Education And Career

Best geography books for UPSC prelims, mains

2020-07-20T10:35:00.001-07:00

This post is intended to clear the confusion that prevails among the aspirants over how to prepare for UPSC geography and the best books for it. I will also share the best source for geography for UPSC prelims and mains.

Geography syllabus for UPSC: Its importance during preparation

Before starting preparation for any subject for UPSC civil services examination, one has to first look at the syllabus.

It will give an overview of the topics that one should focus on not only during the preparation of that particular subject but also while reading a newspaper and other current affairs material.

For example, volcanoes is one of the topics in UPSC syllabus. If any article or news on volcanoes appear in newspaper, it should not be neglected rather aspirant must dig the topic deeper.

>>Click here to continue reading

GATE 2020 Exam, Scholarship: Everything you need to know

2020-07-20T10:26:00.003-07:00

Graduate Aptitude Test in Engineering (GATE) 2020 is being conducted by the Indian Institute of Technology Delhi (IIT-D). This year, it will be held in the month of February.

As per the information brochure provided by the IIT-D, the examination will be held for 25 papers. It will be conducted in an Online Computer Based Test (CBT). Candidates will have to answer 65 questions in 3 hours.

>>Click here to continue reading

TS PGECET web options 2019: Step-by-step procedure

2019-08-10T00:55:00.002-07:00

TS PGECET web options 2019 started. Candidates can login to their account on the official website of TS PGECET to give their web options. Final allotment will be based on the priority of the options given on the website.

TS PGECET web options: Following is the step-by-step procedure

Step 1: Open website of TS PGECET (Click here). You will see this window.

Step 2: Click on ‘Web options login’ link in ‘web counseling’ block. You will get the following window.

Watch the above video to understand the step-by-step process of web option.

Step 3: Enter hall ticket number of PGECET or Registration Id of GATE/GPAT and Rank of PGECET or Score of GATE/GPAT. After entering these details click on 'login' button. You will get the following window.

Step 4: You need to select either registered email or phone number where you will get PIN which can be used for future logins too. After selecting the registered email id or phone number, hit on ‘Send PIN’ button. Then you will get the below window.

Step 5: After entering the PIN received on email or phone number, hit on ‘verify‘ button.

After successful login, check the information displayed on the website.

Finally, click on the ‘continue‘ button to add web options.

Step 6: Add options by clicking on 'Add option' button on the window as shown below.

Step 7: Option can be selected by clicking on checkbox shown on the left-hand side of the option.

Step 8: Priority of the option can be adjusted by clicking on the check box and then on ‘up’ or ‘down’ button.

Step 9: After getting satisfied with the options and its priority, click on ‘Save and continue‘ button and then on the ‘confirm’ button.

Step 10: Final submission can be done by clicking on ‘final submit‘ button.

Note: Once the final submission is done, the candidate cannot change the option.

Bookmark the blog and subscribe the above youtube channel to get a notification once final allotment is done.

GATE 2018: Procedure for requesting Change of Examination City

2017-10-31T02:59:00.002-07:00

GATE 2018: Procedure for requesting Change of Examination City

Below video will guide you step-by-step. View it to get information to change examination city for GATE 2018.

Important information for M.Tech admission seeker (Personal View)

2014-08-20T09:07:00.000-07:00

Many students takes their career decisions without thinking, first find out what is your interest before taking any decision. Think every year many Engineers are getting passed but question is how many are actually satisfied with what they are doing after Engineer. Think twice what I mean to say don't just go for Engineering without thinking what you want to do in future, market status and please don't be in myth that when you take admission in Engineering or M.Tech there faculty will teach you every thing if even though your interest is not Engineer, Practically speaking they will only help you less than 3% in your studies rest more than 97% depends on you. As M.Tech admission are nearer before taking admission into M.Tech ask yourself this important question

“Why I am taking admission into M.Tech?”

Now your answer may be one of the following

1. I dint get placed in B.Tech as my college doesn’t have placements and hoping for placement in M.Tech

2. Want to go for Teaching profession

3. Want to relax for a while as I was struggling for job and my parent's pressure to get job is increasing

4. For Stipend (in case of GATE students)

5. While studying M.Tech I want to prepare for Govt. job

6. To enter into Research sector

If your answer is out of this then my suggestion for every option

1. If your aim is to get placed then let me clear your chances to get recruited in Software company will be very less in 'on campus' or 'off campus recruitment' once you are done with M.Tech from university or private college apart from NITs or IITs or IIIT as companies will always prefer for B.Tech not M.Tech (chances are 99:1 for B.Tech and M.Tech) and finally you will regret for your decision to take admission in M.Tech. If willing for placements in M.Tech then take admission only in NITs, IITs, IIIT.

