## Random Thoughts on a Neat Puzzleby robert

Here is a puzzle from Quanta magazine that I spent wayyy too many spare cycles on (and then a lot of cycles afterwards telling everybody the answers to the puzzle and also all the variations that I thought of). (EDIT: I saw that the Quanta article had a “solution” section after I wrote this, but I think it does a horrible job of explaining why randomness helps. Maybe this will be better.). Quote:

I write down two different numbers that are completely unknown to you, and hold one in my left hand and one in my right. You have absolutely no idea how I generated these two numbers. Which is larger? You can point to one of my hands, and I will show you the number in it. Then you can decide to either select the number you have seen or switch to the number you have not seen, held in the other hand, as your final choice. Is there a strategy that will give you a greater than 50 percent chance of choosing the larger number, no matter which two numbers I write down?

Rather than immediately give you the solution, which is technical, let’s introduce two easier versions of this puzzle to get the juices flowing. First, the easiest version:

Puzzle 1. Consider the following game. There are two players, Alice and Bob. Privately, Alice chooses two integers uniformly at random between 1 and 6 (say, by rolling two fair die), places one in each hand randomly, and shows her closed hands to Bob. Bob’s goal is to find the larger number, and he is allowed the following two actions. First, he chooses one of Alice’s hands, and Alice reveals her number to Bob. Then, he can either choose to take the number he has seen, or he can switch to the other hand. If he switches, he must take the number in the other hand. If the two numbers are the same then Bob always wins. Is there a strategy for Bob that will allow him to find the larger number with probability greater than 50%?

Here is a “dumb” solution. Bob picks a hand, ignores the number, and stays. The probability that he wins is the probability that the numbers are different and he chose the larger number, plus the probability that both numbers were the same (note that these events are disjoint!). Summing up, we get the probability that Bob wins is

$\displaystyle \Pr[\text{Numbers are different and Bob chooses larger}] + \Pr[\text{Numbers are the same}]$

$\displaystyle = \frac{30}{36}\cdot \frac{1}{2} + \frac{1}{6} = \frac{42}{72} \approx 57\%$

However, he can do much better. Here is one of many possible strategies: Bob chooses a hand uniformly at random, and receives a number L. If 1 <= L <= 3, then Bob switches. If 4 <= L <= 6, then Bob stays.

To see that this works better, suppose that Bob chooses the hand and receives a value L. The probability that Bob wins is the probability that he chose the smaller number first and switched, or he chose the larger number first and stayed, or both numbers were the same. Clearly, the probability that he gets the smaller number first or the larger number first is 1/2 in both cases. The probability that the smaller number is no more than 3 is 2/3, and symmetrically the probability that the larger number is no less than 4 is 2/3. Finally, the probability that the numbers are the same is 1/6, in which case he always wins. Thus the probability of winning is

$\displaystyle \frac{1}{2}\cdot\frac{2}{3} + \frac{1}{2}\cdot\frac{2}{3} + \frac{1}{6} = \frac{5}{6} \approx 83\%$

This strategy is both intuitive and surprisingly effective. If Alice instead chooses her random integers to be between 1 and k, for some positive k (let’s assume it is even, for simplicity), the above strategy generalizes in the obvious way, with Bob’s probability of success jumping to (exercise if you’re bored)

$\displaystyle \frac{3}{4} + \frac{1}{4k}$

which is guaranteed to be above 3/4 = 75%.

While this game is tilted towards Bob even from the outset, the second strategy above reveals the remarkable power that randomization can have. In Computer Science, randomness tends to have two major applications:

1. In any sort of optimization or problem solving, when you are confronted with a problem for which computing the best answer is hard, but you (via some other knowledge) know that most answers are high quality, then generating an answer uniformly at random will give you a good approximation most of the time. You can often improve the random answers through repeated independent trials: for example, you can take the best answer (i.e. the Monte-Carlo method), take a “majority vote” (in which case you need a concentration of measure phenomenon such as the Chernoff bounds), or perhaps you can somehow breed the results together and get a new answer for which the whole is greater than the sum of its parts (this type of argument often appears in extremal combinatorics, for example in the Rödl nibble). Of course, there is no need to sample answers uniformly: if instead there is an efficiently computable distribution of good answers you can sample from that. Let us call this the sampling regime.
2. In cryptography, randomness is used almost exclusively for unpredictability, as not even the most powerful computer in the world can predict the outcome of a truly random bit. In this guise, randomness is used to help parties perform a computation correctly while maintaining some data privately either from each other, or from some adversaries who have compromised the method of communication. This is the privacy regime.

