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    • CommentAuthorSamael
    • CommentTimeNov 30th 2006
     
    I finished reading the book "How Would You Move Mount Fuji" on Tuesday- a great book, if you get a chance. It's basically about the way that some business, like Microsoft, use puzzles and riddles as part of the interview process. It talks about how the practice probably started, what you can really learn from a puzzle interview (how effective it is, and what makes for a good/bad puzzle question in an interview) and gives some background about IQ/Puzzle tests, in general. Anyway, I'm a huge, huge fan of riddles, brain teasers, and logic puzzles, and I loved that the book included a bunch of the questions and riddles that companies have been using. I was able to solve most of them (not all) without looking at the answers.

    So, I thought I'd share a few of them as well.

    1. There are 8 pool balls on a table in front of you. 7 of the pool balls weigh exactly the same, but 1 of the balls is slightly heavier than the others. Using only a beam balance (like the one Lady Justice holds), you are to determine which of the balls is heavier than the others. What is the minimum number of measurements needed to determine the heavier ball 100% of the time (You could, for example, find the heavier ball on the first try by putting one ball on each side of the balance if you were lucky, and picked the heavier ball by chance. This would not work 100% of the time, though- you only stand a 1 in 8 chance of selecting the heavy ball when you pick the first one, and a 1 in 7 chance of selecting it as your secod ball)?

    2. There are five bottles of pills. 4 of the bottles are full of 10 gram pills, but the fifth bottle contains contaminated pills. They weigh only 9 grams. You're not sure which bottle contains which pills, and the pills all look the same. Using an electronic scale, how could you determine which bottle contains the tainted pills in exactly one measurement, 100% of the time?

    3. You have two jars, 50 red marbles, and 50 blue marbles. A blind man is going to select one of the jars at random, and pull one random marble out of the selected jar. If he pulls out a blue marble, you win $5,000. All of the marbles must be placed inside of the jars. Once they're inside the jars, they will be stirred up, to mix the marbles together. Once you've placed the marbles inside of the jars, you will no longer have any influence over what happens- what strategy will give you the best chances for winning the $5,000, and- roughly- what are your odds of winning?
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      CommentAuthortallman
    • CommentTimeNov 30th 2006
     
    Lets see here, without thinking tooo much (thinking hurts when I'm sick):

    1. I believe 3. Put 4 balls on each side. One side should be heavier. Remove the other side, then take two of the balls from the heavier side and put them on the other side. Again, one side should be heavier, narrowing the choices down to 2 balls. Then you weigh those two balls against each other, thus revealing which one is the heavy ball.

    2. I'm guessing that the "electronic" scale can measure multiple bottles at once, but I'm pretty sure that's wrong.

    3. 50% Right. Right?

    Sometimes I hate these. I always feel like I'm missing something (and I'm pretty sure that I am, especially with that second one... but the third one also seemed way too easy).

    ~tallman
    • CommentAuthorSamael
    • CommentTimeDec 3rd 2006
     
    I'm sorry to hear that you're sick- anything in particular, or just a general illness?

    1. It can be done in fewer than 3 measurements.
    2. The electronic scale is just a regular weight scale. It could weigh multiple bottles at once, but it's only going to tell you the combined weight. =)
    3. You can get much better than 50%.

    For what it's worth- according to the book, those are pretty much the first answers people come up with.
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      CommentAuthortallman
    • CommentTimeDec 3rd 2006
     
    Just a cold, really. But I'm really congested and coughing a lot. Getting a little better.

    So... I figured I'd try to fire up a few neurons and come up with an answer to a few of them

    1. Ok, so I think I can do it in two. I'll put 3 on each side of the scale, and leave 2 off. If the scale is level, it means the heavier ball is one of the two not on the scale, at which point you can put each of those balls on the scale to figure out which is heavier. However, if one side of the scale is heavier, the ball we're looking for is on that side. At this point, you take the three balls (1 of which is the heavier one) and place one on each side of the scale, leaving one off. If the scale is level, it means the heavier ball is the one not on the scale. Otherwise, it's one of the ones on the scale. Score.

    2. You have to take the pills out of the bottles to figure out which bottle is tainted. Let's say we take the 5 bottles and give them letters ABCDE. We'll put 1 pill from A, 2 from B, 3 from C, 4 from D, and 5 from E. The total weight should then give away the tainted bottle. For instance, if the total weight was 48 grams, it would be bottle B (because there are two pills from B, and if they're tainted, we'll be missing 2 grams). When I first read this, I thought the solution had something to do with taking pills out of the bottle, but being in the non-thinking mode, I went for the easy "magic" answer.

