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Endianess – When it matters ?

We all knew what is an endianess in computer architectures , the big endian and little endian architectures works different. Big endian where most significant bit stored at smallest address where as in little endian least significant bit stored at smallest address.

The world’s most popular processor Intel is little endian machine, where world’s most oldest motorola processors are all big endian. All the network protocols , routers works in big endian architecture.

Most of the time endianess doesn’t come into picture , because programmers doesn’t deal with very low level bits. Ideally when I say int number = 4; the program runs properly in both kind of architecture. The reason is when the value is processed , it will be always in processor registers, not directly into memory. Endianess comes into picture when we deal with memory bits.

For example – If we are applying some effects or transform on an image by using shift operators or some standard APIs, then the things might go wrong. OpenGL APIs internally deal with shifting the bits in memory to perform certain operations in such cases the entire image may look inverted.

The common convention followed in networking is big endian , all TCP/IP protocols and other network protocols deal with big endian data. Those who work on networking should be aware of endianness. 


Let us take an example how endianness might be a problematic

Assume that a System 1 is sending an ip address to System 2 , If system 1 is 80×86 processor which is a little endian and System2 is SPARC machine which is a big endian. An ip address sent from System 1 , will be received by SPARC machine and converts it to big endian format.


If I consider , let us convert this to Hex value

ip address to Hex a logic =  take each octet , multiply it by 256 ^ n, where n is zero based index

= (1 * 256 ^ 0) + (2 * 256 ^ 1) + ( 168 * 256 ^ 2) + (192 * 256 ^ 3)  

= 1 + 512 + 11010048 + 3221225472

= 3232236033 ( decimal )

= C0A80201 ( Hex )

System 1 is sending value C0 A8 02 01 , but System 2 will interpret it as 01 02 A8 C0

This kind of issues happens due to endianness.




Recursion VS Iteration – An Analysis with fibonacci and factorial

Recursion VS Iteration – An Analysis with fibonacci and factorial

It is always difficult to choose one over the other , but recursive and iterative methods can be chosen wisely by analysing the algorithm with certain input values. Let us study the usage of recursive methods and let us analyse how recursive call works internally.

How Recursion  Works ?

The concept of stack frame comes into picture when we deal with recursion, for each call a stack frame will be created and the values will be pushed into stack. Once the recursive call reach the final condition, the program will start popping out from the stack frame and returns the values . A C program to find factorial of given number and analyse it using recursive method. 

Let us take an example –

Finding Factorial of a Number

long int count = 1;
long int factorial(long int n){
  count++;//to count how many times the method getting called
  if (n == 0) {
   return 1;
  return n * factorial(n - 1);

int main(int argc, const char * argv[])
   int n = 13;
   printf("Factorial of %d = %lu , number of time called %lu\n",n,factorial(n),count);
   return 0;

Let Analyse how it works ? 

When factorial(n) will be called every time , a stack frame will be created and the call will be pushed to stack, the entire call stack looks like below. The below image shows stack operations for a given recursive call.


If I give input as 10, the recursive method will be called 12 times,

If I give input as 14, the recursive method will be called 16 times,

If I generalise the number time of call then it will be – n + 2 times.

This looks ok, but let us take different example and try to compare both. A program to find fibonacci number and analyse it with respect to call stack.

Finding Fibonacci number 

long int count = 1;
long int fibonacci(int n){
    if (n <= 1) {
        return n;
        return fibonacci(n-1) + fibonacci(n - 2);
long int fibonacciIteration(int n) {
    int x = 0, y = 1, z = 1;
    for (int i = 0; i < n; i++) {
        x = y;
        y = z;
        z = x + y;
    return x;
int main(int argc, const char * argv[])
    int n = 4;
    printf("fib value recursive %lu ",fibonacci(n));
    n = 100; // let us give 100 as input for iterative method
    printf("\nIterative method %lu ",fibonacciIteration(100));
    printf("\nnumber of time called %lu\n",count);
    return 0;

Now the call stack looks as follows, memory layout of call stack for a given recursive call of fibonacci program. 

Call stack of Fibonacci

Call stack of Fibonacci

Let us analyse this example ,

If we observe the callstack , fib(2) , Fib(1) and Fib(0) called multiple times, which is not necessary as the value will be calculated .

For input n = 4 , the total number of times the fibonacci method called = 10.

