This is an attempt to format the subversion log of a one-month period in the following way:

• Beautiful HTML output.
• Compact representation of lots of information
• Usable with a not-so color rich beamer.
• Fully automatic.

Usage

1. Install Ruby (both 1.8 or 1.9 should work).
2. Download svn-shortlog.rb.
3. Open svn-shortlog.rb with your favourite text editor, and configure the config section according to your needs.
4. Doubleclick svn-shortlog.rb
5. Open the generated changelog_....html file with your favourite browser.

Sample Output

Here is a sample output of one month of boost commits into trunk, taken from the public repository. The output is quite information dense, a quick description is in the screenshot:

Issues

Ideas, suggestions, problems? Please post them as a comment here, at the bug tracker.

Credits

This tool is based on the idea from my colleague Christoph Heindl and inspired by Linus’ Kernel shortlog and Gmail.

Thats why I have created The Online Password Encrypter. Here users can enter their desired username and password, and the encrypted key is automatically generated online, without transmitting anything to any server.

Here is an iframe of the file. Click here for full screen.

The Online Password Encrypter is just one single HTML page, it does not depend on any other files. So it is easy to download it, modify and send it around. Feel free do whatever you want with it.

Have fun,
Martin

Thanks to Johan Kivinemi I have just found out how to bring back the excellent legacy subpixel hinting engine. This has a much more crisp hinting, and uses subpixels only where it really is an improvement:

Just open these files in your home directory, and copy the content into them:

~/.fonts.conf

<?xml version='1.0'?>
<!DOCTYPE fontconfig SYSTEM 'fonts.dtd'>
<fontconfig>
  <match target="font">
    <edit name="antialias" mode="assign">
      <bool>true</bool>
    </edit>
    <edit name="hinting" mode="assign">
      <bool>true</bool>
    </edit>
    <edit name="hintstyle" mode="assign">
      <const>hintfull</const>
    </edit>
    <edit name="lcdfilter" mode="assign">
      <const>lcdlegacy</const>
    </edit>
    <edit name="rgba" mode="assign">
      <const>rgb</const>
    </edit>
  </match>
</fontconfig>

~/.Xresources

Xft.antialias:  true
Xft.hinting:    true
Xft.hintstyle:  hintfull
Xft.lcdfilter:  lcdlegacy
Xft.rgba:       rgb

This should work in Ubuntu 8.04, 8.10, and 9.04 too, and makes all fonts much more crisp. Of course, your mileage may vary.

UPDATE: Comparison Screenshots

As promised on reddit, I got back from an awesome snowboard trip so I am able to put up extensive comparison screenshots of the two subpixel hinting engines. Move your mouse over the following images to see the differences. Watch especially out for letters like “m” where the spacing between the lines is very small. You might have to wait a bit for the image to load.

I have used all of the most important fonts that I usually use, and just for fun I have added “Dijkstra”, which just looks cool.

Sans Fonts

Mouse to see the same fonts with the legacy hinter.

Mono Fonts

Mouse to see the same fonts with the legacy hinter.

Zoomed Comparison Screenhots

Here is an excerpt with 400% magnifications. Mouse over the pictures to see the legacy hinter.

Zoomed Sans

Zoomed Mono

Java 1.5 Collections Hierarchy

As a sidenote, have written this post almost a year ago and just found it in my drafts. I have completely forgotten about it! So here it is anyways. And now I remember why I did not post it: becauseit contains only a minor subset of the collections! Thanks Artur Biesiadowski for reminding me about that…

Sourcecode

I use floating point tricks based on my pow() approximation. Basically I just took the pow() formula and for a^b I substitued b with 0.5, then simplified this as much as possible. As it turns out the result is very simple and short. This initial approximation can be easily made more precise with Newton’s method:

    public static double sqrt(final double a) {
        final long x = Double.doubleToLongBits(a) >> 32;
        double y = Double.longBitsToDouble((x + 1072632448) << 31);

        // repeat the following line for more precision
        //y = (y + a / y) * 0.5;
        return y;
    }

Here is a comparison of the performance and accurancy versus the number of repetitions:

With 2 repetitions, the trouble is not worth the effort, as the approximation is already slower than the original Math.sqrt() which is more precise.