2. If you want to go for teaching job then doing M.Tech is perfect decision but in this case M.Tech should be done from State (OU, JNTU) or Central Universities (NITs, IITs, IIIT) only.

3. Then you are just trying to relax in war field.

4. Good, provided you must also think about your future career plan as you will be having financial support provided you must utilize this financial resource and time you get during M.Tech for better career.

5. This is wonderful option especially for GATE student as they will be getting M.Tech degree, Financial support and good career.

6. If your goal is to enter into Research sector then take only admission in IITs or NITs.

"Think before taking admission!! Best of Luck for bright Career"

Notification for PGEC / PGECET-2013 Final Phase Web Counselling

2013-08-31T09:46:00.002-07:00

All the registered GATE/GPAT candidates and PGECET qualified and eligible candidates of 2013 are informed to attend PGEC / PGECET Web Counseling for admission into M.E./M.Tech./M.Arch./M.Planning/ M.Pharmacy/Pharm.D (P.B.) courses for the academic year 2013-2014.

For more details

Click here

Web Allotments for PGEC/PGECET 2013

2013-08-24T09:40:00.001-07:00

Web Allotments for PGEC/PGECET 2013

Click to know where you got admission

PGEC/PGECET-2013 ADMISSIONS Notification released

2013-07-23T02:54:00.000-07:00

The GATE qualified and eligible candidates of 2012 and 2013, GPAT and PGECET qualified and eligible Candidates of 2013 are informed to attend PGEC / PGECET Web Counseling for admission into M.E./ M.Tech./ M.Arch / M.Pharmacy/Pharm.D (P.B.) courses for the academic year 2013-2014 as per the
schedule given below at any one of the Help Line centers.

Candidates are informed that they should satisfy the admission rules prescribed in the related G.O. besides satisfying the eligibility criteria of the respective universities and should be prepared to pay the tuition fee as applicable.

The left over seats after allotment for GATE / GPAT Candidates are alone available for PGECET candidates This notification itself is a call for counseling.

No separate call letters are sent to the individual candidates. Mere attending counseling does not guarantee a seat. Candidates have to pay non-refundable processing fee of Rs.300/- (SC/ST) and Rs.600/-(others) at
the time of certificate verification.

For More details Click here

RMI implementation program

2013-06-26T08:08:00.000-07:00

Hello.java

import java.rmi.*;
public interface Hello extends Remote
{
String sayHello(String s) throws RemoteException;

}

HelloImpl.java

import java.rmi.*;
import java.rmi.server.*;
public class HelloImpl extends UnicastRemoteObject implements Hello
{
private int i;
public HelloImpl() throws RemoteException
{
System.out.println("Object created");
}
public String sayHello(String s) throws RemoteException
{
i++;
return "Hello " + s + " Yours is " + i + " request.";
}

}

HelloServer.java

import java.rmi.*;
import java.rmi.registry.*;
public class HelloServer
{
public static void main(String[] args)
{
try
{
LocateRegistry.createRegistry(1112);
HelloImpl r = new HelloImpl();
Naming.rebind("//127.0.0.1:1112/value", r);
System.out.println(" Object ready");
}
catch (Exception e)
{
System.out.println(e);
}
}

}

HelloClient.java

import java.rmi.*;
public class HelloClient
{
public static void main(String[] args)
{
try
{
Remote r = Naming.lookup("//127.0.0.1:1112/value");
Hello i = (Hello) r;
String result = i.sayHello(args[0]);
System.out.println( "From Server: " + result);
}
catch (Exception e)
{
System.out.println(e);
}
}

}

Osmania University : UG Degree (BA, BSC, BCOM) Results 2013.

2013-06-07T00:09:00.002-07:00

Osmania University : UG Degree (BA, BSC, BCOM) Results 2013.

