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You are not logged in. #1 20121124 02:48:23
Longest increasing subsequenceLet Hi forum! Using the Erdos' lemma I can only deduce that I would appreciate any further ideas! Thanks for your help, Michael #2 20121124 07:19:03
Re: Longest increasing subsequenceHi Naumberg; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #3 20121124 07:36:05
Re: Longest increasing subsequenceHi bobby, #4 20121124 07:41:22
Re: Longest increasing subsequenceYou want a sequence that is next to each other or can we skip? In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #6 20121124 07:47:06
Re: Longest increasing subsequenceAs far as I know the Erdos bound is for integers that are permutations it does not apply to continuous data. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #7 20121124 08:15:49
Re: Longest increasing subsequenceSure. In the lecture we had the following version of Erdos/Szerkeres which applies to continuous data as well: Now define another sequence In our exercise we need to prove that #8 20121124 08:28:01
Re: Longest increasing subsequenceHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #9 20121124 08:39:06
Re: Longest increasing subsequenceHi #10 20121124 08:42:42
Re: Longest increasing subsequenceHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #11 20121124 09:15:17
Re: Longest increasing subsequenceUnfortunately not. But I could scan you the respective page from the notes. Is there a possibility to upload it here? #12 20121124 09:21:05
Re: Longest increasing subsequenceHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #13 20121124 09:30:17
Re: Longest increasing subsequenceMmh, I am really a newbie #14 20121124 09:38:05
Re: Longest increasing subsequenceHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #16 20121124 10:30:38
Re: Longest increasing subsequenceHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #18 20121124 16:06:09
Re: Longest increasing subsequenceHi Naumberg; was a major achievement. I could not find anyway to tie in what they were doing with the sharper bound you are interested in. I would very much like to see your answer when you find it. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #19 20121127 11:16:29
Re: Longest increasing subsequenceHi bobbym! Code:input: sequence x(1),...x(n) output: length Y of an increasing subsequence y(1)<=...<=y(Y) Y = 0 \\ counting the length of the subsequence s = zero array \\ storing the subsequence here \\ go through intervals elements of L_j for j = 1 to m { \\ boolean helper to implement stopping time, i.e. breaking condition for the loop success == 0 while (success == 0) do { \\ go through elements of L_j for k = (j1)*m+1 to j*m { \\ find a larger element which is still small enough if (s(Y) <= x_k <= s(Y)+1/m) { Y = Y+1 \\ length of subsequence ++ s(Y) = x_k \\ store element success == 1 \\ stop searching in L_j \\ and go to next interval } } } } return(Y) Now let us estimate the expectation of $Y$, #20 20121127 18:17:56
Re: Longest increasing subsequenceHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #21 20121128 01:27:21
Re: Longest increasing subsequenceHi naumberg The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment 