随机算法 (Fall 2011)/Graph Coloring and Graham's number: Difference between pages

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=Graph Colorings=
{{Orphan|date=December 2010}}
A '''proper coloring''' of a graph <math>G(V,E)</math> is a mapping <math>f:V\rightarrow[q]</math> for some integer <math>q</math>, satisfying that <math>f(u)\neq f(v)</math> for all <math>uv\in E</math>.
'''Graham's number''' is a very, very big [[natural number]] that was defined by a man named Ronald Graham. Graham was solving a problem in an area of mathematics called [[Ramsey theory]]. He proved that the answer to his problem was smaller than Graham's number.


We consider the problem of sampling a uniformly random proper coloring of a given graph. We will later see that this is useful for counting the number of proper colorings of a given graph, which is a fundamental combinatorial problem, having important applications in statistic physics.
Graham's number is one of the biggest numbers ever used in a [[mathematical proof]]. Even if every digit in Graham's number were written in the tiniest writing possible, it would still be too big to fit in the [[observable universe]].


Let's first consider the decision version of the problem. That is, given as input a graph <math>G(V,E)</math>, decide whether there exists a proper <math>q</math>-coloring of <math>G</math>. Denote by <math>\Delta</math> the maximum degree of <math>G</math>.
==Context==
* If <math>q\ge \Delta+1</math>, there always exists a proper coloring. Moreover, the proper coloring can be found by a simple greedy algorithm.
* If <math>q=\Delta</math>, <math>G</math> has a proper coloring unless it contains a <math>(\Delta+1)</math>-clique or it is an odd cycle. ([http://en.wikipedia.org/wiki/Brooks'_theorem Brooks Theorem])
* If <math>q<\Delta</math>, the problem is NP-hard.


Sampling a random coloring is at least as hard as deciding its existence, so we don't expect to solve the sampling problem when <math>q<\Delta</math>. The decision problem for the case <math>q=\Delta</math> is also nontrivial. Thus people are interested only in the case when <math>q\ge \Delta+1</math>.
Ramsey theory is an area of mathematics that asks questions like the following:


The following is a natural Markov chain for sampling proper colorings.
{{quote|<p>Suppose we draw some number of points, and connect every pair of points by a line. Some lines are blue and some are red. Can we always find 3 points for which the 3 lines connecting them are all the same color?}}


{{Theorem|Markov Chain for Graph Coloring <math>\mathcal{M}_c</math>|
It turns out that for this simple problem, the answer is "yes" when we have 6 or more points, no matter how the lines are colored. But when we have 5 points or fewer, we can color the lines so that the answer is "no".
:Start with a proper coloring of <math>G(V,E)</math>. At each step:
# Pick a vertex <math>v\in V</math> and a color <math>c\in[q]</math> uniformly at random.
# Change the color of <math>v</math> to <math>c</math> if the resulting coloring is proper; do nothing if otherwise.
}}


For a fixed graph <math>G(V,E)</math>, the state space of the above Markov chain is the set of all proper colorings of <math>G</math> with <math>q</math> colors.
Graham's number comes from a variation on this question.


{{Theorem|Lemma|
{{quote|<p>Once again, say we have some points, but now they are the corners of an ''n''-dimensional [[hypercube]]. They are still all connected by blue and red lines. For any 4 points, there are 6 lines connecting them. Can we find 4 points that all lie on one [[Plane (mathematics)|plane]], and the 6 lines connecting them are all the same color?}}
The followings hold for <math>\mathcal{M}_c</math>.
# Aperiodic.
# The transition matrix is symmetric.
# Irreducible if <math>q\ge \Delta+2</math>.
}}


The followings are the two most important conjectures regarding the problem.
By asking that the 4 points lie on a plane, we have made the problem much harder. We would like to know: for what values of ''n'' is the answer "no" (for some way of coloring the lines), and for what values of ''n'' is it "yes" (for all ways of coloring the lines)? But this problem has not been completely solved yet.
{{Theorem|Conjecture|
#<math>\mathcal{M}_c</math> has mixing time <math>O(n\ln n)</math> whenever <math>q\ge\Delta+2</math>.
# Random sampling of proper graph colorings can be done in polynomial time whenever <math>q\ge\Delta+1</math>.
}}


These two conjectures are still open. People approach them by relax the requirement for the number of colors <math>q</math>. Intuitively, the larger the <math>q</math> is, the more freedom we have, the less dependency are there between non-adjacent vertices.
In 1971, Ronald Graham and B. L. Rothschild found a partial answer to this problem. They showed that for ''n''=6, the answer is "no". But when ''n'' is very large, as large as Graham's number or larger, the answer is "yes".


