Rethinking my Facebook friend acceptance policy

My Facebook-friend-acceptance policy is this.
Actually more restricted than that. If I don't share my personal detail with you in real life, I don't want to share it with you on Facebook.

The Facebook data team recently published this study which basically tells that if you want to receive interesting diverse news from your social network you should add people you hardly talk to. People with whom we have strong ties are similar minded, so they share things that we'd have shared. If you social network only contains similar minded people it'd become an "echo chamber." The more diverse your network is the more diverse and novel information you'd get from it.

I find the study very interesting. People with weak ties are good for getting interesting information, but if I have too many of them then I'd have to add extra privacy settings on my photo albums and status updates or be extra cautious about things I share. I'm not sure if people use facebook for getting news, it's an added benefit. News or no news, with my friends literally staying all over the world, I'd use Facebook anyway.

latex on mac

http://guides.macrumors.com/Installing_LaTeX_on_a_Mac#First_.28easier.29_option
And I love texmaker, so awesome!

I want to know if you can see Beauty
even when it is not pretty
every day.
And if you can source your own life
from its presence.

I watch the stars from my window sill
The whole world is moving, and I'm standing still

How far I am from home!

How far I am from home, originally uploaded by sheetal_shundori.

Feeling homesick

Surprisingly, all my worst days are actually my best days in disguise.

"Would you tell me, please, which way I ought to go from here?"
"That depends a good deal on where you want to get to," said the Cat.
"I don’t much care where--" said Alice.
"Then it doesn’t matter which way you go," said the Cat.
"--so long as I get SOMEWHERE," Alice added as an explanation.
"Oh, you’re sure to do that," said the Cat, "if you only walk long enough."

I couldn't open file name `*.aux'

If you have more than one bib file, add them like this:
\bibliography{bibliography_1,bibliography_2,bibliography_3}

Open up the .tex file and compile it using Latex (Shift+Apple+L) then compile it using Bibtex (Shift+Apple+B)

Gender Guesser

Gender Guesser Script: http://www.hackerfactor.com/GenderGuesser.php
This script can guess gender from writing sample.
It's not 100% accurate but it predicted me as a female!!
It's so weird that how much information we reveal about ourselves without even thinking about it.

Project Implicit

Project Implicit (https://implicit.harvard.edu/implicit/) is about our conscious and unconscious preferences and biases. For example, most people connote blondes with "dumbness", and tall and handsome with "intelligence"; we don't do it consciously, or we don't want other people to know about our true biases. But even when we try to hide our preferences, we end up revealing them through our attitudes or actions.

The Implicit project tries to measure these unconscious biases. In the test you have to categorize words into different groups, generally two. For example, the gender-science test will ask you to categorize different words into two groups: male/science and female/arts. Then it will ask to categorize words into other two groups: female/science and male/arts. If you are biased about man and science, then you'll take more time to classify science words as female/science than to classify them as male/science. The test will measure your response time and figure out your true preference.

I read about this project in "Blink." Then I took several tests and got very weird results. Like, I don't have any preference for female and science over male and science. But my test suggested that I do. I took another test about preferring self over others, whose result was contradictory as well. I feeling a little confused about myself now.

But it was fun!

Those that I fight I do not hate,
Those that I guard I do not love.

Full poem

Crypto codes: GGH public key cryptosystem

Download code from here
I only did the decryption part, using Babai's algorithm (for homework :D)

Crypto codes: Elliptic ElGamal

Download code from here

Crypto codes:Miller Rabin Primality test

Download code from here

Output: (exercise 3.14 in the Introduction of Mathematical Cryptography)

1105 is composite with witness 2
294409 is composite with witness 2
118901509 is prime
118901521 is composite with witness 2
118901527 is prime
118915387 is composite with witness 2

Crypto codes: Elliptic curve cryptography

Download code from here
Code includes:
Elliptic curve addition,
Point in Elliptic curve over finite fields
double-and-add algorithm for Elliptic curves

Output:

Points in E(F13) are :
(-Infinity,-Infinity)
(1.0,5.0)
(1.0,8.0)
(2.0,3.0)
(2.0,10.0)
(9.0,6.0)
(9.0,7.0)
(12.0,2.0)
(12.0,11.0)
Addition test->
(9.0,7.0) + (1.0,8.0) =(2.0,10.0)
double-and-add algorithm for elliptic curve->
947*(6.0,730.0) mod 3623= (3492.0,60.0)

Crypto codes: Pollard’s p − 1 factorization algorithm

Code here
It uses prime factor code from square root

Output shows the steps of the algorithm with results:

Find prime factors of 48356747:
2^19! -1 = 13944673 (mod 48356747)
gcd(2^19!-1,48356747) = 6917
factors of 48356747, p=6917, and q=48356747/6917=6991
The prime factorization of 6916 is: 2, 2, 7, 13, 19

The prime factorization of 6990 is: 2, 3, 5, 233

Day `haven't-count-for-a-while:' Congratulations! You Won!

All the emails congratulate me, turn out to be spam.
I love you spammers, you keep me going.

Crypto codes: Square roots modulo composite moduli

Chinese remainder code: Here
Fast squaring code: Crypto codes 1


Square roots modulo composite moduli: Download from here

Crypto codes (3)

Download from here
For the inverse and fast squaring code, see Crypto codes: 1

Finds eth root modulo p:

public class RSA {

 //find x from x^e = c (mod p)
  public static long findRoot(long e, long c, long p)
  {
   long x = 0;
   
   //find e^-1 (mod p-1)
   long eInverse = crypto.inverse(p-1, e);
   
   //x = c^d (mod p)
   x = crypto.fastSq(p, c, eInverse);
   
   return x;
  }
       //test: answer = 6059
  public static void main(String[] args) {
   
      long e = 1583;
      long c = 4714;
      long p = 7919;
      
   long x = RSA.findRoot(e,c, p);
   System.out.println("Solution to x^"+e+" = "+c+" (mod "+p+"): x= "+x);
  }

}

This test will return:

Solution to x^1583 = 4714 (mod 7919): x= 6059

Matlab codes: http://www-users.math.umd.edu/~lcw/MatlabCode/

Crypto codes: Chinese remainder theorem

For gcd and inverse code, see Crypto codes

Java implementation of Chinese remainder theorem, download from here
Matlab code here

For this particular example, output will look like the following:

Step:0->x=2+3*5=17
Step:1->x=17+21*7=164
Solution:164