B.A

Click Here

B.Sc

Click Here

B.Com

Click Here

Pattern Matching implementation in JAVA implementing Directory

2013-02-07T19:15:00.000-08:00

Pattern Matching implementation in JAVA implementing Directory

import java.io.*;
class Mat
{
public String patt;
public int flg=0;
public Mat(String p)
{
      patt=p;
}
public void chk(File f) throws IOException
{
       if (f.isDirectory())
       {
           File[] lf = f.listFiles();
           for (int i = 0; i < lf.length; i++)
           {
           chk(lf[i]);
           }
       }
       else
           srch(f);
}
public void srch(File f) throws IOException
{
       FileInputStream fstream = new FileInputStream(f);
       DataInputStream in = new DataInputStream(fstream);
       BufferedReader brd = new BufferedReader(new InputStreamReader(in));
       String strLine;
       int l=0;
       while ((strLine = brd.readLine()) != null)
           {
           l=l+1;
               int n=strLine.length();
               int m=patt.length();
               char p;
               char t;
               int s=0;
               int   i=0;
               int j=0;
               while(i<n && s!=1)
               {
                   t=strLine.charAt(i);
                   p=patt.charAt(j);
                   if(p==t)
                   {
                       i++;
                       j++;
                       if(j>=m)
                       {
                           s=1;
                       }
                   }
                   else
                   {
                       j=0;
                       i++;
                   }
               }
               if(s==1)
               {
                   System.out.println("Found Pattern at Line "+ l+" in file" +f);
                   flg=flg+1;
               }
               s=0;
            }
}
}

class Pat
{
   public static void main(String[] args) throws IOException
   {
       String patt;
       File f=new File("D:\\Sameer\\Pat.java");
       InputStreamReader isr = new InputStreamReader(System.in);
       BufferedReader br = new BufferedReader(isr);
       System.out.println("enter pattern");
       patt=br.readLine();
       File f1=f.getParentFile();
       Mat m=new Mat(patt);
       m.chk(f1);
       System.out.println("Pattern Found "+m.flg+" times");
   }
}

Pattern Matching using Java

2013-02-04T07:12:00.001-08:00

import java.io.*;
class Pat
{
    public static void main(String[] args) throws IOException
    {
        String patt,strLine;
        int l=0,f=0,count=0,fc=1;
        InputStreamReader isr = new InputStreamReader(System.in);
        BufferedReader br = new BufferedReader(isr);
        System.out.println("enter pattern");
        patt=br.readLine();
        while(fc<6)
        {
        FileInputStream fstream = new FileInputStream("textfile"+fc+".txt");
                DataInputStream in = new DataInputStream(fstream);
                BufferedReader brd = new BufferedReader(new InputStreamReader(in));
        while ((strLine = brd.readLine()) != null)
            {
            l=l+1;
                int n=strLine.length();
                int m=patt.length();
                char p;
                char t;
                int s=0;
                int    i=0;
                int j=0;
                while(i<n && s!=1)
                {
                    t=strLine.charAt(i);
                    p=patt.charAt(j);
                    if(p==t)
                    {
                        i++;
                        j++;
                        if(j>=m)
                        {
                            s=1;
                        }
                    }
                    else
                    {
                        j=0;
                        i++;
                    }
                }
                if(s==1)
                {
                    count++;
                    System.out.println("Found Pattern in File textfile"+fc+" at Line "+ l+" at position "+(i-j+1));
                    f=1;
                }
                s=0;
            }
            if(f==0)
                    System.out.println("Pattern not found ");
            else
                    System.out.println("Pattern count= "+count);
            fc++;
        }
    }
}

Note: Before executing this program create 5 text files with name textfile1.txt, textfile2.txt, textfile3.txt, textfile4.txt, textfile5.txt.

PCA Assignment for my friends

2013-01-19T02:54:00.003-08:00

DM Assignment for my friends

2013-01-19T02:47:00.001-08:00

DM Assignment..

How to Prepare for GATE 2013 Exam?

2013-01-11T20:16:00.000-08:00

The Graduate Aptitude Test in Engineering (GATE) is an All-India Examination conducted by the Seven IITs and IISC, Bangalore, on behalf of the Ministry of Human Resources Development (MHRD), Government of India. GATE is undertaken to admit students in Postgraduate Engineering Programs at the National level. One of the IIT or IISc is selected every year to conduct the test.

GATE is basically an exam that assesses your fundamentals. The questions asked in the exam are mostly derivations of the fundamentals. While preparing for GATE, you should go step-by-step. Firstly, you need to clear your basics and then test yourself if you have got them right. After this, you should take a mock test among your competitors to see where you stand.

