.:: roKB ::.

Today's Trend: Baba Ramdev's Satyagrah v/s Salman's movie "Ready"

2011-06-03T10:45:00.001+05:30

After reading news articles about "Baba Ramdev's 4th june Satyagrah against Corruption" and Salman Khan's next flick "Ready" releasing today, i am wondering how PR teams work in india nowadays. While Govt. of India is doing the most of the PR work (unintentionally) for Baba Ramdev's Satyagrah, PR team of Ready is trying very hard to keep salman in limelight just before release of movie.

As the above Google Trend shows, generally salman has better "News reference volume (NRV)" then Baba Ramdev, but surpassing of Baba's NRV just before the release of movie "ready" might effect movie's boxoffice collection. But "Search Volume Index (SVI)" of salman is much higher than Baba Ramdev's SVI. It might be a some relief for PR team of "Ready".

PS:
1. Satyagrah haven't yet started. Sources are expecting around 10 Million people joining baba at ramleela maidan at new delhi for Satyagrah.
2. Both are 1965 model of Homosepian (Man the wise) product.
3. My expectation: Satyagrah will surely affect movie's box office collections.

Wee Willy Widget Shop ( Discrete Event System Simulation )

2011-03-15T13:17:00.001+05:30

Problem: The Wee Willy Widget Shop overhauls and repair all types of widgets. The shop consists of five work stations, and the flow of jobs through the shop is as depicted here:

Regular jobs arrive at station A at the rate of one every 15 ± 13 minutes. Rush jobs arrive every 4 ± 3 hours and are given a higher priority except at station C, where they are put on a conveyor and sent through a cleaning and degreasing operation along with all other jobs. For jobs the first time through a station, processing and repair times are as follows:

The times listed above hold for all jobs that follow one of the two sequences A → B → C → D → E or A → B → D → E. However, about 10% of the jobs coming out of station D are sent back to B for further work (which takes 30 ± 10 minutes) and then are sent to D and finally to E.
Every 2 hours, beginning 1 hour after opening, the degreasing station C shuts down for routine maintenance, which takes 10 ± 1 minute. However, this routine maintenance does not begin until the current widget, if any, has completed its processing.
(a) Make three independent replications of the simulation model, where one replication equals an 8-hour simulation run, preceded by a 2-hour initialization run. The three sets of output represent three typical days. The main performance measure of interest is mean response time per job, where a response time is the total time a job spends in the shop. The shop is never empty in the morning, but the model will be empty without the initialization phase. So run the model for a 2-hour initialization period and collect statistics from time 2 hours to time 10 hours. This "warm-up" period will reduce the downward bias in the estimate of mean response time. Note that the 2-hour warm-up is a device to load a simulation model to some more realistic level than empty. From each of the three independent replications, obtain an estimate of mean response time. Also obtain an overall estimate, the sample average of the three estimates.
(b) Management is considering putting one additional worker at the busiest station (A, B, D, or E). Would this significantly improve mean response time?
(c) As an alternative to part (b), management is considering replacing machine C with a faster one that processes a widget in only 14 minutes. Would this significantly improve mean response time?

Solution: Code is written in C++ and there is also a pdf file containing the analysis for all three parts of this problem. It is available here.

Catenary - shape of a chain hanging from two points

2011-03-15T10:44:00.004+05:30

Problem: Given a chain (string) or length L and it is hanged somewhere by fixed its both endpoints, but the euclidean distance between endpoints in 3D is less than L. And the chain have μ mass per unit length. Determine the shape of the resulting curve.

Commentary: Shape formed is known as Catenary in literature. Solution of this problem is precisely the shape of garlands hanging in weddings or temples. Mathematical treatment of this problem leads to some very interesting insight into the physical characteristics of these types of curves.