Now let us take few more inputs and check how many times the function is getting called.
n = 5 => 16 times,
n = 6 => 26 times ,

If I generalise the number of time the function called will give us an astonishing result ,
number of times fib function calls = (2^n – C), Where C is constant, and very less ~ 2^n

Let us take some interesting inputs 🙂
n = 30 => 2692538 times
n = 31 => 4356618 times (just by increasing n by 1 , the number of times function called became so huge)

n = 42 => 866988874 times , Unbelievable !!! These results shows how recursive method is dangerous in this case.

n = 43 => my program not even stopping , even after 5 minutes .

Let us look on iterative method 

long int fibonacciIteration(int n) {
    int x = 0, y = 1, z = 1;
    for (int i = 0; i < n; i++) {
        x = y;
        y = z;
        z = x + y;
    return x;

This method will return , fibonacci value within few seconds even if input n = 100 .

Conclusion is , selecting between iterative and recursive methods should be decided based on the problem statement, the problem looks simple with simple inputs, but analysing the problem with bigger inputs will actually shows the big picture .

Update: Complexity of recursive Fibonacci –

The recursive equation for this algorithm is T(n)=T(n1)+T(n2)+Θ(1)

For this it will be T(n)=Θ(ϕn)

where ϕ is the golden ratio (ϕ=(1+5) / 2).

I strongly recommend to read Introduction to Algorithms 3rd edition : Purchase in FlipKart

A Video Tutorial on Analysis Of Recursive Function For Fibonacci is here

Understanding of computer Memory With a C Program

Memory is one of the important resource , we can say its a crucial resource in computers. In a limited amount of memory an operating system tries to satisfy all the applications and the need of memory.

Let us try to understand how a memory looks – the theory and an example simple program to understand the concept in a better way.

Memory –


 Understanding Of Memory

When we say a memory is allocated for an instance ! How exactly you visualise the memory and allocation ?

A memory is array of smallest unit of space which is nothing but a byte of space location. If a variable holds an integer, float or any type of data, it occupies certain set of such memory locations. Typically it looks like below.

Array of memory locations

In the above memory structure we can see how data is stored in memory.

The String “XYZ” , the number 15 , it looks like at location 12, number 15 is stored, starting from location 21, string XYZ is stored.

If I dig into the depth of it, every thing in computers will be converted into binary , so ‘X’ will be converted into binary which occupies 8 bits , similarly each character will be converted to its ASCII and the ASCII value will be converted to binary ( Ref – http://sticksandstones.kstrom.com/appen.html )

Let us consider one example and try to fill the memory locations 

char subjects[3][2][25] ;

The above character multidimensional array will holds 2 subjects of 3 semesters each of length 25 characters.

subjects[0] -> will give you 2 subjects of first sem, each subject of 25 characters.

subjects[1] -> will give you 2 subjects of second sem, each subject of 25 characters.

If I assign the subjects to this array , the subject names in memory looks like below


For better understanding just check the below diagram of memory layout

Memory Layout


How string copy becomes costly – An Analysis

Let us try an interesting example , a simple but an interesting analysis at the end.  Let me start with a requirement by taking an example .

Assume that you are getting ‘ N ‘ number of strings from an xml file , where ‘ N ‘ is really huge. Each string is again having too many characters. A library is being used to parse the strings from the xml file, library will return the strings one by one.

The library returns string and the length of the string in a callback method, our program has to copy the string to keep a copy of it.

Let me write a simple code for this , to make the code simple for better understanding I will avoid library usage and all other stuff. Just let us focus of core problem , that is copying ‘ N ‘ strings.

int main(){
  return 1;

void getStrings(){
   int numberOfCharacters = 2014; // this length will be returned by Library after parsing xml file
   char *string1 = (char*)malloc(numberOfCharacters); // to hold string
   strcpy(string1,"This string we will get  from library, for this example we are hard coding it");
   char *string2 = (char*)malloc(numberOfCharacters); // to hold string
   strcpy(string2,string1); // have a  copy of string returned by library .

The above problem looks very simple, but the last strcpy() call will become costly for you since number of strings are too big. 

How strcpy is costly ? 

strcpy typically looks like below –

char* strcpy(char * string1, const char * string2){
  char * originalStringPointer = string1;
  while(*string2 != '\0'){
    *string1++ = *string2++;
 *string1 = '\0';
 return originalStringPointer;

So behind the screen for every string copy ( for all your N strings, where N is too big ) , strcpy method iterates each character and copies .

Let us take one example and analyse this – 

Analysis I –

If I have 100 strings all of same length say 10

then the number of times the loop in strcpy runs is 

100 * 10 = 1000  times, this doesn’t make much difference .