Check the below links also to prepare better for GATE 2013.

--> -->
Start your preparation from the first chapter. You need to separate the books that include the fundamental concepts and the ones that include problem oriented questions. You should make a note of probable concepts, important theories, formulae etc. Try to use different tricks and solve as many problems as possible. Take chapter-wise self-tests to improve your score. In the end, take self-tests that include the whole syllabus. It is imperative to analyse your test results as this will help you improve more. You should see your strong and weak points. Check your accuracy level in different topics and maintain a topic-wise datasheet to check your performance level. Visit the library often to get hold of some good books that can help you get through GATE.Also, get some good guide books for GATE and question papers of previous years. Ebooks help alot in GATE Examination.

Dijkstra algorithm's implementation in java (From one source to all the other nodes in Graph)

2013-01-03T08:57:00.001-08:00

import java.io.*;
class Tnode
{
int prev;
int len;
char label;
boolean vis;
public Tnode(char lab)
{
label = lab;
prev=-1;
len=100000;
vis=false;
}
}

class graph
{
public final int MAX = 20;
public int nverts,i,min;
public Tnode vlist[];
public int adj[][];
public graph()
{
nverts = 0;
adj = new int[MAX][MAX];
vlist = new Tnode[MAX];
for(int i=0;i<MAX;i++)
{
for(int j=0;j<MAX;j++)
adj[i][j] = 0;
}
}
public void addver(char lab)
{
vlist[nverts++] = new Tnode(lab);
}
public void addedge(int start,int end,int cost)
{
adj[start][end] = cost;
}
public int getind(char l)
{
for(int i=0;i<nverts;i++)
if(vlist[i].label==l)
return i;
return (MAX+1);
}
public void brfs(int src)
{
int k,cnt=0;
int path[]=new int[20];
vlist[src].len=0;
vlist[src].vis=true;
k=src;
do{
cnt++;
for(i=0;i<nverts;i++)
{
if((adj[k][i]!=0)&&(vlist[i].vis==false))
{
if(vlist[k].len+adj[k][i]<vlist[i].len)
{
vlist[i].prev=k;
vlist[i].len=vlist[k].len+adj[k][i];
}
}
}
k=0;
min=100000;
for(i=0;i<nverts;i++)
{
if((vlist[i].vis==false)&&(vlist[i].len<min))
{
min=vlist[i].len;
k=i;
}
}
vlist[k].vis=true;
}while(cnt<nverts);
System.out.println("shortest path from source to Other Nodes is \n");
for(int l=0;l<nverts;l++)
{
i=0;
k=l;
int c=0;
do{
path[i++]=k;
c++;
k=vlist[k].prev;
}while(k>=0);
System.out.println("shortest path from " +vlist[src].label+" to " +vlist[l].label);
for(i=c-1;i>=0;i--)
System.out.print("-->"+vlist[path[i]].label);
System.out.println("\n shortest distance from " +vlist[src].label+" to " +vlist[l].label+" \t\t"+vlist[l].len);
}
}
}
class Dijra
{
public static void main(String args[])throws IOException
{
graph gr = new graph();
char c;
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(isr);
System.out.println("Enter the number of vertices");
int n = Integer.parseInt(br.readLine());
String t;
System.out.println("Enter the labels for the vertices");
for(int i=0;i<n;i++)
{
t = br.readLine();
c = t.charAt(0);
gr.addver(c);
}
System.out.println("Enter the number of edges");
int edg = Integer.parseInt(br.readLine());
System.out.println("Enter the vertices which you need to connect");
for(int j=0;j<edg;j++)
{
System.out.println("Enter the first vertex");
t = br.readLine();
c = t.charAt(0);
int start = gr.getind(c);
System.out.println("Enter the second vertex");
t = br.readLine();
c = t.charAt(0);
int end = gr.getind(c);
System.out.println("Enter the cost");
int cost = Integer.parseInt(br.readLine());
gr.addedge(start,end, cost);
}
System.out.println("Enter the source vertex");
t = br.readLine();
c = t.charAt(0);
int src = gr.getind(c);
gr.brfs(src);
}
}

EAMCET Counselling procedure

2013-01-01T18:11:00.000-08:00

Stages in AP EAMCET Web Counselling 2013

EAMCET Notification:

APSCHE will be published the EAMCET web Counselling notification during June-July in all the newspapers. Verification is done in help line center. Candidates who fail to report for Certificates verification will not be considered for allotment of seats in any of the colleges.

