Comments for ChapterZero http://www.tangentspace.net/cz somewhere near the beginning. Sat, 22 Nov 2008 00:31:09 +0000 http://wordpress.org/?v=2.5 Comment on Testing graphs for connectivity by John http://www.tangentspace.net/cz/archives/2008/11/testing-graphs-for-connectivity/#comment-431465 John Fri, 21 Nov 2008 20:24:21 +0000 http://www.tangentspace.net/cz/?p=1035#comment-431465 Another somewhat related result. In 1980, Babai-Erdos-Selkow used these kinds of properties to develop a graph isomorphism testing algorithm that works almost always and runs in quadratic time. This result should be in any of the books on Random Graphs, but here is the paper where it was originally proved. László Babai, Paul Erdös, Stanley M. Selkow: Random Graph Isomorphism. SIAM J. Comput. 9(3): 628-635 (1980) Another somewhat related result.

In 1980, Babai-Erdos-Selkow used these kinds of properties to develop a graph isomorphism testing algorithm that works almost always and runs in quadratic time.

This result should be in any of the books on Random Graphs, but here is the paper where it was originally proved.

László Babai, Paul Erdös, Stanley M. Selkow: Random Graph Isomorphism. SIAM J. Comput. 9(3): 628-635 (1980)

]]> Comment on Testing graphs for connectivity by John http://www.tangentspace.net/cz/archives/2008/11/testing-graphs-for-connectivity/#comment-431460 John Fri, 21 Nov 2008 20:15:44 +0000 http://www.tangentspace.net/cz/?p=1035#comment-431460 Random Graphs (each edge has probability [tex]p[/tex] of appearing, all events independent) have been studied since the 50s. See some of the books from http://mathworld.wolfram.com/RandomGraph.html For connectivity, the threshold is [tex]p \sim \log n / n [/tex] That is, if [tex]p = c \log n / n[/tex] and [tex]c<1[/tex], then as [tex]n[/tex] goes to infinity the probability [tex]G(n,p)[/tex] has an isolated vertex goes to 1. If [tex]c>[/tex], then as [tex]n[/tex] goes to infinity the probability [tex]G(n,p)[/tex] has a spanning cycle goes to 1. Testing graphs for connectivity is in the domain of Property Testing, and is a recent area of study. See for example Chapter 17 of Alon's book http://www.amazon.com/Probabilistic-Wiley-Interscience-Discrete-Mathematics-Optimization/dp/0470170204/ref=pd_bbs_sr_1?ie=UTF8&s=books&qid=1227298300&sr=8-1 One of the big recent results is the Alon, Shapira paper http://portal.acm.org/citation.cfm?id=1060590.1060611 Random Graphs (each edge has probability p of appearing, all events independent) have been studied since the 50s.

See some of the books from
http://mathworld.wolfram.com/RandomGraph.html

For connectivity, the threshold is p \sim \log n / n

That is, if p = c \log n / n and c&lt;1, then as n goes to infinity the probability G(n,p) has an isolated vertex goes to 1.

If c>, then as n goes to infinity the probability G(n,p) has a spanning cycle goes to 1.

Testing graphs for connectivity is in the domain of Property Testing, and is a recent area of study. See for example Chapter 17 of Alon’s book
http://www.amazon.com/Probabilistic-Wiley-Interscience-Discrete-Mathematics-Optimization/dp/0470170204/ref=pd_bbs_sr_1?ie=UTF8&s=books&qid=1227298300&sr=8-1

One of the big recent results is the Alon, Shapira paper
http://portal.acm.org/citation.cfm?id=1060590.1060611

]]> Comment on Testing graphs for connectivity by Alex http://www.tangentspace.net/cz/archives/2008/11/testing-graphs-for-connectivity/#comment-431350 Alex Fri, 21 Nov 2008 08:24:24 +0000 http://www.tangentspace.net/cz/?p=1035#comment-431350 No, I wasn't. It looks pretty enticing. Thanks! No, I wasn’t. It looks pretty enticing. Thanks!

]]> Comment on Testing graphs for connectivity by Rod Carvalho http://www.tangentspace.net/cz/archives/2008/11/testing-graphs-for-connectivity/#comment-431307 Rod Carvalho Fri, 21 Nov 2008 02:51:58 +0000 http://www.tangentspace.net/cz/?p=1035#comment-431307 Are you acquainted with this survey paper? <a href="http://arxiv.org/abs/cond-mat/0106096" rel="nofollow">Statistical mechanics of complex networks</a> It's not rigorous, but it's an interesting read, and it contains tons of references to prior work on random graph theory and other cool stuff. Are you acquainted with this survey paper?

Statistical mechanics of complex networks

It’s not rigorous, but it’s an interesting read, and it contains tons of references to prior work on random graph theory and other cool stuff.

