When I find something interesting and new, I post it here - that's mostly programming, of course, not everything.

Saturday, October 16, 2010

Sunday, September 19, 2010

Running with SBT

So, you are tired of your slow and half-dumb ides, ok, what can you do to make your scala dev cycle more agile? Use sbt.

First go to that website, and download the jar. Store it somewhere. Write a shell script or a bat file with the only command:

java -Xmx512M -jar wherever you store the jar/sbt-launch-0.7.4.jar

Go to your project directory and run your sbt there. It will ask you some questions regarding what version of Scala you want to order, the name of your project, the like. Then it will create an sbt project in that directory. Now the project expects the directory structure to follow maven rules of the game. You don't have to. Go to the directory named project, create a directory named build, and within that directory create a scala file Project.scala, that looks something like this:

import sbt._

class CategoriesProject(info: ProjectInfo) extends DefaultProject(info)
// lazy val hi = task { println("Categories in Scala"); None } // that's for fun
override def mainScalaSourcePath = "src" // your source directory
override def mainResourcesPath = "resources" // fiik what is this for

override def testScalaSourcePath = "tests" // your tests directory
override def testResourcesPath = "test-resources" // fiik what is this for
override def outputDirectoryName = "bin" // your output directory

Additionally, check out lib directory in your project; it probably contains already some stuff that sbt believes it needs - that is, scala-compiler.jar, and scala-library. Copy over there the other jars you believe you need. There should be a way to reference them where they are, but I have not figured it out yet.

Now start sbt again.

If you give it a command ~test, it will cycle through compiling the code and running the tests, automatically detecting what to compile and guessing what tests to run. As soon as you save a source file, sbt wakes up and does the job.

So... I love it... as do many other scala people.

Sunday, September 12, 2010

Dealing with Infinities in Your Code - 3

Here I had introduced countable sets, and here I had demonstrated how we can build a union of two countable sets. Now is time for Cartesian products.

Suppose we have two sets, {'A', 'B'} and {1, 2, 3}. Their product will look like this: {('A',1), ('A',2), ('A',3), ('B',1), ('B',2), ('B',3)}. It is pretty easy to build: iterate over the first set, and for each element iterate over the second set. In Scala it is pretty easy:

for (a <- setA;
b <- setB) yield (a, b)

The problem is that if the second component (setB) is infinite, we will never be able to iterate over all elements. Let setB be, for instance, the set of all natural numbers, N. In this case the product setA x setB will be yielding ('A',0), ('A',1), ('A',2),..., and will never switch to pairs like ('B',i). We have to do something about it.

Say, in the case when only one component is infinite, we could think about changing the order of the loop, first iterating over the infinite component.

That won't work actually. We cannot always know for sure if a set is finite or not. Halting problem, you know. So we have to figure out how to do it in a general way, without losing common-sense efficiency for the finite case.

The solution has been known for over 100 years, and is called Kantor Pairing Function.
In this picture: you see this function enumerating pairs of natural numbers in the following order: (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), (3,0)... - you got it.

In our case we have two essential differences: first, the sets we have are not necessarily finite, and second, elements are not retrievable by their "sequential numbers"... meaning, it's possible, but it is horribly ineffecient: to reach element #1000, we need to rescan all the elements from #1 (or rather #0, since natural numbers and array indexes start with 0). So we also do not need a fancy formula for Kantor Pairing Function (see link above), the formula that calculates product elements by their sequential numbers. All we need is a way to iterate over elements in the right order.

Okay, let's first write the factory method for Cartesian products:

def product[X, Y](xs: Set[X], ys: Set[Y]): Set[(X, Y)] = {
val predicate = (p: (X, Y)) => xs.contains(p._1) && ys.(p._2)
kantorIterator(xs, ys),
xs.size * ys.size,

The predicate definition is extracted just for readability purposes. What's important here is kantorIterator that is supposed to work with any combination of finite and infinite countable sets.

