CaptainEd
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Everything posted by CaptainEd
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also, are the values required to be integer?
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These two lines seem a little out of pattern H+I > A+B+C+D+E+F+G+H, H+J > A+B+C+D+E+F+G+H+I, Given the pattern above, I would expect H+I > A+B+C+D+E+F+G H+J > A+B+C+D+E+F+G+I However, it's your puzzle, so I'm simply asking, is this discrepancy intentional, or is it a typographical error?
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It seems like an incredible coincidence. Maybe the 0s have been replaced with 9s?
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Yeah, I think you guys nailed the unscrambling into the sequence Barbapapa--I think that's all the letters and the colors are for. Was it Molly Mae who found some patterns? What have you learned?
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That's a lot of interesting thoughts, Molly Mae. The notion that A through F could be numbers (not restricted to single digits) for long/lat (or lat/long, sigh) is interesting. Certainly, the OP states a restriction to single digits, but the posted document does not. If they are single digits, the geocache is in a pretty cold place...But your suggestion proposes how a mere 6 numbers could suffice as a geocache locator. Yes, the denominator of 7 is what comes to mind looking at those first row numbers. I remember when I learned about the magic of the repeating digits of 1/7 = .142857... where multiplying simply rotates the string (e.g. 3/7=428517... and 4/7=571428...). Of course, they aren't in order. (as you said, "Shrug")
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Like RESHAW, I'm looking for more context--you say that once the values to A-F are found, you'll be able to determine the location...Does that mean (a) ABCDEF are pieces of location information, or (b) you have additional clues to the location of the geocache, and ABCDEF are merely just the next hurdle you're supposed to jump through? (I'm not trying to steal your geocache, I'm just trying to squeeze any more context out of this amazingly empty-looking puzzle.)
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I'm with Maurice:You asked for expected value. If you were to bet a fixed fraction f of your holdings, your expected value is (1 + f*.045)^n Clearly the largest value comes when f = 1, the Let-It-Ride strategy. For n = 365, this amounts to 9 Million bucks and of course, some outcomes are much larger. Yes you could go broke. But so what? Suppose you use f = .01, at the end of 365 days, you've got $1.18. That's nearly broke. So, if you'd like to add a proviso that the player guarantee not to go broke, we'd need the definition of "go broke". We could make f = .99, you could easily go to very tiny fractions of a dollar, but the expected value is still over 8 million. Would you, instead, like to maximize mean time to bankruptcy? Or optimize value while preventing your value going below some threshold T more than probability P? So, are you sure the task is to optimize expected value? Or is it really something more complicated?
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I assume that the broken balance can give accurate readings, and can give any answer, can say heavy is heavy, even, or light, and the same for equal.
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Quite right, Swapnil, my assumptions aren't even true, much less conclusions. I'll be back later, I hope...
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Sorry, bad math.
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Superprismatic, it's amazing to me that (1) this claim is true and (2) someone knows it. This is interesting, as always. Where do you come up with these things?
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Thanks for the puzzle, Anza Power!
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Finally, I wrote a Python program, benefiting from the iterators over combinations. there are none with the lowest speed < 1000 for eight cars
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Anza Power, what if the best answer is a single product? For example, if the list is $24, 10 hours $24, 11 hours $49, 50 hours, the best combination would be the singleton for 50 hours, rather than the pair for 21. Do you want the desired answer to be the pair for 21 hours?
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Aparichit, your P(3) = 1, but the OP requires that P(3)=2. Superprismatic, I enjoyed the "foraging", as I manually explored the landscape around these integer coefficients. I would have posted a parity-based answer, but stigge posted a clearer answer sooner. But thank you for the journey!
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Bhramarraj, look at the 4-car case 2:3:4:6. Clearly the 2, 4, and 6 represent the 1:2:3, and we've inserted one between the first two. When looking at a pair of cars, I like to divide their speeds by the smaller. So, looking at the 3:4 case, I think of that as 1:1.33. This means, when the slower one has reached the intersection for the first time (1), the faster has gone one third of the way around the loop beyond. When the slower one has reached for the second time, the faster is now 2.66, and when the slower reaches for the third time, the faster is reaching it for the fourth time. The other cases are familiar from the 1:2:3 situation. As it turns out, to get more than 4 cars with compatible speeds, we have to abandon the 1:3 ratio. Is this a useful explanation?
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Thanks for a good puzzle, plainglazed, I thought this was going to be impossibly complicated. (Notice I'm not touching the question about the value of the game...)
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Yes Plasmid, I figured that out when you posted your answer and accepted 45:42 as being N+2:N, by transforming it into 30:28. For me, it is an example of N+1:N (15:14), (and therefore represents a "cross" rather than an "overtake"--not that this difference matters in the OP; we've both assumed that both crosses and overtakes are permitted at the intersection). And since your suite of six speeds is valid according to my view of the constraints, I figured we would agree on the same solutions. So you'll probably agree with my suite in post #35. This is an interesting puzzle!
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Bhramarraj, As you point out, speed ratios of 4:1 and 6:1 are too big--the faster car overtakes the slower car before the slower one can get to the intersection. However, other speed ratios are possible. Try 3:2 or 4:3 or 5:3. Plasmid and I have different statements of the possibilities, but we agree on what possibilities there are, and there are plenty more than 2:1 and 3:1.
