Tuesday, February 21, 2006

RFC: Mozilla Architecture Changes

Dave Liebreich is looking for people to participate in conversations about proposed major changes in the architecture of Mozilla.

Dave provided the following URLs for current discussions about the issues (garbage collection approach and using exceptions) and asks that interested people jump in and participate:
Mozilla Wiki
Benjamin Smedbergs on Exceptions
Benjamin Smedbergs on Garbage Collection
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Should Arab Media Take Lessons from the West?

An excellent debate among journalists from and working in the Arab world, debating the motion "This house believes that the Arab media need no lessons in journalism from the West."

Saturday, February 18, 2006

Bunchball - put interactive stuff in your blog


The idea is you put a code snippet in your web page, blog etc. and then you can upload photos etc. to bunchballs servers and they show up in your page. Also does games, and various other interactive thingies.

Friday, February 17, 2006

Perl IDE Demonstration and Comparison

The San Francisco Perl Mongers are sponsoring a demo of three Perl IDE's on Tuesday, February 28, 2006:

When & Where:
When: Tuesday, February 28, 2006, 8:00 p.m.
Where: Perpetual Entertainment, Fifth floor;
149 New Montgomery Street, San Francisco, CA
RSVP: qw@sf.pm.org - please let us know if you are attending, and if you want thai food, bring some $$

If you would like to attend, please contact Quinn Weaver at the address above - and bring cash if you want Thai food!

Sunday, February 12, 2006

Lets Make A Deal Game Theory

(originally posted 10:08 AM, Feb 11, 2006)

A couple days ago (yes, in a bar) I became intrigued with a fairly well known game problem, here's my version:

  1. There are three doors, one of which is a winner.
  2. You pick one door, but it isn't opened yet.
  3. At this point at least one of the unpicked doors (maybe both) are losers.
  4. A losing unpicked door is opened.
  5. You now have the option of either sticking with your original choice, or switching to the remaining unopened door. Which has a better chance of being the winner?
Here is a Perl script I wrote which will play the game using both the "switch" and "stay" strategies: http://eigenstate.net/misc_scripts/make_a_deal.pl

What do you think?

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12 Feb 2006 - addition:

Here is one way of explaining what is going on:

When the losing choice is revealed, you gain some new information.

At the start, you have equal knowledge of all the choices, so at first every door has 1/n chance (1/3 if there are three doors).

You then divide doors into two groups: one door in group A (your first pick) and the rest of the doors in group B. With three doors group A carries 1/3 of the chances and group B carries 2/3.

When one of the doors in group B is revealed to be a loser you know something new:

You know that the group of chances that was in group B at the start must now be redistributed among the remaining doors in group B.

So if group B had 2/3 of the chances at the start, it *still has* 2/3 of the chances, but those all ride on the one remaining door in group B. So, the remaining door in group B has a 2/3 chance fo winning, whilst the door you originally picked, in group A still has only a 1/3 chance.

What has happend is that the revelation of the loser in group B has told you that all the other doors in group are more likely to be a winner.

If you try it with 10 doors it is still better to switch - to pick any remaining door in group B - than to stay with the door in group A.

Where N is the number of doors, the chance of the first pick winning is 1/N, and the chance of a door in group B after eliminating a loser from group B is: (N-1)/N

It is the -1 that makes swicthing worth it.

We can generalize the problem further by saying that T is the Total number of doors, A is the number of doors in group A, and B is the number in group B before a loser is revealed. Then the chance of a door in group A is A/T (for example, 1/3) and the chance of a door in group B, after the elimination is B - 1 / T.

So if there are 10 door total, and you put 3 doors in group A at the start:

T = 10
A = 3
B = 7

Each group A door has a 1/10 chance (3/10 for the group as a whole.) After eliminating a loser from group B, the group as a whole still has 7/10 chances and thus each remaining group B door has a 6/10 11 2/3% chance of being the winner. (thanks to keith for the correction)

Thursday, February 09, 2006

Unit tests for Mozilla / Firefox coming soon

Dave Liebreich of Mozilla.com has posted the source code for jssh-driver - a unit test framework for use in Mozilla browsers, (e.g. Firefox). This little framework will allow you to have a directory of HTML files, and test if the browser renders them correctly. Basically the test loads the "golden master" versions into the browser and compares the rendered version to the version on disk. (When the browser renders an HTML page it takes it apart and puts it back together - hopefully correctly.)

Thursday, February 02, 2006

US Internet biz profits from oppression

From the Washington Post:

House: Internet Companies Give in to China

The Associated Press
Wednesday, February 1, 2006; 10:37 PM

WASHINGTON -- Lawmakers on Wednesday accused U.S.-based Internet companies of
giving in to pressure from China and helping to censor Web users in violation
of American principles of free speech.

I think it's disgusting that these companies are seeking to profit from cooperation with oppression while simultaneously benefiting from the freedoms afforded them here in the US.

Wednesday, February 01, 2006

More tests, fewer bugs

Mozilla is looking for a few good test-driven engineers...

I visited the Mozilla Corporation offices yesterday, at the invitation of Dave Liebreich, who is heading up the effort to bring more testing to the Mozilla code base. There is a wiki page: http://wiki.mozilla.org/SoftwareTesting describing some projects and ideas. Dave is a good guy doing good work - let's help!