Since we do not have direct access to the elements of the sets by their indexes, we will, instead, keep the rows in iterators. Note that we can build a queue of such iterators. Let's call them rows. What the algorithm does is this:
- push row0 to the iterator queue
- yield (row0.next, y0)
- push row1 to the iterator queue
- yield (row0.next, y0) (first iterator from iterator queue)
- yield (row1.next, y1) (second (last) iterator from iterator queue)
- push row2 to the iterator queue
- yield (row0.next, y0) (that is, (x2,y0))
- yield (row1.next, y1) (that is, (x1,y1))
- yield (row2.next, y2) (that is, (x0,y2))

and so on: every time we reach the end of the queue, we push the next row iterator into the queue, reset the vertical iterator, and start all over.

The problem arises when one of the iterators is out of elements.

We have two possible cases.

Case 1. The vertical iterator halts. In this case we stop pushing new iterators into the queue, so that the size of the queue is the size of the vertical component.
Case 2. the horizontal iterator halts. The vertical one may be infinite or not, we have to shift it. E.g. if we build {'A', 'B','C'} x N, we will be producing ('A',0), ('B',0), ('A',1), ('C',0), ('B',1), ('A',2), ('C',1), ('B',2), ('A',3).... Note that after ('C',1) row0 is out of elements, and this row should be dropped from the queue. At the same time the beginning element of the vertical iterator should be shifted, so that the next loop will be not 0,1,2, but 1,2,3. In such a way we go all along the vertical iterator.

Here's the code:
def kantorIterator[X, Y](xs: Iterable[X], ys: Iterable[Y]): Iterator[(X, Y)] =
new Iterator[(X, Y)] {
var iterators: Queue[Iterator[Y]] = Queue()
var xi = xs.iterator
var yi: Iterator[Iterator[Y]] = Iterator.empty
var shift = 0

def next = {
if (!yi.hasNext) {
if (xi.hasNext) {
iterators enqueue ys.iterator
yi = iterators.iterator
xi = xs.iterator.drop(shift)

val yii = yi.next
val y = yii.next
val res = (xi.next, y)

if (!iterators.isEmpty && yii.isEmpty) {
shift += 1


def hasNext = !xs.isEmpty && !ys.isEmpty &&
(xi.hasNext || (!iterators.isEmpty && iterators(0).hasNext))

hasNext covers all the cases of empty components.

I probably have to mention that you won't be always able to iterate over all the subsets of a countable set. If the set is finite, it's okay, its powerset is finite; but if it is not, here you can find Kantor's proof that the powerset of an infinite countable set is not countable.

Dealing with Infinities in Your Code - 2

In this post I had introduced the idea of having countable sets in your code. Next I want to show how we can combine countable sets producing new countable sets.

We have two operations, union and Cartesian product. Let's start with disjoint union - the union of non-intersecting sets.

If we expect our sets to be both finite, there's nothing special here. The predicate checks whether an element belongs to one set or another; the iterator iterates over the first set, then the second.

What if one or both of our sets is/are infinite? The predicate seems to be still okay: we just check whether the element belongs to one of the sets. But the iterator... if the first set is infinite, our naive iterator will never reach the second set. I mean, not in this universe; we will need transfinite numbers readily available, which is well beyond the power of modern computers or most of modern humans. What can we do then? Well, we can alternate over two iterables.

Let's look at the code:
def union[X, X1 <: X, X2 <: X](set1: Set[X1], set2: Set[X2]): Set[X] = {
lazy val parIterable = new ParallelIterable(set1, set2)
lazy val size = if (set1.size == Integer.MAX_VALUE ||
set2.size == Integer.MAX_VALUE) {
} else {
set1.size + set2.size
(x: X) => (x.isInstanceOf[X1] && (set1 contains x.asInstanceOf[X1]))||
(x.isInstanceOf[X2] && (set2 contains x.asInstanceOf[X2]))

What we have here? A new iterable, a new size evaluator, a new predicate. We can ignore the size evaluator; nothing is calculated here until someone requests it. The predicate is obvious; the only new thing is ParallelIterable. We need it to iterate over the union of two countable sets (rather, over two iterables) in a smart way:

class ParallelIterator[X, X1 <: X, X2 <: X](
iterator1: Iterator[X1],
iterator2: Iterator[X2]) extends Iterator[X] {
var i2 : (Iterator[X], Iterator[X]) = (iterator1, iterator2)
def hasNext = iterator1.hasNext || iterator2.hasNext
def next = {
i2 = (i2._2, i2._1)
if (i2._1.hasNext) i2._1.next else i2._2.next

Note that we flip i2, so it changes from (iterator1, iterator2) to (iterator2, iterator1) and back every cycle.

