Missing square in a triangle puzzle:
Both of these triangles are made up of the exact same shapes, just reconfigured. So where does that gap in the lower one come from?
It's already been sort of brought up, but I want to be more explicit about it: those shapes are not triangles by any definition. Calling them such is just misinformation and (in my mind) kind of ruins the brain teaser.
It's a quadrilateral. What appears to be a hypotenuse is two sides with a vertex where the red and blue triangles meet. I strongly suspect, though I'm not going to bother formally proving, that the riddle wouldn't work if the triangles had the same slope and formed a continuous side.
A sees 3 people with blue eyes, and is therefore thinking, "If I have brown eyes, B sees 2 people with blue eyes, and is therefore thinking, "If I have brown eyes, C sees 1 person with blue eyes, and is therefore thinking, "If I have brown eyes, D sees nobody with blue eyes."""
Without the base case in there, no conclusion may be drawn.
Uh, no. Want to think that through again?
Uh, no. I could elaborate further, if it'd help:
Let's assume the Guru has said her line. Now, the reasoning goes as such:
No, I was pointing out that you wrote that the deduced situation was
A sees 3 blue, therefore he thinks:
B sees 2 blue, therefore he thinks:
C sees 1 blue, therefore he thinks:
D sees 0 blue
A sees 3 people with blue eyes, and is therefore thinking, "If I have brown eyes, B sees 2 people with blue eyes, and is therefore thinking, "If I have brown eyes, C sees 1 person with blue eyes, and is therefore thinking, "If I have brown eyes, D sees nobody with blue eyes."""
Without the base case in there, no conclusion may be drawn.
Uh, no. Want to think that through again?
Uh, no. I could elaborate further, if it'd help:
Let's assume the Guru has said her line. Now, the reasoning goes as such:
No, I was pointing out that you wrote that the deduced situation was
A sees 3 blue, therefore he thinks:
B sees 2 blue, therefore he thinks:
C sees 1 blue, therefore he thinks:
D sees 0 blue
Which is very obviously totally wrong.
Yes. That is totally wrong. It also is not what I wrote.
A sees 3 people with blue eyes, and is therefore thinking, "If I have brown eyes, B sees 2 people with blue eyes, and is therefore thinking, "If I have brown eyes, C sees 1 person with blue eyes, and is therefore thinking, "If I have brown eyes, D sees nobody with blue eyes."""
Without the base case in there, no conclusion may be drawn.
Uh, no. Want to think that through again?
Uh, no. I could elaborate further, if it'd help:
Let's assume the Guru has said her line. Now, the reasoning goes as such:
No, I was pointing out that you wrote that the deduced situation was
A sees 3 blue, therefore he thinks:
B sees 2 blue, therefore he thinks:
C sees 1 blue, therefore he thinks:
D sees 0 blue
Which is very obviously totally wrong.
Yes. That is totally wrong. It also is not what I wrote.
a and b are two numbers (integers) in the interval [2,99]. Mr. P and Mr. S are told a*b and a+b respectively. They have the following conversation:
Mr. P: I don't know the two numbers.
Mr. S: I already knew you didn't.
Mr. P: Ah, now I know them!
Mr. S: And now, so do I.
Assuming they aren't lying or incorrect, what are a and b?
Note: Both Mr. P and Mr. S know that the numbers are in the interval [2,99]. That's important, if I recall correctly. Also, this can be done on pencil and paper but some very basic programming helps. If you think you have a method of getting the solution and don't want to program it or do it by hand, just post a description of the method and I'll tell you if it's correct or not.
I got hooked on this one and finally spoiled it for myself. It's meta-neat to me how many variants they are and how subtly this one can be perverted. Seeing implementations of the solution algorithm in different languages (especially Haskell) was neat too.
I guess what I'm saying is that I'm a horrible person but thanks for sharing that one.
A sees 3 people with blue eyes, and is therefore thinking, "If I have brown eyes, B sees 2 people with blue eyes, and is therefore thinking, "If I have brown eyes, C sees 1 person with blue eyes, and is therefore thinking, "If I have brown eyes, D sees nobody with blue eyes."""
Without the base case in there, no conclusion may be drawn.
Uh, no. Want to think that through again?
Uh, no. I could elaborate further, if it'd help:
Let's assume the Guru has said her line. Now, the reasoning goes as such:
No, I was pointing out that you wrote that the deduced situation was
A sees 3 blue, therefore he thinks:
B sees 2 blue, therefore he thinks:
C sees 1 blue, therefore he thinks:
D sees 0 blue
Which is very obviously totally wrong.
Yes. That is totally wrong. It also is not what I wrote.
The important bit is, "if I have brown eyes".
I'm going to glare at you until you stop this.
I'm trying to provide an explanation you'd understand, but I really can't think of one simpler than what I've already written. So, instead, I invite you to prove how what I'm saying is wrong, in the hopes that you'll learn from that.
Here's one to think about - what is the absolute maximum amount of money that someone could win in a single day of competition in Jeopardy? And what, exactly, would it take to reach this maximum?