Analysis: A nearly complete analysis of this problem is given at http://en.wikipedia.org/wiki/Catenary#Alternative_analysis

Coding: Even though the equations are given on wikipedia page, it is unclear how to code to get the shape of curve given two endpoints and length of the string. But this problem was thoroughly discussed in NSDE lecture 04 (12-01-2011) by Prof. Atanu Mohanty. He gave some hints regarding how to implement it in C.
If you are interested in seeing the code it is available here.

Result:

Plotting a basic 2D graph using C + gnuplot (Tutorial)

2010-09-07T20:54:00.002+05:30

In this tutorial we will learn how to draw a graph using ANSI C language with the help of gnuplot utility.

Step 1 write a c program using any editor. (here file name is plot.c)

/*
 *      file: plot.c
 *
 *      Copyright 2010 Rooparam Choudhary <rooparam@rishi.serc.iisc.ernet.in>
 *
 *      Date : 07.09.2010
 *
 *      This program is free software; you can redistribute it and/or modify
 *      it under the terms of the GNU General Public License as published by
 *      the Free Software Foundation; either version 2 of the License, or
 *      (at your option) any later version.
 *     
 *      This program is distributed in the hope that it will be useful,
 *      but WITHOUT ANY WARRANTY; without even the implied warranty of
 *      MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
 *      GNU General Public License for more details.
 *     
 *      You should have received a copy of the GNU General Public License
 *      along with this program; if not, write to the Free Software
 *      Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston,
 *      MA 02110-1301, USA.
 *
 */

#include <stdio.h>

double func ( double x ) {
    if ( x < 0 )
        return 0 ;
    if ( x < 2 )
        return x * x ;

    return 16.0 / ( x * x ) ;
}

/*
 * output will be in 2 coloumns
 * 1 column contains x-axis values
 * 2 column contains y-axis {or f(x) } values corresponding to x-value
 */
int main ( ) {
    double x = 0.0;     // initial value
    double x_max = 5.0;
    double step = 0.01;
    
    printf ( "# x \t f(x) \n" );

    while ( x <= x_max ) {
        printf ( "%.3f \t %.3f \n", x, func(x) );
        x += step;
    }

    return 0;
}

Step 2 compile it and run it and grab the output in plot.txt file

$ cc plot.c
$ ./a.out >plot.txt

Step 3 Draw using gnuplot

$ gnuplot
gnuplot> plot "./plot.txt" with lines

OK folks.
Tutorial is over.

This was just an introduction for plotting basic graphs. For more plotting tutorials and advanced plotting, just visit gnuplot .

See ya later.

Facts of Lagrange Interpolation

2010-08-27T01:18:00.001+05:30

"Lagrangian interpolation is praised for analytic utility and beauty but deplored for numerical practice." This heading, from the extended table of contents of one of the most enjoyable textbooks of numerical analysis [1],expresses a widespread view.

[...] Given (x0, f0), (x1, f1), . . ., (xn, fn) with arbitrary spaced xj, Lagrange had the idea of multiplying each fj by a polynomial that is 1 at xj and 0 at the other n nodes and then taking the sum of these n + 1 polynomials. Clearly, this gives the unique interpolation polynomial of degree n or less. [...]

( Erwin Kreyszig, Advanced Engineering Mathematics )

Figure : Lagrange Interpolation of function 1/(1+x*x)

[...] Lagrange and other interpolation at equally spaced points, as in the example above, yield a polynomial oscillating above and below the true function. This behaviour tends to grow with the number of points, leading to a divergence known as Runge's phenomenon; the problem may be eliminated by choosing interpolation points at Chebyshev nodes. [...]

( Wikipedia )

Although their are superior interpolation methods than lagrange interpolation method. But it is quite easy to understand and it is superior than Taylor Series Approximation of a function. It can be seen easily by following diagram.

Here I choose f(x) = e^x in the interval [0, 2]

[1] F. S. Acton, Numerical Methods That [Usually] Work, AMS, Providence, RI, 1990.

