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.

• Text over the colored background should be easily readable
• Colors should be very distinct
• The number of required colors is not initially known

Naïve Approach

The first and simplest approach is to create random colors by simply using a random number between [0, 256[ for the R, G, B values. I have created a little Ruby script to generate sample HTML code:

# generates HTML code for 26 background colors given R, G, B values.
def gen_html
  ('A'..'Z').each do |c|
    r, g, b = yield
    printf "<span style=\"background-color:#%02x%02x%02x; padding:5px; -moz-border-radius:3px; -webkit-border-radius:3px;\">#{c}</span> ", r, g, b
  end
end

# naive approach: generate purely random colors
gen_html { [rand(256), rand(256), rand(256)] }

The generated output looks like this:

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

As you can see this is quite suboptimal. Some letters are hard to read because the background is too dark (B, Q, S), other colors look very similar (F, R).

Using HSV Color Space

Let's fix the too dark / too bright problem first. A convenient way to do this is to not use the RGB color space, but HSV (Hue, Saturation, Value). Here you get equally bright and colorful colors by using a fixed value for saturation and value, and just modifying the hue.

Based on the description provided by the wikipedia article on conversion from HSV to RGB I have implemented a converter:

# HSV values in [0..1[
# returns [r, g, b] values from 0 to 255
def hsv_to_rgb(h, s, v)
  h_i = (h*6).to_i
  f = h*6 - h_i
  p = v * (1 - s)
  q = v * (1 - f*s)
  t = v * (1 - (1 - f) * s)
  r, g, b = v, t, p if h_i==0
  r, g, b = q, v, p if h_i==1
  r, g, b = p, v, t if h_i==2
  r, g, b = p, q, v if h_i==3
  r, g, b = t, p, v if h_i==4
  r, g, b = v, p, q if h_i==5
  [(r*256).to_i, (g*256).to_i, (b*256).to_i]
end

Using the generator and fixed values for saturation and value:

# using HSV with variable hue
gen_html { hsv_to_rgb(rand, 0.5, 0.95) }

returns something like this:

Golden Ratio

Using just rand() to choose different values for hue does not lead to a good use of the whole color spectrum, it simply is too random.

Here I have generated 2, 4, 8, 16, and 32 random values and printed them all on a scale. Its easy to see that some values are very tightly packed together, which we do not want.

Lo and behold, some mathematician has discovered the Golden Ratio more than 2400 years ago. It has lots of interesting properties, but for us only one is interesting:

[...] Furthermore, it is a property of the golden ratio, Φ, that each subsequent hash value divides the interval into which it falls according to the golden ratio!
-- Bruno R. Preiss, P.Eng.

Using the golden ratio as the spacing, the generated values look like this:

Much better! The values are very evenly distributed, regardless how many values are used. Also, the algorithm for this is extremly simple. Just add 1/Φ and modulo 1 for each subsequent color.

# use golden ratio
golden_ratio_conjugate = 0.618033988749895
h = rand # use random start value
gen_html {
  h += golden_ratio_conjugate
  h %= 1
  hsv_to_rgb(h, 0.5, 0.95)
}

The final result:

You can see that the first few values are very different, and the difference decreases as more colors are added (Z and E are already quite similar). Anyways, this is good enough for me.

And because it is so beautiful, here are some more colors
s=0.99, v=0.99, s=0.25, h=0.8, and s=0.3, v=0.99

Have fun!
Martin

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

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.

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!

Even though there is not much about this search, I have done quite a bit of user interface tweaks to make them easier to use. Here are the new features:

Remember Latest Selection

When you use e.g. the Open Source Search, the previously used operating system (selected by clicking on the button) is automatically used when you just enter the search term end press Return. The selection is highlighted with bold text.

Select Text On Click

Clicking into the text field automatically selects the whole text, so it is easier to reuse the textfield. I have also entered a default value, because from the webanalyzer logs I found out that many people just click on the buttons without entering any text, so now it should be more obvious how this thing works.

No Textfield Changes

The previous version added a text like more:Linux into the search box that was used by the custom search. This clutter is now hidden from the user.

I think the widgets are much more useful now. If you like them, add them to Netvibes, iGoogle, Apple Dashboard, Opera, Windows Live, or Windows Vista here:

• Open Source Software Search Widget
• Java Developer Kit Documentation Widget

I think the widgets are now much more useful. Post if you have any problems!

Try Them

For example, enter Word and click Linux to find open source alternatives for your favorite operating system. The widget remembers what your choice, and the next time you use the search you just have to enter the search term and press enter to search with the same operating system again.

Type concurrent and click 6.0 in the JDK search to find out about the concurrency package in JDK 1.6.0.

Add Them

You can add the engines to Netvibes, iGoogle, Apple Dashboard, Opera, Windows Live, or Windows Vista here:

• Open Source Software Search Widget
• Java Developer Kit Documentation Widget

Are the widgets useful to you? Please tell me!

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!

• Go here: javadoc.ankerl.com

It’s still a draft, so the documentation is surely subject to change. You can also use the search directly from here:

happy hacking!