For those who haven't seen the show, here's the format:
The show consists of three rounds.
In the first round, there are 6 sets of 5 trivia questions each, valued from $200 to $1000 in $200 increments. In addition, one question, instead of being worth a set value, allows the player to bet all or a portion of their winnings, recieving the amount bet on a successful answer.
The second round has the same format as the first, except the values are doubled, and there are two questions allowing the player to bet their winnings.
The third round consists of a single question that the player bets all or a portion of their winnings on.
If the player gets a question wrong, they lose money equal to the value of the question they missed.
That's all the info you need to answer this puzzle.
Correctly answer every question in the first round minus the daily double, then get the daily double, make it a true daily double, wager everything, get it -> 2 * (3000 *6 - 400)
Add to that correctly answering every question in the second minus the two daily doubles, which would have to be in the 800 dollars -> 6000 * 6 - 1600
Then true daily double twice, and then double again in Final Jeopardy.
Of course, you're an obnoxiously cocky jackass if you do that, but theoretically.
enlightenedbum on
Self-righteousness is incompatible with coalition building.
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VariableMouth CongressStroke Me Lady FameRegistered Userregular
A classic one: the infinite hotel. You can dress it up as you want to make it a nice story for your family, but the basic point is: you have a hotel with an infinite number of rooms, all occupied, and you want to fit in more people. How do you do it?
You ask each person in each room to move to the room with the double of their current room number. That way, every odd-numbered room will be free.
I love this one, got it in some after school program I did.
A sees 3 people with blue eyes, and is therefore thinking, "If I have brown eyes, B sees 2 people with blue eyes, and is therefore thinking, "If I have brown eyes, C sees 1 person with blue eyes, and is therefore thinking, "If I have brown eyes, D sees nobody with blue eyes."""
Without the base case in there, no conclusion may be drawn.
Uh, no. Want to think that through again?
Uh, no. I could elaborate further, if it'd help:
Let's assume the Guru has said her line. Now, the reasoning goes as such:
No, I was pointing out that you wrote that the deduced situation was
A sees 3 blue, therefore he thinks:
B sees 2 blue, therefore he thinks:
C sees 1 blue, therefore he thinks:
D sees 0 blue
Which is very obviously totally wrong.
Yes. That is totally wrong. It also is not what I wrote.
The important bit is, "if I have brown eyes".
I'm going to glare at you until you stop this.
I'm trying to provide an explanation you'd understand, but I really can't think of one simpler than what I've already written. So, instead, I invite you to prove how what I'm saying is wrong, in the hopes that you'll learn from that.
I understand the problem perfectly. However, you're stating the situation incorrectly and taking the induction past where it makes any kind of logical sense.
Let's examine your 4 monk example.
A sees 3 blue. He assumes that B sees 2 blue, C sees 2 blue, and D sees 2 blue.
Since he knows that B sees two blue, he believes that C and D will both see 1 blue.
That's it.
That's as far as A's induction can logically progress. You can't recurse on that again. He knows for certain that
B sees (at least) 2.
C sees (at least) 2.
D sees (at least) 2.
From there, he cannot logically reach the conclusion that C would think D sees none. It's even more obvious with larger numbers.
A sees 3 people with blue eyes, and is therefore thinking, "If I have brown eyes, B sees 2 people with blue eyes, and is therefore thinking, "If I have brown eyes, C sees 1 person with blue eyes, and is therefore thinking, "If I have brown eyes, D sees nobody with blue eyes."""
Without the base case in there, no conclusion may be drawn.
Uh, no. Want to think that through again?
Uh, no. I could elaborate further, if it'd help:
Let's assume the Guru has said her line. Now, the reasoning goes as such:
No, I was pointing out that you wrote that the deduced situation was
A sees 3 blue, therefore he thinks:
B sees 2 blue, therefore he thinks:
C sees 1 blue, therefore he thinks:
D sees 0 blue
Which is very obviously totally wrong.
Yes. That is totally wrong. It also is not what I wrote.
The important bit is, "if I have brown eyes".
I'm going to glare at you until you stop this.
I'm trying to provide an explanation you'd understand, but I really can't think of one simpler than what I've already written. So, instead, I invite you to prove how what I'm saying is wrong, in the hopes that you'll learn from that.
I understand the problem perfectly. However, you're stating the situation incorrectly and taking the induction past where it makes any kind of logical sense.
Let's examine your 4 monk example.
A sees 3 blue. He assumes that B sees 2 blue, C sees 2 blue, and D sees 2 blue.
Since he knows that B sees two blue, he believes that C and D will both see 1 blue.
No. He believes that B will believe that C and D will see 1 blue. He also believes (actually, assumes is a better description) that B will believe that C and D will each believe that the other will see no blues.
That's it.
That's as far as A's induction can logically progress. You can't recurse on that again. He knows for certain that
B sees (at least) 2.
C sees (at least) 2.
D sees (at least) 2.
From there, he cannot logically reach the conclusion that C would think D sees none. It's even more obvious with larger numbers.