Certificate Verification:

Document to be produced:

EAMCET 2013 Hall ticket
AP EAMCET 2013 Rank Card
SSC Original Certificate or Marks Memo
Intermediate Certificate
Study Certificates for 10th Class to Intermediate.
Caste Certificate of SC / ST / BC / Minorities (Except OC Students)
Income Certificate issued by MRO
PH/NCC/CAP/Sports and Games Certificates (if applicable)
Transfer Certificate of Intermediate

Exercising of Options:

After completion of certificate verification, helpline center officials will give a scratch card which is used to exercising options in EAMCET official website.

Based option provided seat will be allotted.

Kruskal's algorithm's implementation in java

2012-12-30T05:14:00.000-08:00

Kruskal's algorithm is a greedy algorithm in graph theory that finds a minimum spanning tree for a connected weighted graph. This means it finds a subset of the edges that forms a tree that includes every vertex, where the total weight of all the edges in the tree is minimized. If the graph is not connected, then it finds a minimum spanning forest (a minimum spanning tree for each connected component).

Example run

Image	Description
	AD and CE are the shortest edges, with length 5, and AD has been arbitrarily chosen, so it is highlighted.
	CE is now the shortest edge that does not form a cycle, with length 5, so it is highlighted as the second edge.
	The next edge, DF with length 6, is highlighted using much the same method.
	The next-shortest edges are AB and BE, both with length 7. AB is chosen arbitrarily, and is highlighted. The edge BD has been highlighted in red, because there already exists a path (in green) between B and D, so it would form a cycle (ABD) if it were chosen.
	The process continues to highlight the next-smallest edge, BE with length 7. Many more edges are highlighted in red at this stage: BC because it would form the loop BCE, DE because it would form the loop DEBA, and FE because it would form FEBAD.
	Finally, the process finishes with the edge EG of length 9, and the minimum spanning tree is found.

import java.io.*;
class Tnode
{
char label;
//boolean vis;
int prev;
public Tnode(char lab)
{
label = lab;
//vis=false;
prev=-1;
}
}

class graph
{
public final int MAX = 20;
public int nverts,i,min;
public Tnode vlist[];
public int adj[][];
public graph()
{
nverts = 0;
adj = new int[MAX][MAX];
vlist = new Tnode[MAX];
for(int i=0;i<MAX;i++)
{
for(int j=0;j<MAX;j++)
adj[i][j] = 10000;
}
}
public void addver(char lab)
{
vlist[nverts++] = new Tnode(lab);
}
public void addedge(int start,int end,int cost)
{
adj[start][end] = cost;
adj[end][start] = cost;
}
public int getind(char l)
{
for(int i=0;i<nverts;i++)
if(vlist[i].label==l)
return i;
return (MAX+1);
}
boolean find(int i, int j)
{
int a,b,k=0,l=0;
a=i;b=j;
while(a!=-1)
{
k=a;
a=vlist[a].prev;
}
if(a==b)
return false;
a=i;b=j;
while(b!=-1)
{
l=b;
b=vlist[b].prev;
}
if(a==b || k==l)
return false;
return true;
}
public void brfs()
{
int k,c=0,minicost=0,j,mini=100000,m=0,p=0,ct=1;
while(ct<nverts)
{
for(i=0;i<nverts;i++)
{
for(j=0;j<nverts;j++)
{
if((mini>adj[i][j])&&(find(i,j)))
{
mini=adj[i][j];
p=i;
m=j;
}
}
}
System.out.print("\n vertex" +vlist[p].label+"is connected to"+vlist[m].label);
minicost=minicost+mini;
mini=100000;
adj[p][m]=100000;
adj[m][p]=100000;
if(vlist[m].prev==-1)
vlist[m].prev=p;
else
vlist[p].prev=m;
//vlist[m].vis=true;
//vlist[p].vis=true;
ct++;
}
System.out.println("\nMinimum cost=" +minicost);
}
}
class Krus
{
public static void main(String args[])throws IOException
{
graph gr = new graph();
char c;
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(isr);
System.out.println("Enter the number of vertices");
int n = Integer.parseInt(br.readLine());
String t;
System.out.println("Enter the labels for the vertices");
for(int i=0;i<n;i++)
{
t = br.readLine();
c = t.charAt(0);
gr.addver(c);
}
System.out.println("Enter the number of edges");
int edg = Integer.parseInt(br.readLine());
System.out.println("Enter the vertices which you need to connect");
for(int j=0;j<edg;j++)
{
System.out.println("Enter the first vertex");
t = br.readLine();
c = t.charAt(0);
int start = gr.getind(c);
System.out.println("Enter the second vertex");
t = br.readLine();
c = t.charAt(0);
int end = gr.getind(c);
System.out.println("Enter the cost");
int cost = Integer.parseInt(br.readLine());
gr.addedge(start,end, cost);
}
gr.brfs();
}
}

