Code on Zio Ivan

Coffee-Shit

Thu, 02 Jun 2011 12:24:10 +0000

I had it. I’m sick and tired of this nonsense. People, Devs in the specific, should pull their head out of their ass right now and act like adults.

What am I on about? I’m about re-factored, recompiled languages. Stuff like CoffeShit (yeah, Coffee-SHIT - so what?).

Who the fuck cares?

Come on people, are we really saying that the grammar of a language makes you a better/smarter/faster developer? This is BULLSHIT! Grammar has nothing to do with the quality of your code or with your development speed. NOTHING!

Thinking out aloud: a JavaScript based console?

Thu, 05 May 2011 21:45:56 +0000

In the last month or two I started contributing to PhantomJS:

PhantomJS logo

PhantomJS is a headless WebKit with JavaScript API. It has fast and native support for various web standards: DOM handling, CSS selector, JSON, Canvas, and SVG. PhantomJS is an optimal solution for headless testing of web-based applications, site scraping, pages capture, SVG renderer, PDF converter and many other use cases.

The project is really interesting and we are seeing a steady growth in terms of:

From Wordpress to Bloggart

Mon, 11 Oct 2010 22:32:25 +0000

This post has been in my “TODO list” long enough. It’s time to put it in words. Let’s go!

As you probably know, I decided to give up Wordpress.org in favour of Bloggart on App Engine. A great choice so far I must say.

Migration, a proper one

What I wanted to achieve

My aim was to migrate more than 6 years of content (posts and comments) from my previous blog at detronizator.org into a bloggart installation. Why bloggart? Nick Johnson, the original author of Bloggart on Appengine, wrote a series of article on how to build a nice, little, smart blogging app on App Engine. At the same time, work on txty.mobi started to get more and more engaging, making me a lover (and advocate) of Google App Engine. Plus, I was interested in “starting from scratch”: Wordpress.org is a great project, but I just didn’t feel it “mine enough”. My blog must feel really mine: it’s about me after all.

gredirector - 'HTTP 301' through App Engine

Wed, 29 Sep 2010 00:00:29 +0000

When I decided to migrate to this new blog I was confronted by a very important issue: how do I make sure that the (already modest) traffic going to http://www.detronizator.org/* would be redirected to this new URL?

301 Redirection

I started searching and I found this article by Danny Tuppeny on how to use a Google App Engine application to do the trick.

So, what I did was to take his code and started putting it in place for me at http://redirector.ivandemarino.me. But because I’m a Software Developer that likes to make elegant stuff, I noticed that quite few things were missing:

3 o'clock javascript

Sun, 28 Mar 2010 00:00:00 +0000

I was writing some code to react at a textarea.onKeyUp. I take the size of the current textarea.val().length, update an element and do some other stuff.

yawn

The first version of the code looked like:

$('#message').keyup(function(e){
   $('#chars_num').html( new_len );
   $('#sms_num').html( Math.floor($('#message').val().length / 161) +1) );
});

Working good, but was clearly slow: every keystroke was “giving back the cursor” too slowly for a fast typer like me. I went to take a look at twitter, and their text box was WAY FASTER.

Binary Tree Rebuilder

Fri, 05 Feb 2010 00:00:00 +0000

Imagine you have a Binary Tree, with those characteristics:

* Nodes do not respect any order relation - In other words: it's <strong>not</strong> a [Binary Search Tree](http://en.wikipedia.org/wiki/Binary_search_tree) of any kind

* Every node appears <em>once and only once</em> within the tree

A nice Binary Tree

Then, your little brother passes by your desk and, to upset you, deletes the tree from your computer memory/HD (yeah, I know, I’m pathetic at inventing hypothetical situations :-P ).

Fortunately though, you previously did a Pre-Order and an In-Order visit of your tree, and stored the result in an 2 nice array.

Can you rebuild the original tree structure out of this 2 array?

How are you going to rebuild it?

Yes, you can! (Sorry, I couldn't resist). And it's quite easy as well. What you have to do, is the following:

Take the first element of the PreOrder Array and use it as root of a new tree
Find the position of this New Node in the InOrder Array, scanning it from 0 to n-1 (n is the number of Nodes)
IF next element in the PreOrder Array is on the left of the New Node in the InOrder array: call RECURSIVELY this procedure, this time taking into account the portion of InOrder array that goes from 0 to the position of the New Node in the InOrder Array -1.
IF next element in the PreOrder Array is on the right of the New Node in the InOrder array: call RECURSIVELY this procedure, this time taking into account the portion of InOrder array that goes from the position of the New Node in the InOrder Array +1 to n-1.
Return the New Node

By the way, this doesn’t work. To fix it we should be more generic, specifying things a little bit better. Things like:

* Every recursive calls takes into account a portion of the InOrder array; in the case of the first call it's the entire array

* There is going to be as much recursive calls as the number of elements in the PreOrder array

Of course, is a tree what we are talking about here: recursion is a MUST.