He's not forming firm conclusions, but rather describing uncertainties. A is uncertain of his own eye color. Consequently, he can be certain only that B sees two blues. A knows that B is likewise uncertain of his own eye color, so A believes that B believes that C can only be certain of seeing one blue. The uncertainty of A with respect to his own eye color carries over into his mental projection of B's uncertainty, and so on and so on. That's why no one leaves without a declaration from the guru.
From there, he cannot logically reach the conclusion that C would think D sees none. It's even more obvious with larger numbers.
That's the thinking that's gonna get you in trouble. As with many logic puzzles, this one is deliberately obfuscated. It is trivial to solve for two islanders, just as the Monty Hall problem is quite obvious if you imagine a hundred doors. The problem as stated is not logically any different from the base case, but it is made to seem so by the magic of large numbers.
Any time you're approaching a logic puzzle with the idea that it's obvious if you look at the problem as stated, you can be pretty sure you're on the wrong track.
Adrien on
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SmasherStarting to get dizzyRegistered Userregular
A sees 3 people with blue eyes, and is therefore thinking, "If I have brown eyes, B sees 2 people with blue eyes, and is therefore thinking, "If I have brown eyes, C sees 1 person with blue eyes, and is therefore thinking, "If I have brown eyes, D sees nobody with blue eyes."""
Without the base case in there, no conclusion may be drawn.
Uh, no. Want to think that through again?
Uh, no. I could elaborate further, if it'd help:
Let's assume the Guru has said her line. Now, the reasoning goes as such:
No, I was pointing out that you wrote that the deduced situation was
A sees 3 blue, therefore he thinks:
B sees 2 blue, therefore he thinks:
C sees 1 blue, therefore he thinks:
D sees 0 blue
Which is very obviously totally wrong.
Yes. That is totally wrong. It also is not what I wrote.
The important bit is, "if I have brown eyes".
I'm going to glare at you until you stop this.
I'm trying to provide an explanation you'd understand, but I really can't think of one simpler than what I've already written. So, instead, I invite you to prove how what I'm saying is wrong, in the hopes that you'll learn from that.
I understand the problem perfectly. However, you're stating the situation incorrectly and taking the induction past where it makes any kind of logical sense.
Let's examine your 4 monk example.
A sees 3 blue. He assumes that B sees 2 blue, C sees 2 blue, and D sees 2 blue.
Since he knows that B sees two blue, he believes that C and D will both see 1 blue.
That's it.
That's as far as A's induction can logically progress. You can't recurse on that again. He knows for certain that
B sees (at least) 2.
C sees (at least) 2.
D sees (at least) 2.
From there, he cannot logically reach the conclusion that C would think D sees none. It's even more obvious with larger numbers.
Four people have blue eyes, and we label them A, B, C, and D.
I see no people with blue eyes. If I have brown eyes, then nobody on the island has blue eyes. If we learn that somebody on the island has blue eyes, then I will leave that night.
If I have blue eyes, then D will not leave that night, and so I will leave the second night.
If I have blue eyes, then C and D will not leave the second night, and so I will leave the third night.
If I have blue eyes, then B, C, and D will not leave the third night, and so I will leave on the fourth night.
Note that the quote tree does not (necessarily) describe what B, C, and D actually think, merely what A can suppose about their recursive thoughts about each others' thoughts.
Also, hopefully this makes it more apparent why the Guru is necessary; it eliminates any chain of supposition that results in someone recursively thinking anybody else thinks there are no people with blue eyes.
I'll probably be repeating what other people said earlier, but I'm going to write down my answer to the island puzzle:
Imagine there are four blue-eyed people on the island and a random number of brown-eyed people. The blue eyes are Alice, Bob, Carol, and Derek. Someone announces that he sees one blue-eyed person.
Alice sees three people with blue eyes. She knows that Bob, Carol, and Derek will see at least two people; possibly three, if she herself has blue eyes. She wants to know the actual number of blue-eyed people on the island, so she starts a thought experiment.
If her eyes are not blue, Bob would see two people with blue eyes.
She knows, of course, that the imaginary Bob will be starting his own thought experiment, in which Bob will also calculate the odds, same as Alice is doing right now. She knows that, in this second level thought experiment, Bob would imagine a universe in which Carol sees only one person with blue eyes.
The thing about Alice is that she’s a perfect logician, just like everyone else on the island. She has to take her thought experiment to its logical conclusion. So, the imaginary Carol would create yet another level of reality in which Derek sees absolutely no one with blue eyes.
Alice continues. Derek would realize that he has blue eyes, since no one else has them. As such, he has to leave at midnight. If, in this farfetched thought experiment, Derek does not leave, there have to be at least two people on the island with blue eyes.
Alice knows that Derek will do no such thing. After all, she can count the number of people with blue eyes and it's higher than one. In fact, at least three people have blue eyes.
So what would happen if Derek does not leave? Well, Carol would look at Derek and realize Derek expected Carol to leave, since Carol is herself unsure about her eye color. As such, they must both have blue eyes.