Analysis II – 

Let us take bigger value , 

If I have 3000 strings of length 100 each 

then your loop in strcpy runs 3000 * 100 = 3,00,000 times 

3,00,000 times is big, also we assumes all strings are of same size and that is just 100, but in realworld example that might be different again.

So in such a scenario your strcpy becomes very costly for you, your application performs really bad that too if you didn’t handle this case properly. If by chance you are reading all the strings during the app launch, your application takes several seconds to show the first screen.

A clever programmer finds a better way to achieve this with better way.

Better approach for better performance –

If you observe the problem statement, just notice that we have a string and its length both. so instead of using strcpy we can go for memcpy which actually just transfers chunk of data from one location to another location rather than copying it character by character.

so just replacing strcpy() by memcpy() avoids 3,00,000 loop executions in your second analysis .

your getStrings method looks like below now

void getStrings(){
  int numberOfCharacters = 2014; // this length will be returned by Library after parsing xml file
  char *string1 = (char*)malloc(numberOfCharacters); // to hold string
  strcpy(string1,"This string we will get  from library, for this example we are hard coding it");
  char *string2 = (char*)malloc(numberOfCharacters); // to hold string
  memcpy(string2,string1,numberOfCharacters); // have a  copy with memcpy.


It is always a very good practice to find best possible way to do something, even a one line code which runs hundreds, thousands of times will become bottleneck. 

Introduction to Struts

Struts is an open source web application framework which will be used to develop the web-
based applications. Struts framework was implemented based on 3 popular design patterns.

  • MVC design pattern.
  • Front Controller design pattern.
  • Application Controller design pattern.

MVC design pattern :-

There are 3 layers in MVC Design pattern

I.  Presentation Layer

II.  Controller Layer

III.  Model Layer


I Presentation Layer :-

It contains the presentation logic and can be implemented with JSP or with any other presentation frameworks like flex, velocity etc. Struts Presentation Layer contains following components.

1)     JSPs/HTMLs
2)      Struts custom tags
3)      Form Beans
4)      Message Bundles

1)     JSPs/HTMLs:

In struts presentation layer JSPs will be used to display the data to the client and also used to receive the input data from the client.

2)      Struts custom tags:

Apache has implemented various custom tags for the struts based application development and are categorized as follows.

  • Html tag libraries
  • Bean tag libraries
  • Logic tag libraries
  • Tiles tag libraries
  • Template tag libraries
  • Nested tag libraries
  1. Form Beans:

Form bean is a simple java bean styled class which stores the data. When we are writing the form bean java class, the bean class must be a subclass of one of the following.

  • ActionForm
  • DynaActionForm
  • ValidatorForm
  • DynaValidatorForm
  • ValidatorActionForm
  • DynaValidatorActionForm


  1. Message Bundles:

Message Bundles are nothing but property files which contain key value pair. Mainly message bundles will be used to achieve i18n (internationalization).

II Controller Layer :-

The Controller portion of the application is focused on receiving requests from the client (typically a user running a web browser), deciding what business logic function is to be performed, and then delegating responsibility for producing the next phase of the user interface to an appropriate View component. Following are the various components which we use in struts controller layer.   

   i)   ActionServlet

   ii)  RequestProcessor

   iii)  Action class

 i) ActionServlet:

ActionServlet is the one and only servlet for the entire struts application which is implemented based on Front controller design pattern and this servlet is loaded during server startup time. ActionServlet is responsible for receiving all incoming request and delegating to RequestProcessor. It also initialize the struts-config.xml file.

 ii)  RequestProcessor

RequestProcessor is responsible for processing all incoming request. It is implemented on Application Controller Design pattern. RequestProcessor is responsible for identifying the incoming action class and its form bean, it is also responsible for managing life cycle of form bean. After getting response, it manages to return to corresponding jsp by using ActionForward.

iii)  Action class

Action Class is the beginning of application’s business logic.

