Prime
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Everything posted by Prime
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There was so much space devoted to the subject and such a vivid description... I got confused. Sorry. But you must agree, bosom has to play some role there.
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Small correction: I said that if Waldo does not see the cards he must find one out 52, whereas he can still narrow down his choice to 48.
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There are many simple ways to make a code: I say where magician's assistant saw cards -- it is not a trick.
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I am not sure… Sounds like what you say is, the problem can be solved if it can be solved. Isn’t the set of mutually exclusive tests with collectively exhaustive outcomes, the very thing that the problem asks you to find? Regard 4 coins, where either only one can be either heavier, or lighter, or all four of them may be true. There are 9 combinations in all: 4 cases where one of the coins is heavier, 4 cases where one is lighter, and one case with all four of them true. Yet, I’m convinced that in two weighings (each with 3 distinct outcomes), you are not guaranteed even to find which one is counterfeit, if any, let alone tell whether it is heavier, or lighter. Despite the fact that 2 weighings have a potential to distinguish 3**2=9 cases. Check out my posts in the “Weghing in a harder way” puzzle. That was the topic, which prompted me to come up with this weighing problem.
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An empty set is a perfectly valid notion. You can have an empty set of Fords. (I do.) I think, what causes the outrage here, is that the anwer 2 cars assigns the names to the members of an empty set. I.e., I have an empty set and its members are named Ford.
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My daughter charges that I was not entirely fair with my second problem (40 in 4). To clear the air, but without admitting to any wrongdoing, I will stipulate as following: My "spoiler" in my previous post is neither a solution, nor any part of the solution. Feel free to peek.
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The original statement tells us nothing about how far either of the trains travel -- only the distance travelled for both trains.
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No, you are not allowed to add any coins. I agree that if you pull a true coin out of your pocket and add to the set, you can solve the problem. You are trying to reuse the same trick, that you have found when solving 39 coins. Namely, when you had 3 weighings left and a set of 13 coins with undetermined fault (heavy or light) and you used a reference coin to isolate 9 coins instead of 8. (The very thing that separated my problem from the one where you solve 12 coins in 3 weighings.) The trick for solving 40 is different. I’ll give a small hint here, after which you should solve it with ease. Check it out, if you like: After that we shall revisit your proposition that you can always solve 81 or less possible cases with 4 weighings.
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I'm in a particular nasty mood today. So I'm going to take that puzzle apart and show that the answer that the author has in mind is not the correct one. Cute problem, but the statement needs a bit more work.
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Suicide has the solution, toughing it through the last variation, where everything balances to the end in the last post. Another challenge is to express the solution in simpler terms. I will try that a little later. Now Suicide is ready to solve 40. Don't bother describing the cases where one group of 13 is heavier than the other. You've done that already. Just go directly into the path where everything balances.
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When you directed to find the difference of the two numbers, you forgot to specify to subtract smaller from larger. Without that stipulation your formula only guarantees to determine the lower number every time.
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Well, who is the author?
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Or, perhaps, there are 78 cases for 39 coins. There are 39 distinct possibilities for which coin is a counterfeit and there are two possibilities for each -- heavy or light. But I don't see that as a guarantee that you can arrange your 4 weighings in a way to determine the counterfeit. Furthermore, I see a variation of the problem, where you can not find such an arrangement.
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I don't think you have the answer. The description of your method is incomplete. Also, I think you are mistaking about 78 cases. I did ask to find the counterfeit coin and whether it is lighter or heavier. That makes it 117 cases for 39 coins, as I see it. Thanks for trying.
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I studied my math back when they did not come up with calculators yet. So the integral of t**2 is (1/3)* t**3. If you plug in the boundaries and subtract lower from upper, you get 2/3. Cosine of 60 degrees is 1/2, and I guess you have figured out the logarithm. This puzzle is hard to solve unless, you heard that limmerick before.
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After seeing so many variations of the counterfeit coin weighing problem on this forum, I am compelled to take that puzzle to the top and slightly over. Here is my own variation: There are 39 coins, one of them counterfeit and weighs either more, or less than a true coin. In 4 weighings, or less on a balance-type scale, find the counterfeit and determine whether it weighs more or less than a true coin. And for an encore… When you have solved the above problem (or not), find the only counterfeit coin (which weighs more or less) out of 40 coins with 4 weighings, or less.
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First, I’d like to complement the clever redundant use of the factorial sign. (2!!!! = 2). However, your answer is wrong. The original question was: “… how many minimum ways …”. Whereas, you pointed out only one such way. Indeed, the minimum number of weighings to determine counterfeit coin is 2, if you get lucky. But there are 27!/(25!*2!)=351 ways to compare 2 coins out of 27. And then there are 25*2 ways to compare one of those two to the 25 on the side. Or you could choose to drop 13 coins onto one side of the scale and 13 onto another, and get lucky finding that they weigh the same. And there are C(27,13)*C(14,13) ways, or 27!/(14!*13!)*14 of doing that. Then there are 26 ways to compare the counterfeit coin with one of the true ones. Simplified, the correct answer is: 27! / 24! + 27! * 26 / (13! * 13!). That seems like a very large number, which may not fit into the space allotted by this forum. (Also, see my previous post for further discussion of the puzzle as it was meant.)
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I’ll take up an issue with that statement. I’d say it depends on the kind of possibilities that you are trying to distinguish. The puzzle required to find the counterfeit coin AND tell whether it is heavier or lighter. While you can definitely solve 27 in 4 weighings, I don’t see how you can solve 40 in 4. Although, there are only 80 possibilities (less than 3**4). You can find one counterfeit coin out of 40 in 4 weighings, but not necessarily tell whether it is heavier, or lighter. For the sake of saving space, solve 13 coins in 3 weighing. I can see how you can find counterfeit coin there, but I can’t ensure determining whether it is heavier, or lighter.
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It evaluates: (2/3)*(1/2)=1/3, so there is nothing special there. On the other hand, it's Intergral square tea times. Three pie over nine equal log cubic root. Yeee... Or something like that. Sounds like something from Alice in Wonderland.
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The solution you have in mind may lose you your bet. There are draws, you know... And draw is not win. Also, it would not work in practice. When giving a simul, your opponents make their move when you come to their board, and you are expected to make your move before going to the next board.