What if, Alice wonders - and this is the point where she gets excited - both Derek and Carol fail to leave on that second day? Bob expected them to do so, if there were only two people with blue eyes. The fact that they didn't means he has blue eyes too.
Alice starts to see a pattern, and she ponders the implications. She realizes that every blue-eyed person would leave on the same day, and that day is based on how many there are. For one person, it's the first day. For two people, it's the second day. For three people, it's the third day.
Since Alice counts three people with blue eyes, she does not expect anyone to leave until the third day. If, however, no one leaves on the third day, she would know that she herself has blue eyes.
So she waits.
And she starts packing, just in case.
The reason the Guru, or any outsider, is necessary is because everyone needs to think the same thing at the same time for this to work. I think. I'm not sure on that last point.
Missing square in a triangle puzzle:
Both of these triangles are made up of the exact same shapes, just reconfigured. So where does that gap in the lower one come from?
It's already been sort of brought up, but I want to be more explicit about it: those shapes are not triangles by any definition. Calling them such is just misinformation and (in my mind) kind of ruins the brain teaser.
That's a good point, actually. Given the solution, they're not actually triangles. It's best to not be explicit, as it does kind of count as a cheat.
You are blindfolded in front of a lazy Susan. There are four identical shot glasses arranged in a square along the perimeter of the lazy Susan. Each shot glass is either sitting right side up or upside down. I'm going to rotate the lazy Susan so one of the shot glasses, selected at random, is directly in front of you. You, still blindfolded, can then reach out and grab any two of the shot glasses, feel how they are oriented, then place them back in the same locations on the lazy Susan oriented however you wish. I'll then rotate the lazy Susan at random again, and let you feel and rotate two shot glasses again. This process repeats until all four shot glasses are oriented the same way (either right side up or upside down), at which point I will announce that you have won and remove your blindfold. What strategy do you use to get all four shot glasses oriented the same way in as few steps as possible, and how many steps does this strategy take in the worst case?
Placed in front of you is a "Lazy Suzan" with four shot glasses placed equally apart around the outer edge. They have been randomly placed either right side up or upside down and it is your challenge to get them all oriented the same way; that is, either all four up, or all four down, as quickly as possible.
Of course you will be blindfolded! You're not surprised...
Each turn the Lazy Suzan will be spun and you may then touch any two glasses, and then decide to keep each as it is or turn either one, or both, over. OK?
You may keep both that you touch the same, turn them both over, or turn just one of them over (your choice, either one). Obviously, as you reach down to grab two of the glasses you will be able to tell which two of the four you are touching - both in front of you; both away from you; the two on the left; the two on the right; or either of the two corner to corners.
You will start with a $10,000 Prize Fund which will be decreased by $1,000 after each turn. I will stand next to you and announce after each of your efforts whether you have solved the puzzle. Of course, there's no chance in the whole world that you will be so lucky as to start with all four in a winning position, but of course you already knew that. And also, if you decide to rely on luck to get you to the final solution you should understand that once the $10,000 Prize fund is gone, you will then have to start paying $1,000 for each wrong guess you make from then on out - No Stopping until you get it right!
Question: What process can you use to guarantee success?
The solution is not posted on that website, but I have found a strategy which guarantees success in at most
That was quite an interesting exercise, I was writing up this big long post about how the eye color problem didn't make sense when I finally understood it.
So the spin on the lazy susan isn't random? It's just rotating one position clockwise/ccw?
The spin on the lazy Susan is random. I framed the problem so that a shot glass always ends up directly in front of you, and the website I linked framed the problem so that the shot glasses always end up [two in front, two in back], but that detail isn't really important. There are four discrete rotations the lazy Susan can take, and the specific rotation chosen each turn is random.
So the spin on the lazy susan isn't random? It's just rotating one position clockwise/ccw?
The spin on the lazy Susan is random. I framed the problem so that a shot glasses always ends up directly in front of you, and the website I linked framed the problem so that the shot glasses always end up [two in front, two in back], but that detail isn't really important. There are four discrete rotations the lazy Susan can take, and the specific rotation chosen each turn is random.
So, how can you definitively solve it in a maximum number of moves then? There's no guarantee that you ever even touch a given glass, and the only way to account for that would be to swap the other three all one way or another, and you can't verify that you've ever accomplished that, either.
If it can only rotate a quarter turn either direction I bet you can do it, but I don't see how you can if it can adopt any of four positions.
Eat it You Nasty Pig. on
it was the smallest on the list but
Pluto was a planet and I'll never forget
So the spin on the lazy susan isn't random? It's just rotating one position clockwise/ccw?
The spin on the lazy Susan is random. I framed the problem so that a shot glasses always ends up directly in front of you, and the website I linked framed the problem so that the shot glasses always end up [two in front, two in back], but that detail isn't really important. There are four discrete rotations the lazy Susan can take, and the specific rotation chosen each turn is random.
So surely you can't guarantee succeeding in any number of turns. I mean there's always a (small) chance that you'll just see the same two glasses in front of you for the first 2000 turns.