Imagine you have some Java code that does lots and lots of computation. All the time intensive calculations is performed by the class SlowCalculator which implements the interface Calculator:

public static interface Calculator {
    public String calculate(int a, String b);
}

public static void main(String[] args) {
    Calculator c = new SlowCalculator();
    // call c.calculate() a lot of times here...
}

You notice that calculate() is often called with the same parameters which lead to the exact same result (SlowCalculator is stateless). This means it is possible to cache values so there’s no need to recompute. Using the generic CachingProxy™ described below, you can create a cached proxy for any class with just one single line of code:

// ...

public static void main(String[] args) {
    Calculator c = new SlowCalculator();
    c = CachedProxy.create(Calculator.class, c);
    // call c.calculate() a lot of times here...
}

That’s it, and the application is blazingly fast again.

UPDATE: Support for null values, transparently handles exceptions, better hash, nullpointer-bugfix.

UPDATE: Here is an article “Memoization in Java Using Dynamic Proxy Classes” that does (almost) exactly the same as this code.

How To Do This

All this sounds nice, but can you do this in java? Turns out you can and it is not that difficult either. The feature that makes it all possible is Dynamic Proxy. With it you can implement interfaces at runtime. You take an interface, create a proxy for it with Proxy.newProxyInstance(...), supply an InvocationHandler that implements the invoke() method, and you are done.

The code for the CachedProxy.create() method is this:

    /**
     * Creates an intermediate proxy object that uses cached results if
     * available, otherwise calls the given code.
     *
     * @param <T>
     *            Type of the class.
     * @param cl
     *            The interface for which the proxy should be created.
     * @param code
     *            The actual calculation code that should be cached.
     * @return The proxy.
     */
    @SuppressWarnings("unchecked")
    public static <T> T create(final Class<T> cl, final T code) {
        // create the cache
        final Map<Args, Object> argsToOutput = new HashMap<Args, Object>();

        // proxy for the interface T
        return (T) Proxy.newProxyInstance(cl.getClassLoader(), new Class<?>[] { cl }, new InvocationHandler() {

            @Override
            public Object invoke(final Object proxy, final Method method, final Object[] args) throws Throwable {
                final Args input = new Args(method, args);
                Object result = argsToOutput.get(input);
                // check containsKey to support null values
                if (result == null && !argsToOutput.containsKey(input)) {
                    // make sure exceptions are handled transparently
                    try {
                        result = method.invoke(code, args);
                        argsToOutput.put(input, result);
                    } catch (InvocationTargetException e) {
                        throw e.getTargetException();
                    }
                }
                return result;
            }
        });
    }

1. First I create a HashMap that is the cache for the return values. I have written a class Args (omitted here) that is as the key to map from the method and the parameters to the cached output.
2. A Proxy is created for the interface cl, with an InvocationHandler that does all the magic.
3. The magic is actually very simple: If there is no cached result already available (line 25), perform the computation (line 26) and store the result in the map, then return the result.

Download

You can get the full code at my github repository:

• CachedProxy

Benchmarks

In my benchmark I can run about 8 million calls per secons via the cached proxy. That’s not too bad, given all the additional overhead with reflection and the HashMap.

Do you know of any way to improve this? Any ideas or suggestions are welcome!

Write a program that generates all two-word anagrams of the string “documenting”. Here’s a word list you might want to use: wordlist.zip.

When you’re done, send the results to selfdocumenting@hotmail.com.

Good luck!

So this caught my interest and i wrote a little entry in Ruby that is 23 lines long with whitespace and very nice to read. But I won’t show you this code until the contest is over, and this is not the reason for this post. The reason is, that the nice version takes about 2 seconds, and somebody else has coded a Python solution that takes only 1 second (I have no idea what his code looks like). This post is about a fast anagram finding algorithm, and how I developed this algorithm. The final result takes about 0.11 seconds.

Algorithm

The most basic algorithm has two phases:

1. Read in the file
2. Build all combinations of two words and compare the letter count with the query.

Building the combinations is usually done with two nested loops and takes O(n^2) runtime. This is slow, so I have added another step in between:

Idea #1: Filter out Candidate Words

The second step is really slow, but it would be a lot faster if it has to handle less words. So I wrote a little filtering step that lets only words through which are made out of the same letters as the query word.

For example when the query is documenting, the word men or go and even too are extracted, even if the number of letters might not match. But that’s not important, what is important is that the number of possible words are reduced a lot, and so the next phase is faster.