Damien Walters

2010-06-16T20:43:00.003+05:30

Damien Walters is a British gymnast specialising in tumbling. In 2003, the British Men's Tumbling team won the World Team title. Walters is also known for his participation in the Trampoline World Championships in 2007, finishing fourth in the team events. and 22nd in the individual events. Walters has also had the role of stuntman in the films Hellboy II and Kick-Ass. In 2010, he won the Taurus World Stunt Awards for "Best Fight" in the film Ninja Assassin.

SMVDU CSE Batch 2006

2010-06-03T15:13:00.281+05:30

Name : Rameez Ahmad 2006ECS01
DOB : March 10, 1988
Email ID : rameezahmad@gmail.com, tycoon_raja@yahoo.co.in
Phone : 06542-251503 (home), 9431744776 (Abbu)
Address : c/o M D Badruddin, Millat Nagar, Siwandih,
Bokaro Steel City, Jharkhand

Name : Gaurav Bhardwaj 2006ECS02
DOB : May 27, 1988
Email ID : gsbhardwaj27@gmail.com
Phone : 01494-246220, 21 (home), 9414637779 (papa),
9829158404 (self)
Address : s/o Satish Bhardwaj, Old Market, Neemrana,
Alwar, Rajasthan 301705

Name : Ankit Gupta 2006ECS03
DOB : April 03, 1987
Email ID : cyb3r40@gmail.com
Phone : 9795056049
Address : s/o Mr. Anil Gupta, Gandhi Nagar, Mahoba
U. P. 210427

Name : Praven Singh 2006ECS04
DOB : July 31, 1987
Email ID : pravinsmvdu@gmail.com
Phone : 9928476967
Address : E - 18, Friends Colony, Imliphatak, Jaipur,
Rajasthan 302015

Name : Vishal Natani 2006ECS05
DOB : November 16, 1988
Email ID : natanivishal@gmail.com, onenonlyone88@gmail.com
Phone : 8003422044 (self), 9414718742, 0141-2322327 (home)
Address : 1104, Tikkadmal ka Rasta, Kishanpol Bazar, Jaipur,
Rajasthan 302001

Name : Romi Kumar 2006ECS06
DOB : not yet provided
Email ID : romikumar7@gmail.com
Phone : 9430476253
Address : Near Welham English School, New Godown,
Gaya, Bihar 823001

Name : Amit Priyadarshi 2006ECS07
DOB : not yet provided
Email ID : amit.langratyagi@gmail.com, amit.ecs07@gmail.com,
amit.ecs07@yahoo.com, amit.ecs07@live.com
Phone : 9431483350 (papa), 9031585859 (papa), 9031585869 (maa)
Address : c/o Shri Shio Shankar Rai, S - 330, Dalmiyanagar,
Rahtash, Bihar 821305

Name : Jitendra Yadav 2006ECS08
DOB : April 04, 1988
Email ID : jitu0404@gmail.com, jitu0404@rediffmail.com
Phone : 9412659352, 9286878253, 9119162514
Address : Shivpuri Shala, Pachawali Road, Etawah,
U. P. 206001

Name : Nilamber Kumar 2006ECS09
DOB : not yet provided
Email ID : bittuSavarn@gmail.com, bittuSavarn_2287@yahoo.com
Phone : 9835809393, 9934708102
Address : c/o Ramnaresh Chaudhary, vill - Bandhan,
Distt - Samastipur, Bihar

Name : Narendra Singh Choudhary 2006ECS10
DOB : October 24, 1986
Email ID : ncr.cse@gmail.com, love0012007@gmail.com
Phone : 9929191740 (self), 9413088666 (papa), 9413839830 (Bhaiya)
Address : 1. Plot No - 406, Lion'slane coloney, near sirsi road,
   Jaipur, Rajasthan
   2. vill - Mundly, post - Renwal, distt - Jaipur
   Rajasthan 303603

Name : Santosh Kumar 2006ECS11
DOB : October 25
Email ID : sankumarsingh@gmail.com, s.smvdu@gmail.com
Phone : 9852860226 (Bhaiya), 9504353054 (self)
Address : Chawk Bazar, near Shyam Market, Mokama,
Patna, Bihar 803302

