The Great Mind Challenge (TGMC) is a programming contest by IBM, India to help engineering students in testing their skills. Student Engineers interested to join software industry can get lot of exposure from this.
How this can help Engineering/programming students ?
This contest is made for students interested in programming only. Biggest opportunity for final year students as they can use this for final year project and i believe this can change your future. You can grab an opportunity to work with IM (Information Management software) team of IBM.
Launched in 2004, TGMC is an innovative program that addresses the need to better educate millions of students for a more competitive information technology (IT) workforce by partnering with colleges and universities.
Every year, the growth and reach of TGMC has increased, seeing a marked increase in the number of students who are keen to participate in such an endeavor. In fact, TGMC is now listed in the Limca Book of Records as the largest technological contest of its kind – testament to the fact that there is a need in today’s competitive world for such an initiative that supports, guides, and challenges students to higher goals.
Universities provide excellent learning environments, but find it difficult to teach students about an IT transformation in progress unless those driving the transformation – IT companies at the helm of transformation – provide real-time input and support. TGMC is an out reach of the IBM Academic Initiative, a global program that facilitates the collaboration between IBM and educators to teach students the IT skills they need to be competitive and keep pace with changes in the workplace.
The IBM Academic Initiative includes an online portal that provides access to software, training, and course materials, most at no charge. Available within the Academic Initiative web site, IBM offers hundreds of resources for integration into college curricula to help teach students how to master the fast-growing market of open technologies. Through this initiative, IBM is working with students to build technology and science skills.
Have questions? Call us toll-free at 1800-425-9366.
We are available from 8 a.m. to 6 p.m. Monday through Friday.
You can also visit our FAQs, e-mail us at tgmc@in.ibm.com or post a question on our group message board.
Hello Friends,
Have you ever been in a situation where you as a developer knows that your data is stored in your table rows, but you would like to present the data as a column ??
There are many way to handle this situation. Here I am sharing few of them which I know.?
Below is the sample table and data :
————————————————————
[sql]CREATE TABLE yr_table (
Roll_No NUMBER,
Subject VARCHAR2(20),
Marks NUMBER
);
[/sql]
[sql]insert into yr_table VALUES (1212324,’MTH’,90);
insert into yr_table VALUES (1212324,’PHY’,72);
insert into yr_table VALUES (1212324,’CHE’,85);
insert into yr_table VALUES (1212324,’BIO’,78);
insert into yr_table VALUES (1212334,’MTH’,85);
insert into yr_table VALUES (1212334,’PHY’,65);
insert into yr_table VALUES (1212334,’CHE’,74);
insert into yr_table VALUES (1212672,’MTH’,88);
insert into yr_table VALUES (1212672,’PHY’,42);
insert into yr_table VALUES (1212672,’BIO’,12);
COMMIT;
[/sql]
——————————————————————-
Now I would like to represent above data into below format :
We can achieve above output in different way.
# Method 1 -> Using DECODE function
[sql]SELECT ROLL_NO,
MAX( DECODE( SUBJECT , ‘MTH’ , MARKS ) ) AS MTH ,
MAX( DECODE( SUBJECT , ‘PHY’ , MARKS ) ) AS PHY ,
MAX( DECODE( SUBJECT , ‘CHE’ , MARKS ) ) AS CHE ,
MAX( DECODE( SUBJECT , ‘BIO’ , MARKS ) ) AS BIO
FROM yr_table
GROUP BY ROLL_NO ORDER BY 1 ;?
[/sql] To understand the above query, first see the below output of above query without using MAX function then you can understand how above query works.
# Method 2 -> Using CASE? statement
[sql]SELECT ROLL_NO,
MAX( CASE WHEN SUBJECT = ‘MTH’ THEN MARKS END) MTH ,
MAX( CASE WHEN SUBJECT = ‘PHY’ THEN MARKS END) PHY ,
MAX( CASE WHEN SUBJECT = ‘CHE’ THEN MARKS END) CHE ,
MAX( CASE WHEN SUBJECT = ‘BIO’ THEN MARKS END) BIO
FROM yr_table
GROUP BY ROLL_NO ORDER BY 1 ;?
[/sql] Here CASE statement works same as DECODE function.
