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Topic: Expected length of longest increasing subsequence (Read 11354 times) 

Eigenray
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Expected length of longest increasing subsequence
« on: Feb 26^{th}, 2009, 2:51am » 
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Let E(n) be the expected value of the length of the longest increasing subsequence of a random permutation of {1,2,3,...,n}. E.g., E(1) = 1, E(2) = 3/2, E(3) = 2. Show that there exist constants 0 < A < B < such that for all n, An < E(n) < Bn

« Last Edit: Feb 26^{th}, 2009, 3:13am by Eigenray » 
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towr
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Re: Expected length of longest increasing subseque
« Reply #1 on: Feb 26^{th}, 2009, 3:10am » 
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Would you prefer something more interesting for A than 0?


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Eigenray
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Re: Expected length of longest increasing subseque
« Reply #2 on: Feb 26^{th}, 2009, 3:14am » 
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...yes.


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Aryabhatta
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Re: Expected length of longest increasing subseque
« Reply #3 on: Mar 14^{th}, 2009, 11:44am » 
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Interesting problem. An attempt at partial solution (> Asqrt(n) part). If F(n) is the expected length of the longest decreasing subsequence, then by symmetry E(n) = F(n) (I hope). Now consider E+F. We can use the following fact: If there are n^{2} + 1 distinct integers, then there is a subsequence of n+1 integers which is monotonic. Thus for every permutation, E + F > A sqrt(n) for some A > 0. This shows that E(n) + F(n) > A sqrt(n) for some A > 0. Upper bound, will have to think about it.


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Eigenray
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Re: Expected length of longest increasing subseque
« Reply #4 on: Mar 20^{th}, 2009, 12:35pm » 
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For the upper bound, it suffices to bound the probability that there exists an increasing subsequence of length k.


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TenaliRaman
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Re: Expected length of longest increasing subseque
« Reply #5 on: Mar 20^{th}, 2009, 6:05pm » 
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*Spoiler* A discussion I had at another forum sometime back  AI

« Last Edit: Mar 20^{th}, 2009, 6:06pm by TenaliRaman » 
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Eigenray
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Re: Expected length of longest increasing subseque
« Reply #6 on: Mar 20^{th}, 2009, 6:50pm » 
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Here's what got me thinking about this: Suppose we have n rectangles, each with a width and height chosen uniformly at random between 0 and 1. Now suppose we stack them, without rotation, into as few piles as possible, with the restriction that if (w,h) is above (W,H), then w < W and h < H. Then the expected number of piles we will need is E(n). Extension 1: What if rotations are allowed? Extension 2: What if we had boxes (i.e., rectangular parallelepipeds) instead, and we wanted to nest them into as few pieces as possible. Then how many would we need?


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ThudnBlunder
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Re: Expected length of longest increasing subseque
« Reply #7 on: Mar 20^{th}, 2009, 10:36pm » 
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on Mar 20^{th}, 2009, 6:50pm, Eigenray wrote: Suppose we have n rectangles, each with a width and height chosen uniformly at random between 0 and 1. ? 
 And suppose further that this were possible. LOL

« Last Edit: Mar 21^{st}, 2009, 6:18am by ThudnBlunder » 
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Eigenray
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Re: Expected length of longest increasing subseque
« Reply #8 on: Mar 21^{st}, 2009, 8:19am » 
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Eh? They're mathematical objects. What does "possible" have to do with anything?

« Last Edit: Mar 21^{st}, 2009, 8:20am by Eigenray » 
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ThudnBlunder
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Re: Expected length of longest increasing subseque
« Reply #9 on: Mar 21^{st}, 2009, 2:04pm » 
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on Mar 21^{st}, 2009, 8:19am, Eigenray wrote:Eh? They're mathematical objects. What does "possible" have to do with anything? 
 But impossible mathematical operations should not be a means to an end.


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towr
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Re: Expected length of longest increasing subseque
« Reply #10 on: Mar 21^{st}, 2009, 2:30pm » 
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on Mar 21^{st}, 2009, 2:04pm, ThudanBlunder wrote:But impossible mathematical operations should not be a means to an end. 
 It's possible up to an arbitrarily close approximation. What's the fuss? Picking a real number uniformly from a finite interval is well defined mathematically. Besides, I never hear you complain about picking a number from a normal distribution, which is just as 'impossible'.


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Eigenray
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Re: Expected length of longest increasing subseque
« Reply #11 on: Mar 21^{st}, 2009, 3:15pm » 
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They're mathematical means to a mathematical end. A sequence of n rectangles is a point in the probability space [0,1]^{2n}. The number of piles required is a function on this space. I was curious about its expected value. I guess I shouldn't have wasted my time thinking about this impossible problem. Instead I will think about plows that move infinitely fast when there's no snow on the ground.


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Obob
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Re: Expected length of longest increasing subseque
« Reply #12 on: Mar 21^{st}, 2009, 6:27pm » 
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on Mar 21^{st}, 2009, 3:15pm, Eigenray wrote:Instead I will think about plows that move infinitely fast when there's no snow on the ground. 
 Zing!


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ThudnBlunder
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Re: Expected length of longest increasing subseque
« Reply #13 on: Mar 28^{th}, 2009, 11:03am » 
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on Mar 21^{st}, 2009, 3:15pm, Eigenray wrote:They're mathematical means to a mathematical end. A sequence of n rectangles is a point in the probability space [0,1]^{2n}. The number of piles required is a function on this space. I was curious about its expected value. I guess I shouldn't have wasted my time thinking about this impossible problem. Instead I will think about plows that move infinitely fast when there's no snow on the ground. 
 No need to be so touchy. You should know I would never knowingly kee separate you from your mathematical universe. If you like your ploughs to have a demonstrable top speed, what model would you use then?

« Last Edit: Jun 30^{th}, 2009, 1:02am by ThudnBlunder » 
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Eigenray
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Re: Expected length of longest increasing subseque
« Reply #14 on: Mar 30^{th}, 2009, 11:15pm » 
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on Mar 28^{th}, 2009, 11:03am, ThudanBlunder wrote: No need to be so touchy. You should know I would never knowingly keep separate you from your mathematical universe. If you like your ploughs to have a demonstrable top speed, what model would you use then? 
 I don't really care how fast the ploughs move; let them go backwards in time if you want. I just think it's funny that you find a probability question objectionable because it involves, of all things, random numbers. Suppose instead that we flip a fair coin (not that those exist either) to determine the binary digits of the lengths and widths. With probability one, after a finite number of flips we will have enough information to be able to tell which of any two given numbers is larger.


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