stigant
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Actually, since the kings are in on the strategy meeting, that strategy will result in a 50/50 chance that 99 wives are killed.

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lopsidation
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How could 99 wives possibly be killed? The first two wives' responses put together are essentially the same as the first wife's response in the original strategy. This means everyone after them will be certain of their hat color.

Here's an example with 6 people:

B W B W W B

The first wife randomly chooses *flips a coin* black. The second wife counts the black hats in front of her. There are 2, which is even, so Wife #2 would say "white"- except the first wife said "black", so she switches her response to "black".

Wife #3 thinks "Okay, Wife #2 said "black", but Wife #1 also said "black", which means Wife #2's original response would have been "white". This means there's an even number of black hats left, but I see an odd number of black hats in front of me. Therefore, my hat is black."

This method continues down the line, so all of the remaining wives will live.

EDIT: By the way, this method won't work in King Erik's court, because 1) using it would give the first two wives a 5/6 chance of death, which is worse than one certain death, and 2) King Erik's wives are presumably honest and trustworthy, so he can trust the first wife to give a correct response. The best King Erik can do is to just randomly choose someone to go first.

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[Last edited by lopsidation at 01-25-2009 05:20 PM]

Kwakstur
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IMO, this just simply makes it uncertain who and how many will die. Here's a bunch of example rows, the wives responses, and who dies depending on what the first wife says.

B W B B W       B W B W W
---------Dead   ---------Dead
B B B B W  2    W B B W W 1,2
W W B B W  1    B W B W W
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B B B B W       W W B W W
---------Dead   ---------Dead
B B B B W       B W B W W 1
W W B B W 1,2   W B B W W 2

To me, it seems like it always averages out to 1 wife dying. Since his other wives hate Celine, I don't think they'd like her strategy that gets them killed.

"Yes, I know you already have a plan that just kills me, but I have something better: Let's take 50% off of my chance of death and place it on one of you! What do you say? Please?"

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[Last edited by Kwakstur at 01-25-2009 07:38 PM]

lopsidation
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True, King Bill's other wives wouldn't want a 50% chance of death for one of them...
However, they have a natural reason not to trust Celine at all, given that she's married to King Bill too. Choosing this method would ensure that Celine won't kill them all to save herself.

Although, maybe King Bill's wives are psychology majors, so they would know Celine wouldn't have betrayed them...

Enough discussion about this, let's just move on to the next puzzle!

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stigant
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Ok, a teacher wants to give his class a pop quiz, and he wants to make sure that:

1. They know it's coming so that the experience as much dread as possible

2. They are totally surprised when it finally happens.

So on a Friday, he tells the class that they will have a pop quiz sometime next week, but they will be totally surprised when it comes. After a few moments of thought, one of the more mathematically inclined students says, "But Mr. Smith, you can't give us the quiz on Friday since we'll know that after not getting a quiz on Thursday, it must be on Friday. But that means you can't give us a quiz on Thursday either since we know it's not going to be on Friday, and so when we haven't had one after Wednesday, we'll know it's coming on Thursday. But then you can't give it to us on Wednesday either, or Tuesday, or Monday. So you can't possibly give us a pop quiz next week that we will be totally surprised by"

When did the teacher give the quiz?

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Nuntar
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Ah, this one is a classic I'll follow the rules and give people who don't already know it a chance to answer.

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Dex Stewart
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I think I *may* see a solution. As the student pointed out, having the test on any day would be illogical. Since the students expect the professor to act logically, they expect them to not give them a test after all. Therefore, any day would surprise them.

I could also be totally wrong, as this doesn't seem a... "clean" solution to me.

lopsidation
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Saturday.

Well, I'd be completely surprised...

Alternatively, he could just flip a coin every day, and give them the test if it's heads.

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[Last edited by lopsidation at 01-27-2009 10:06 PM]

stigant
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I told this story to my students (I enjoy popping quizzes on them, and refuse to tell them if we have a quiz until I tell them to clear their desks) who immediately asked, "So when was the quiz" My response is always "I don't know, but boy were the students surprised!"

Dex got it, so the puzzle stick is in his lap.

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RoboBob3000
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stigant wrote:
Dex got it, so the puzzle stick is in his lap.

I don't know if I like that answer, though. If the students assume that it's not going to happen, then the teacher doesn't fulfill goal #1.

How do we maximize dread? How do we define surprise? Are we maximizing surprise? Are students more surprised if they take the quiz on a day when it had a 33% chance of occurring than a 50% chance of occurring, or do we just care that they were surprised at all?

This question is poorly defined. To maximize dread, I'd say you'd have to minimize surprise by giving the quiz on the last day of the window. To maximize surprise, you'd have to minimize dread, by giving the quiz out of the blue without any warning at all.

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Nuntar
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For anyone who wasn't convinced by Dex's answer, there's a very good discussion of this puzzle in Raymond Smullyan's Forever Undecided. Maybe I'll post a summary of it if I have time later.

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stigant
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How do we maximize dread? How do we define surprise? Are we maximizing surprise? Are students more surprised if they take the quiz on a day when it had a 33% chance of occurring than a 50% chance of occurring, or do we just care that they were surprised at all?

This question is poorly defined. To maximize dread, I'd say you'd have to minimize surprise by giving the quiz on the last day of the window. To maximize surprise, you'd have to minimize dread, by giving the quiz out of the blue without any warning at all.

Well, the student's logic is a bit, I don't know, nerve wrecking in itself. I mean it seems to come to a ludicrous conclusion, but doesn't appear to have any obvious holes. I would be pretty nervous that there was something amiss. However, that feeling might start to subside after a couple of days of no quiz. So now the object is to give the quiz just as the nervousness starts to subside thus maximizing both dread and surprise at the same time.

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Garlonuss
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You know, this reminds me of Star Trek II.

[Spoilers ahead, but if you haven't seen it by now, I don't particularly care to protect you about this. However I've placed the following in secret tags just to be a good sport.]



But that's a bit off topic.

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Sometimes they don't make much sense.
Refrigerator.

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[Last edited by Garlonuss at 01-28-2009 05:54 PM]

Dex Stewart
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It was a huge surprise that I actually got it right; I myself am not convinced by my answer. I therefore do not have any puzzle in store, anyone can take the turn.

Nuntar
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OK, here's quite a simple one. It is mathematical, but doesn't require any formulas or programming.

How many different ways are there to tile the plane with just squares and equilateral triangles of the same side length, if adjacent tiles are required to touch along a complete edge (so you can't get infinitely many variants of the tesselation of just squares by shifting one row by x) and rotations of the same tiling do not count as distinct? Of course, you must prove that your answer is correct.

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stigant
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That's funny, I was playing around with tiling the plane with squares and equilateral triangles just the other day. But I think, to answer your question, there are an infinite number of distinct tilings.

You can tile a strip 1 wide with an infinite line of squares. And you can tile a strip sqrt(3)/2 wide with a line of triangles alternating 1 up, 1 down.

It would take an infinite number of strips to tile the whole plane, and strips are pretty much interchangeable (ie you could do 2 strips of triangles, followed by a strip of squares, followed by 3 strips of triangles, followed by 5 stips of squares etc.), so there's one tiling for every infinite string of bits (substitue strips of squares for 1's and strips of triangles for 0's). ANd there are probably other, non-strippy tilings, or more complicated ways to make strips to boot. So basically, there are an infinite number of ways to tile the plane with squares and triangles.

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Dex Stewart
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I was thinking this way: There's at least one way to do it (using both triangles and squares). Stretch that tiling by factor x, and split every square and every triangle in x squares and triangles and you get a new tiling.

stigant
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Not necessarily... One such tiling is to use no triangles, only squares (like a piece of graph paper). If you stretch and subdivide as you said, you end up with the same tiling. Now, that method may work on other tilings to create new tilings, but you can't say it works on every tiling. Further, there may be cycles T1 -> (stretch and subdivide) -> T2 -> (stretch and subdivide) -> T1 etc.

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Nuntar
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You're nearly there, but "an infinite number" isn't a complete answer. Is it a countable or uncountable infinity?

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stigant
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Since the number of infinitely long binary strings is uncountable (diagonal proof), the number of tilings is uncountable.

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Nuntar
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Almost. The standard diagonal proof is for strings with a definite start that extend to infinity in only one direction, so you need one more step to prove that it can be applied here.

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[Last edited by Nuntar at 01-28-2009 10:22 PM]

stigant
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Well, you can choose a strip as a starting place (say the strip that includes the x-axis), and just consider the sequence of strips above it. That's enough to get to uncountable via the diagonal proof. Alternatively, you can let every odd bit represent the next strip in the positive y-direction and every even bit represent the next strip in the negative y-direction. (ie let the x-axis strip be your first bit, and then consider the strips going up as one sequence of bits, and the strips going down as another, and interleave them to form the final sequence.) Both of those options are as standard as the diagonal proof itself. Although, with the second, you might have to consider that one sequence might be a translation of another. It's enough, though, to note that letting the negative sequence be all square-strips will eliminate this possibility (this of course reduces to the first option).

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[Last edited by stigant at 01-28-2009 10:17 PM]

Nuntar
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Correct! Over to you.

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stigant
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TonyPa recently published a new game called Pushori. It's pretty easy to learn, has groovy music, and is oddly addictive. I won't explain all the rules. They're pretty easy to figure out, except for the rule where you fill the board with no repeating tiles (this moves you on to the next level rather than ending the game).

The puzzle: Does the scoring system make sense? Why or why not? ie can you find a way, given enough time, to score an arbitrarily large number of points.

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[Last edited by stigant at 02-02-2009 06:40 PM]

Nuntar
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It's not possible to play indefinitely.

Sooner or later it will eventually happen that you get seven consecutive tiles with no matches and the displayed next tile also doesn't match. Then you have six tiles on the board -- let's take the best case and say two each in three rows -- and you have no information as to where to put the seventh tile. You must fill up one row. But then if the ninth tile matches the middle tile of the full row, you cannot continue.

Of course you could keep playing for ever if you always got lucky, but mathematically the probability of being able to play for ever is zero.

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stigant
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Good! But totally wrong.

To be sure, your analysis is correct (and mirrors my own). However, you haven't thought of another possibility.

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TripleM
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Hint? It's hard to know what to look for when you say the analysis is correct (and I've tried and agree with it).

stigant
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You can't last forever when there are so many tiles in the pot...

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TripleM
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Ah. Now this is turning into a very interesting problem..

ballzie
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I think the key is that you try to stay on level 1 forever. There are only 5 unique tiles (that I have seen) on level 1. I have not spent the time trying to come up with a method to prove that you can keep from matching 2 tiles on a specific square indefinitely (although I'm sure a proof is what you're looking for). It does however "feel" like it is possible. Maybe somebody else can prove it.

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[Last edited by ballzie at 02-17-2009 08:06 PM]