Name : Saurabh Sharma 2006ECS12
DOB : March 16, 1988
Email ID : mossesb43@yahoo.com
Phone : not yet provided
Address : c/o Mr. Ramavtar Sharma, 3 - Naval Nagar,
Hathras, U. P. 204101

Name : Vipin Kumar Sharma 2006ECS13
DOB : not yet provided
Email ID : vipinsharma786@yahoo.in, vipin.sharma2009@hotmail.com
Phone : 9027452142, 9468809240 (brother)
Address : 1. byepass road, krishna colony, kheragarh,
   Agra, U. P. 283121
   2. 17-B, State Bank Colony, Jaipur house,
   Agra, U. P.

Name : Vivek Sharma 2006ECS14
DOB : not yet provided
Email ID : not yet provided
Phone : not yet provided
Address : not yet provided

Name : Sanjeev Kumar 2006ECS15
DOB : not yet provided
Email ID : not yet provided
Phone : 9835648353
Address : south gali of devi sthan, bihar sharif, nalanda,
Bihar 803101

Name : Shakeel Ahmad 2006ECS17
DOB : July 01, 1987
Email ID : shakeel2006ecs17@gmail.com
Phone : 9695391701 (self)
Address : vill- tulsiyapur,PO - Aundahiklan, Siddharth Nagar,
U. P. 272201

Name : Shashi Kumar Pal 2006ECS18
DOB : November 01, 1986
Email ID : shashi403@gmail.com
Phone : 05265-239446, 276562, 9628514195
Address : B - 403, Indian Telephone Industries Ltd.
Mankapur, Gonda, U. P. 271308

Name : Pulkit Gupta 2006ECS19
DOB : November 18, 1988
Email ID : pulkit2008.us@gmail.com
Phone : 0122-2300857, 9917323215
Address : 1311-A, New Shivpuri, Hapur, Ghaziabad,
U. P. 245101

Name : Rooparam Choudhary 2006ECS20
DOB : September 15, 1987
Email ID : rrc.iisc@gmail.com, rooparambhakar@gmail.com
Phone : 9461314938 (self), 9414484438 (bhaiya)
Address : s/o Shri Mohan Ram Bhakar, Bhakaro ki dhani,
Makrana, Rajasthan 341505

Name : Sumit Kumar Bola 2006ECS21
DOB : April 11, 1988
Email ID : sumitkumarcse@gmail.com
Phone : 9828737807 (mother), 9828571475 (mama)
Address : s/o Parmeshwar lal, vill - Bola ki dhani, via - Mandrella, post - Bajawa,
Jhunjhunu, Rajasthan 333025

Name : Jayant Pathak 2006ECS22
DOB : December 25, 1986
Email ID : charlie0456@gmail.com, jayant0456@indiatimes.com
Phone : 011-28365029 (home), 9999208161 (self)
Address : C-53, nilothi mode, shivram park ext. II, nangloi,
New Delhi 110041

Name : Devendra Kumar Upadhyay 2006ECS23
DOB : not yet provided
Email ID : devendra_567@yahoo.co.in, devendraupadhyay567@gmail.com
Phone : 9907008057 (home), 07662-299626 (home)
Address : C - 2, MPSEB colony, silpara, rewa,
M. P. 486550

Name : Dileep Parashar 2006ECS24
DOB : not yet provided
Email ID : dileepparashar06@gmail.com, dileep_parashar87@yahoo.co.in
Phone : 05641-230344 (home), 9780943677
Address : Betal Mohalla, Maszid ke piche, New Road,
Deeg, Bharatpur, Rajasthan 321203

Name : Narendra Kumar Agrawal 2006ECS25
DOB : May 24, 1986
Email ID : narendrabtechcse@hotmail.com, narendrabtechcse@gmail.com
Phone : 07766-277771, 9893735925
Address : nawapara - dhal, Raigarh, C. G.