Prim's algorithm's implementation in java

2012-12-30T04:10:00.000-08:00

Prim's algorithm is a greedy algorithm that finds a minimum spanning tree for a connected weighted, undirected graph. This means it finds a subset of the edges that forms a tree that includes every vertex, where the total weight of all the edges in the tree is minimized.

Example run

U	possible edges	V \ U	Description
{}		{A,B,C,D,E,F,G}	This is our original weighted graph. The numbers near the edges indicate their weight.
{D}	{D,A} = 5 V {D,B} = 9 {D,E} = 15 {D,F} = 6	{A,B,C,E,F,G}	Vertex D has been arbitrarily chosen as a starting point. Vertices A, B, E and F are connected to D through a single edge. A is the vertex nearest to D and will be chosen as the second vertex along with the edge AD.
{A,D}	{D,B} = 9 {D,E} = 15 {D,F} = 6 V {A,B} = 7	{B,C,E,F,G}	The next vertex chosen is the vertex nearest to either D or A. B is 9 away from D and 7 away from A, E is 15, and F is 6. F is the smallest distance away, so we highlight the vertex F and the edge DF.
{A,D,F}	{D,B} = 9 {D,E} = 15 {A,B} = 7 V {F,E} = 8 {F,G} = 11	{B,C,E,G}	The algorithm carries on as above. Vertex B, which is 7 away from A, is highlighted.
{A,B,D,F}	{B,C} = 8 {B,E} = 7 V {D,B} = 9 cycle {D,E} = 15 {F,E} = 8 {F,G} = 11	{C,E,G}	In this case, we can choose between C, E, and G. C is 8 away from B, E is 7 away fromB, and G is 11 away from F. E is nearest, so we highlight the vertex E and the edge BE.
{A,B,D,E,F}	{B,C} = 8 {D,B} = 9 cycle {D,E} = 15 cycle {E,C} = 5 V {E,G} = 9 {F,E} = 8 cycle {F,G} = 11	{C,G}	Here, the only vertices available are C and G.C is 5 away from E, and G is 9 away from E.C is chosen, so it is highlighted along with the edge EC.
{A,B,C,D,E,F}	{B,C} = 8 cycle {D,B} = 9 cycle {D,E} = 15 cycle {E,G} = 9 V {F,E} = 8 cycle {F,G} = 11	{G}	Vertex G is the only remaining vertex. It is 11 away from F, and 9 away from E. E is nearer, so we highlight G and the edge EG.
{A,B,C,D,E,F,G}	{B,C} = 8 cycle {D,B} = 9 cycle {D,E} = 15 cycle {F,E} = 8 cycle {F,G} = 11 cycle	{}	Now all the vertices have been selected and the minimum spanning tree is shown in green. In this case, it has weight 39.

Implementation:

import java.io.*;
class Tnode
{
int prev;
int len;
char label;
boolean vis;
public Tnode(char lab)
{
label = lab;
prev=-1;
len=100000;
vis=false;
}
}