Idea #2: Use a Commutative Hashing Function

String comparisons are slow. To common way to find out if the strings coming with tuned is an anagram of the word documenting is to sort the letters and make a comparison, like this:

irb(main):003:0> "documenting".unpack("c*").sort.pack("c*")
=> "cdegimnnotu"
irb(main):004:0> ("coming" + "tuned").unpack("c*").sort.pack("c*")
=> "cdegimnnotu"

The strings are equal, so we have a match. But this comparison is terribly slow! What’s worse, the computations have to be redone for each match. It would be much better to just compare hash values, and find a hash function to quickly check if we might have a match, and only do the string comparison when the hash check matches. The hash has to be good enough that we don’t have too much false positives (hashes are equal but the real comparisons not) to get a speed advantage. So why not just sum up all the letters bytes?

irb(main):005:0> "documenting".sum
=> 1181
irb(main):006:0> "coming".sum + "tuned".sum
=> 1181

Ruby’s String#sum does exactly this. we can now precalculate the sum for each word, and to find a match we just add the two hashes and compare the result to the query’s hash:

irb(main):007:0> query="documenting"; first="coming"; second="tuned"
=> "tuned"
irb(main):008:0> first.sum + second.sum == query.sum
=> true

When this very quick check returns true, we have to do the string comparison to be absolutely sure it is a match. This considerably speeds up the whole program, but it is still O(n^2).

Idea #3: Reformulate Problem

Now here comes the trickiest and coolest part. Since Idea #2 the slowest part is matching the numbers, with still quadratic complexity. But the hard task is not anagram finding any more, we have reduced it to finding two hashes that combined have the same hash as the query. We can reformulate this problem into something completely detached from the anagram problem:

Given a list of numbers, find all combination of two numbers that add up to a given number

When we concentrate on just this problem and ignore the rest, we might come up with a better way of doing things.

I came up with a fast solution, described below. Somebody posted a better solution that is both faster and simpler, if you want just this final solution skip ahead to Idea #4 as the following description is outdated.

It clearly looks stupid to just try all combinations to add the numbers.
So lets sort them first. Quicksort is fast, especially with numbers, so no worries here. Now consider a list of numbers like this example:

1   3   7   10   10   12   17   20   22   23   24   24   25   26   30

Find all the combinations of two number that add up to 27. They are

• 1 + 26 = 27
• 3 + 24 = 27
• 7 + 20 = 27
• 10 + 17 = 27
• 10 + 17 = 27 (a second time)

You can detect a pattern here: the first number always increases, the second number always decreases! We can now formulate an algorithm for this:

We can have two pointers to the array, one starting from the left side, the other starting from the right side. When the numbers behind the pointers add up to a bigger result than the query (e.g. 1 + 30 = 31), we decrease the right pointer to find a smaller combination (1 + 26 = 27). When the sums are too small (1 + 25 = 26), we move the left pointer to the right (3 + 25 = 28).

This way we walk through the whole array in O(n) time and the sum of the pointers is always kept as close the the desired result as possible. When the pointers meet each other, we can stop the whole process or otherwise we would just reverse the words.

This algorithm gets a bit more complicated when you consider that we might have lots of numbers in it that are equal, whenever this happens you have to fall back into an O(n^2) matching algorithm for just this section.

Idea #4: Use Hash directly

UPDATE Scrap the implementation in idea #3. A blog post here from a reader of this article posted a way to do this really in O(n), without any sorting which is O(n*log(n)). The idea is to use a hashmap that maps from the hash key of the word to its matches:

M = {}
S = the target sum
for each element e in the list
      if M[S-e] exists? (e,S-e) is a pair
      add e to the M

Just use a Hashmap that maps from the cummulative hash of a word to a list of words that have the same hash. Whenever a new word is added, get the list of words that is stored under query.sum - current_word.sum. When the hashes are the same we just have to create a list of all the matches under this key, and check each of the matches sequentially for equality. This is just normal hash collision handling through a linked list. That’s very simple and works like a charm.

I have revised the code, it got both simpler and faster. That’s a win-win situation, wohoo!

The Sourcecode

I hope the code is understandable now with the above explanation. If you have any questions or ideas, please share them here!

#!/usr/bin/ruby

# created by Martin Ankerl http://martin.ankerl.com/

class String
	# creates an array of characters
	def letters
		unpack("c*")
	end
end

class Array
	# converts an array of letters back into a String
	def word
		pack("c*")
	end
end

query = "documenting"
query_letters_sorted = query.letters.sort
txt = File.read('wordlist.txt').downcase

# to quickly check if a letter is part of the query word
used_letters = Array.new(256, nil)
query_letters_sorted.each do |letter|
	used_letters[letter] = true
end