The strategy outlined above is an excellent example of the efficacy of the sampling regime of randomization. In the rest of this post we will show how randomization can be used to give a strategy to solve the puzzle at the top better than randomly guessing.

Before we get there, however, let’s first make the previous puzzle a bit harder. Introducing

Puzzle 2. Consider the following modification of the previous game. Now, Alice generates her numbers in some way known only to her (perhaps she always generates them randomly, or, perhaps she always picks the numbers 1 and 2 and places them in the left and right hand, respectively). Assume also that Alice always chooses numbers which are distinct. Is there now a strategy for Bob that will allow him to find the larger number with probability greater than 50%?

Notice that because Bob now has no information on how Alice is generating her numbers, the strategy from the previous puzzle will no longer work. However, by a suitable modification of the previous strategy, he can (surprisingly) do better than 1/2.

The key observation is this. Suppose Alice has generated two numbers, and let S be the smaller number and let L be the larger number. Then there are more numbers between S and 6 then there are between L and 6; symmetrically, there are more numbers between 1 and L than there are between 1 and S. Considering the sampling regime above, we arrive at the following strategy:

Strategy: Bob chooses a hand uniformly at random and receives a number N. He then chooses a uniformly random number R between 1 and 6. If R <= N, he stays. Otherwise, he switches.

We can calculate the probability that the above strategy succeeds as follows. The probability that Bob chooses the smaller number, S, first is 1/2, and the probability that he switches is (6 – S)/6. Similarly, the probability that Bob chooses the larger number first is 1/2, and the probability that he stays is L/6. So, the probability that Bob succeeds is

$\displaystyle \frac{1}{2}\cdot \frac{6-S}{6} + \frac{1}{2}\cdot \frac{L}{6} = \frac{1}{2} + \frac{L-S}{6}$

Since L and S are integers and L > S we know L – S is at least one, and so Bob succeeds with probability greater than 1/2 + 1/6 = 4/6. This strategy similarly generalizes: if Alice chooses a number between 1 and k instead of 1 and 6, we get that Bob succeeds with probability at least 1/2 + 1/k > 1/2.

So, despite the fact that Bob has no idea how Alice is choosing her numbers, by cleverly using randomization he can still strictly improve on randomly guessing (it’s not a great improvement, but the fact of it is still fascinating).

Of course, there are caveats. You have to play the game multiple times in order for Bob’s strategy to really quantitatively improve on random guessing — but, this is the same grain of salt that comes with any randomized strategy. In other words, the “gain” from using such a randomized strategy will really only start showing itself once you are allowed to do repeated and independent trials, as described in the sampling regime above.

Now, what about the puzzle that we first discussed? We will state it again, but using our language:

Puzzle 3: Consider the following modification of Puzzle 2. Now, Alice can choose any two distinct real numbers, by way of whatever strategy she wants. Is there a strategy for Bob that will allow him to find the larger number with probability greater than 50%?

The strategy from Puzzle 2 would dictate that Bob selects a hand uniformly at random, chooses a uniformly random real number R, and if R is less than the number he chose he stays, otherwise he switches. Unfortunately, there is no uniform distribution on the real numbers (there are “too many of them” in a certain sense: any attempt to create a distribution on the reals which has the properties of the uniform distribution will violate either the second or third axiom of probability).

To get around this, we will use a bijective mapping from the reals to the real interval [0, 1] that is monotonically increasing — i.e. if x < y then f(x) < f(y) —- then, we choose a random number on [0, 1] (this, fortunately, can be done to any degree of accuracy). There are many possible choices: an obvious one is the logistic function

$\displaystyle f(x) = \frac{e^x}{e^x + 1}$.

