Monthly Archive for October, 2011

Mistake of the Day : rm -rf ~

Published on 4 months ago in Anu. 0 Comments

Kids these days huh! They never exercise caution before executing the rm -rf command. It’s gotten ingrained into their habit to use it just as commonly as ls. Careless people. And that is exactly what I did! Damn it! So here’s what happened.

I use tmux in which ` is set as the control character. So it’s pretty common for me to reach out to the ` keystroke. Now I was editing a file in vi. I wanted to save the file and move to the previous tmux screen.

:wq followed by “

Now : requires the SHIFT key be pressed. So usually you get used to having a finger over the SHIFT key. I entered the :wq command. But then somehow my fingers also pressed the SPACE and ~ keystrokes. So I ended up doing

:wq ~

Pff. Vi created a new file named ~. Bah. And then by habit, even before my brain could process anything, my fingers typed in

rm -rf ~

:D . In their defense their intention was just to delete the ~ file. But – Disaster! :’(

At this point I must mention that I was in the root directory /. Luckily enough I did not create a file named *. :D .

Popularity: 1% [?]

Project Euler : Frequency Analysis For Message Decryption

Published on 4 months ago in Problems. 0 Comments

Problem 59 of Project Euler brought back memories of Sherlock Holme’s Dancing Men. Holmes had to decrypt a series of messages – the characters of which were stick men figures in varying dancing positions. He computed the frequencies of each dancing figure. Then since ‘e’ is the most used letter in English, he replaced the most frequent dancing figure with ‘e’.

I did the same thing with a difference. The given text in the problem did not have any spaces. This meant that even the spaces had been encrypted. And the most frequent number occurred almost twice as much as the next one. So I replaced that with space. That’s it. The problem was solved.

The password used to encrypt the text is – ‘god’ and the following code decrypts the text and finds the answer in the count variable.


    private List<Distribution<Integer>> parseInputFile(String fileName) throws FileNotFoundException {
        List<Distribution<Integer>> frequencies = Lists.newArrayList(new Distribution<Integer>(), new Distribution<Integer>(), new Distribution<Integer>());

        Scanner scanner = new Scanner(new File(fileName)).useDelimiter(",");
        int count = 0;
        for(int i = 0;  scanner.hasNext(); i++) {
            int num = scanner.nextInt();
            frequencies.get(i % 3).add(num);

            if(i % 3 ==0) num ^= 103;
            else if(i%3 == 1) num ^= 111;
            else if(i%3 == 2) num ^= 100;
            count += num;
            System.out.printf("%c", num);
        }

        System.out.println("Sum = " + count);

        return frequencies;
    }

Popularity: 5% [?]

Groovy : Parse All Soccer Players Info

Published on 4 months, 2 weeks ago in Coding. 0 Comments

I am new to groovy and am still getting used to the scripting way of thing coming from Java. So as a learning exercise I wrote up the following lines to parse information of all the soccer players from ESPN Soccernet. I have used the Jsoup library to get the document and parse it.

def leagues = [
        "http://soccernet.espn.go.com/clubs/_/league/eng.1/english-premier-league?cc=4716",
        "http://soccernet.espn.go.com/clubs/_/league/esp.1/spanish-la-liga?cc=4716",
        "http://soccernet.espn.go.com/clubs/_/league/ita.1/italian-serie-a?cc=4716",
        "http://soccernet.espn.go.com/clubs/_/league/ger.1/german-bundesliga?cc=4716",
        "http://soccernet.espn.go.com/clubs/_/league/fra.1/french-ligue-1?cc=4716",
]

leagues.each {leagueUrl ->
    Utils.getDocument(leagueUrl).select("table[class=tablehead]").get(0).select("td:eq(2)").select("a[href]").each {teamStatsUrl ->
        Utils.getDocument(teamStatsUrl.attr("abs:href")).select("tbody").each {playerGroup ->
            playerGroup.select("td:eq(1)").select("a[href]").each {playerLink ->
                Element playerProfile = Utils.getDocument(playerLink.attr("abs:href")).select("div.profile").get(0)
                String playerName = playerProfile.select("h1").text()

                def profilePrperties = [:]
                playerProfile.select("li").each {item ->
                    String[] itemProperties = item.text().split(":")
                    if(itemProperties.size() == 1) profilePrperties.get("teams", []).add(itemProperties[0])
                    else profilePrperties[itemProperties[0]] = itemProperties[1]
                }
                println playerName + " " + profilePrperties
            }
        }
    }
}

All that Utils.getDocument(url) here does is to call Jsup.connect(url).get() within a loop with number of retries set to 5. The script produces output as follows:

Ramires [Full Name: Ramires, Squad No: 7, Position: Midfielder, Age: 24, Birth Date: Mar 24, 1987, Birth Place: Barra do Piraí, Rio de Janeiro, Brazil, Height: 5' 11'' (1.80m), Weight: 73 kg, teams:[Brazil, Chelsea]]
Frank Lampard [Squad No: 8, Position: Midfielder, Age: 33, Birth Date: Jun 21, 1978, Birth Place: Romford, Height: 6' 0" (1.83m), Weight: 174 lbs (78.7 kg), teams:[England, Chelsea]]
Fernando Torres [Full Name: Fernando Torres, Squad No: 9, Position: Forward, Age: 27, Birth Date: Mar 20, 1984, Birth Place: Fuenlabrada, Madrid, Height: 6' 1'' (1.85m), Weight: 174 lbs (78.7 kg), teams:[Spain, Chelsea]]
John Mikel Obi [Squad No: 12, Position: Midfielder, Age: 24, Birth Date: Apr 22, 1987, Birth Place: Jos, Nigeria, Height: 5' 11'' (1.80m), Weight: 179 lbs (81.3 kg), teams:[Chelsea]]
Raul Meireles [Full Name: Raul Meireles, Squad No: 16, Position: Midfielder, Age: 28, Birth Date: Mar 17, 1983, Birth Place: Porto, Portugal, Height: 1.79m, Weight: 65 kg, teams:[Chelsea, Liverpool, Portugal]]
Branislav Ivanovic [Squad No: 2, Position: Defender, Age: 27, Birth Date: Feb 22, 1984, Birth Place: Sremska Mitrovica, Yugoslavia, Height: 6' 2" (1.88m), Weight: 86 kg, teams:[Serbia, Chelsea]]
Juan Mata [Full Name: Juan Mata, Squad No: 10, Position: Forward, Age: 23, Birth Date: Apr 28, 1988, Birth Place: Burgos, Spain, Height: 1.70m, Weight: 61 kg, teams:[Spain, Valencia, Chelsea, Spain U21]]

Popularity: 2% [?]

Facebook Media Server Failing

Published on 4 months, 3 weeks ago in Miscel. 0 Comments

We’ve been seeing lots of Facebook static file requests failing.

And this issue is slowing down the other websites that have deep Facebook integration. Soccernets css is getting screwed up and even cricinfo is damn to slow.

Here’s a screenshot from my Chrome web-developer bar:

fbImageServerFailure

[UPDATE]

Well, within minutes of this post getting live the things were back to normal. But the fact that there was a failure remains. Discussing among ourselves we came to the conclusion that perhaps the Akamai cdn was having some issues. We suspect s because Twitter was failing as well and the links in question were those of Akamai. The ak prefix in the links above suggest that even Facebook uses Akamai.

Popularity: 1% [?]

Project Euler : Expansion Of Continued Fraction For Square Root Of Two

Published on 4 months, 3 weeks ago in Problems. 0 Comments

Problem 57 of Project Euler is pretty simple forward. It is very easy to see the following relation:

Numerator = Numerator + 2 * Denominator

Denominator = Numerator + Denominator

Simple code and was done. Here’s that:

    public int getCount(int numberOfExpansions) {
        int count = 0;

        BigInteger numerator = BigInteger.valueOf(3);
        BigInteger denominator = BigInteger.valueOf(2);
        while(--numberOfExpansions > 0) {
            BigInteger a = denominator.multiply(NumberUtil.BIG_TWO).add(numerator);
            BigInteger b = numerator.add(denominator);
            numerator = a;
            denominator = b;
            if(NumberUtil.numberOfDigits(numerator) > NumberUtil.numberOfDigits(denominator)) count++;
        }

        return count;
    }

Popularity: 5% [?]

Project Euler : n Digit Integers That Are Also nth Power

Published on 4 months, 3 weeks ago in Problems. 0 Comments

Problem number 63 of Project Euler gives the expression a = b^c and imposes the condition that the number of digits in a is equal to c. An example would be 16807 = 7^5. a, b and c are all positive integers. The problem asks one to find the number of such possibilities.

Solving this problem made me happy. So, I’ll go into the details.

Given the problem, the first thing that comes to mind is the special relationship between the log of a number and the number of digits. The hint is to use logarithms to simplify the problem.

a = b^c

\Rightarrow \log(a) = c \log(b)

But \log(a) = c - f, where f \in (0, 1]

\Rightarrow c - f = c \log(b)

\Rightarrow c \log(\frac{10}{b}) = f

\Rightarrow 0 < \log(\frac{10}{b}) \leq \frac{1}{c}, dividing by c \in Z^+

That’s it! Our problem has been solved. I hope you have already noted that b \in [1, 9]. This means that we can precompute the values of \log(\frac{10}{b}) and store the results in an array. Then for all c \geq 1, just count how many of the precomputed values are smaller than or equal to \frac{1}{c}. The iteration completes when the smallest precomputed value is greater than \frac{1}{c}.

Below is the piece of code I wrote in Java to solve the task.

    public int getCount() {
        m_precomputedValues = new double[9];
        for(int i = 1; i < 10; i++) {
            m_precomputedValues[i-1] = Math.log10(10.0 / i);
        }

        int count = 0;

        int numberOfDigits = 1;
        double val = 1.0 / numberOfDigits;
        while (val > m_precomputedValues[8]) {
            for(int idx = 8; idx >= 0 &amp;amp;&amp;amp; m_precomputedValues[idx] <= val; idx--) count++;
            val = 1.0 / ++numberOfDigits;
        }

        return count;
    }

Popularity: 5% [?]