III Model Layer : –

Since there are no predefined components for model layer in struts like presentation layer and controller layer, so struts will not deal with model layer. But struts allows us to make use of any of the following model layer.

i)    Simple JDBC component

ii)   DAO + Hibernate

iii)  DAO + JDBC

iv)  Spring

v)   Web services     etc…


Capture1flow diagram showing struts flow

Steps for the flow         

  • When we submit the request after entering value in jsp form, request will be submitted to ActionServlet which is configured in web.xml.
  • This ActionServlet is responsible for initializing struts-config.xml and this struts-config.xml file is     configured in web.xml.
  • ActionServlet delegates the request to RequestProcessor
  • RequestProcessor takes the incoming request uri and tries to find the matching <action> tag in struts-config.xml. If no matching action is found, then client will get “cannot retrieve mapping for action”.
  • Once matching action is found for the incoming request uri, then corresponding “name” will be taken and tries to find the corresponding <form-bean> under <form-beans> configuration. If no matching <form-bean> is found then client will get the error called “cannot retrieve definition for form beans”.
  •  Once matching is found, RequestProcessor is responsible for managing form-bean life cycle.
  •  RequestProcessor takes the <action> “type” which is Action java class and create the instance    for the first time. With that Action class instance, execute() method  will be called.
  •  Once execute() method is completed ActionForward will be returned to RequestProcessor with      the forward name associated with the ActionForward object and RequestProcessor will try to  identify corresponding <forward> inside struts-config.xml file.
  •  Once the forward is found then its path will be taken and will be forwarded to the client.
  •   RequestProcessor takes the <action> “type” which is Action java class and create the instance     for the first time. With that Action class instance, execute() method  will be called.
  •  Once execute() method is completed ActionForward will be returned to RequestProcessor with  the forward name associated with the ActionForward object and RequestProcessor will try to  identify corresponding <forward> inside struts-config.xml file.
  •  Once the forward is found then its path will be taken and will be forwarded to the client.


A Bird eye on Education System

This article is neither to judge quality of education nor to conclude on any plans or steps that have been taken by Government of India(GOI) to improve the literacy rate in the country.

The purpose of this article is just to provide certain facts and I would like to leave analysis of these facts to reader. Report on your analysis is welcomed through comments; do share your thoughts related to this article. All the data is collected from the reports which have been published by Government of India for public domain.

India is struggling with many issues, issues with respect to caste, religion, politics, languages , the mindset of people across the country varies a lot with respect to all the above factors. Despite of being such a diverse society, I would say India has achieved and reached many milestones in every sector. Considering education sector for this article, let us see what the report says.

Find out best realtime examples, interesting facts on concepts and all the formulas. The best mobile application for 10+ students (PUC I and II year students)


Android Mobile Application for 10+ students

Android Mobile Application for 10+ students

Achievements during 11th Five-year plan period

The result of 2011 census reveals that despite an impressive decadal increase of 9.2 percent points in literacy, national literacy levels have risen to no more than 74.0 percent (from 64.8 percent in 2001) Only 15 States/Union Territories, namely Kerala, Lakshadweep, Mizoram, Tripura, Goa, Daman & Diu, Puducherry, Chandigarh, Delhi, Andaman & Nicobar Islands, Himachal Pradesh, Maharashtra, Sikkim, Tamil Nadu and Nagaland could achieve 80 percent or above literacy rate.

The 2011 Census has shown that female literacy has increased much more than male literacy. While male literacy rate increased by 6.86 percent 279 points from 75.26 percent in 2001 to 82.14 in 2011, the female literacy increased by 11.79 percent points from 53.67 to 65.46 percent during the same period. Thus, by the end of the 11th Five Year Plan in 2012, the three Plan Targets would not have been achieved: overall literacy rate being short by five percent points.

Increase in literacy between 2001-2011 at National level:

There has been a substantial progress in literacy with the planned intervention and sustained effort from 64.83 per cent in 2001 to 74.04 per cent in 2011, an increase of 9.21 percentage points.


Sakshar Bharat

The Prime Minister launched Saakshar Bharat, a centrally sponsored scheme of Department of School Education and Literacy (DSEL), Ministry of Human Resource Development (MHRD), Government of India (GOI), on the International Literacy Day, 8th September, 2009.

Despite significant accomplishments of the Mission, illiteracy continues to be an area of national concern. 2001 census had revealed that there were still 259.52 million illiterate adults (in the age group of15 +) in the country.

Fund utilization and capacity building to achieve the milestone in education

During the financial year 2011-12, a total amount of Rs.488.50 crores was provided for Sakshar Bharat programme as Central share. An amount of Rs.429 crores was released till December 2011 to State Level Mission Authorities (SLMAs). A total of 1098.33 crores was released by NLMA to SLMAs during the last three years.

Government is keep trying and funding to increase the literacy rate within the country, in this process government has moved from SSA(Sarva Shiksha Abhiyan) Program to RTE ( Right to Education ).

