techgig.com Code Wizards 2015 Contest

Organisation : Techgig
Announcement : Code Wizards 2015 Contest

Code Wizards 2015 Contest :

Concept :
Code Wizards 2015 is the biggest tech contest designed to identify the best programmers and tech enthusiasts Students across India.

The unique idea has been conceptualized by India’s No. 1 platform for IT people – TECHGIG.com.

Code Wizards 2015 gives a platform for all code junkies to showcase their coding talent. It gives coders:
“The Best Coder” recognition at the national level.
Sharpen your edge by competing against the best of the league and win attractive prizes worth more than 4.2 Lakhs.
The best Students programmers get an opportunity to be interviewed by Top companies.

The contest is open to coders for 10 different languages. Take your pick from Java, C, C++, C#, VB.net. Java Script, Python, Perl, Ruby and PHP.

Who can register :

The contest is open to :
(a) Participant should be the final year (2015-16 batch) student of either Engineering or MCA College approved by AICTE and UGC respectively.

(b) The contest is valid for student who is a resident of India only.

Dates and Contest Details :

Key dates to watch out:
Round 1 : 12th Oct – 06th Nov 2015
Round 2 : 06th Nov – 21st Nov 2015
Results : 23rd Nov 2015

How to participate?
You can register by completing the user registration form here – Click Here.

Contest Rules :

Code Wizard contest will have 2 rounds – Round1 & Round2
There are 2 types of participation in Code Wizard 2015 –
Intra College round
Open Round

In Intra College, participating Colleges will compete against each other to reach the Round 2. Participants who study in the respective registered participant Colleges need to register themselves via Intra College round registration.
In Open Round, participants from all other Colleges battle it out to reach the round 2.
If any participant is found having multiple profile while participating in the contest for any kind of reasons, he/she will be disqualified from the event.

How to Play the Code Wizards :

The individual will be able to access the problem statement once s/he clicks “Participate” button on the Coding Challenge box in the application. The application has a Live Coding Environment where the individual can write his/her code and compile the code.
Once the individual completes writing the code, a default test case will be used to check the code. To submit the code the individual must pass the default test case, however, individual does not score for passing the default test case.
To score marks, participants must clear default test cases which are built to check code. Each test case carries equal marks with equal weightage. The total of all the test cases is 100 marks.
Maximum two attempts are given in each problem. Higher score of the two attempts will be considered as final score.

How do you score in the Code Wizards Contest :

Each round will have 1 level.
Each level will have a mix of Multiple Choice Questions (MCQs) and a coding problem.
There will be 20 Multiple Choice Questions (MCQs).
In case of the MCQs, each correct answer will fetch the user 5 marks.
There will be a maximum of 2 attempts.
A coding problem will consist of a certain number of test cases. Each test case will carry 10 marks.

Scoring logic of level is as follows:

(Number of MCQs answered correctly x 5) = A
Total number of test cases answered correctly x 10 = B
Total score = A + B
The total score will be the cumulative score of the level.
If 2 or more users score the same, tie break logic would apply (Detailed below)
Participant will get 2 attempts per level to improve their scores. The higher score from both the attempts will be considered as the final score for that level.
There is no negative marking.

Tie breaker logic for coding contest :

Let’s take an example in order to understand tie breaker logic –

We have two users who have got equal marks –

LEVEL 1 CHECK – The user with higher marks in coding problem will be ranked higher. For example ‘A’ and ‘B’ has got following marks in subsequent level –
A B
MCQ 80 100
CODING 100 80
TOTAL 180 180

For instance A and B both have got equal marks in level

In this case following levels of checking will be done in order to determine the higher ranker –

LEVEL 2 CHECK (Attempt level check) – For instance ‘A’ has scored 90 marks in first attempt and ‘B’ has scored 90 marks in second attempt. In this case ‘A’ will be the higher ranker As ‘A’ got 90 marks in one attempt and ‘B’ got 90 marks in two attempts.

LEVEL 3 CHECK (Code execution time) – In case there is a tie in both LEVEL 1 CHECK and LEVEL 2 CHECK i.e. both A and B has scored 90 marks in same number of attempt. Now execution time of the code will be taken into consideration. For instance ‘A’s’ code takes 10 seconds for execution and ‘B’s’ code takes 12 seconds for execution. In this case ‘A’ will be the higher ranker.

LEVEL 4 CHECK (Time taken in submission) – In case there is a tie in LEVEL 1, LEVEL 2 and LEVEL 3 CHECK. Now total time taken by user in completing round 1 will be taken into consideration. For instance ‘A’ takes 25 minutes for completing his level and ‘B’ takes 30 minutes for completing his level. In this case ‘A’ will be the higher ranker.

LEVEL 5 CHECK (Participation time) – In case there is a tie in LEVEL 1, LEVEL 2, LEVEL 3 and LEVEL 4, the rule of early bird winner will be taken into consideration. For instance ‘A’ and ‘B’ fulfils all the criteria of Level 1,2,3,4. But ‘A’ has participated and completed his level before ‘B’, then in this case ‘A’ will be the higher ranker.