The new strategy is as follows: we choose a hand at random, and receive a number N. Compute f(N), and choose a uniformly random number R from the interval [0, 1]. If f(R) <= f(N), then stay. Otherwise, switch. By a similar argument as before, the probability that Bob wins the new game will be

$\displaystyle \frac{1}{2} + \frac{f(L) - f(S)}{2}$,

and since L > S we get that this will always be bounded away from 1/2 (although, it can be bounded away by an arbitrarily small fraction, and so success will crucially depend on the degree of accuracy by which we calculate the random number R).

We can interpret the role of the function f above in a similar way as the role of the uniform distributions in the previous two puzzles. Recall that the intuition in the other puzzles was as follows: Bob chooses a random hand, and if he gets the smaller number S first there are “more” numbers between the number he sees and the highest number, 6, so by choosing a random number R between 1 and 6 and switching if R > S he guarantees that more of the time he will be switching to the larger number (and symmetrically if he gets the larger number first). The problem in the case of the real numbers is that if x is any real number, then there are infinitely many numbers greater  and less than x. The function f, by mapping the entire real line to [0, 1] monotonically, gives us a “measure” for S and L that allows us to re-use this intuition: there are still not more numbers greater than S then there are numbers greater than L, but the numbers greater than S have larger measure (according to f) then the corresponding measure for L.

## Infinity is Weirdby robert

Infinity is weird.

This post is about an odd little thing I learned about involving infinite sets quite recently. First, let’s introduce some notation. We let N = {0, 1, 2, … } denote the set of natural numbers, and Q denote the set of rational numbers (recall a number is rational if we can write it as a fraction of two integers in lowest terms. So 1/2 is rational while  2 is not). If A and B are sets then a function f mapping A to B is one-to-one if everything in B is mapped to by a unique thing in A (so, for every y in B there is at most one element x in A such that f(x) = y), and it is onto if everything in B is mapped to by something in A (so for every y in B there is some element x in A such that f(x) = y). Finally, f is bijective if it is one-to-one and onto. In other words, f is bijective if everything in B is mapped to by a unique element in A.

In elementary set theory we use bijections to define what we mean by the “size” of a set. In other words, two sets A and B have the “same size” (now called cardinality) if there is some bijection f mapping A to B. For example, if A = {1, 2, 3} and B = {a, b, c}, then we can say that A and B have the same size since the function f mapping f(1) = a, f(2) = b, f(3) = c is a bijection.

What do we get by defining “size” in this manner? Well, clearly we recover “size” in the “regular” sense. If A and B are finite sets with a bijection between them and A has 10 elements, then B certainly has 10 elements as well. The nice thing about this definition of size is that it generalizes to infinite sets in a clean way. Once you realize this you can get some remarkable observations. Here is a nice one:

Theorem 1: Let N be the set of natural numbers and Q be the set of rational numbers. Then there is a bijection f from N to Q.

So, even though there are “clearly more” rational numbers than natural numbers, really the two sets have the same size. Do all infinite sets have the same size? Cantor showed that this is false via, famously, the diagonal argument.

Theorem 2 (Cantor’s Theorem): Let N be the set of natural numbers and R be the set of real numbers. Then there is no bijection from N to R.

Now, Theorem 1 tells us that there is a bijection f from N to Q. This gives us a natural way of ordering the set of rational numbers: just order them according to f! That is, we can define Q by

Q = {f(0), f(1), f(2), …}.

In particular, for every natural number x, this gives us a finite set Q(x) defined by

Q(x) = {f(0), f(1), …, f(x-1), f(x)}.

Note that for any x < y we have Q(x) is strictly contained in Q(y), and the union of Q(x) over all natural numbers x gives us every rational number! In the language of order theory the sequence of sets

{}, Q(0), Q(1), …, Q(n), …

yields an infinite chain in the lattice of all subsets of rational numbers. Moreover, this infinite chain is countable: there is a bijection between it and the natural numbers (just map any natural x to the set Q(x)).

Now, let us consider different subsets of rational numbers. For any real number t, define the set P(t) to be the collection of all rational numbers x < t.