# Maps from cummulative hash of a word to a list of words that have this hash code.
hashToWords = Hash.new do |hash, key|
	hash[key] = Array.new
end

query_hash = query.sum

prev = 0
txt_size = txt.size
separator = 10
idx = txt.index(separator, prev)
while prev < txt_size

	letter_idx = prev

	# no need to check end of word because it is \n
	# which is not part of the word anyways
	while used_letters[txt[letter_idx]]
		letter_idx += 1
	end

	# ignore word if the above quick check fails
	if letter_idx == idx
		word = txt[prev, idx-prev]

		# check all key matches
		key = word.sum
		hashToWords[query_hash - key].each do |other_word|
			if (word.letters + other_word.letters).sort == query_letters_sorted
				puts "#{word} #{other_word}"
				puts "#{other_word} #{word}"
			end
		end

		# insert word
		hashToWords[key] << word
	end

	prev = idx + 1

	# no need to check end of file because we have to end with new line
	idx = txt.index(separator, prev)
end

When you rewrite the algorithm in C++ or Java or Python I am sure it will be faster than this one. But this is not the point of this post. The point is, “The Best Optimizer is between Your Ears” (Michael Abrash, Graphics Programming Black Book).

Have fun!

You are asked to sort a collection of records. The records are stored in a file on the disk, and the size of the file is much bigger than available memory you can work with. Assume a record size is 10 bytes and file size is 1GB. you read and write data to/from disk in units of pages where each page is 1KB big. You have few (less than 10) available pages of physical memory. Assume you are given a method to compare two records.

Here is how I would approach this problem. It was just a quick direct writeup of the thoughts that popped up in my mind:

1. Easiest solution: use Unix’s sort tool, with -S option to specify max memory usage.
2. If this is not allowed, the second easiest solution is to implement a simple Merge sort, and use multiple passes with one temporary file until everything is sorted.
3. If speed is an issue, you can have a pre-pass before the merge sort that loads as much data junks into memory as possible, sorts this junk with Quicksort, and write each sorted junk back into the temporary file.
4. If you want to go even faster, you can go crazy with 10 passes at once and use a sorting network to merge these 10 streams. Here comes the point where you should benchmark before coding, because caching, harddisk properties etc. might have an enormous influence.

I believe this is a quite reasonable approach. It is important to have multiple solutions for a problem to select from, then you can choose the simplest one that works. This is the Do The Simplest Thing That Could Possibly Work (DTSTTCPW) – Principle. This approach is very important in test driven development, and in general a big part of the scientific method (see also Occams’ Razor). If the simplest solution is enough, that’s great! If this is not enough, you can try again with the second simplest solution. Repeat until the problem is solved

I have use this approach for a long time now, and its really productive. Never go for cool and interesting but complex solutions just because they are cool and interesting. These kind of solutions are just for your own ego. There are even 10 commandments for egoless programming.

Screenshot

Unfortunately I don’t really have the time nor the interest to continue development for XDCC-Fetch. Please contact me if you are interested to continue development.

Approximation of pow() in Java

public static double pow(final double a, final double b) {
    final int x = (int) (Double.doubleToLongBits(a) >> 32);
    final int y = (int) (b * (x - 1072632447) + 1072632447);
    return Double.longBitsToDouble(((long) y) << 32);
}

This is really very compact. The calculation only requires 2 shifts, 1 mul, 2 add, and 2 register operations. That’s it! In my tests it usually within an error margin of 5% to 12%, in extreme cases sometimes up to 25%. A careful analysis is left as an exercise for the reader. This is very usable for in e.g. metaheuristics or neural nets.

I use Linux, Java 1.6.0-b105 with the server VM, and execute the benchmark with this command:

sudo nice -n -20 java -cp . -server PowTest

The approximation is 27 times faster than Math.pow() on my Pentium-M. On a Pentium 4 it is 41 times faster. Unfortunately, microbenchmarks are difficult to do in Java, so your mileage may vary. You can download the benchmark PowTest.java and have a look, I have tried to prevent overoptimization while still having a low overhead.

Approximation of pow() in C and C++

double pow(double a, double b) {
    int tmp = (*(1 + (int *)&a));
    int tmp2 = (int)(b * (tmp - 1072632447) + 1072632447);
    double p = 0.0;
    *(1 + (int * )&p) = tmp2;
    return p;
}

Compiled on my Pentium-M with gcc 4.1.2:

gcc -O3 -march=pentium-m -fomit-frame-pointer -fno-strict-aliasing

This version is 7.8 times faster than pow() from the standard library.

WARNING! you HAVE to use the -fno-strict-aliasing option, or this does not work!

