随机算法 (Fall 2011)/Problem set 2 and Graham's number: Difference between pages

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==Problem 1==
{{Orphan|date=December 2010}}
Some known facts:
'''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.
* Balls-into-bins: <math>m</math> balls are uniformly and independently thrown to <math>n</math> bins. If <math>m=\Theta(n)</math>, then the maximum load is <math>\Theta\left(\frac{\ln n}{\ln\ln n}\right)</math> with high probability.
* Power of two choices: <math>m</math> balls are sequentially thrown to <math>n</math> bins. For each ball, uniformly and independently choose random bins <math>i</math> and <math>j</math> (might be the same bin), and throw the ball to the currently less loaded bin among bin <math>i</math> and bin <math>j</math>. Assuming that <math>m=\Theta(n)</math>, after all <math>m</math> balls are thrown to bins, the maximum load is <math>\Theta\left(\ln \ln n\right)</math> with high probability.


Questions: Assume <math>n</math> balls and <math>n</math> bins. We throw the <math>n</math> balls sequentially.
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]].
# If we throw the first <math>\frac{n}{2}</math> balls uniformly and then throw the rest balls as the way of power-of-two-choices, what is the asymptotic maximum load with high probability?
# For the <math>k</math>th ball, if <math>k</math> is even, we throw the ball to a uniform bin; and if <math>k</math> is odd, we throw the ball as the way of power-of-two-choices. What is the asymptotic maximum load with high probability?


== Problem 2==
==Context==
* Generalize the LazySelect algorithm for the general selection problem: finding the <math>k</math>th smallest element in an array. Choose appropriate parameters to do the job.
* Use the Chernoff bounds instead of Chebyshev's inequality in the analysis of the above algorithm and try to use a smaller number of random samples.


== Problem 3 ==
Ramsey theory is an area of mathematics that asks questions like the following:
;The Monte Carlo method for computing <math>\mathrm{e}</math>


Suppose there are <math>n</math> peoples, each wearing a hat. We collect all their hats and randomly distribute the hats back to them, with each people receiving one hat. We ask for the probability that none of the people receive his/her own hat. We denote this probability by <math>p(n)</math>
{{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?}}


Formally, a '''permutation''' is a bijection <math>\pi:[n]\xrightarrow[\text{1-1}]{\text{onto}}[n]</math>. A '''fixed point''' for permutation <math>\pi</math> is an <math>i\in[n]</math> such that <math>\pi[i]=i</math>. Then <math>p(n)</math> is the probability that a uniformly random permutation has no fixed point. Such permutations are called '''derangement'''.
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".


This derangement problem is taught in undergraduate class Combinatorics at Nanjing University. See the [[组合数学 (Fall 2011)/Sieve_methods#Derangements|lecture notes]] for details. By the Principle of Inclusion-Exclusion (容斥原则), we know that <math>p(n)=\sum_{k=0}^n\frac{(-1)^k}{k!}</math>.
Graham's number comes from a variation on this question.


Recall that due to Taylor's expansion <math>\mathrm{e}=\sum_{k=0}^\infty\frac{(-1)^k}{k!}=p(\infty)</math>.
{{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?}}


Now suppose you are given access to a balckbox <math>derange(n)</math> which given as input a number <math>n</math>, samples a uniform and independent permutation <math>\pi</math> of <math>[n]</math>, and returns true if <math>\pi</math> has no fixed point and false if otherwise.
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.


Develop a randomized algorithm <math>randE(\epsilon,\delta)</math>, whose inputs are <math>0<\epsilon<1</math> and <math>0<\delta<\frac{1}{2}</math>. The returned value of the algorithm should satisfy that
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".
:<math>\Pr[\mathrm{e}-\epsilon\le randE(\epsilon,\delta)\le\mathrm{e}+\epsilon]\ge1-\delta</math>.


* Your algorithm should call <math>derange(n)</math> as subroutine. Try to make both <math>n</math> and the number of times calling <math>derange(n)</math> as small as possible.
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.
* Please given detailed analysis of your algorithm.
 
==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>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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