RTE-SSA Financial Allocation

The Governmentn of India approved an outlay of Rs 71,000 crore for SSA in the 11th Five year plan. Considering the increased requirements, ,the Government of India approved an outlay Rs 2,31,233 crore for the combined RTE-SSA programme. This increased outlay was for a five year period from 2010-11 to 2014-15 to be shared between the Center and the State on a 65:35 ratio (90:10 for North Eastern States). The Finance Ministry allotted Rs 25,555 crore for the RTE-SSA programme for 2012-13. Just check the report chart below, government has spent 19,103,22cr upto January 2012.


Considering all the facts, percentile of literacy between 2001 – 20011, a 10 year period, the amount of money spent on it, the results may not be satisfactory, the reason is that focusing only on spending and opening schools doesn’t give the expected result, but focusing and executing the program towards quality education does bring the change. The path towards achieving really quality education is not an easy, more and more educated people has to come front to bring the change, it is not just the effort of government.

Interesting way of explaining Pythagoras Theorem

Pythagoras Theorem Explained – Mathemagic with Bawa

Sources of data




An yearly report



Importance of DB Design – An analysis

The most important step in any big application development life cycle is database design, If it requires any data storage. We have come across many developers who will never think much before finalising the database design. Some people will simply add fields into the table , some people simply define the data type without considering the type of data stored in real time. Let’s take a simple example and do analysis which shows big picture of database design importance with small set of data.

Consider the following example , a database designed for a simple requirement and then the analysis of the same.

Example – Requirement

Design a database for a web application where registered users will post articles. Any one can post any article, each article might fall under some category, A Category is again defined under a department and a stream. There will be Maximum 100 departments and 100 streams in the system.

For example , an article on “ Analysis of computer simulated data on wheels of a plane “, this article might fall under “Aerospace” Department, Physics stream .

Another article on “ Calculation of Voltage variations within the cooling system “ ,this article might fall under “Aerospace” Department , Electrical stream.

If we consider the above requirement and two examples , the following design sounds good – ( Note the database designed here is simple and just taken for simple analysis )

DB Design

Considering the above design, and observe the department Id and Stream id fields in respective tables are defined as INT ,  The values of these two fields will never go beyond 100 as per the requirement. The first mistake is the datatype defined INT is not right. It can be declared as TINYINT(1) which holds value upto +127 , since both department id and stream id are declared as INT , extra 3 bytes is wasted for each record, because INT takes 4 bytes, and TINYINT takes 1 byte memory.

Let us analyse this scenario with simple real time example –

Analysis I –

If by chance each table ( department and stream ) holds 100 rows , then considering above storage details of each data type , for each row 3 bytes is wasted and hence if there are 100 records , for each table 300 bytes of data wasted.

So overall 600 bytes wasted from these two tables for just 100 records. This doesn’t make any difference because it is just a matter of few hundred bytes.

But If I change the above ArticePost table as follows ( that kind of mistakes usually happens when the database is bigger which will be having 100+ tables , since here only few tables are there then there is very less chance of such mistakes ).

Analysis II –

Modified ArticlePost table


For a quick access if I put both department id and stream id in ArticlePost table itself , the things will go worse !

Now consider 100 articles for each department and for each stream then total number of rows in ArticlePost = 10 * 100 * 100 = 100,000 articles, this is a simple math total ArticlePosts = (Articles for one dept/stream) * number of dept * number of streams 

If we consider 100,000 number of rows in articles then the things will be worse ! For each row 600 bytes of memory waste as per Analysis I , so now for 100,000 rows how much memory waste ?

100,000 * 600 = 600,00,000 bytes  = 57.220 MB memory wasted.

In real time , 100,000 of rows is not a big one , We should consider one thing very seriously here , 57.220 MB memory waste just because of two extra fields in a table and with improper usage of datatype .

Have a look on Data types and storage capacity from Mysql official documentation



Update –

Just observe how the data types in RDBMS are , INT(4) , INT(10), VARCHAR(25)

in general INT(N) , where N decides how much space it should consume ?.

Remember what we studied in C ? Can we relate this to C bit fields concept ?

Assume that I wanted to hold a status of the request in ma C program written for managing a government office complaint management system.

struct complaintStatus{
  int isComplaintRaised;
  int isComplainReceived;
  int isComplaintResolved;
} status;

Since We are holding only status , it not required to have an int of 4 bytes ! only one byte is sufficient to hold either 0 or 1 , where 0 indicates NO , 1 indicates YES.

Let me modify the above Struct 

struct complaintStatus{
 int isComplaintRaised : 1;
 int isComplainReceived : 1;
 int isComplaintResolved : 1;
} status;

By above structure your compiler will restrict the amount of memory used by status .