"This story only inspired to get the Coordinator post for DalalBull next year. No matter what happens or who coordinates the event, I will make sure next year, Dalal Bull gets the spotlight right from day one. I will write about it next year."
-Sujeetha (Myself) a year ago.
I am glad that I could get to coordinate the onsite event this year. This year's experience was the pinnacle of all my extracurricular activities. I am talking about Dalalbull, an event in Kurukshetra which is about simulation of stock market. There is both online and onsite event of Dalalbull. The online event was coordinated by shanketh, one of my classmate.I organized the onsite event. For the onsite, we had 5 organizers just like last year. 2 girls and 3 boys from same class. But this year it was from IT dept. Last year it was myself, usha, thanu, rajesh and pradep who did the organizing work. This year it was prathyusha, gayathri, ramki, rahul and ganesh. The work started by first reviewing our last year software, then fixed some bugs. The organizers this time did a wonderful job. The ultimate experience was the volunteer meet. Dalalbull, being an event which involves the maximum number of volunteers and the largest hall in college for venue, solely depends on volunteer's enthu for a good show. We arranged at 2 the meeeting in 111. But only after 2.30 some volunteers entered the hall. One person who I really should mention is dharani of second year computer science. This junior had super contacts and arranged like 100 volunteers(first and second years, both girls and boys) for the event in just one phone call(slightly exaggerated!). In the meet, I really understood the difficulty in making all 100 people listen to you. I understood the difficulites of the faculty. After the meet, I completely lost my voice but I gained confidence that I am still the same good old speaker who used to get prizes for debates and state level speech competitions. The hall 13 where the event was supposed to take place was just getting ready for the computer installation after the Contraption event. When my hostel curfew got over, my work still continued. Myself, kalpu and abi went to first year block and conducted an informal meeting there. Then finally on first day of the event, I heard as usual the night was a nightmare for the guys.
So the day started by first setting the computers, this time we had 30 systems like last year. Some juniors helped to set up the LAN to all computers. This time there was no problem with LAN wire at all. Thanks to vishnu and the juniors. I forgot to mention about some of my friends who helped me the most in conducting this event. Abi and kalpu spent their time with me by helping me in every aspect I could think of. The onsite event had a great help from thanumalayan. For record, he is not the coordinator, but off the record he is one of the coordinator. He helped me by staying in the hall the day before the event. Thanks to vicky for arranging the lunch at the hall.
My voice was completely out by the end of the day. Finally we had a successful event. No fight, no complaints.
to be contd.........
Thursday, February 5, 2009
Sunday, December 21, 2008
KURUKSHETRA - BATTLE BEGINS
One more year has gone by.. I could say this because the next big thing in my calendar is KURUKSHETRA'09. I never believed in adage. According to me its all crap. But one thing I learnt from Kuruk is "united we stand, divided we fall". All students of our college work together to make this great event. Do check out the website. Lot of online events are open.
the website address is : http://kurukshetra.org.in/
The battle begins on Jan 21st. But registerations open now !!!
the website address is : http://kurukshetra.org.in/
The battle begins on Jan 21st. But registerations open now !!!
Wednesday, August 20, 2008
Quine programs
A “quine” (or “selfrep”) is a computer program which prints its own listing.
complex yet trivial code to print the own source file is
#include
char *p="#include%c%cchar *p=%c%s%c; %c%cint main(){%c%cprintf(p,13,10,34,p,34,13,10,13,10,13,10,13,10); %c%cgetch();%c%c }";
int main(){
printf(p,13,10,34,p,34,13,10,13,10,13,10,13,10);
getch();
}
Lot more about this type of programming in
http://www.madore.org/~david/computers/quine.html
complex yet trivial code to print the own source file is
#include
char *p="#include
int main(){
printf(p,13,10,34,p,34,13,10,13,10,13,10,13,10);
getch();
}
Lot more about this type of programming in
http://www.madore.org/~david/computers/quine.html
Wednesday, June 25, 2008
Googly Google
Well, just like everybody I was there doing my business, yes browsing through technical blogs for passing time. Trust thats one good freetime job. So I came across this blog, it has informations to help you chat well and also to write Project Management Exams.
http://sanjaykattimani.blogspot.com/
From that blog I came across this Google's yet another invention or what ever it is, Toogle. Nice and bubbly.
Just like Google image search, here the slight difference is the image is drawn with the search string.
Speaking of searches and Google, this feature really helps you if the website you are looking for has no custom Google search.
http://www.google.com/advanced_search
And trust me, it gives better result than custom search.
I will sure update more about Google's products.
http://sanjaykattimani.blogspot.com/
From that blog I came across this Google's yet another invention or what ever it is, Toogle. Nice and bubbly.
Just like Google image search, here the slight difference is the image is drawn with the search string.
Speaking of searches and Google, this feature really helps you if the website you are looking for has no custom Google search.
http://www.google.com/advanced_search
And trust me, it gives better result than custom search.
I will sure update more about Google's products.
Intriguing Puzzles
As I am preparing for my placements, I was on the lookout for books with puzzles to work out and to train the mind to think. I came across this book name in many forums I saw. So when I searched for this book, viola I got a pdf copy of the book online. The book name is Shakuntala- Puzzles to puzzle you. This book is recommended for anybody who is looking forward to Infosys interviews.
