Monday, March 11, 2019

Quick Quiz - 1
BRIGHTEN BRAINS
3rd SEM 
COMPUTER ORGANIZATION
DATE:06/10/2018












Prepared and Executed by
Kotreshi SN
Assistant Professor,
Department of CSE,GMIT, Davangere 577006
Students Participation in quiz images



Winning Teams:
I - Prize
Pooja Walishettar
Bhoomika Bhat
Samatha A
Amreen banu
Nikshitha                                                           Inside Computer

II-Prize
Akash
Manjanagouda
Pradeep
Akshatha
Bhavya











                                                           


Thursday, March 7, 2019


Innovative Teaching Method : Pass the “Mic”
Subject: Machine Learning (15CS073)                               
Semester: 7th 
Faculty: Ms. Rachana N B.                                       
Date:  13/11/2018




                                       Students involved in the discussion.

As an instructor, it’s amazing how much information you can gather from a student centered review session. Specifically, if you leave the review in the hands of your students, you can get an easy and thorough assessment of what is being absorbed, and what is being left by the wayside. The more you encourage participation, the more you’ll see where your class is struggling and the more comfortable students will become with course material. 
Summary:


            Students will have a tendency to pick the topics that they are most comfortable speaking about and those left consistently untouched will give you a clear assessment of the subjects in which your class is struggling, and where comprehension is lacking. Once your class has narrowed down the list to just a few terms, you can switch gears into a more classic review session. Bringing a bit of interaction and fun into a review can help loosen things up during exam time, when students and teachers alike are really starting to feel the pressure.

Wednesday, March 6, 2019

                                                                ROLE PLAY


Subject: C Programming for Problem Solving


TOPIC: BINARY SEARCH
FACULTY INCHARGE: SANDEEP SIR
DATE:05/01/2019
                                                        Presented by:
                                                                                                        NIKHIL MADWACHAR V K
                                                                                                  MANJUNATH SONNAD
                                                                         SANDESH
                                                                                          ABHIRAM MITHUR
                                                                     CHETAN
                                                                             SIDDARTHA 
                                                                 VIJAY
                                                                      ABUZER
                                                                                   DATE:5/01/2019
DESCRIPTION:
Binary search is a process of finding an element in the list of ‘n’ elements and are in sorted order.it works on the principle of finding the middle position,compare the key element with the middle element,if it matches then search is successful otherwise the key element may be present on the left or right side of the array by separating the middle position and repeat the procedure until you get a single element then conclude key element is found or not
Algorithm:
Step1:Start
Step2:Read the number of elements
Read the array elements
Read the key elements to be searched
Step3:Assign low=0
High=n-1
Mid=low+high/2
Step4:Check if(key==a[mid])
Assign flag=1
Break statement
Check if(key>a[mid])
Low=mid+1
else
high=mid-1
Step5: Check if(flag==1)
Display search successful and the position of the element
else
Display search unsuccessful
Step6: Stop
LOGIC:
Low=0;
high=n-1;
while(low<=high)
{
Mid=low+high/2;
If(key==a[mid])
{
Flag=1;
Break;
}
If(key>a[mid])
Low=mid+1;
Else
High=mid-1;
}
Example:
Enter the number of elements
9
Enter the 9 elements in ascending order
3 7 13 22 35 42 50 58 66
Enter the key element to be searched
22
Search successful and element found at the position 3
Explanation:
First low and high are assigned to 0th and 8th position.
Low=0
High=8
First iteration:


Mid=4

Key!= a[mid]
high =4 by high=mid-1
Second iteration:
Mid=1
Key!=a[mid]
Low=2 by low=mid+1
Third iteration:
Mid=2
Key!=a[mid]
Low=3 by low=mid+1
Fourth iteration:
Mid=3
Key= a[mid]

Search successful element found position at 3

                                            Software Installation
Course Name: Computer Graphics Laboratory with Mini Project
Course Code: 15CSL68
     
    Resource Person                                                                                                                                                    Staff In-Charge
Mr. Thippeswamy G N                                                                       Mr. Shivanna K



                


Friday, March 1, 2019





No. Activity Name Sem Link
1 Towers of Honoi 3 Link
2 Software installation 6 link
3 X-game 6 Link
4 C Programming for Problem Solving 1 Link



Activity : Solving Tower of Hanoi Problem

Sem: 3
Subject: Data Structures and Applications
Subject Code: 17CS33
Faculty Name:   Deepak D J
 Asst. Professor, Dept of CSE, GMIT
Date : 11-10-2018

The Towers of Hanoi Problem:
In the game of Towers of Hanoi, there are three pegs labeled A, B, and C. Tower A contains n disks with decreasing size.

Objective : Move disks from peg A to peg C using peg B as an auxiliary.

Rules: The rules to be followed in moving the disks from peg A to peg C using peg B are as follows:
• Only one disk can be moved at a time.
• Only the top disk on any tower can be moved to any other tower.
• A larger disk cannot be placed on a smaller disk.

Recursive Solution:
Tower of Hanoi can be reduced to following subproblems for n>1 disks
1.       Move the top (n-1) disks from peg A to peg B.
2.       Move the top disk from peg A to peg C.
3.       Move the top (n-1) disks from peg B to peg C

Activity Description:
·       Props were prepared for 4 disks and 3 pegs.
·       Each team participating in activity had 4 students.
·       Every student is assigned a specific disk. It is the responsibility of the that student to move the disk based on the recursive solution. Student assigned with a particular disk should not move other disks.
·       Total of 12 teams participated in the activity.
·       The team which solved the problem fastest was considered as the winning team.

Sl no
Team Head
Time taken (sec)
Winners
1
Kavana
30

2
Uma
37

3
Sneha
29

4
Priyanka
38

5
Shruthi
30

6
Vidyasagar
19

7
Prajwal
12
Anand, Manjana Gowda, Pradeep
8
Vinayak
25

9
Nirmala
18

10
Sharath
18

11
Namaratha
15

12
Tejaswini
28