class graph
{
public final int MAX = 20;
public int nverts,i,min;
public Tnode vlist[];
public int adj[][];
public graph()
{
nverts = 0;
adj = new int[MAX][MAX];
vlist = new Tnode[MAX];
for(int i=0;i<MAX;i++)
{
for(int j=0;j<MAX;j++)
adj[i][j] = 0;
}
}
public void addver(char lab)
{
vlist[nverts++] = new Tnode(lab);
}
public void addedge(int start,int end,int cost)
{
adj[start][end] = cost;
adj[end][start] = cost;
}
public int getind(char l)
{
for(int i=0;i<nverts;i++)
if(vlist[i].label==l)
return i;
return (MAX+1);
}
public void brfs(int src)
{
int k,c=0,minicost=0;
vlist[src].len=0;
vlist[src].vis=true;
k=src;
while(c<nverts){
for(i=0;i<nverts;i++)
{
if((adj[k][i]!=0)&&(vlist[i].vis==false))
{
if(adj[k][i]<vlist[i].len)
{
vlist[i].prev=k;
vlist[i].len=adj[k][i];
}
}
}
k=0;
min=100000;
for(i=0;i<nverts;i++)
{
if((vlist[i].vis==false)&&(vlist[i].len<min))
{
min=vlist[i].len;
k=i;
}
}
vlist[k].vis=true;
c++;
}
for(i=0;i<nverts;i++)
{
if(vlist[i].prev!=-1)
{
System.out.print("\n vertex" +vlist[i].label+"is connected to"+vlist[vlist[i].prev].label);
minicost=minicost+vlist[i].len;
}
}
System.out.println("\nMinimum cost=" +minicost);
}
}
class Prims
{
public static void main(String args[])throws IOException
{
graph gr = new graph();
char c;
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(isr);
System.out.println("Enter the number of vertices");
int n = Integer.parseInt(br.readLine());
String t;
System.out.println("Enter the labels for the vertices");
for(int i=0;i<n;i++)
{
t = br.readLine();
c = t.charAt(0);
gr.addver(c);
}
System.out.println("Enter the number of edges");
int edg = Integer.parseInt(br.readLine());
System.out.println("Enter the vertices which you need to connect");
for(int j=0;j<edg;j++)
{
System.out.println("Enter the first vertex");
t = br.readLine();
c = t.charAt(0);
int start = gr.getind(c);
System.out.println("Enter the second vertex");
t = br.readLine();
c = t.charAt(0);
int end = gr.getind(c);
System.out.println("Enter the cost");
int cost = Integer.parseInt(br.readLine());
gr.addedge(start,end, cost);
}
System.out.println("Enter the source vertex");
t = br.readLine();
c = t.charAt(0);
int src = gr.getind(c);
gr.brfs(src);
}
}

GATE 2013 Preparation tips

2012-12-27T17:22:00.002-08:00

Important tips for Preparation:

Foremost thing is to prepare right from the day you want to plan for it. As it is a saying beginning is half success. If you wait for some good day to start, that good day will never come and eventually you will loose/waste your time gratuitously.

Candidate must collect GATE 2013 syllabus copy and find the subjects asked during examination and segregate the subjects based on past questions and mark weightage of each subjects.

GATE paper consists of that questions that are connected to specialization in M. Tech and M.E. therefore, when you start your preparation, just focus on concept not on theory derivation. Have preceding year questions papers handy to know which type of questions can be asked in the exam and contemplate on the topics that you skipped or you do not know in syllabus during your graduation.

Do not revise the topics that you already know as this will waste your time. Make a separate note of equations with 2 marks and grasp it along with its concepts.

Important dates

Last date for Submission of Online Application- September 30, 2012

Last date for request of change of city- November 20, 2012

Availability of admit card on Online Application Interface- December 5, 2012

GATE 2013 Online Examination- January 20, 2013

GATE 2013 Offline Examination- February 10, 2013

Announcement of results on Online Applicant Interface- March 15, 2013

Dijkstra algorithm's implementation in java (From one source to one destination in Graph)

2012-12-21T03:39:00.000-08:00

import java.io.*;
class Tnode
{
int prev;
int len;
char label;
boolean vis;
public Tnode(char lab)
{
label = lab;
prev=-1;
len=100000;
vis=false;
}
}
class graph
{
public final int MAX = 20;
public int nverts,i,min;
public Tnode vlist[];
public int adj[][];
public graph()
{
nverts = 0;
adj = new int[MAX][MAX];
vlist = new Tnode[MAX];
for(int i=0;i<MAX;i++)
{
for(int j=0;j<MAX;j++)
adj[i][j] = 0;
}
}
public void addver(char lab)
{
vlist[nverts++] = new Tnode(lab);
}
public void addedge(int start,int end,int cost)
{
adj[start][end] = cost;
}
public int getind(char l)
{
for(int i=0;i<nverts;i++)
if(vlist[i].label==l)
return i;
return (MAX+1);
}
public void brfs(int src, int dest)
{
int k;
int path[]=new int[20];
vlist[src].len=0;
vlist[src].vis=true;
k=src;
do{
for(i=0;i<nverts;i++)
{
if((adj[k][i]!=0)&&(vlist[i].vis==false))
{
if(vlist[k].len+adj[k][i]<vlist[i].len)
{
vlist[i].prev=k;
vlist[i].len=vlist[k].len+adj[k][i];
}
}
}
k=0;
min=100000;
for(i=0;i<nverts;i++)
{
if((vlist[i].vis==false)&&(vlist[i].len<min))
{
min=vlist[i].len;
k=i;
}
}
vlist[k].vis=true;
}while(k!=dest);
i=0;
k=dest;
int c=0;
do{
path[i++]=k;
c++;
k=vlist[k].prev;
}while(k>=0);
System.out.println("shortest path from source to destination is \n");
for(i=c-1;i>=0;i--)
System.out.print("-->"+vlist[path[i]].label);
System.out.println("\n shortest distance from source to destination is \n"+vlist[dest].len);
}
}