# Method 3-> Using PIVOT Operator?
The PIVOT and the UNPIVOT operators were introduced in Oracle version 11g. The PIVOT operator takes data in separate rows, aggregates it and converts it into columns. The following query will give the same result as the query above, just by using the PIVOT operator.
[sql]SELECT * FROM yr_table
PIVOT (
MAX ( MARKS ) FOR (SUBJECT) IN (‘MTH’ AS MTH, ‘PHY’ AS PHY, ‘CHE’ AS CHE, ‘BIO’ AS BIO)
)
ORDER BY 1??
[/sql] You can check below link for more clarification on PIVOT Operator.http://www.oracle.com/technetwork/articles/sql/11g-pivot-097235.html
# Method 4-> Using WITH clause and PIVOT Operator? ?
[sql]WITH TMP AS (
SELECT ROLL_NO,SUBJECT , MARKS FROM yr_table
)
SELECT * FROM TMP
PIVOT (
MAX(MARKS) FOR (SUBJECT) IN (‘MTH’ AS MTH, ‘PHY’ AS PHY, ‘CHE’ AS CHE, ‘BIO’ AS BIO)
)
ORDER BY 1
[/sql]
The WITH clause, was added into the Oracle SQL syntax in Oracle 9.2 . The WITH clause can be used to reduce repetition and simplify complex SQL statements. Here don’t get confuse with WITH Clause; just think that it create a temporary table which we can use it in select statement.
# Method 5-> Using WITH clause and Sub-query
[sql]WITH TMP AS (
SELECT ROLL_NO,SUBJECT , MARKS FROM yr_table
)
SELECT Y.ROLL_NO ,
( SELECT TMP.MARKS FROM TMP WHERE TMP.ROLL_NO = Y.ROLL_NO AND TMP.SUBJECT = ‘MTH’) AS MTH ,
( SELECT TMP.MARKS FROM TMP WHERE TMP.ROLL_NO = Y.ROLL_NO AND TMP.SUBJECT = ‘PHY’) AS PHY ,
( SELECT TMP.MARKS FROM TMP WHERE TMP.ROLL_NO = Y.ROLL_NO AND TMP.SUBJECT = ‘CHE’) AS CHE ,
( SELECT TMP.MARKS FROM TMP WHERE TMP.ROLL_NO = Y.ROLL_NO AND TMP.SUBJECT = ‘BIO’) AS BIO
FROM (SELECT DISTINCT ROLL_NO FROM yr_table ) Y
ORDER BY 1 ;
[/sql] # Method 6-> Using Multiple Joins
[sql]SELECT DISTINCT
A.ROLL_NO ,
B.MARKS AS MTH ,
C.MARKS AS PHY ,
D.MARKS AS CHE ,
E.MARKS AS BIO
FROM yr_table A
LEFT JOIN yr_table B
ON A.ROLL_NO = B.ROLL_NO AND B.SUBJECT = ‘MTH’
LEFT JOIN yr_table C
ON A.ROLL_NO = C.ROLL_NO AND C.SUBJECT = ‘PHY’
LEFT JOIN yr_table D
ON A.ROLL_NO = D.ROLL_NO AND D.SUBJECT = ‘CHE’
LEFT JOIN yr_table E
ON A.ROLL_NO = E.ROLL_NO AND E.SUBJECT = ‘BIO’
ORDER BY 1;
[/sql]
Friends, please share your knowledge as well, if you know any other method.
If drupal is not working in your Windows(like if you have tried installing drupal but it’s giving some errors). Try these (worked for me) –
1. Increase PHP’s memory settings by doing–
a) Click on the WAMP icon in your system tray (or wherever it is just click on it). It will show you all the options like Apache, Mysql, PHPadmin, Localhost etc.
b) Select Config files (in newer versions ) or PHP (in old versions) —-à then click on PHP.ini
c) Go to line “upload_max_filesize=2M”, change it to 32 M or more
d) Save and restart WAMP
2. Got an ERROR : C:\WAMP\www\includes\file.inc on line 911 – Do this :Go to the line where it is written “elseif ($depth>=$min_depth && ereg($mask, $file))” change ereg to mb_ereg
3. Setup Database problem
a) First create a database named drupal or whatever you want.
b) Enter the database username as root and password “”[null]
Keep in mind : When you are renaming the default_settings.php to settings.php then make sure you are copying the content of default_settings.php & renaming it to settings.php.
