvious way" is FORBIDDEN. Speculation as to the less-obvious way is grudgingly permitted as I'm only allowed to FORBID one thing.

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Skeptic's Circle #68

The latest Skeptic's Circle is now up over at Aardvarchaeology. Read. Learn. Skepticize.

As a side note, I'm going to be moving up to grad school tomorrow, and I won't have internet connected until Wednesday. So, don't go calling the cops if you don't see me commenting on your posts in that time.

Open thread as usual, though pointing out that "Skepticize" isn't a verb is FORBIDDEN - it's a word now.

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Skeptic's Circle #67

Ask yourself this: What's the coolest plot concept in Japanese animation and manga?

No, cooler than ninjas.

Cooler than pirates, too.

Magical girls? Are you kidding me?

There ya go: Giant robots. And the Skeptic's Circle has decided to harness their awesome power to combat weaponized woo. You'd better check it out; what's a giant robot without a prodigious teen pilot?

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Skeptic's Circle #66

Ook.

Ook! Eek ook-ook OOK!

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Skeptic's Circle #64

The latest Skeptic's Circle is now up at The Skeptical Alchemist. Read. Learn. Pass it on.

Open thread as usual, but mentioning how this is so late it's kind of pointless as you already knew about this Skeptic's Circle is FORBIDDEN.

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Skeptic's Circle #63

The latest Skeptics Circle is now up, courtesy of Relatively Science.

Open thread as usual, but stealing ideas from Bronze Dog is FORBIDDEN, unless, that is, you're sure to point out that imitation is a form of flattery.

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Skeptic's Circle #62

The latest Skeptics Circle is now up, courtesy of Polite Company.

Open thread as usual, but pointing out how I use this same format for announcing every Skeptics Circle but never bother to copy and paste it is FORBIDDEN.

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Skeptic's Circle #61

The latest Skeptic's Circle is now up, courtesy of Memoirs of a Skepchick, so go check it out when you have a chance.

Open thread as usual, though giving any credence to Rebecca's findings unsubstantiated claims about me is FORBIDDEN.

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The 60th Skeptic's Circle

In a flurry of activity, the blog has gone from completely empty to being set up for an elaborate presentation. A stage is set up with a podium just to the left of center. To the right of and behind it is a large screen, presumably for displaying highlighted posts for the current Skeptic's Circle. Soon enough, the lights dim and a spotlight appears, tracking Infophile as he walks to the podium.

Welcome one and all to the latest edition of the Skeptic's Circle, where we highlight the best posts from the last two weeks in critical thinking and in debunking pseudoscience, quackery, denialism, pseudohistory, and frauds. I am particularly honored to be able to host the 60th edition of this since, as many of you are likely aware, the number 60 is quite an interesting number.

The number 60 is the lowest number to be divisible by the first 5 natural numbers (and the 6th as a bonus). This leads it to having a plethora of divisors, making it the most relatively abundant number below 100. It's also notably a unitary perfect number and a semiperfect number.

For an instant, the shadows off to the side of the stage appear to be take on a human form, but it quickly fades. Probably just pareidolia.

Additionally, the number 60 and related numbers show up in numerous places in human culture. The most obvious would be that it's the number of seconds in a minute and the number of minutes in an hour. It's also the number of degrees in each angle of an equilateral triangle. Also, it's divisors sum up to 108, which shows up in many more places in religion and literature.

There's that human-like shadow again. Wait... if it were just irrelevent pareidolia, would it be pointed out in the post like this? Or maybe it's just a test of your skepticism.

But enough about the number 60, onto the circle. While I'd hoped I could get 60 posts for this, I wasn't quite that fortunate. Nevertheless, I doubt you'll be disappointed. Our first post comes from the organizer of the circle himself, Orac from Respectful insolence...

Infophile appears to be triggering a remote in his hand, likely to turn on the screen and start the slide show. The screen switches to show a "Loading..." display for a few moments, and then promptly switches to the dreaded Blue Screen of Death, stating that the system has performed an illegal operation and must be shut down.

Great, just great. I apologize for the inconvenience people, but this should only take a moment. I really should have made sure the company I hired to organize this had upgraded at least to Windows XP, but it's too late for that now. While waiting, why don't you talk amongst yourself. I'll even give you a subject: Why did it take so long for Microsoft to fix this problem.

Infophile strikes a key on the podium to commence the shutdown, and at that moment every electrical device in the room simultaneously shuts off, plunging the auditorium into darkness.

Okay, whose idea was it to run the entire room's electrical functions from the same computer, and not even install a backup? You can't tell me this hasn't happened before. Trust me, I'm going t...

Infophile is abruptly cut off, as if his speech were overwritten by silence. This is followed by a disturbing thud. After a few hectic seconds of worried audience members scrambling to get up to the stage and many others asking about what had happened, the lights switch back on.

Somehow, you managed to get pushed to front, and you can see that Infophile is now lying prone on the stage, showing no signs of life. Looking over his body, you see no obvious signs indicating what might have happened to him, but you're not a doctor (for the purposes of this story), so you can't be sure. While inspecting him, you also find a crumpled note which appears to be listing various websites - perhaps the list of posts submitted for this circle.

Call out, "Is anyone here a doctor?"

Flatten out the note and use the computer to check out the sites for the latest Skeptic's Circle. Yes, it's heartless, but you're a busy person, and this is all fictional anyways.

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Fade to Black

Time is fast running out. The deadline for contributing to the next Skeptic's Circle is tomorrow at 9 PM Eastern time, so get your posts written and e-mail a link to TheInfophile {at} gmail {dot} com before time runs out.

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