Placed in front of you is a "Lazy Suzan" with four shot glasses placed equally apart around the outer edge. They have been randomly placed either right side up or upside down and it is your challenge to get them all oriented the same way; that is, either all four up, or all four down, as quickly as possible.
Of course you will be blindfolded! You're not surprised...
Each turn the Lazy Suzan will be spun and you may then touch any two glasses, and then decide to keep each as it is or turn either one, or both, over. OK?
You may keep both that you touch the same, turn them both over, or turn just one of them over (your choice, either one). Obviously, as you reach down to grab two of the glasses you will be able to tell which two of the four you are touching - both in front of you; both away from you; the two on the left; the two on the right; or either of the two corner to corners.
You will start with a $10,000 Prize Fund which will be decreased by $1,000 after each turn. I will stand next to you and announce after each of your efforts whether you have solved the puzzle. Of course, there's no chance in the whole world that you will be so lucky as to start with all four in a winning position, but of course you already knew that. And also, if you decide to rely on luck to get you to the final solution you should understand that once the $10,000 Prize fund is gone, you will then have to start paying $1,000 for each wrong guess you make from then on out - No Stopping until you get it right!
Question: What process can you use to guarantee success?
The solution is not posted on that website, but I have found a strategy which guarantees success in at most
five
steps, and I think it's the best you can do.
This problem, I haven't seen it before. It's great. I think I have a solution that matches yours.
For shorthand, I'll be using the abbreviation 'U' for an upturned glass, and 'D' for a downturned glass.
In the beginning of the game, you have one of the following four arrangements: DDDU, DDUU, DUDU, DUUU. Otherwise you'd have already won. Start off the game by picking up two adjacent glasses (1st turn).
If they are UU, flip them to DD (consider this as case 1). If they are DD, flip them to UU (case 2). If they are DU, flip them to DD (case 3). Now, the game might end here, but then again it might not.
Let's assume the game isn't over and start with case 1. Since the game hasn't ended, then the other two glasses must be DU. They can't be DD, since the game would have ended before it started. So the current arrangement must be DDDU. Grab two adjacent glasses (2nd turn).
If they are DU, then flip them to DD - that ends the game (the other glasses must be DD). If they are DD, then flip them to DU. The other two glasses are DU, so you must either have DDUU (case a) or DUDU (case b). To distinguish between them, grab two diagonal glasses (3rd turn).
If they are DD or UU, then you are in case b. Flip both glasses over and you're done. If you get DU, then you are in case a. Keep the glasses the same to stay in the DDUU state. On your fourth turn, grab adjacent glasses. If they are DD or UU, flip both over to complete the game. Otherwise, if you have DU, flip them to UD to get to a DUDU state. Grab diagonal glasses and flip them on the fifth turn to solve the problem.
Case 2 is exactly the same as case 1, except with U and D reversed. Done in five turns as well.
Case 3 is trickier, since we need to differentiate between two different possibilities: the glasses arrangements could be DDUU (case x) or it could be DDDU (case y). To distinguish between them, grab two diagonal glasses (2nd turn).
If you grab DD, you are in case y. Flip the glasses to DU and you will be in a DDUU state. On your third turn, grab adjacent glasses. If they are DD or UU, flip both over to complete the game. Otherwise, if you have DU, flip them to UD to get to a DUDU state. Grab diagonal glasses and flip them on the fourth turn to solve the problem.
If you grab DU, you could still be in case x or case y. Flip the glasses to DD. If you were in case y, then the game will end. Otherwise, you will know that you are in case x, and your current state is DDDU. Take diagonal glasses on the third turn. If you get DU, then flip them to DD to finish. If you get DD, then flip them to DU to get to a DDUU state. Pick up adjacent glasses on turn four. If you get DD or UU, flip both over to finish. Otherwise, with DU, flip them to UD to get to a DUDU state, and on your fifth turn, grab diagonal glasses and flip them both to obtain a solved state.
EDIT: My first solution was wrong (see posts 159 and 161)! Edited with a corrected solution.
For shorthand, I'll be using the abbreviation 'U' for an upturned glass, and 'D' for a downturned glass.
In the beginning of the game, you have one of the following three arrangements: DDDU, DDUU, DUUU. Otherwise you'd have already won. Start off the game by picking up two adjacent glasses (1st turn).
If they are UU, flip them to DD (consider this as case 1). If they are DD, flip them to UU (case 2). If they are DU, flip them to DD (case 3). Now, the game might end here, but then again it might not.
Let's assume the game isn't over and start with case 1. Since the game hasn't ended, then the other two glasses must be DU. They can't be DD, since the game would have ended before it started. So the current arrangement must be DDDU. Grab two adjacent glasses (2nd turn).
If they are DU, then flip them to DD - that ends the game (the other glasses must be DD). If they are DD, then flip them to DU. The other two glasses are DU, so you must either have DDUU (case a) or DUDU (case b). To distinguish between them, grab two diagonal glasses (3rd turn).
If they are DD or UU, then you are in case b. Flip both glasses over and you're done. If you get DU, then you are in case a. Flip both glasses to UD. That makes the glasses set up DUDU. Grab two diagonal glasses on your 4th turn and flip them both over. You're done!