class Dijr
{
public static void main(String args[])throws IOException
{
graph gr = new graph();
char c;
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(isr);
System.out.println("Enter the number of vertices");
int n = Integer.parseInt(br.readLine());
String t;
System.out.println("Enter the labels for the vertices");
for(int i=0;i<n;i++)
{
t = br.readLine();
c = t.charAt(0);
gr.addver(c);
}
System.out.println("Enter the number of edges");
int edg = Integer.parseInt(br.readLine());
System.out.println("Enter the vertices which you need to connect");
for(int j=0;j<edg;j++)
{
System.out.println("Enter the first vertex");
t = br.readLine();
c = t.charAt(0);
int start = gr.getind(c);
System.out.println("Enter the second vertex");
t = br.readLine();
c = t.charAt(0);
int end = gr.getind(c);
System.out.println("Enter the cost");
int cost = Integer.parseInt(br.readLine());
gr.addedge(start,end, cost);
}
System.out.println("Enter the source vertex");
t = br.readLine();
c = t.charAt(0);
int src = gr.getind(c);
System.out.println("Enter the destination vertex");
t = br.readLine();
c = t.charAt(0);
int dest = gr.getind(c);
gr.brfs(src,dest);
}
}

GATE 2013 useful preparation tips

2012-12-20T07:24:00.001-08:00

GATE 2013 useful preparation tips

Get your fundamentals right:

It is important that you get your fundamentals right. GATE is one such exam which tests you on your fundamentals.

Take exercise:

After you gain full confidence in your fundamentals then go for the exercise for the each topic you studied, that will boast your confidence.

Solve previous year's papers:

To understand the pattern of the exam and get familiar with it, aspirants should solve the previous year's papers of GATE exam. It will also brushes up your basic concepts and exposes you to real exam pattern.

Solve mock test papers:

While preparing for the exam it is really important that you should solve as many papers as you can. Solve mock papers to understand the basic concepts. Mock test will give you the image of real paper and make you aware about what type of question will be asked in the exam. So try to solve as much papers as you can.

Books to refer:

Step 1: Go for the academic books (B.Tech/B.E books) first that will brush up your fundamentals.

Step 2: G.K. Publication for more than fundamentals preparation. Solve all the problems in the G.K. publication.

Step 3: After this try to solve the mock test from G.K. publication C.D.

Step 4: Try to solve as many previous papers as possible.

Note: The most important, have a balance on your mind in the exam hall, even if you prepare well but if you get tense in exam hall then you will sure loose.

PGECET-2012 Admissions Vacancy Position in SVU region as on 17-12-2012

2012-12-17T18:18:00.002-08:00

Left over seat in M.Tech Colleges in SVU region.
Student who are not qualified in PGECET or who are not allotted in PGECET Counselling can get seat in left over seats in spot admission.

To know the left over seat in the respective colleges Click here

PGECET-2012 Admissions Vacancy Position in AU region as on 17-12-2012

2012-12-17T18:17:00.002-08:00

Left over seat in M.Tech Colleges in AU region.
Student who are not qualified in PGECET or who are not allotted in PGECET Counselling can get seat in left over seats in spot admission.

To know the left over seat in the respective colleges Click here

PGECET-2012 Admissions Vacancy Position in OU region as on 17-12-2012

2012-12-17T18:16:00.002-08:00

Left over seat in M.Tech Colleges in OU region.
Student who are not qualified in PGECET or who are not allotted in PGECET Counselling can get seat in left over seats in spot admission.

To know the left over seat in the respective colleges Click here