=Coupling: <math>q\ge 4\Delta+1</math>=
One of the reasons this partial answer is important is that it means that the answer is eventually "yes" for at least some large ''n''. Before 1971, we didn't know even that much.
We couple the two copies of <math>\mathcal{M}_c</math>, <math>X_t</math> and <math>Y_t</math>, by the following coupling rule:
* At each step, let <math>X_t</math> and <math>Y_t</math> choose the same vertex <math>v</math> and same color <math>c</math>.


Note that by this coupling, it is possible that <math>v</math> is recolored to <math>c</math> in both chains, in one of the two chains, or in no chain. We may separate these cases by looking at the Hamming distance between two colorings.
==Definition==


Let <math>d(X_t,Y_t)</math> be the number of vertices whose colors in <math>X_t</math> and <math>Y_t</math> disagree. For each step, for any choice of vertex <math>v</math> and color <math>c</math>, there are three possible cases:
Graham's number is not only too big to write down all of its digits, it is too big even to write in [[scientific notation]]. In order to be able to write it down, we have to use [[Knuth's up-arrow notation]].
* '''"Good move"''' (<math>d(X_t,Y_t)</math> decreases by 1): <math>X_t</math> disagree with <math>Y_t</math> at <math>v</math>, and <math>c</math> is not used in the colors of <math>v</math> in both <math>X_t</math> and <math>Y_t</math>.
* '''"Bad move"''' (<math>d(X_t,Y_t)</math> increases by 1):
* '''"Neutral move"''' (<math>d(X_t,Y_t)</math> is unchanged):


=Path Coupling: <math>q\ge 2\Delta+1</math> =
We will write down a [[sequence]] of numbers that we will call '''g1''', '''g2''', '''g3''', and so on. Each one will be used in an equation to find the next. '''g64 is''' Graham's number.
 
First, here are some examples of up-arrows:
 
* <math>3\uparrow3</math> is 3x3x3 which equals 27. An arrow between two numbers just means the first number multiplied by itself the second number of times.
* You can think of <math>3 \uparrow \uparrow 3</math> as <math>3 \uparrow (3 \uparrow 3)</math> because two arrows between numbers A and B just means A written down a B number of times with an arrow in between each A. Because we know what single arrows are, <math>3\uparrow(3\uparrow3)</math> is 3 multiplied by itself <math>3\uparrow3</math> times and we know <math>3\uparrow3
</math> is twenty-seven. So <math>3\uparrow\uparrow3</math> is 3x3x3x3x....x3x3, in total 27 times. That equals 7625597484987.
* <math>3 \uparrow \uparrow \uparrow 3</math> is <math>3 \uparrow \uparrow (3 \uparrow \uparrow 3)</math> and we know <math>3\uparrow\uparrow3</math> is 7625597484987. So <math>3\uparrow\uparrow(3\uparrow\uparrow3)</math> is <math>3\uparrow \uparrow 7625597484987</math>. That can also be written as <math>3\uparrow(3\uparrow(3\uparrow(3\uparrow . . .(3\uparrow(3\uparrow(3\uparrow3)</math> with a total of 7625597484987 3s. This number is so huge, its digits, even written very small, could fill up the observable universe and beyond.
** Although this number may already be beyond comprehension, this is barely the start of  this giant number.
* The next step like this is <math>3 \uparrow \uparrow \uparrow \uparrow 3</math> or <math>3 \uparrow \uparrow \uparrow (3 \uparrow \uparrow \uparrow 3)</math>. This is the number we will call '''g1'''.
 
After that, '''g2''' is equal to <math>3\uparrow \uparrow \uparrow \uparrow \ldots \uparrow \uparrow \uparrow \uparrow 3</math>; the number of arrows in this number is '''g1'''.
 
'''g3''' is equal to <math>3\uparrow \uparrow \uparrow \uparrow \uparrow \ldots \uparrow \uparrow \uparrow \uparrow \uparrow 3</math>, where the number of arrows is '''g2'''.
 
We keep going in this way. We stop when we define '''g64''' to be <math>3\uparrow \uparrow \uparrow \uparrow \uparrow \ldots \uparrow \uparrow \uparrow \uparrow \uparrow 3</math>, where the number of arrows is '''g63'''.
 
This is Graham's number.
 