Name : Mahesh Kumar 2006ECS26
DOB : January 15, 1987
Email ID : er.mahesh26@gmail.com
Phone : 9430739055
Address : Shri Govind Prasad Modi, Chandwara, Koderma,
Jharkhand

Name : Dheeraj Kumar 2006ECS27
DOB : January 01, 1986
Email ID : dheerajagrahari86@gmail.com
Phone : 9794335238 (papa)
Address : vill & post - vijaypur, dist - mirzapur
U. P. 231303

Name : Kumud Ranjan 2006ECS28
DOB : not yet provided
Email ID : catch_kranjan@yahoo.co.in
Phone : 9934037495 (papa), 9234268392 (mummy)
9199407192 (self)
Address : 504, Sakuntalam Place, Boring Road, Patna
Bihar 800013

Name : Anant Agarwal 2006ECS29
DOB : February 20, 1989
Email ID : anantrams2005@gmail.com
Phone : 9211243593 (self)
Address : s/o Shri Shyam Kr. Agarwal, 246, opp. Jagat Cinema,
Naveen Threads, Bara Bazar, Bareilly
U. P. 243003

Name : Shakti Singh Chauhan 2006ECS30
DOB : not yet provided
Email ID : anshushakti@gmail.com
Phone : 9451803297, 9452876323
Address : c/o Daveta Deen Singh Chauhan, vill - chak pancham,
post - lalganj, distt - Raebareli
U. P.

Name : Saurav Savarn 2006ECS31
DOB : not yet provided
Email ID : sauravsavarn@gmail.com, sauravsavarn@yahoo.co.in
Phone : 9308655763 (maa)
Address : c/o K. B. Sharma, Vivekanand Park, Road no 2,
House no IV, Patliputra, Patna
Bihar 800013

Name : Shreepriya Dogra 2006ECS32
DOB : January 21, 1988
Email ID : not yet provided
Phone : not yet provided
Address : not yet provided

Tour de Bangaluru

2010-04-16T20:48:00.000+05:30

life is good here, and this city lot less noisy than north india. Nobody is shouting here at public places (railway station, bus station) unlike north india. Food is good, cheap, and tasty. Just one problem here, and thats of language. Every written word on walls, papers seems like different varieties of jalebi. Apart from it, its a very nice place. And people are nice and friendly here.

... BYE ....

rrc @ cybercafe near ganesh hotel

twitter account

2010-03-30T14:39:00.003+05:30

just removed 2 redundant accounts from twitter.com and now i have only 1 twitter acct and its http://twitter.com/rrc_iisc

moment

2010-03-29T17:16:00.000+05:30

sitting in net lab thinking "upto which episode of HIMYM season 5 we have on LAN in hostel? it might be 15 or upto 16" but not sure, so delaying download of further episodes....

As u can guess from current situation, after gate results We have no more surprises left in BTech at SMVDU, so its much booring life now at SMVDU and after 8 or 10 days here we will have RESU#$#$%# { should be a youth fest } so more booring dayz are comming. Even though ArchiFest was nice entertainment just before anouncement of GATE results.

PS: currently reading Mark Allen Weiss and listening to Lucky Ali.

Happy New Year

2010-01-27T12:17:00.000+05:30

First of all, HAPPY NEW YEAR to all readers (if any) [i now it's too late for that, but its the first post in this year]

and yesterday night around 3:00 AM, i finally solved the last problem in my new Minimum Spanning Tree calculating algorithm. All credit goes to Dr. Naveen Garg (IITD) who teaches me BCD (Boy, Cat, Doll) of Data Structures and Algorithm (although through videos :) ).

So, today i am pretty excited, cause now i just need to calculate the time complexity of algorithm. (implementation will only take one night stand) And i have saved my full 2 months for other projects as it will be my final year major project and it is 67% done without much effort.

(if it will produce good statistical graphs in testing with Prim's and kruskal's algo, then it might lend in some journal and might also help me in securing job).