Keep this file in sites\default. After doing all this I got my drupal site , hoping the same for you. Cheers!
Normalization Resolved Normalization is one of the favorite topics of interviewee. It does not matter whether you have mentioned DBMS in your resume or not .This question is going to come and the funny part is that all of us know
what is normalization?
What are the different types of normalization?
So when this question on being asked the interviewer who have already prepared for it start with the history of normalization and end with the geography of normalization but when the next question for which they have not prepared i.e. apply normalization in real case scenario.
Now here comes the real part of normalization and just because of not proper concepts, people end up confusing themselves. So the idea is to not only to get familiar with normalization but also how to apply it in real time scenario.
What is Normalization?
Database designed based on ER model may have some amount of inconsistency, ambiguity and redundancy. To resolve these issues some amount of refinement is required. This refinement process is called as Normalization. I know all of you are clear with the definition, let’s go with :
what is the need of normalization?
What are the problems we can face if we proceed without normalization?
What are the advantages of normalization?
Asking question to oneself is the best way to get familiar with all the concepts.
The need of Normalization
I am going to show you one simple E-R model database.
Student Details
Course Details
Result details
1001 Ram 11/09/1986
M4 Basic Maths 7
11/11/2004 89 A
1002 Shyam 12/08/1987
M4 Basic Maths 7
11/11/2004 78 B
1001 Ram 23/06/1987
H6 4
11/11/2004 87 A
1003 Sita 16/07/1985
C3 Basic Chemistry 11
11/11/2004 90 A
1004 Gita 24/09/1988
B3 8
11/11/2004 78 B
1002 Shyam 23/06/1988
P3 Basic Physics 13
11/11/2004 67 C
1005 Sunita 14/09/1987
P3 Basic Physics 13
11/11/2004 78 B
1003 Sita 23/10/1987
B4 5
11/11/2004 67 C
1005 Sunita 13/03/1990
H6 4
11/11/2004 56 D
1004 Gita 21/08/1987
M4 Basic Maths 7
11/11/2004 78 B
In first look the above table is looking so arranged and well in format but if we try to find out what exactly this table is saying to us , we can easily figure out the various anomalies in this table . Ok let me help you guys in finding out the same.
Insert Anomaly: We cannot insert prospective course which does not have any registered student or we cannot insert student details that is yet to register for any course.
Update Anomaly: if we want to update the course M4’s name we need to do this operation three times. Similarly we may have to update student 1003’s name twice if it changes.
Delete Anomaly: if we want to delete a course M4 , in addition to M4 occurs details , other critical details of student also will be deleted. This kind of deletion is harmful to business. Moreover, M4 appears thrice in above table and needs to be deleted thrice.
Duplicate Data: Course M4’s data is stored thrice and student 1002’s data stored twice .This redundancy will increase as the number of course offerings increases.
Process of normalization:
Before getting to know the normalization techniques in detail, let us define a few building blocks which are used to define normal form.
Determinant : Attribute X can be defined as determinant if it uniquely defines the value Y in a given relationship or entity .To qualify as determinant attribute need NOT be a key attribute .Usually dependency of attribute is represented as X->Y ,which means attribute X decides attribute Y.
Example: In RESULT relation, Marks attribute may decide the grade attribute .This is represented as Marks->grade and read as Marks decides Grade.
Marks -> Grade
In the result relation, Marks attribute is not a key attribute .Hence it can be concluded that key attributes are determinants but not all the determinants are key attributes.
Functional Dependency: Yes functional dependency has definition but let’s not care about that. Let’s try to understand the concept by example. Consider the following relation :
IName- Name of the instructor who delivered the course
Room#-Room number which is assigned to respective instructor
Marks- Scored in Course Course# by student Student #
Grade –Obtained by student Student# in course Course #
Student#,Course# together (called composite attribute) defines EXACTLY ONE value of marks .This can be symbolically represented as
Student#Course# Marks
This type of dependency is called functional dependency. In above example Marks is functionally dependent on Student#Course#.
