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#1
September 9th 07, 05:05 PM
posted to rec.bicycles.racing
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Don't subscribe to CYCLING.TV because they are thieves.
Don't subscribe to CYCLING.TV because they are thieves.
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#2
September 9th 07, 05:32 PM
posted to rec.bicycles.racing
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Don't subscribe to CYCLING.TV because they are thieves.
"Carmine Clamenza" wrote in message
... Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Don't subscribe to CYCLING.TV because they are thieves. Oh, I get it, it's a contest to see who can figure out the minimum number of cut-and-pastes for that many identical lines! Are the blank lines at the end a clue? Or is that to throw us off? --Mike-- Chain Reaction Bicycles www.ChainReactionBicycles.com
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#3
September 9th 07, 06:02 PM
posted to rec.bicycles.racing
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Don't subscribe to CYCLING.TV because they are thieves.
Mike Jacoubowsky wrote:
Oh, I get it, it's a contest to see who can figure out the minimum number of cut-and-pastes for that many identical lines! Well, there's always copy-paste, starting with one line, doubling each iteration, yielding log2(N) * 2, where N is the final number of lines 1 copy 2 paste (2) 3 copy 4 paste (4) 5 copy 6 paste (8) 7 copy 8 paste (16) 9 copy 10 paste (32) 11 copy 12 paste (64) But one can paste twice, yielding log3(N) * 3, improving on this: 1 copy 2 paste (2) 3 paste (3) 4 copy 5 paste 6 paste (9) 7 copy 8 paste 9 paste (27) 10 copy 11 paste 12 paste (81) pasting 3 times, however, yields log4(N) * 4, = (log2(N) / 2) * 4 = log2(N) * 2, so no better than the first method: 1 copy 2 paste 3 paste 4 paste (4) 5 copy 6 paste 7 paste 8 paste (16) 9 copy 10 paste 11 paste 12 paste (64) pasting 4 times, yields log5(N) * 5, which is worst of all. Analytically, one can calculate the optimal n, where n is the number of pastes per cut: d/dn logn(N) * n = d/dn (ln N / ln (n + 1)) * (n + 1) = ln N d/dn [ (n + 1) / ln (n + 1) ] = (ln N) [ 1 / ln (n + 1) - 1 / [ln (n + 1)]^2 ] = 0 solve for n: ln (n + 1) = 0 = n = e - 1 So e - 1 pastes per cut is optimal, but the closest integer is 2, which turns out to be the correct answer. Dan
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#4
September 9th 07, 08:34 PM
posted to rec.bicycles.racing
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Don't subscribe to CYCLING.TV because they are thieves.
Dan Connelly wrote:
solve for n: ln (n + 1) = 0 = n = e - 1 So e - 1 pastes per cut is optimal, but the closest integer is 2, which turns out to be the correct answer. Congratulations, you've won the 2007 rbr Nobel prize for Prolix Mathematics.
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#5
September 10th 07, 04:25 AM
posted to rec.bicycles.racing
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Don't subscribe to CYCLING.TV because they are thieves.
In article
, Dan Connelly wrote: Mike Jacoubowsky wrote: Oh, I get it, it's a contest to see who can figure out the minimum number of cut-and-pastes for that many identical lines! Well, there's always copy-paste, starting with one line, doubling each iteration, yielding log2(N) * 2, where N is the final number of lines 1 copy 2 paste (2) 3 copy 4 paste (4) 5 copy 6 paste (8) 7 copy 8 paste (16) 9 copy 10 paste (32) 11 copy 12 paste (64) But one can paste twice, yielding log3(N) * 3, improving on this: 1 copy 2 paste (2) 3 paste (3) 4 copy 5 paste 6 paste (9) 7 copy 8 paste 9 paste (27) 10 copy 11 paste 12 paste (81) pasting 3 times, however, yields log4(N) * 4, = (log2(N) / 2) * 4 = log2(N) * 2, so no better than the first method: 1 copy 2 paste 3 paste 4 paste (4) 5 copy 6 paste 7 paste 8 paste (16) 9 copy 10 paste 11 paste 12 paste (64) pasting 4 times, yields log5(N) * 5, which is worst of all. Analytically, one can calculate the optimal n, where n is the number of pastes per cut: d/dn logn(N) * n = d/dn (ln N / ln (n + 1)) * (n + 1) = ln N d/dn [ (n + 1) / ln (n + 1) ] = (ln N) [ 1 / ln (n + 1) - 1 / [ln (n + 1)]^2 ] = 0 solve for n: ln (n + 1) = 0 = n = e - 1 So e - 1 pastes per cut is optimal, but the closest integer is 2, which turns out to be the correct answer. If you want the optimal schedule for a final count of n, I think the problem remains unsolved. It goes under the appellation _addition_chains_. -- Michael Press
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