Finally, Happy New Year Again

[Keep watching for more code]

Area Filling Algorithms (seed filling, ScanLine conversion)

2009-12-11T09:48:00.000+05:30

#include <graphics.h>

#include <iostream>

#define ROUND(x) ((int)(x+0.5))

typedef int Point[2];           // P[0] - x coordinate P[1] - y coordinate

void seedFill(int x, int y, int color) {

     if(x > 640 || x < 0 || y > 480 || y < 0)

          return;

     if(getpixel(x, y) == BLACK) {

                       putpixel(x, y, color);

                       seedFill(x+1, y, color);

                       seedFill(x-1, y, color);

                       seedFill(x, y+1, color);

                       seedFill(x, y-1, color);

     }

}

void scanlineFill(Point *polygon, int nodes, int color) {

     setcolor(color);

     Point array[480];

     for(int i=0; i<480; ++i) {

             array[i][0] = 640;

             array[i][1] = 0;

     }

     for(int i=0; i<nodes; ++i){

             Point p1, p2;

             p1[0] = polygon[i][0];   p1[1] = polygon[i][1];

             p2[0] = polygon[(i+1)%nodes][0];               p2[1] = polygon[(i+1)%nodes][1];

             if(p1[1] > p2[1]) {

                      p2[0] = polygon[i][0];   p2[1] = polygon[i][1];

                      p1[0] = polygon[(i+1)%nodes][0];               p1[1] = polygon[(i+1)%nodes][1];

             }

             double m = (double)(p2[1]-p1[1])/(p2[0]-p1[0]);

             double xd = p1[0] - 1/m;

             for(int y=p1[1]; y<=p2[1]; ++y) {

                     xd += 1/m;

                     int x = ROUND(xd);

                     if(array[y][0] > x)

                                    array[y][0] = x;

                     if(array[y][1] < x)

                                    array[y][1] = x;

             }

     }

     for(int i=0; i<480; ++i)

             if(array[i][0] < array[i][1])

                            line(array[i][0]+1, i, array[i][1]-1, i);

}

void drawPolygon(Point *polygon, int nodes) {

     setcolor(WHITE);

     for(int i=0; i<nodes; ++i){

             Point p1, p2;

             p1[0] = polygon[i][0];   p1[1] = polygon[i][1];

             p2[0] = polygon[(i+1)%nodes][0];               p2[1] = polygon[(i+1)%nodes][1];

             line(p1[0], p1[1], p2[0], p2[1]);

     }

}

int main(){

    int gd = DETECT, gm;

    initgraph(&gd, &gm, "C:\\");

    Point polygon[20];

    int size = 0;

    while(true) {

                std::system("cls");

                std::cout << "Enter size of polygon : (-1 for exit) : ";

                std::cin >> size;

                if(size == -1)

                        return 0;

                if(size > 20){

                        std::cout << "Enter size less than 20" << std::endl;

                        continue;

                }

                for(int i=0; i < size; ++i) {

                        std::cout << "Enter vertex : (x y) : ";

                        std::cin >> polygon[i][0] >> polygon[i][1];

                }

                drawPolygon(polygon, size);

                FILL:

                std::system("cls");

                std::cout << "Choose an algorithm to fill." << std::endl;

                std::cout << "\t1. seed fill" << std::endl;

                std::cout << "\t2. scan line fill" << std::endl;

                std::cout << "choice: ";

                int choice;

                std::cin >> choice;

                switch(choice) {

                               case 1:

                                    {

                                        std::cout << "Enter a point inside polygon: (x y) : ";

                                        Point p;

                                        std::cin >> p[0] >> p[1];

                                        seedFill(p[0], p[1],RED);

                                        break;

                                    }

                               case 2:

                                    scanlineFill(polygon, size, BLUE);

                                    break;

                               default:

                                       std::cout << "I can't interpret your choice." << std::endl;

                                       goto FILL;

                                       break;

                }

    }

}