Prime Numbers Generator

Sat, 23 Jan 2010 00:00:00 +0000

I believe I don’t have to describe what primes are, what are their properties and what not. This post is more a tribute to geek-ness of 2 friends-and-colleagues (@lucabox) that have fun thinking of algorithms to solve stupid (or less stupid), and always useless problems ;-) .

Optimus Prime :-P - yeah yeah, a stupid joke

Briefing

This code is based on the assumption that we want to generate very very large primes, so it uses unsigned long long to store the values, instead of classical unsigned int. Live with that.

"Bidirectionally multiplied" array

Fri, 22 Jan 2010 00:00:00 +0000

Another small problem before I go to sleep tonight:

There is an array A[N] of N numbers. You have to compose an array Output[N] such that Output[i] will be equal to multiplication of all the elements of A[N] except A[i]. For example Output[0] will be multiplication of A[1] to A[N-1] and Output[1] will be multiplication of A[0] and from A[2] to A[N-1]. Solve it without division operator and in O(n).

Use Backtracking to print all Subsets

Fri, 22 Jan 2010 00:00:00 +0000

I’m “studying” this algorithmic technique: Backtracking. This is an algorithmic approach to problem solution that I trust every good Computer Scientist already uses; but taking a good theo-practical look at it is something better.

A Backtracking Tree

I believe you can find enough informations about it online (the 2 links I provided are more than enough), so I’ll go straight to the problem.

Problem definition

Given an integer n, and the set S = { 1 … n }, calculate (print) all the possible subsets of S. For example, for n = 1, all the possible subsets are { 1 } and { } (empty set). For n = 2, all the possible subsets are: { 1 2 } { 1 } { 2 } { }. In general, for the set made of the first n Integers, the number of possible subsets is 2ⁿ.

Money change problem: Greedy vs. Dyn.Pro.

Sun, 17 Jan 2010 00:00:00 +0000

This is a classical problem of Computer Science: it’s used to study both Greedy and Dynamic Programming algorithmic techniques.

I hate having my pocket full of copper!!! -_-

Definition

Given:

A set of n Denominations D[0…n-1] in ascending order, representing a Monetary Coin System
An money amount A, as input

calculate a solution:

* <strong><code>S[0...n-1]</code></strong>, with <code>0 &lt;= S[i] &lt;= (A/S[i])</code> and <code>0 &lt; i &lt; n-1</code>

where:

* <strong><code>A = Sum<sub>[i=0 -> n-1]</sub> { D[i] * S[i] }</code></strong>

* <strong><code>Min{ Sum<sub>[i=0 -> n-1]</sub> { S[i] } }</code></strong>

In other words

Find the smallest amount of coins to make the given amount.

Calculate abs(int) without branching

Wed, 13 Jan 2010 00:00:00 +0000

For this you need someone to teach it to you: if you made it yourself, then you are a very good Comp-Sci, and you should send your CV to Google ASAP. ;)

Without branching O_o?

Yes, without using any “if ( a < 0 )…”. To do that, you need to refresh how Two’s Complement works, then come back.

What we really need to focus on is that, given a signed int A, the negative of that number is: B = ~A + 1. BUT, we are trying to calculate the Absolute Value, not the negative. So, something like:

Swap the value of two integers without temporary storage

Wed, 13 Jan 2010 00:00:00 +0000

Someone says this is an old lame trick. I think it’s simple and clever use of XOR.

How/Why does it work?

It’s built around the properties of the XOR ^ operator, who has the following properties:

A ^ B = B ^ A (commutative)
A ^ 0 = A
A ^ 1 = ~A
A ^ A = 0

So, you can see how it get’s applied here:

#include <stdio .h>
int main(void)
{
   unsigned int a, b;
   // ... populate somehow "a" and "b"...
   
   printf("a = %d - b = %d\n", a, b);

   a ^= b; // store in "a" the value of "a XOR b"
   b ^= a; // store in "b" the value of "a XOR b XOR b" = "a XOR 0" = "a"
   a ^= b; // store in "a" the velue of "a XOR b XOR a" = "b XOR 0" = "b"

   printf("a = %d - b = %d\n", a, b);
}

Neat.

Pascal's Triangle generator

Mon, 11 Jan 2010 00:00:00 +0000

What’s Pascal’s Triangle? That’s what it is (Wikipedia has all the theory, if you need).

Pascal’s Triangle first 6 rows

The thing I wrote here is a generator of the n-th row of the triangle, that doesn’t use more then the memory needed to store the solution. Instead of allocating a Triangular Matrix, and building every row based on the one above, solution is built in place.

How does it work

The result is generated “filling the row from right to left”. I start initiating the element on the right hand side to ‘1’. Then, I run something like:

String search in O(n + m)

Fri, 08 Jan 2010 00:00:00 +0000

Searching a string inside another string is a very easy task. Given a string A and a string B, start a loop over A from left to right and, for every letter, start an internal loop to see if there is a match with the letters of B.