    3. Yeah, if you put only 1 blue marble in 1 of the jars, and the rest of the marbles in the other, your chances should increase significantly (somewhere around 74-75%).

    There might be more to number 3, but now it's late and my neurons wish to stop firing so quickly.

    ~tallman
    • CommentAuthorSamael
    • CommentTimeDec 4th 2006
     
    Nope, those are all exactly right.

    You want the hard one, now? ;)

    Imagine that you're dead. You're walking along the Golden Path, when you come to a fork. One path leads to heaven, the other to hell. Obviously, you want to avoid hell.
    Standing at the fork are three guardians- the red guardian, the blue guardian, and the green guardian.
    Further, you know that one of the guardians is Satan's, and will only tell lies.
    One guardian is God's, and will only tell the truth.
    One guardian is the guardian of chaos, and will lie or tell the truth randomly.
    Unfortunately, you have no way of knowing which guardian is which.
    You may address any two of the guardians with one question each (Two questions, one to each of two guardians). No guardian will respond to a question not asked of him, and you may only address a guardian once.
    How can you make sure that you take the path to heaven?
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      CommentAuthortallman
    • CommentTimeDec 4th 2006
     
    Score. It's nice to know that me noggin' still functions semi-properly.

    The hard one sounds like a variation on one I've heard of before. It's the "random" guardian that's tricky here. I do have one question regarding the setup: When you say "No guardian will respond to a question not asked of him" does that mean I can't ask the red guardian what the blue (or green) guardian would say? This is the "ah ha" of the original version of this puzzle, but I'm not sure if this variation requires it or not...

    ~tallman
    • CommentAuthorSamael
    • CommentTimeDec 4th 2006
     
    Ah, I blew it. Heh.

    The deal is that you get two "yes/no" questions. You can ask any yes/no question you want, but you have to addresss it to one of them.
    So, you could, for example, ask the Red Guardian "Does the Blue Guardian like cheese." But, no guardian is going to answer a question that he wasn't asked. The Blue Guardian isn't going to pipe up with "Yes, I like cheese" when you ask the Red Guardian. And you can't ask two questions of the same Guardian.
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      CommentAuthortallman
    • CommentTimeDec 6th 2006
     
    I thought about it a bit, but nothing in particular came to mind and I don't feel like revving up the brain right now. Care to post the answer?

    ~tallman
    • CommentAuthorSamael
    • CommentTimeDec 12th 2006
     
    Sure.
    The first question is addressed to any of the three. For our example, we'll ask Red.
    You ask Red "Is Blue more likely to tell the truth than Green?"
    If the answer is "yes" then you address the next question to Green.
    If the answer is "no" then you address the next question to Blue.
    The next question you ask is this: "Do I assume right, that if I ask whether the left way is the way to Heaven, you would answer 'Yes'?"

    This will always work because the second question will always end up being addressed to either Heaven's guardian or Hell's guardian. The first question ensures this because you want to ask the second question to whomever is indicated to be more likely to *lie*. Why?
    If Heaven's guardian is the one you're addressing with the first question, you want to end up talking to Satan's guardian, so you take whichever guardian is indicated to be the liar. On the other hand, if your first question is addressed to Satan's guardian, he's going to lie to you and tell you that Chaos' guardian is more likely to tell the truth, so you want to take the person indicated to be a liar anyway (thus, ending up with Heaven's guardian." If you're talking to Chao's guardian with the first question, it doesn't matter, because you're going to end up with either Satan's or Heaven's guardian, anyway.
    And, as long as the second question is addressed to one of those two, you will always know the correct path, because of how the second question is worded. It's obvious that the Heaven guardian will tell you the truth, but the Hell's guardian will respond the same way because of this: You're embedding an assumed question. Assume that Heaven's path is on the left. His answer is reasoned like this. If you asked him if the left was the right way, he would lie and say "no." But, the question you're actually asking is what he would say *if* you asked him. Since he'd actually say "no," he has to lie and say "yes" to the question of whether he would say "yes."
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      CommentAuthortallman
    • CommentTimeDec 12th 2006
     
    Ahhhh, very clever. Thankee.
    • CommentAuthorSamael
    • CommentTimeDec 15th 2006
     
    You're much welcome. I did not correctly solve that particular riddle without looking up the answer. It makes sense, but damn if it's not tricky.