CaptainEd
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I am boggled both by the interestingness of the problem and the clarity of the solution. Way to go!
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Yes, once I learned to read English, I saw and removed the ones with 3 1's per row. But, it does seem that you could make N as large as you want with a 2 and a 1 in each row, doesn't it? (Or even a single 3). Maybe I still don't understand the OP.
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If I understand the goal right, you want to generate A with probability 2/3 and B with probability 1/3, using coin flips Flip two coins. If both heads, declare B If HT or TH, declare A if both tails, repeat. let E be the expectation of getting A When you flip two coins, half the time you get A, one quarter of the time you get B, and the other quarter you get E. E = 1/2 + E/4 3*E / 4 = 1/2 E = 2/3 Probably don't have to flip a whole lot.
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Don't forget reducing to lowest terms.
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once again, I doubt 4 and 10 are compatible. Also, what does it mean to have a length of 8? The speeds we have are just relative to each other.
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Thank you, plainglazed. Once I finally had the ratio constraints, I was able to enumerate candidates, so the optimality proof is by exhaustion--I tried all smallest speeds from 1 to 39
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Thanks, again, plainglazed and everyone who doesn't let wrong answers get by.
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worse yet, neither 36 nor 39 goes with 20. Neither does 50. Sigh! pesky constraints...
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I love these simply stated problems that keep showing weird interactions. I'm finding it easy to be a critic, and hard to see how to generate "the next candidate" solution.
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I question 15:24 = 5:8. I think the fast one passes the slow one before the slow crosses the fourth time.
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3:5 does work. The ratio is 1 : 1 2/3. Let's speak of the track as two loops. When slow car reaches the intersection first time, the fast car has gone 2/3 around the next loop, so he's ahead of the slow car, in the same loop. WHen slow reaches the second time, the fast guy has gone 3 1/3 times, so he's in the other loop When slow reaches the third time, the fast guy just overtakes him. As far as I can tell, any successful ratio, reduced to lowest terms, must be one of: 2n : 2n + 1 2n : 2n - 1 2n -1 : 2n + 1 In the last case, the fast car overtakes the slow car at the intersection, in the other cases, the fast car crosses the slow car at the intersection (both drivers closing their eyes, I'm sure)
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are you sure 12 and 30 can coexist for 5 cars? Similarly, I doubt the 2nd and 6th car.
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Well, part of what I said was true--"every step brought surprises". Valic, if I understand your objection, they can only avoid rear-ending if they arrive at the intersection at precisely that moment, and the faster overtakes the slower. (We're ignoring the ludicrous situation of all six arriving at the same moment, each overtaking the next...) Phillip, thanks for waking me up. Now that you point out 7 and 10 fail, I"m at a loss again. It looks to me now that, expressed as integers with common factors taken out, any two compatible speeds have to differ by one or two. So 3:5 works, 5:7 works, 7:8 and 7:9 work, but 7:10 does not. (also 9 vs. 16 or 8 vs 18). I think...(and I don't have the six car speeds now, either)
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I'm sure your concern is valid, Valic. I've been making an assumption--that two cars can "cross" as well as "pass" at the intersection. Under that assumption, What a marvelous puzzle, plainglazed! Each step brought surprises, thank you. (Sure hope my answer is right...at least it's better than my first one)
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Thank you thoughtfulfellow. Woops!
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Thank you, Bushindo, for another surprise. I would not have imagined this result!
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ok Swapnil, and your answer crossed the line at 37, so the remaining moves weren't even necessary.
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intersecting the point (18,19) qualifies as crossing the line. So, a move from (13,18) to (23,20) qualifies. Stopping on the point would qualify if both moves (it. the entering and leaving moves) move in the required X direction.
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Superprismatic, please forgive my unintentional rudeness for failing to express my enjoyment with your racing challenge. I DID INDEED like your challenge. I intend interpretation 1--If the line segment of the move does not intersect the line segment of the mark, it does not count as a traversal. So, the move from (16,20) to (19,17) does not round the mark from (18,19)to (18,100). ----- Separate issue: Re-reading my verbose statement of the problem, I see that I made a confusing statement, corrected here: You can specify an acceleration vector (a,b), where a and b can independently be selected from the set (-1,0,1). Then it makes sense to say that you add the acceleration to the velocity in the first part of a move. I made the edit fix in the original post -- SP