Case 2 is exactly the same as case 1, except with U and D reversed. Done in four turns.
Case 3 is trickier, since we need to differentiate between two different possibilities: the glasses arrangements could be DDUU (case x) or it could be DDDU (case y). To distinguish between them, grab two diagonal glasses (2nd turn).
If you grab DU, you are in case x. Flip them to DD, and you will be in the DDDU situation. This is the same as case 1! Follow the steps for case 1, and you will complete the task within five turns. If you grab DD, then you are in case y, which is case 1. Again, follow the case 1 steps and you will complete the task within five turns.
This is very close, but:
If you are in case a (DDUU), grabbing two diagonal glasses and flipping both will keep you in case a (you'll still have DDUU).
I was also under the impression that you couldn't tell which way the glasses were oriented when you grabbed them.
You can tell how they are oriented by feeling them. You're blindfolded, not numb.
EDIT: To clarify, you can tell how the glasses you grab are oriented after you grab them, and before you decide how to put them back. You don't see the orientations of the glasses before you choose which glasses to grab.
For shorthand, I'll be using the abbreviation 'U' for an upturned glass, and 'D' for a downturned glass.
In the beginning of the game, you have one of the following three arrangements: DDDU, DDUU, DUUU. Otherwise you'd have already won. Start off the game by picking up two adjacent glasses (1st turn).
If they are UU, flip them to DD (consider this as case 1). If they are DD, flip them to UU (case 2). If they are DU, flip them to DD (case 3). Now, the game might end here, but then again it might not.
Let's assume the game isn't over and start with case 1. Since the game hasn't ended, then the other two glasses must be DU. They can't be DD, since the game would have ended before it started. So the current arrangement must be DDDU. Grab two adjacent glasses (2nd turn).
If they are DU, then flip them to DD - that ends the game (the other glasses must be DD). If they are DD, then flip them to DU. The other two glasses are DU, so you must either have DDUU (case a) or DUDU (case b). To distinguish between them, grab two diagonal glasses (3rd turn).
If they are DD or UU, then you are in case b. Flip both glasses over and you're done. If you get DU, then you are in case a. Flip both glasses to UD. That makes the glasses set up DUDU. Grab two diagonal glasses on your 4th turn and flip them both over. You're done!
Case 2 is exactly the same as case 1, except with U and D reversed. Done in four turns.
Case 3 is trickier, since we need to differentiate between two different possibilities: the glasses arrangements could be DDUU (case x) or it could be DDDU (case y). To distinguish between them, grab two diagonal glasses (2nd turn).
If you grab DU, you are in case x. Flip them to DD, and you will be in the DDDU situation. This is the same as case 1! Follow the steps for case 1, and you will complete the task within five turns. If you grab DD, then you are in case y, which is case 1. Again, follow the case 1 steps and you will complete the task within five turns.
This is very close, but:
If you are in case a (DDUU), grabbing two diagonal glasses and flipping both will keep you in case a (you'll still have DDUU).
Ah, indeed. (That's what I get for trying to solve brainteasers at the end of an all-nighter.) Mixed up my adjacencies and diagonals. So let's rewrite that section.
... If they are DD or UU, then you are in case b. Flip both glasses over and you're done. If you get DU, then you are in case a. Keep the glasses the same - you are in a DDUU state. On your fourth turn, grab adjacent glasses. If they are DD or UU, flip both over to complete the game. Otherwise, if you have DU, flip them to UD to get to a DUDU state. Grab diagonal glasses and flip them on the fifth turn to solve the problem.
Of course, this now requires a refinement of the case 3 strategy, but now that I've corrected case 1 (and case 2), I can see how it's done. Gonna edit my previous post with the correct solution, then...
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It's already been sort of brought up, but I want to be more explicit about it: those shapes are not triangles by any definition. Calling them such is just misinformation and (in my mind) kind of ruins the brain teaser.
the "no true scotch man" fallacy.
It's a quadrilateral. What appears to be a hypotenuse is two sides with a vertex where the red and blue triangles meet. I strongly suspect, though I'm not going to bother formally proving, that the riddle wouldn't work if the triangles had the same slope and formed a continuous side.
Oh hey you're right. I never noticed that before.
the "no true scotch man" fallacy.
No, I was pointing out that you wrote that the deduced situation was
A sees 3 blue, therefore he thinks:
B sees 2 blue, therefore he thinks:
C sees 1 blue, therefore he thinks:
D sees 0 blue
Which is very obviously totally wrong.
Yes. That is totally wrong. It also is not what I wrote.
The important bit is, "if I have brown eyes".
Bonus: the bottom "triangle" is an octogon. I think that's cool for some reason.
I'm going to glare at you until you stop this.
I got hooked on this one and finally spoiled it for myself. It's meta-neat to me how many variants they are and how subtly this one can be perverted. Seeing implementations of the solution algorithm in different languages (especially Haskell) was neat too.