==Related pages==
* [[Knuth's up-arrow notation]]
 
[[Category:Mathematics]]
[[Category:Hyperoperations]]
[[Category:Integers]]

Latest revision as of 20:06, 6 April 2017

Template:Orphan Graham's number is a very, very big natural number that was defined by a man named Ronald Graham. Graham was solving a problem in an area of mathematics called Ramsey theory. He proved that the answer to his problem was smaller than Graham's number.

Graham's number is one of the biggest numbers ever used in a mathematical proof. Even if every digit in Graham's number were written in the tiniest writing possible, it would still be too big to fit in the observable universe.

Context

Ramsey theory is an area of mathematics that asks questions like the following:

Template:Quote

It turns out that for this simple problem, the answer is "yes" when we have 6 or more points, no matter how the lines are colored. But when we have 5 points or fewer, we can color the lines so that the answer is "no".

Graham's number comes from a variation on this question.

Template:Quote

By asking that the 4 points lie on a plane, we have made the problem much harder. We would like to know: for what values of n is the answer "no" (for some way of coloring the lines), and for what values of n is it "yes" (for all ways of coloring the lines)? But this problem has not been completely solved yet.

In 1971, Ronald Graham and B. L. Rothschild found a partial answer to this problem. They showed that for n=6, the answer is "no". But when n is very large, as large as Graham's number or larger, the answer is "yes".

One of the reasons this partial answer is important is that it means that the answer is eventually "yes" for at least some large n. Before 1971, we didn't know even that much.

Definition

Graham's number is not only too big to write down all of its digits, it is too big even to write in scientific notation. In order to be able to write it down, we have to use Knuth's up-arrow notation.

We will write down a sequence of numbers that we will call g1, g2, g3, and so on. Each one will be used in an equation to find the next. g64 is Graham's number.

First, here are some examples of up-arrows:

  • [math]\displaystyle{ 3\uparrow3 }[/math] is 3x3x3 which equals 27. An arrow between two numbers just means the first number multiplied by itself the second number of times.
  • You can think of [math]\displaystyle{ 3 \uparrow \uparrow 3 }[/math] as [math]\displaystyle{ 3 \uparrow (3 \uparrow 3) }[/math] because two arrows between numbers A and B just means A written down a B number of times with an arrow in between each A. Because we know what single arrows are, [math]\displaystyle{ 3\uparrow(3\uparrow3) }[/math] is 3 multiplied by itself [math]\displaystyle{ 3\uparrow3 }[/math] times and we know [math]\displaystyle{ 3\uparrow3 }[/math] is twenty-seven. So [math]\displaystyle{ 3\uparrow\uparrow3 }[/math] is 3x3x3x3x....x3x3, in total 27 times. That equals 7625597484987.
  • [math]\displaystyle{ 3 \uparrow \uparrow \uparrow 3 }[/math] is [math]\displaystyle{ 3 \uparrow \uparrow (3 \uparrow \uparrow 3) }[/math] and we know [math]\displaystyle{ 3\uparrow\uparrow3 }[/math] is 7625597484987. So [math]\displaystyle{ 3\uparrow\uparrow(3\uparrow\uparrow3) }[/math] is [math]\displaystyle{ 3\uparrow \uparrow 7625597484987 }[/math]. That can also be written as [math]\displaystyle{ 3\uparrow(3\uparrow(3\uparrow(3\uparrow . . .(3\uparrow(3\uparrow(3\uparrow3) }[/math] with a total of 7625597484987 3s. This number is so huge, its digits, even written very small, could fill up the observable universe and beyond.
    • Although this number may already be beyond comprehension, this is barely the start of this giant number.
  • The next step like this is [math]\displaystyle{ 3 \uparrow \uparrow \uparrow \uparrow 3 }[/math] or [math]\displaystyle{ 3 \uparrow \uparrow \uparrow (3 \uparrow \uparrow \uparrow 3) }[/math]. This is the number we will call g1.

After that, g2 is equal to [math]\displaystyle{ 3\uparrow \uparrow \uparrow \uparrow \ldots \uparrow \uparrow \uparrow \uparrow 3 }[/math]; the number of arrows in this number is g1.

g3 is equal to [math]\displaystyle{ 3\uparrow \uparrow \uparrow \uparrow \uparrow \ldots \uparrow \uparrow \uparrow \uparrow \uparrow 3 }[/math], where the number of arrows is g2.

We keep going in this way. We stop when we define g64 to be [math]\displaystyle{ 3\uparrow \uparrow \uparrow \uparrow \uparrow \ldots \uparrow \uparrow \uparrow \uparrow \uparrow 3 }[/math], where the number of arrows is g63.

This is Graham's number.

Related pages

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