Other Functional dependencies in above examples are:
Course# -> CourseName
Course#-> IName(Assuming one course is taught by one and only one instructor )
IName -> Room# (Assuming each instructor has his /her own and non shared room)
Marks ->Grade
Formally we can define functional dependency as: In a given relation R, X and Y are attributes. Attribute Y is functional dependent on attribute X if each value of X determines exactly one value of Y. This is represented as :
X->Y
However X may be composite in nature.
Full functional dependency: In above example Marks is fully functional dependent on student#Course# and not on the sub set of Student#Course# .This means marks cannot be determined either by student # or Course# alone .It can be determined by using Student# and Course# together. Hence Marks is fully functional dependent on student#course#.
CourseName is not fully functionally dependent on student#course# because one of the subset course# determines the course name and Student# does not having role in deciding Course name .Hence CourseName is not fully functional dependent on student #Course#.
Student#
Marks
Course#
Formal Definition of full functional dependency : In a given relation R ,X and Y are attributes. Y is fully functionally dependent on attribute X only if it is not functionally dependent on sub-set of X.However X may be composite in nature.
Partial Dependency: In the above relationship CourseName,IName,Room# are partially dependent on composite attribute Student#Course# because Course# alone can defines the coursename, IName,Room#.
Room#
IName
CourseName
Course#
Student#
Formal Definition of Partial dependency: In a given relation R, X and Y are attributes .Attribute Y is partially dependent on the attribute X only if it is dependent on subset attribute X .However X may be composite in nature.
Transitive Dependency: In above example , Room# depends on IName and in turn depends on Course# .Here Room# transitively depends on Course#.
IName
Room#
Course#
Similarly Grade depends on Marks,in turn Marks depends on Student#Course# hence Grade
Fully transitively depends on Student#Course#.
Key attributes : In a given relationship R ,if the attribute X uniquely defines all other attributes ,then the attribute X is a key attribute which is nothing but the candidate key.
Ex: Student#Course# together is a composite key attribute which determines all attributes in relationship REPORT(student#,Course#,CourseName,IName,Room#,Marks,Grade)uniquely.Hence Student# and Course# are key attributes.
Types of Normal Forms
First Normal Form(1NF)
A relation R is said to be in first normal form (1NF) if and only if all the attributes of the relation R are atomic in nature
Student Details
Course Details
Result details
1001 Ram 11/09/1986
M4 Basic Maths 7
11/11/2004 89 A
1002 Shyam 12/08/1987
M4 Basic Maths 7
11/11/2004 78 B
1001 Ram 23/06/1987
H6 4
11/11/2004 87 A
1003 Sita 16/07/1985
C3 Basic Chemistry 11
11/11/2004 90 A
1004 Gita 24/09/1988
B3 8
11/11/2004 78 B
1002 Shyam 23/06/1988
P3 Basic Physics 13
11/11/2004 67 C
1005 Sunita 14/09/1987
P3 Basic Physics 13
11/11/2004 78 B
1003 Sita 23/10/1987
B4 5
11/11/2004 67 C
1005 Sunita 13/03/1990
H6 4
11/11/2004 56 D
1004 Gita 21/08/1987
M4 Basic Maths 7
11/11/2004 78 B
Table shown above Student Details ,Course Details and Result Details can be further divided. Student Details attribute is divided into Student#(Student Number) , Student Name and date of birth. Course Details is divided into Course# ,Course Name,Prerequisites and duration. Similarly Results attribute is divided into DateOfexam,Marks and Grade.
Second Normal Form (2NF)
A relation is said to be in Second Normal Form if and only If :
It is in the first normal form ,and
No partial dependency exists between non-key attributes and key attributes.
Let us re-visit 1NF table structure.
Student# is key attribute for Student ,
Course# is key attribute for Course
Student#Course# together form the composite key attributes for result relationship.
Other attributes are non-key attributes.
To make this table 2NF compliant, we have to remove all the partial dependencies.
StudentName and DateOfBirth depends only on student#.
CourseName,PreRequisite and DurationInDays depends only on Course#
DateOfExam depends only on Course#.
To remove this partial dependency we need to split Student_Course_Result table into four separate tables ,STUDENT ,COURSE,RESULT and EXAM_DATE tables as shown in figure.