As a result, in the output we just interleave members from the first component with the members of the second component. What is important here is that we do not make any assumptions regarding which iterator is finite and which is infinite; if one component does not have more elements, fine, we continue scanning the other one.

This was probably pretty simple; now we have to deal with Cartesian products, that is, the set of all pairs (x: A, y: B) where x is from the first set and y is from the second set; with no assumptions regarding the finiteness of the said sets.

That would be the next part.

Thursday, September 09, 2010

Dealing with Infinities in Your Code - 1

These days it is not unusual to have an infinite sequence in a computer system. The idea is that while the sequence grows, the system upgrades, and there's no limit unless we make certain assumptions like accepting non-standard arithmetics. Then even sequences that we expect to be finite may turn out to be infinite - like Goodstein's sequence.

Still, it seems like in a foreseeable future we will only encounter countable sets. They are not necessarily infinite; any finite set is traditionally considered countable, that is, equal in size to a subset N, the set of natural numbers. The set of all real numbers, R, is a typical example of an uncountable set (it is a continuum). (Whether there is a set of intermediate size, between N and R, depends on what axiom we assume. If we assume it exists, it exists; if we assume it does not exist, it does not. Quite an eye-opener for people that use to treat Mathematics as a provider of absolute truths, right?)

What's good in a countable set is that (assuming set-theoretical axioms) one can list its element in such a way that any arbitrary element may be eventually reached. (That's due to another axiom, the Axiom of Choice - it says that we can introduce linear order in any set. If we don't assume this axiom - then we cannot.)

When I worked in Google, one of my favorite interview questions was "given an alphabet, write a code that would produce all words that сan be built based on this alphabet". Then I had to explain that when words are built out of alphabet, it does not mean each letter is used at most once. It does not. Then I had to explain that by "producing all words" I mean writing a code that could reach any word that can be built out of the alphabet. For instance, given {A,B}, the sequence "A", "AA", "AAA", etc would be infinite, but it won't constitute a solution: it will never reach letter 'B'.

In my work I limit myself with sets as they are presented in Java or Scala.

One particular problem is that, usually, an implementation of Set is supposed to have a method that returns the set's size. Not sure I know why. Probably it has something to do with array reallocations... some technical stuff. Because see, if you know that your set is infinite, you are safe, you can always return, by the rules of Java, Integer.MAX_VALUE; but what if you do not know? Say you have a function that takes another function, calls it subsequently until it halts (e.g. fails), and records the times it takes each call. What would be the size? We have to solve Halting Problem to know the answer.

On the other hand, isEmpty is a pretty decent method; it could easily belong to Iterable: to check if an Iterable is empty, one has to just get its Iterator and call its hasNext.

Also, it would be good to supply a set with a predicate, similar to what they do in Zermelo-Fraenkel Set Theory: according to the Extensionality Axiom, two sets are equal iff they consist of the same elements. Not having to count to infinity (or futher) to check if an element belongs would be a good advantage. E.g. for the set of all odds the predicate would not have to scan through all odd numbers from 1 to the given one; just check its last bit.

So that's why I came with a set that is defined by its

  • iterable that enumerates the elements;

  • predicate that "efficiently" checks if a value belongs to the set

  • size evaluator that, when called, tries to evaluate the size of the set
  • .

