Prime
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Everything posted by Prime
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For the question (2) the average number of picks of random numbers to add up to 1 or more, the intuitive answer 2 is wrong. Think of it this way, the probability to pick exactly 1.0 out infinite number of points is zero for any finite number of attempts. Thus, for the purpose of average, getting 1 with just one pick does not come into play. The minimum number of picks that would count in calculation of average is 2. And then there are cases when it takes more than 2 picks to add up to 1. So the average must be greater than 2.
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The #0 task appears to be similar to what we've done in Bonanova's infinite exponential, where EventHorizon explained stable equilibrium points.
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Is it a common knowledge?
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Question (1) has recently been done here.
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I don't quite see how increasing number of states can win anything in speed. Computers don't do their process one bit at a time. On each clock pulse the state changes. If you had a simple 1000-state machine you'd need a 3-digit (bit) register to represent that state in decimal and a 10-digit (bit) register to represent state in binary. So you'd use less space, but no gain in speed. The engineering part of a 10-state element would be a technological nightmare. Parallel processor is not an exact analogy here. Each processor has its own clock. Anything you can do in decimal, you can do in the same amount of time in binary. Takes more space, but the design is much simpler on all levels.
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Against...against? Double negative? Does it mean the probability of winning is 1/3?
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In case people are still working on this, I'll throw this suggestion into the pot. First, try solving 4 rocks ranging from 1 to 4 lb with one reference weight.
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Either way the problem does not have a unique solution. Or are we are missing some other possible interpretation here?
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I pondered on this one for a while and I don't see a unique solution, but rather many...
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Since this puzzle is my own invention based on the previous "Weighty thoughts" topic, the solution I have, may not be so simple and elegant. However, ...
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I've made several typos and must clarify. I meant one "reference weight" of your choosing. You can make any number of rock weighings, until you find all individual rocks' weights. Thanks for pointing that out. Also, I meant rock-weighing firm in the first sentence. The word "firm" is missing (which is not rellevant to the solution, though.)
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OK I'll try to answer that making as little giveaway as I can manage:
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Condition-wise it is the same problem as your original "Weighty thoughts". You have to excuse my grammar. I meant there is a balance-type scale, where you can only weigh one side against another and see if one weighs more, or another, or they are equal. My variation of the problem explores some logical inferences that we passed in your original post. I believe, I solved this problem. If I point out what's wrong with your proof above, it would be a dead giveaway. And I think you might enjoy finding the solution on your own.
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Disregard that. I made a mistake. You need more than 2 rocks to solve that. See my new topic.
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Having disposed of one weighing job, same rock-weighing issued requirements for another challenging task. The task involves a balance-type scales. 1). Due to government regulations you can use only one measuring weight. It can be any weight you want. There are 7 rocks each weighing a whole number of pounds between 1 and 5 inclusively. The rocks are not necessarily distinct in weight from one another. Using just one weigh and balance scales, figure out the weight of each rock.
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So you did. While I went on rambling about why, what, and how.
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Thanks. So I won the bid and stand to earn $20. However, I don't feel that we have exhausted the potential of this puzzle. For one thing, I'm still looking for a 2-weigh solution. And although, I haven't found one yet, neither have I proved it impossible. Consider a follow up weighing job: There are just two rocks, which can weigh a whole number of pounds from 1 to 4. You are asked to use only one reference weight of your choosing to determine the weight of the two rocks.
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So, now let the OP say ... To me the OP clearly implied 13 different colors present in the bag. For this new problem statement the minimal solution is:
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That wouldn't work if all 26 stones weighed, say 23 lb each. (They don't have to weigh different per OP). However, I can lower my bid to $50 (using only 3 whole number weighs). Actually, Imran already used same idea, but didn't optimize the solution.
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I thought of that. However, since you bid before taking the contract, I assume, you must count on the worst case scenario.