Now things are getting interesting. Like before, for any pair of real numbers t, u with t < u we have that P(t) is strictly contained in P(u). In the language of order theory the collection

{P(t) : t >= 0}

is an infinite chain in the lattice of all subsets of rational numbers. Also, like before, if we take the union of every set P(t) for every positive real number t, we get Q again. So, we get that there is a natural bijection from non-negative real numbers to this new infinite chain (just map t to P(t)).

But, by combining Theorem 1 and Theorem 2, there is no bijection from the set of rational numbers to the set of real numbers. Why is this weird? Well, if we consider the set of all subsets of rational numbers, we get that

• There is an infinite chain {{}, Q(0), Q(1), …} starting from the empty set that (in the limit) covers all rational numbers, which is also countable — it can be ordered (for all i, j with i < j, Q(i) is strictly contained in Q(j)), and the elements of the chain are bijective with the natural numbers.
• There is another infinite chain {P(t) : t >= 0}, starting from the empty set that (in the limit) covers all rational numbers, which is provably not countable — it can be ordered (for all real numbers t, u with t < u we have P(t) is strictly contained in P(u)), but there is no bijection with the natural numbers.

It is not as paradoxical as it first seems, once you realize that the rational numbers are dense in the real numbers. Moreover, these sets P(t) can (in a loose sense) be used to define the real numbers (this uses the notion of a Dedekind cut). A similar construction was used by John Conway to define the surreal numbers, which is detailed in a fairly entertaining novel by Donald Knuth. But, it is late and I have already ranted for too long. Math is cool.

G’night!

## Moirèby scirvir

Something I always get excited about is when simplistic functions have really interesting side effects, this is more a op art thing, but it’s caused by aliasing from transferring a series of concentric rings from some continuous space to discreet space with really dumb sampling. I’ll get around to actually embedding the shader here but for now paste this link into your browser, because wordpress is failing me right now.

## VBS2 Notepad++ Language (Syntax Highlight)by Steven

This for anyone else unfortunate enough to be using the ARMA or VBS2 scripting language. Cheers.

syntax highlight

To use:

• copy the text below into an xml file
• select import, and import the xml file you created
• the style tag for each group can be edited
• I recommend changing at least the default style in the “folder&default” tab if you do not use a dark background.
• And by recommend, I mean, change it or you probably wont even see plaintext.

## Better Late Than Never!by Steven

FLAWLESS PHOTOSHOP.

## Giant Title That Broke My Formattingby Emily

Remember the time I worked at a gym. Who could have seen that in my future, eh? So, I’ve picked up some shifts on the front desk to pad my cheque because, contrary to popular opinion, personal training is not a lucrative business. Unfortunately. So I am currently bored out of my tree because it’s 6am at the gym and I’m sitting on my ass serving my only purpose, which is smiling at people as they walk in… thank God for wikipedia. Sooooo, since I’ve thoroughly exhausted my daily wiki (Happy Polish Mother’s Day) I figured writing to youse guys would be more productive than picking my teeth (although I did have loose-leaf tea this morning…). Who knows, maybe it’ll become a regular thang.

So the other day we had an interesting…encounter…at work. Guy comes in to the front desk, seems to be having trouble articulating his problem but from what I can gather he stopped authorizing payments for his gym membership a few months ago. Yet he kept coming to the gym…not much, but nonetheless… Anyways, turns out he wanted us to wave his service charges. Probably not going to happen but our Assistant Manager is rad so she was at least humouring the poor guy. She asks him why he thought we would be able to do that, to which he responds along the lines of, “Well, I don’t have any money. I’m obliged to give everything away.” “Oh?” “Yes. You see, I’m the Lord and Savior. I was sent by God to save man from the asteroids.”

I shit you not. You would think a false prophet would be more concerned about his health…

And this was a good day.

Had a woman come in this morning for a tan (comes in every second morning, like clockwork, 5:30am). She always asks for a towel to put over her face so she has this super tanned body and ghostly white face. I saw her come in this morning so I said, “Good Morning! Nine minutes in the stand up?” She smiled and said, “You’ve got such a good memory!” Of course I remember you, you look like a goblin.