I use Scala for coding these days; hope it is reasonably pretty:

private def setForIterator[X](sourceIterator: => Iterator[X], sizeEvaluator: => Int, predicate: X => Boolean): Set[X] = {
new Set[X] {
override def contains(x: X) = predicate(x)
override def size = sizeEvaluator
override def iterator = sourceIterator filter predicate
override def isEmpty = !iterator.hasNext
override def -(x:X) = requireImmutability
override def +(x:X) = requireImmutability
def map[Y] (f: Functions.Injection[X,Y]) : Set[Y] = f.applyTo(this)
def map[Y] (f: Functions.Bijection[X,Y]) : Set[Y] = f.applyTo(this)
override def filter(p: X => Boolean) = filteredSet(sourceIterator, (x: X) => predicate(x) && p(x))

def setOf[X](source: Iterable[X], sizeEvaluator: => Int, predicate: X => Boolean) =
setForIterator(source.iterator, sizeEvaluator, predicate)

Had to override a bunch of methods declared in trait Set, to avoid any immediate calls of size evaluator or assuming that the size is known.

Here's how, using this class, we can define the set of natural numbers:
val N = setOf(new Iterable[Int] {
def iterator = new Iterator[Int] {
private var i: Int = -1
def hasNext = true
def next = {i += 1; i}
(x: Any) => x.isInstanceOf[Int] && x.asInstanceOf[Int] >= 0)

That's it for now. In the next part I'll talk about how to combine two countable sets. It is an old fact that a union and a product of two countable sets are countable; we will need a code that behaves according to this knowledge.

Tuesday, July 20, 2010

iterable, tree, monad, foundation axiom

Everybody knows now that linear data are bad for performance. But in Java everything is based on Iterable, with the idea, deeply rooted in math, that we can only visit the content of X if we iterate over it (enumerable sets etc), that is, have an epimorphism from N to that X.

Well, that's stupid. All we need is foundation axiom to hold.

In plain words, a tree is good as well, but a tree can be scanned in parallel.

But programmers of the world were trained to write a loop every time they see a plurality of something.

But the loop does not actually mean it should be interpreted sequentially.

If you write

for (x <- container) {

it only means that the job is done inside a monad.

That's why you should think monad, eh. That's why enumerables are XX century. Foundation axiom is way more powerful.

I actually can even prove it, but only for a Boolean base topos.

(oneof the sources)

Sunday, April 18, 2010

How I Installed a TV Antenna

I live in South San Jose; recently we have discovered that now that we have fast Internet, WII and Netflix, we already have access to hundreds of movies, so what do we need a tv for? Local news. Okay, but they are being broadcast for free, in hdtv, just install an antenna on the roof and go.

Started with looking up our area on antennaweb: entered my address and it gave me directions and this map.

The site also say that frequency ranges have colors, and that I have to choose an antenna that has the right colors.

Since there's over 60 miles from my home to San Francisco, the only reasonable choice was ChannelMaster 4228D - they sell it at Fry's.

Turned out that to install it I need more than just this box. I need a pole and the stuff to attach the pole to the roof. Below I explain what I did.

I bought a 5-foot piece of "black pipe" at Lowe's (here on the picture the end of the pipe), a cap for the pipe and the piece which I want to attach to the roof.

You need a cap on top so that rain won't penetrate the pipe.
Before attaching the bottom piece to the roof, I had patched the area below with roof coating, to avoid leaks:

Now we need these things from OSH (electric department and bolts and nuts department) to attach to the pole the guy wire that will hold it.

This is how the top of the pole looks like with all the stuff attached:

I have attached two eye screws to the sides of my roof, and two went straight into the roof:

After I drilled the hole for the screw, I patched it so the rain won't get in:

Now I have to attach the guy wire to the pole and to the eye screws; for this I use clams:

The guy wire does not go all the way through from the top of the pole to the screws. I cut the wire and attached turnbuckles that allow me to tighten the wires later:

This is how it looks with antenna attached to the pole using brackets, and the guy wires tightened with turnbuckles.

Now it's time to get the cable into the house. I bought a 25-foot white coaxial cable, connected it to the antenna's amplifier, got it along the roof and the wall.

I had measured the position of an existing tv socket inside the room, and used a long drill to drill through the wall to get to the socket as close as possible.

I used this bushing for the cable to get through. Bought it at Lowe's.

Had to cut the bushing, or else how would I get the cable through?

When I got the cable through the wall, I applied a good amount of clear caulk to make sure no water gets through.