A ball of string

Pseudo code would look something like this:

function NaiveSearch(string s[1..n], string sub[1..m])
   for i from 1 to n-m+1
      for j from 1 to m
         if s[i+j-1] ≠ sub[j]
            jump to next iteration of outer loop
      return i
   return not found

Time Complexity? O(n * m), where n is the size of A and m is the size of B.

Count bits set in parallel

Thu, 07 Jan 2010 00:00:00 +0000

This time it’s not something I make myself. Indeed, I still can’t “see” it 100%: I got it, but it’s a bit complex.

A cute little lady counting (bits? ;-) )

It’s a method to count the number of bits in a number in O(1), in just 5 lines of code. INHUMAN.

The “human” solutions

Of course, there are methods that look way more easy and, given that the size of a number in memory is “fixed”, the O(1) still stands. For example: 0. Based on the “evenness/oddness” of the number

unsigned int bits_counter_v0(unsigned int x) {
    unsigned int count = 0;
    while ( x != 0 ) {
        // If odd, add 1
        count += (x % 2 == 0) ? 0 : 1;
        x >>= 1;
    }
    
    return count;
}

1. Counting one bit at a time (always the least significant one)

unsigned int bits_counter_v1(unsigned int x) {
    unsigned int count = 0;
    while ( x != 0 ) {
        // If least-significant bit is 1, add 1
        count += (x & 1) ? 1 : 0;
        x >>= 1;
    }
    
    return count;
}

2. Counting 4 bit at a time with max 8 shifts, using an “hashmap” with precalculated results The fact that it can count the bits in “max 8 shifts” has the trade off of the memory used by the hashmap.

unsigned int bits_counter_v2(unsigned int x) {
    unsigned int count = 0;
    // "Hashmap" of the values for the least significant 4 bits
    unsigned int int_to_bits_count[16] = {
        0, // 0  00
        1, // 1  01
        1, // 2  10
        2, // 3  11
        1, // 4  100
        2, // 5  101
        2, // 6  110
        3, // 7  111
        1, // 8  1000
        2, // 9  1001
        2, // 10 1010
        3, // 11 1011
        2, // 12 1100
        3, // 13 1101
        3, // 14 1110
        4  // 15 1111
    };
    
    while ( x != 0 ) {
        // Add the bits count of the least significant 4 bits
        count += int_to_bits_count[ x & 15 ];
        x >>= 4;
    }
    
    return count;
}

Let’s see what some insane people made.

Largest square array of same integers

Thu, 07 Jan 2010 00:00:00 +0000

Tonight it’s a challenging one. Or, better, a problem of which is really difficult to find a good solution in < O(n³). Indeed, it’s a question that an ex-colleague was asked during an interview with Big-G.

The guy, a part from being a REALLY smart guy, is also very humble, and he doesn’t want to be mentioned by name. So, sorry for girls looking for a young, smart, promising young man: you need to find who he is yourself. ;)

Find the non repeating char in O(n) time and O(1) space - v2

Fri, 18 Dec 2009 00:00:00 +0000

My colleague and friend Luca (@lucabox) described a better solution to the problem of "Finding the first non repearing char in a string in O(n) time and O(1) space". It uses smartly the space, making the solution nicer and slicker. Or we are just 2 geeks that need to give more attention to their girlfriends :P

Luca’s solution description

The logic of this solution is based on the usage of an array of unsigned chars. Every char (assumed to be lowecase) has an associated small byte (1 char = 8 bits), where the bits 0x1 and 0x2 (the 2 least significant) represents, respectively, “present once in the input string” and “present multiple times in the input string”. After the input is “scanned” once, and every letter is marked with the correspondent “presence bit” (once, multiple or none), it get’s scanned a second time to find the first char of the input which has the bit “present once” set to “1”.

Personalize your $PS1

Mon, 21 Apr 2008 00:00:00 +0000

Fancy $PS1 setting

The code to make the Bash prompt looks like mine is (~/.bashrc):

case $TERM in
	xterm*|rxvt*)
		TITLEBAR='\[\033]0;\u@\h:\w\007\]'
		;;
	*)
		TITLEBAR=""
		;;
esac;

PS1="${TITLEBAR}\n\[\033[0;0m\][\033[0;32m\]time: \
\[\033[0m\]\[\033[1;31m\]\t\[\033[0m\]\[\033[0;0m\]]-\
[\[\033[0m\]\[\033[0;32m\]host: \[\033[0m\]\
\[\033[1;31m\]\h\[\033[0m\]\[\033[0;0m\]]-\
[\[\033[0m\]\[\033[0;32m\]user: \[\033[0m\]\[\033[1;31m\]\u\
\[\033[0m\]\[\033[0;0m\]]-[\[\033[0m\]\[\033[0;32m\]bash: \
\[\033[0m\]\[\033[1;31m\]\v\[\033[0;0m\]]\n\
\[\033[0;0m\][\[\033[0;32m\]cwd: \[\033[0m\]\[\033[1;31m\]\w\
\[\033[0m\]\[\033[0;0m\]]# "

Thanks to Bash Prompt HOWTO.