STUDENT TABLE
Student #
Student Name
DateofBirth
1001
Ram
Some value
1002
Shyam
Some value
1003
Sita
Some value
1004
Geeta
Some value
1005
Sunita
Some value
COURSE TABLE
Course#
CourseName
Duration of days
C3
Bio Chemistry
3
B3
Botany
8
P3
Nuclear Physics
1
M4
Applied Mathematics
4
H6
American History
5
B4
Zoology
9
RESULT TABLE
Student#
Course#
Marks
Grade
1001
M4
89
A
1002
M4
78
B
1001
H6
87
A
1003
C3
90
A
1004
B3
78
B
1002
P3
67
C
1005
P3
78
B
1003
B4
67
C
1005
H6
56
D
1004
M4
78
B
EXAM DATE Table
Course#
DateOfExam
M4
Some value
H6
Some value
C3
Some value
B3
Some value
P3
Some value
B4
Some value
In the first table (STUDENT) ,the key attribute is Student# and all other non-key attributes, StudentName and DateOfBirth are fully functionally dependant on the key attribute.
In the Second Table (COURSE) , Course# is the key attribute and all the non-key attributes, CourseName, DurationInDays are fully functional dependant on the key attribute.
In third table (RESULT) Student#Course# together are key attributes and all other non key attributes, Marks and Grade are fully functional dependant on the key attributes.
In the fourth Table (EXAM DATE) Course# is the key attribute and the non key attribute ,DateOfExam is fully functionally dependant on the key attribute.
At first look it appears like all our anomalies are taken away ! Now we are storing Student 1003 and M4 record only once. We can insert prospective students and courses at our will. We will update only once if we need to change any data in STUDENT,COURSE tables. We can get rid of any course or student details by deleting just one row.
Let us analyze the RESULT Table
Student#
Course#
Marks
Grade
1001
M4
89
A
1002
M4
78
B
1001
H6
87
A
1003
C3
90
A
1004
B3
78
B
1002
P3
67
C
1005
P3
78
B
1003
B4
67
C
1005
H6
56
D
1004
M4
78
B
We already concluded that :
All attributes are atomic in nature
No partial dependency exists between the key attributes and non-key attributes
RESULT table is in 2NF
Assume, at present, as per the university evaluation policy,
Students who score more than or equal to 80 marks are awarded with “A” grade
Students who score more than or equal to 70 marks up till 79 are awarded with “B” grade
Students who score more than or equal to 60 marks up till 69 are awarded with “C” grade
Students who score more than or equal to 50 marks up till 59 are awarded with “D” grade
The University management which is committed to improve the quality of education ,wants to change the existing grading system to a new grading system .In the present RESULT table structure ,
We don’t have an option to introduce new grades like A+ ,B- and E
We need to do multiple updates on the existing record to bring them to new grading definition
We will not be able to take away “D” grade if we want to.
2NF does not take care of all the anomalies and inconsistencies.
Third Normal Form (3NF)
A relation R is said to be in 3NF if and only if
It is in 2NF
No transitive dependency exists between non-key attributes and key attributes.
In the above RESULT table Student# and Course# are the key attributes. All other attributes, except grade are non-partially , non – transitively dependant on key attributes. The grade attribute is dependant on “Marks “ and in turn “Marks” is dependent on Student#Course#. To bring the table in 3NF we need to take off this transitive dependency.
Student#
Course#
Marks
1001
M4
89
1002
M4
78
1001
H6
87
1003
C3
90
1004
B3
78
1002
P3
67
1005
P3
78
1003
B4
67
1005
H6
56
1004
M4
78
UpperBound
LowerBound
Grade
100
95
A+
94
90
A
89
85
B+
84
80
B
79
75
B-
74
70
C
69
65
C-
After Normalizing tables to 3NF , we got rid of all the anomalies and inconsistencies. Now we can add new grade systems, update the existing one and delete the unwanted ones.
Hence the Third Normal form is the most optimal normal form and 99% of the databases which require efficiency in
The Great Mind Challenge (TGMC) is a programming contest by IBM, India to help engineering students in testing their skills. Student Engineers interested to join software industry can get lot of exposure from this.
.jpg)
If drupal is not working in your Windows(like if you have tried installing drupal but it’s giving some errors). Try these (worked for me) –
Visitor Rating: 5 Stars