I guess what I'm saying is that I'm a horrible person but thanks for sharing that one.
I'm trying to provide an explanation you'd understand, but I really can't think of one simpler than what I've already written. So, instead, I invite you to prove how what I'm saying is wrong, in the hopes that you'll learn from that.
For those who haven't seen the show, here's the format:
That's all the info you need to answer this puzzle.
Correctly answer every question in the first round minus the daily double, then get the daily double, make it a true daily double, wager everything, get it -> 2 * (3000 *6 - 400)
Add to that correctly answering every question in the second minus the two daily doubles, which would have to be in the 800 dollars -> 6000 * 6 - 1600
Then true daily double twice, and then double again in Final Jeopardy.
Of course, you're an obnoxiously cocky jackass if you do that, but theoretically.
I love this one, got it in some after school program I did.
It is, I pumped out the math real fast, pretty sure I did it right. If Ken Jennings were really smart, he only would have needed like four days.
he had a weird ocd with his daily double betting.
I understand the problem perfectly. However, you're stating the situation incorrectly and taking the induction past where it makes any kind of logical sense.
Let's examine your 4 monk example.
A sees 3 blue. He assumes that B sees 2 blue, C sees 2 blue, and D sees 2 blue.
Since he knows that B sees two blue, he believes that C and D will both see 1 blue.
That's it.
That's as far as A's induction can logically progress. You can't recurse on that again. He knows for certain that
B sees (at least) 2.
C sees (at least) 2.
D sees (at least) 2.
From there, he cannot logically reach the conclusion that C would think D sees none. It's even more obvious with larger numbers.
That's the thinking that's gonna get you in trouble. As with many logic puzzles, this one is deliberately obfuscated. It is trivial to solve for two islanders, just as the Monty Hall problem is quite obvious if you imagine a hundred doors. The problem as stated is not logically any different from the base case, but it is made to seem so by the magic of large numbers.
Any time you're approaching a logic puzzle with the idea that it's obvious if you look at the problem as stated, you can be pretty sure you're on the wrong track.
Four people have blue eyes, and we label them A, B, C, and D.
Note that the quote tree does not (necessarily) describe what B, C, and D actually think, merely what A can suppose about their recursive thoughts about each others' thoughts.
Also, hopefully this makes it more apparent why the Guru is necessary; it eliminates any chain of supposition that results in someone recursively thinking anybody else thinks there are no people with blue eyes.
Imagine there are four blue-eyed people on the island and a random number of brown-eyed people. The blue eyes are Alice, Bob, Carol, and Derek. Someone announces that he sees one blue-eyed person.
Alice sees three people with blue eyes. She knows that Bob, Carol, and Derek will see at least two people; possibly three, if she herself has blue eyes. She wants to know the actual number of blue-eyed people on the island, so she starts a thought experiment.
If her eyes are not blue, Bob would see two people with blue eyes.
She knows, of course, that the imaginary Bob will be starting his own thought experiment, in which Bob will also calculate the odds, same as Alice is doing right now. She knows that, in this second level thought experiment, Bob would imagine a universe in which Carol sees only one person with blue eyes.
The thing about Alice is that she’s a perfect logician, just like everyone else on the island. She has to take her thought experiment to its logical conclusion. So, the imaginary Carol would create yet another level of reality in which Derek sees absolutely no one with blue eyes.
Alice continues. Derek would realize that he has blue eyes, since no one else has them. As such, he has to leave at midnight. If, in this farfetched thought experiment, Derek does not leave, there have to be at least two people on the island with blue eyes.
Alice knows that Derek will do no such thing. After all, she can count the number of people with blue eyes and it's higher than one. In fact, at least three people have blue eyes.
So what would happen if Derek does not leave? Well, Carol would look at Derek and realize Derek expected Carol to leave, since Carol is herself unsure about her eye color. As such, they must both have blue eyes.
What if, Alice wonders - and this is the point where she gets excited - both Derek and Carol fail to leave on that second day? Bob expected them to do so, if there were only two people with blue eyes. The fact that they didn't means he has blue eyes too.
Alice starts to see a pattern, and she ponders the implications. She realizes that every blue-eyed person would leave on the same day, and that day is based on how many there are. For one person, it's the first day. For two people, it's the second day. For three people, it's the third day.
Since Alice counts three people with blue eyes, she does not expect anyone to leave until the third day. If, however, no one leaves on the third day, she would know that she herself has blue eyes.
So she waits.
And she starts packing, just in case.
The reason the Guru, or any outsider, is necessary is because everyone needs to think the same thing at the same time for this to work. I think. I'm not sure on that last point.
I hate you. Four hours I spent thinking about this.
That's a good point, actually. Given the solution, they're not actually triangles. It's best to not be explicit, as it does kind of count as a cheat.