Chances are only three people might read this (maybe 4 if Soucy’s Dad still looks us up once in a while) so it’s a good thing I know my audience. I miss you guys. Bro-cation needed. Will you guys move to Scotland with me? Before you say no, I have two words for you….Eggplant Lasagna.

And I figure it’s only fitting to wish my three favorite computer scientists a joyous Alan Turing’s Birthday:) and if I had even a moderate amount of technical ability I would have been able to figure out how to insert a picture of him shopped with a birthday hat, but I don’t so I couldn’t. So you’ll just have to use your imagination.

## Updateby Steven

Hello internet, sorry for letting the website become so….DERELICT

## Setting up Xinu Development and Back-end in VirtualBoxby Steven

Posting this for anyone unlucky enough to have to set this up, and for anyone desperate enough to be searching the blogosphere for help…

Setting up XINU in VirtualBox

XINU is a teaching tool, as such, users will want to make changes to it quickly and frequently. Using VirtualBox, and open-source virtual machine sandbox program, development and back-end environments can be created and linked on an internal network. The development machine will act as a standardized programming machine. It is a basic Debian install that contains the XINU source code. The back-end machine will boot from VirtualBox’s internal network and run the compiled version of the code.

Installing VirtualBox (Linux / OSX )

• Alternatively, VirtualBox can be downloaded using a software package manager such as aptitude or apt-get. (sudo apt-get install virtualbox)

Installing VirtualBox (Windows)

Obtaining XINU

XINU is available for many different systems and architectures. The current focus of development is to bring it to even more. This guide focuses on the Virtual Machine implementation.

The VM implementation of XINU is available as two “virtual appliances”, essentially pre configured virtual machine settings, in the public FTP of the University of Purdue.

1. Connect using FTP to ftp.cs.purdue.edu

2. Use anonymous login when prompted

3. execute the ‘bin’ command, for binary transfer

4. get ‘xinu-appliance.tar.gz’ this contains the two VirtualBox appliances

5. untar the package

Importing and Configuring XINU

Once the XINU package have been downloaded and extracted, there should be two appliances, develop-end.ova and back-end.ova . Import these two appliances into the newly installed VirtualBox.

Leave all the settings as their defaults, and repeat the import for the second appliance.

In settings, under serial ports, you will want to set BOTH appliances to the same settings, with the addition of checking “Create pipe” for the development machine.

The path setting will vary depending on the operating system you are using. For instance,

in windows, use “\\. \pipe\xinu_com1”. Linux/OSX users will enter something like “/tmp/xinu_com” . This is the link between the two machines. The back-end machine is by default set to boot from the network. Once these settings are edited, you are ready to compile and run XINU!

Running XINU in VirtualBox

XINU is run by first compiling and “uploading” the resulting binary file to a location that the back-end can boot from.

1. Start the develop-end VM

2. When prompted, enter ‘xinu’ for the username, and ‘xinurocks’ for the password

3. Once the system has completed loading, there will be a folder titled xinu-x86-vm

4. cd xinu-x86-vm/compile

5. make clean

6. make

8. XINU has now been placed in the proper location for the back-end to boot when started. However, in order to see the CONSOLE of XINU, you will have to connect using minicom.

9. sudo minicom

10. back in the VirtualBox menu, start the backend appliance. XINU will start, and run it’s main process, which by default will start the shell module. Congratulations, XINU is now running!

glhf

## Codes and Gödelby robert

SHORT POST TIME. Which means I thought of this off-hand (and it’s certainly not new information, but is kind of fun).

Coding theory is concerned with encoding messages in a way so as to minimize their length for transmission over some sort of channel. The mathematical formalization of this goes all the way back to Shannon’s Information Theory, so I’ll give some basics and then mention the RANDOM CONNECTION.