Approximation of pow() in C#

Jason Jung has posted a port of the this code to C#:

public static double PowerA(double a, double b) {
  int tmp = (int)(BitConverter.DoubleToInt64Bits(a) >> 32);
  int tmp2 = (int)(b * (tmp - 1072632447) + 1072632447);
  return BitConverter.Int64BitsToDouble(((long)tmp2) << 32);
}

How the Approximation was Developed

It is quite impossible to understand what is going on in this function, it just magically works. To shine a bit more light on it, here is a detailed description how I have developed this.

Approximation of e^x

As described here, the paper “A Fast, Compact Approximation of the Exponential Function” develops a C macro that does a good job at exploiting the IEEE 754 floating-point representation to calculate e^x. This macro can be transformed into Java code straightforward, which looks like this:

public static double exp(double val) {
    final long tmp = (long) (1512775 * val + (1072693248 - 60801));
    return Double.longBitsToDouble(tmp << 32);
}

Use Exponential Functions for a^b

Thanks to the power of math, we know that a^b can be transformed like this:

1. Take exponential

   a^b = e^(ln(a^b))

2. Extract b

   a^b = e^(ln(a)*b)

Now we have expressed the pow calculation with e^x and ln(x). We already have the e^x approximation, but no good ln(x). The old approximation is very bad, so we need a better one. So what now?

Approximation of ln(x)

Here comes the big trick: Rember that we have the nice e^x approximation? Well, ln(x) is exactly the inverse function! That means we just need to transform the above approximation so that the output of e^x is transformed back into the original input.

That’s not too difficult. Have a look at the above code, we now take the output and move backwards to undo the calculation. First reverse the shift:

final double tmp = (Double.doubleToLongBits(val) >> 32);

Now solve the equation

tmp = (1512775 * val + (1072693248 - 60801))

for val:

1. The original formula

   tmp = (1512775 * val + (1072693248 - 60801))

2. Perform subtraction

   tmp = 1512775 * val + 1072632447

3. Bring value to other side

   tmp - 1072632447 = 1512775 * val

4. Divide by factor

   (tmp - 1072632447) / 1512775 = val

5. Finally, val on the left side

   val = (tmp - 1072632447) / 1512775

Voíla, now we have a nice approximation of ln(x):

public double ln(double val) {
    final double x = (Double.doubleToLongBits(val) >> 32);
    return (x - 1072632447) / 1512775;
}

Combine Both Approximations

Finally we can combine the two approximations into e^(ln(a) * b):

public static double pow1(final double a, final double b) {
    // calculate ln(a)
    final double x = (Double.doubleToLongBits(a) >> 32);
    final double ln_a = (x - 1072632447) / 1512775;

    // ln(a) * b
    final double tmp1 = ln_a * b;

    // e^(ln(a) * b)
    final long tmp2 = (long) (1512775 * tmp1 + (1072693248 - 60801));
    return Double.longBitsToDouble(tmp2 << 32);
}

Between the two shifts, we can simply insert the tmp1 calculation into the tmp2 calculation to get

public static double pow2(final double a, final double b) {
    final double x = (Double.doubleToLongBits(a) >> 32);
    final long tmp2 = (long) (1512775 * (x - 1072632447) / 1512775 * b + (1072693248 - 60801));
    return Double.longBitsToDouble(tmp2 << 32);
}

Now simplify tmp2 calculation:

1. The original formula

   tmp2 = (1512775 * (x - 1072632447) / 1512775 * b + (1072693248 - 60801))

2. We can drop the factor 1512775

   tmp2 = (x - 1072632447) * b + (1072693248 - 60801)

3. And finally, calculate the substraction

   tmp2 = b * (x - 1072632447) + 1072632447

The Result

That’s it! Add some casts, and the complete function is the same as above.

public static double pow(final double a, final double b) {
    final int tmp = (int) (Double.doubleToLongBits(a) >> 32);
    final int tmp2 = (int) (b * (tmp - 1072632447) + 1072632447);
    return Double.longBitsToDouble(((long) tmp2) << 32);
}

This concludes my little tutorial on microoptimization of the pow() function. If you have come this far, I congratulate your presistence

UPDATE Recently there several other approximative pow calculation methods have been developed, here are some others that I have found through reddit:

• Fast pow() With Adjustable Accuracy — This looks quite a bit more sophisticated and precise than my approximation. Written in C and for float values. A Java port should not be too difficult.
• Fast SSE2 pow: tables or polynomials? — Uses SSE operation and seems to be a bit faster than the table approach from the link above with the potential to scale better when due to less cache usage.

Please post what you think about this!