You are blindfolded in front of a lazy Susan. There are four identical shot glasses arranged in a square along the perimeter of the lazy Susan. Each shot glass is either sitting right side up or upside down. I'm going to rotate the lazy Susan so one of the shot glasses, selected at random, is directly in front of you. You, still blindfolded, can then reach out and grab any two of the shot glasses, feel how they are oriented, then place them back in the same locations on the lazy Susan oriented however you wish. I'll then rotate the lazy Susan at random again, and let you feel and rotate two shot glasses again. This process repeats until all four shot glasses are oriented the same way (either right side up or upside down), at which point I will announce that you have won and remove your blindfold. What strategy do you use to get all four shot glasses oriented the same way in as few steps as possible, and how many steps does this strategy take in the worst case?
Pluto was a planet and I'll never forget
No, it's possible. Here's the problem written out in more detail from http://www.twinbear.com/riddles.html:
The solution is not posted on that website, but I have found a strategy which guarantees success in at most
That was quite an interesting exercise, I was writing up this big long post about how the eye color problem didn't make sense when I finally understood it.
Thanks.
The spin on the lazy Susan is random. I framed the problem so that a shot glass always ends up directly in front of you, and the website I linked framed the problem so that the shot glasses always end up [two in front, two in back], but that detail isn't really important. There are four discrete rotations the lazy Susan can take, and the specific rotation chosen each turn is random.
So, how can you definitively solve it in a maximum number of moves then? There's no guarantee that you ever even touch a given glass, and the only way to account for that would be to swap the other three all one way or another, and you can't verify that you've ever accomplished that, either.
If it can only rotate a quarter turn either direction I bet you can do it, but I don't see how you can if it can adopt any of four positions.
Pluto was a planet and I'll never forget
So surely you can't guarantee succeeding in any number of turns. I mean there's always a (small) chance that you'll just see the same two glasses in front of you for the first 2000 turns.
This problem, I haven't seen it before. It's great. I think I have a solution that matches yours.
In the beginning of the game, you have one of the following four arrangements: DDDU, DDUU, DUDU, DUUU. Otherwise you'd have already won. Start off the game by picking up two adjacent glasses (1st turn).
If they are UU, flip them to DD (consider this as case 1). If they are DD, flip them to UU (case 2). If they are DU, flip them to DD (case 3). Now, the game might end here, but then again it might not.
Let's assume the game isn't over and start with case 1. Since the game hasn't ended, then the other two glasses must be DU. They can't be DD, since the game would have ended before it started. So the current arrangement must be DDDU. Grab two adjacent glasses (2nd turn).
If they are DU, then flip them to DD - that ends the game (the other glasses must be DD). If they are DD, then flip them to DU. The other two glasses are DU, so you must either have DDUU (case a) or DUDU (case b). To distinguish between them, grab two diagonal glasses (3rd turn).
If they are DD or UU, then you are in case b. Flip both glasses over and you're done. If you get DU, then you are in case a. Keep the glasses the same to stay in the DDUU state. On your fourth turn, grab adjacent glasses. If they are DD or UU, flip both over to complete the game. Otherwise, if you have DU, flip them to UD to get to a DUDU state. Grab diagonal glasses and flip them on the fifth turn to solve the problem.
Case 2 is exactly the same as case 1, except with U and D reversed. Done in five turns as well.
Case 3 is trickier, since we need to differentiate between two different possibilities: the glasses arrangements could be DDUU (case x) or it could be DDDU (case y). To distinguish between them, grab two diagonal glasses (2nd turn).
If you grab DD, you are in case y. Flip the glasses to DU and you will be in a DDUU state. On your third turn, grab adjacent glasses. If they are DD or UU, flip both over to complete the game. Otherwise, if you have DU, flip them to UD to get to a DUDU state. Grab diagonal glasses and flip them on the fourth turn to solve the problem.
If you grab DU, you could still be in case x or case y. Flip the glasses to DD. If you were in case y, then the game will end. Otherwise, you will know that you are in case x, and your current state is DDDU. Take diagonal glasses on the third turn. If you get DU, then flip them to DD to finish. If you get DD, then flip them to DU to get to a DDUU state. Pick up adjacent glasses on turn four. If you get DD or UU, flip both over to finish. Otherwise, with DU, flip them to UD to get to a DUDU state, and on your fifth turn, grab diagonal glasses and flip them both to obtain a solved state.
EDIT: My first solution was wrong (see posts 159 and 161)! Edited with a corrected solution.
This is very close, but:
You can tell how they are oriented by feeling them. You're blindfolded, not numb.
EDIT: To clarify, you can tell how the glasses you grab are oriented after you grab them, and before you decide how to put them back. You don't see the orientations of the glasses before you choose which glasses to grab.
... If they are DD or UU, then you are in case b. Flip both glasses over and you're done. If you get DU, then you are in case a. Keep the glasses the same - you are in a DDUU state. On your fourth turn, grab adjacent glasses. If they are DD or UU, flip both over to complete the game. Otherwise, if you have DU, flip them to UD to get to a DUDU state. Grab diagonal glasses and flip them on the fifth turn to solve the problem.
Of course, this now requires a refinement of the case 3 strategy, but now that I've corrected case 1 (and case 2), I can see how it's done. Gonna edit my previous post with the correct solution, then...
EDIT: Post 157 edited with corrections.