]]> Comment on Simple question by Rod Carvalho http://www.tangentspace.net/cz/archives/2008/11/simple-question/#comment-429970 Rod Carvalho Sat, 15 Nov 2008 04:39:23 +0000 http://www.tangentspace.net/cz/?p=1027#comment-429970 Correcting the typo: "initially we have $n$ degrees of freedom, we impose the constraint which steals one DOF and end up with $n-1$ DOF. Not very rigorous, but it kind of works." copy+paste sucks Correcting the typo:

“initially we have $n$ degrees of freedom, we impose the constraint which steals one DOF and end up with $n-1$ DOF. Not very rigorous, but it kind of works.”

copy+paste sucks

]]> Comment on Simple question by Rod Carvalho http://www.tangentspace.net/cz/archives/2008/11/simple-question/#comment-429969 Rod Carvalho Sat, 15 Nov 2008 04:37:43 +0000 http://www.tangentspace.net/cz/?p=1027#comment-429969 The subspace of [tex]$R^n$[/tex] consisting of vectors the sum of whose entries is [tex]$0$[/tex] is the nullspace of the map George proposed. Since the rank of the map is [tex]$1$[/tex], then from the rank-nullity theorem, it follows that the dimension of the nullspace is [tex]$n-1$[/tex]. However, that is not how I thought of it at first. We have [tex]$n$[/tex] variables and impose the constraint that their sum is [tex]$0$[/tex]. Thus, we can write [tex]$x_{n} = -\displaystyle\sum_{i=1}^{n-1} x_i$[/tex], from which we can conclude that the subspace of [tex]$R^n$[/tex] which is mapped into zero has dimension [tex]$n-1$[/tex]. In other words, initially we have [tex]$n-1$[/tex] degrees of freedom, we impose the constraint which steals one DOF and end up with [tex]$n-1$[/tex] DOF. Not very rigorous, but it kind of works. The subspace of $R^n$ consisting of vectors the sum of whose entries is $0$ is the nullspace of the map George proposed. Since the rank of the map is $1$, then from the rank-nullity theorem, it follows that the dimension of the nullspace is $n-1$. However, that is not how I thought of it at first.

We have $n$ variables and impose the constraint that their sum is $0$. Thus, we can write

$x_{n} = -\displaystyle\sum_{i=1}^{n-1} x_i$,

from which we can conclude that the subspace of $R^n$ which is mapped into zero has dimension $n-1$. In other words, initially we have $n-1$ degrees of freedom, we impose the constraint which steals one DOF and end up with $n-1$ DOF. Not very rigorous, but it kind of works.

]]> Comment on Simple question by Alex http://www.tangentspace.net/cz/archives/2008/11/simple-question/#comment-429934 Alex Sat, 15 Nov 2008 00:29:52 +0000 http://www.tangentspace.net/cz/?p=1027#comment-429934 Haha George, that's how I did it. But I wonder if I had not come at it in an environment where it was obvious that the Rank-Nullity theorem was the way to go, would I have solved it using the fact that they satisfy one linear constraint? Because it took me forever to recognize that basic fact. I might be going senile. Haha George, that’s how I did it. But I wonder if I had not come at it in an environment where it was obvious that the Rank-Nullity theorem was the way to go, would I have solved it using the fact that they satisfy one linear constraint? Because it took me forever to recognize that basic fact. I might be going senile.

]]> Comment on Simple question by George http://www.tangentspace.net/cz/archives/2008/11/simple-question/#comment-429931 George Sat, 15 Nov 2008 00:21:51 +0000 http://www.tangentspace.net/cz/?p=1027#comment-429931 Consider the linear map from [tex]$R^n$[/tex] to [tex]$R$[/tex] given by [tex]$f(x_1,x_2,\ldots,x_n) = x_1 + x_2 + \dots + x_n$[/tex]. The result follows from the rank-nullity theorem. Consider the linear map from $R^n$ to $R$ given by $f(x_1,x_2,\ldots,x_n) = x_1 + x_2 + \dots + x_n$. The result follows from the rank-nullity theorem.

]]> Comment on Simple question by Alex http://www.tangentspace.net/cz/archives/2008/11/simple-question/#comment-429920 Alex Fri, 14 Nov 2008 23:29:18 +0000 http://www.tangentspace.net/cz/?p=1027#comment-429920 Correcto! how did you get it? Correcto! how did you get it?

]]> Comment on Simple question by Rod Carvalho http://www.tangentspace.net/cz/archives/2008/11/simple-question/#comment-429896 Rod Carvalho Fri, 14 Nov 2008 21:59:00 +0000 http://www.tangentspace.net/cz/?p=1027#comment-429896 Unless I am missing something embarassingly obvious, the answer is n-1. Right? <abbr><em><abbr><em>Rod Carvalhos last blog post..<a href="http://stochastix.wordpress.com/2008/11/09/representing-complex-numbers-as-2x2-matrices/" rel="nofollow">Representing complex numbers as 2×2 matrices</a></em></abbr></em></abbr> Unless I am missing something embarassingly obvious, the answer is n-1. Right?

Rod Carvalhos last blog post..Representing complex numbers as 2×2 matrices

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