I was lucky, the cable got exactly where I wanted it.

So all I had to do is connect my tv, scan the 56 channel it found and enjoy the show.

And you know what? It sucks. I don't have a dvr on those channels, so there's no way I can pause it, or get any information regarding what is it about... no recording. And the channels... what nonsense people watch, omg.

So, was it all worth it? Probably. I had fun with my antenna.

Here's a photo I took from a digital channel:

Monday, March 29, 2010

Hidden Monads in Scala

Actually, not monads, just functorial stuff. But programmers seems to be unaware of the fact that a monad is a special kind of functor, so...

In Scala, one can define a functor something like this:

  trait Functor[X] { 
def map[Y](f: X => Y): Functor[Y]

All we need here is map. Lists have map, Options have map... The natural use of a map is to apply it to a function. Like this:

def loop[X,Y](f: Functor[X], a: X=>Y) = f.map(a)

In Scala, loops can work as maps. And if you are a Haskell programmer, you'll immediately recognize your monadic notation:

def loop[X,Y](f: Functor[X], a: X=>Y) = for (x <- f) yield a(x)

These two are exactly the same.

Praise Scala!

Update: see recent blog entry on monads by Luc Duponchel for a lot more details.

Wednesday, March 24, 2010

Topics in Coding Style

While at Google, I believed in two things: peer code review and unacceptability of ascii art in the code.

Peer code review means you are supposed to show your stuff to somebody in your team before submitting it. So, when you want to submit the stuff in the end of the day (yes, all unittests pass), you are supposed to hold on, wait for half a day (the next day), not touching the code, get into a discussion, and then, eventually get through with it. Which means the code should be pretty solid. No way you would go ahead submitting something working but halfway through, or something that is in the process of being formed, in transition, work in progress.

That's bad. I found myself spending several days forming my ideas regarding how I can possibly specify binary output format in JSON, so that the format can be sent From Above to the client when requirements change (say, a new format is eventually implemented by busy/lazy server guys). If I had to submit a fully-implemented, fully thought-out, neat-looking code, it would take a couple of weeks. Thinking, prototyping, discussing, including three days of explaining how the stuff works and what JSON is, and why it is okay to use strings as keys, etc.

That's about pair programming too. Imagine pair theorem proving. Two guys are placed together by a senior mathematician, say, Grisha and Misha, and told to prove a theorem by the end of the week.

What I want to say: get the fuck off programmers' backs. If they want to work together, let them work together. If they want to work alone, let them work alone. If they want to be agile, let them be agile; if they want to be locally-convex, let them be locally-convex. On the other hand, if you, the manager, know better how to write code, how come you cannot produce in a week anything comparable to what a junior programmer can easily concoct in 30 minutes? Just kidding.

Now ascii art. Programming is art. Ascii art is a part of programming art. If a programmer thinks that his or her code reads better formatted the way the programmer formatted it, challenge this, the fact that it is, not the fact that it does not follow Coding Style Guide Laws.

Monday, March 08, 2010

A Brief History Of Single Exit

Many many years ago people were writing their code mostly in FORTRAN, but also in PL/I. They were told by their teachers not to use too many functions or subroutines, because each call and return is very expensive, compared to plain goto statement.

So the brave programmers managed to write functions or subroutines several thousand lines long; these people were considered smart and very important, compared to the junior kind who only could write a hundred or two lines in one function or subroutine.

Then something bad started happening: the code did not work. Or it did work until you start changing it. As a result, two philosophies were born:
1. Do not change anything;
2. Write structured code.

Structured programming consisted of using functions of reasonable sizes with reasonable structures. No jumps into the middle of a loop, no jumps out of loops into another loop or into a conditional statement; no exits out of the middle of your subroutine.

Then, later, it was discovered that it is just goto statement that should be blamed. So people started setting limitations, depending on their taste and creativity.

- do not goto inside another subroutine;
- do not goto into a loop or a condition;
- do not goto upstream.

The idea of totally abandoning goto was considered too extremist: how else can we make sure that we have only one exit out of a subroutine or a function?