Here’s the idea. We have two parties, Alice and Bob, who are trying to communicate over some sort of digital channel. (For convenience, let’s assume that the channel communicates every message that is sent across without corruption). Alice has a message M that she wants to send, and the message is drawn from some alphabet $\Sigma$. Concretely, let’s assume the message is drawn from the English alphabet

$\displaystyle \Sigma = \{a, b, c, \ldots, x, y, z, \#\},$

where we use # as a placeholder for a blank space. Let $\Sigma^*$ denote the set of all messages we can compose out of the symbols in the alphabet $\Sigma$. For example,

$what\#up\#dog \in \Sigma^*.$

Now, suppose the channel is binary, so it can only send 0s and 1s. Obviously, Alice needs some way to encode her alphabet $\Sigma$ into the alphabet $\{0,1\}$ to send over pressing messages to Bob.

To bring this about, let’s define a binary code to be a function

$\displaystyle C: \Sigma \rightarrow \{0,1\}^*.$

That is, a binary code is any map from our source alphabet to a sequence of bits. Note that if we have a binary code C we can easily extend it to messages (i.e. to elements of $\Sigma^*$) by defining, for any sequence of symbols $\alpha_1 \alpha_2 \cdots \alpha_n \in \Sigma^*$, the map

$\displaystyle C(\alpha_1 \alpha_2 \cdots \alpha_n) = C(\alpha_1) C(\alpha_2) \cdots C(\alpha_n).$

Now, most codes are useless. Indeed, under our above definition, the map $C(\alpha) = 0$ for every English letter $\alpha$ is a code. Unfortunately, if Alice used this code over their channel Bob would have a tough time decoding it.

So, we need some condition that allows us to actually decode the bloody things! We’ll start with a useful type of code called a prefix-free code.

Definition: A binary code $C: \Sigma \rightarrow \{0,1\}^*$ is prefix-free if, for every pair of symbols $\alpha, \beta \in \Sigma$ neither $C(\alpha)$ is a prefix of $C(\beta)$ nor vice-versa.

An example of a prefix-free binary code (for the first four letters of the English alphabet) could be the following:

$\displaystyle C(a) = 0, C(b) = 10, C(c) = 110, C(d) = 111.$

Let’s encode a message with C: if Alice encoded the message $badcab$ via C and sent it to Bob, Bob would receive

100111110010.

Now, the beautiful property of prefix-free codes is the following: Bob can actually decode this message online. That is, he can do the following: iterate through each of the bits in sequence, and store what order they came in. Once his stored bit sequence matches a sequence in the code, he can automatically decode that character and keep going!

To illustrate, Bob first reads a 1 off the string. He convinces himself that 1 is not the code for anything, so he reads the next bit, a 0. He now has the string “10″, which is a code for b. Now, is it possible that this could be the beginning of a code for another letter? NO! Because “10″ is the code for b and is not the prefix of any other code. So Bob can translate the b, and move on.

We define nonsingular codes to be the set of codes that can actually be decoded. After seeing the above example, it’s clear that prefix-free codes are non-singular. However, is it possible for there to be non-prefix-free, non-singular codes? That is, are there codes that are decodable, but require us to read the entire message before we can decode them? (NOTE: These codes are practically useless, from an efficiency point of view. This is just a thought experiment to test the definition.)

The answer is YES, and a natural example are Gödel numberings! Here is how it works: for each letter $\alpha$ in the alphabet $\Sigma$ choose a distinct positive integer $z_\alpha$. Now, to encode a message

$\displaystyle \alpha_1 \alpha_2 \cdots \alpha_n$

let M be the positive integer defined as

$\displaystyle M = 2^{z_1}3^{z_2}5^{z_3}\cdots p_n^{z_n}$

where $p_n$ is the nth prime number. We then send the binary expansion of M as our message.

How does Bob decrypt it? Easily: he reads ALL of M, factors it, and reads off the powers of the exponents: the order of the message is preserved if we read off in order of lowest prime to highest, where the power of the ith prime is the code of the ith symbol in the message. Bob has to read all of the message (and he has to make sure he’s transcribed it correctly), or else he cannot recover any of it! Marvelously useless.

OR IS IT USELESS? Similar ideas lurk under regular RSA encryption which everyone uses a billion times a day without even realizing it (thank you blaggerwebs). If factoring integers is as hard as complexity theorists believe it is, then Alice has just sent Bob a frustratingly uncrackable message.