Well then, why do we need just one exit? There were two reasons:
- it was really hard to set breakpoints on all the exits scattered over half a thousand lines;
- how about releasing resources, like closing files, etc?

So it was decided, more or less unanimously, that we have to a) maintain some kind of "response" or "result" code, and b) goto the bottom of our subroutine and exit from there, using the "result code" to decide whether we should do something to close up our activity.

Since later goto was more and more discouraged (although you can find a goto in Java libraries source code), the kosher solution was suggested that was keeping some kind of integer shitLevel flag; everywhere in your code you are supposed to check the level, and if it is close to the fan, you should proceed without doing anything.

This ideology penetrated into Pascal, C, C++, and, later, Java. And only later people started thinking about structuring their code better.

Like making it smaller.

Besides, some languages have finally block that can be used to neatly close all the streams that may remain open.

These two changes, much smaller code and finally block make "single exit" ideology meaningless. Why does one need a specific variable to pass around if we can just return it? If the whole method is half a screen, is it really hard to find other exit points? Is not it safer to use finally that will release your resources in any case, as opposed to that bottom of the method that may never be reached. Just look at this:

public final static void doit() {
try {
throw new NullPointerException("npe");
} finally {

Here is basically the same argument, specifically for Java.

Wednesday, January 20, 2010

Dispatch by Return Type: Haskell vs Java

If you are a Haskell programmer, this is all trivial for you; but if you are a Java programmer, you would ask yourself two questions:
  • is it possible?!
  • is it safe?!
Let's start with Java. In Java, we have pretty nice polymorphism which includes method overloading: using the same method name with different lists of parameters:

  • int length(Iterable i);
  • int length(Collection c);
  • int length(String s);
  • int length(T[] array);

Not that we need these specific methods, but for the sake of argument. What happens when compiler/JVM finds the right method to call: it finds the method that matches best the list of parameters, and in the case of confusion will either show an error message or (during runtime) throws an exception.

One thing is impossible, though: defining two methods that differ only in return type, something like

  • int parse(String source);
  • long parse(String source);
  • boolean parse(String source);

In principle, we could, in most cases, deduce which method we want, by the context: if we write

int n = parse("123");

the compiler could guess, of course, what exactly we mean... but no, it is not allowed.

So, a Java programmer, given a method name, expects to always get the same return type (even if signatures are different, we are used to expecting the same return type anyway).

Now, how about generics? What if we define
interface Ubiq {
T parse(String s);

then implement the interface and apply it to the strings?

Now we have a problem. If we define

class UbiqInt implements Ubiq<Integer> {

then we explicitly specify the type, and will be forced to write

int n = ubiqInt.parse("12345");

then all the polymorphisms is gone now.

What if we define a class that is still parameterized by T?

class UbiqImpl<T> implements Ubiq<T> {

Then the question is, how exactly can we have several different implementation of parse(String) in one class? We cannot, and we are stuck.

Now, what is different in Haskell that makes it possible? Let's take a look at Haskell code:

class Ubiq a where
parse :: String → a

instance Ubiq Bool where
parse ('T':cs) = True
parse _ = False

instance Ubiq Integer where
parse ('0':cs) = 2 * parse(cs)
parse ('1':cs) = 2 * parse(cs) + 1
parse _ = 0

check :: Bool → Integer → String
check b n = if b then show n else "no nothing"

tryme = check (parse "Tautology") (parse "1001")

What happens here? We defined a typeclass Ubiq, which is a vague analog of Java interface; and then we have a couple of implementations. But both implementations define known types, Integer and Bool to be implementing Ubiq. So that, from now on, if we have an expression parse(String) in a position where is expected to have an Integer type, then the compiler finds the appropriate implementation and "calls" it when needed. How does it happen? We just know the expected return type, so it is not hard to figure out. In cases when the compiler is confused, like in an expression show(parse("something")), the compiler will need a hint:
*Main> let x = parse("T")

Ambiguous type variable `a' in the constraint:
`Ubiq a' arising from a use of `parse' at :1:8-17
Probable fix: add a type signature that fixes these type variable(s)

(please do not hesitate to write your comments, questions, objections)


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