P.S : solutions are at pg.89
http://www.scribd.com/doc/2052493/Devi-ShakuntalaPuzzles-To-Puzzle-You
P.S : solutions are at pg.89
http://www.scribd.com/doc/2052493/Devi-ShakuntalaPuzzles-To-Puzzle-You
Thursday, June 12, 2008
Puzzles
Interesting how these puzzles make your mind fresh. Anyways, I am right now in a puzzle solving spree.. So I was searching for some websites with puzzles .. and found an excellent website, where I can learn and think,
http://en.allexperts.com/q/Puzzle-Solving-1841/
Trust me, this is good. I am going to blog down the supporting theorems and techniques for later reference.
Bayes Theorem:
P(A|B) = P(AnB)/P(B)
P(A|B) - conditional probability. Probability of A happening given B has happened.
P(A) & P(B) - prior probability.
Multiplication tricks:
Multiplication of two digit number by 11:
The result will be the first digit, sum of the two digit and the second digit of the given number.
Eg: 4_2 is the two digit number. The result would be 462, where 6 is sum of first (4) & second(2) digit.
Given 4_8, the result would be 528, since when we add 4+8= 12 add 1 to first digit, and follow the rule above.
Multiplication of any number by 5:
*Divide the number by 2
*If there is a remainder, add 5 to the answer. Else, add 0 to the remainder.
Eg: 246 * 5 :
246/2 = 123 (no remainder)
So result would be 1230.
Say, given 235 * 5:
235/2 = 117(there is a remainder)
so result will be 1175.
Division Techniques:
Divisible by 4, 8,16:
Any number is divisible by 4, only if the last 2 digits of the number are divisible by 4. Similarly for 8 and 16, the last 3 and 4 digits should be considered.
Eg:
5632 is divisible by 4, since 32 is divisible by 4.
It is also divisible by 8, since 632 is divisible by 8.
It is also divisible by 16, since 5632 is divisible by 16.
Divisible by 3,9:
A number is divisible by 3 only if the sum of its digit is divisible by 3. Similarly for 9, the sum of digits should be divisible by 9.
Eg:
23541 is divisible by 3, since 2+3+4+5+1 = 15 is divisible by 3.
334242 is divisible by 9, since 3+3+4+2+4+2=18 is divisible by 9.
Divisible by 11:
A number is divisible by 11, only if the difference between the sum of its digit in the odd places and the sum of its digit in the even places are either 0 or a number divisible by 11.
Eg:
4832718 is divisible by 11. Since (8+7+3+4) - (1+2+8) = 22 - 11 = 11 divisible by 11.
http://en.allexperts.com/q/Puzzle-Solving-1841/
Trust me, this is good. I am going to blog down the supporting theorems and techniques for later reference.
Bayes Theorem:
P(A|B) = P(AnB)/P(B)
P(A|B) - conditional probability. Probability of A happening given B has happened.
P(A) & P(B) - prior probability.
Multiplication tricks:
Multiplication of two digit number by 11:
The result will be the first digit, sum of the two digit and the second digit of the given number.
Eg: 4_2 is the two digit number. The result would be 462, where 6 is sum of first (4) & second(2) digit.
Given 4_8, the result would be 528, since when we add 4+8= 12 add 1 to first digit, and follow the rule above.
Multiplication of any number by 5:
*Divide the number by 2
*If there is a remainder, add 5 to the answer. Else, add 0 to the remainder.
Eg: 246 * 5 :
246/2 = 123 (no remainder)
So result would be 1230.
Say, given 235 * 5:
235/2 = 117(there is a remainder)
so result will be 1175.
Division Techniques:
Divisible by 4, 8,16:
Any number is divisible by 4, only if the last 2 digits of the number are divisible by 4. Similarly for 8 and 16, the last 3 and 4 digits should be considered.
Eg:
5632 is divisible by 4, since 32 is divisible by 4.
It is also divisible by 8, since 632 is divisible by 8.
It is also divisible by 16, since 5632 is divisible by 16.
Divisible by 3,9:
A number is divisible by 3 only if the sum of its digit is divisible by 3. Similarly for 9, the sum of digits should be divisible by 9.
Eg:
23541 is divisible by 3, since 2+3+4+5+1 = 15 is divisible by 3.
334242 is divisible by 9, since 3+3+4+2+4+2=18 is divisible by 9.
Divisible by 11:
A number is divisible by 11, only if the difference between the sum of its digit in the odd places and the sum of its digit in the even places are either 0 or a number divisible by 11.
Eg:
4832718 is divisible by 11. Since (8+7+3+4) - (1+2+8) = 22 - 11 = 11 divisible by 11.
Wednesday, June 11, 2008
To calculate day of the week
When I was reading through some placement question papers, I found this problem asked quite a few times.
Q: Given a date, we have to find out the day of the week.
Solution:
1) C = 2(3-(cent%4))
where cent = first 2 digits of the century
2) y = year + floor(year/4)
where year = last two digits of the year given
3)m = choose from the table
4)sum = c + y + m + date given
5)val = sum%7
6) sun - 0,mon-1,tue-2,... sat - 6, use value and find the day
Sample program:
#include
int leapyearMonValue[] = {0,3,3,6,1,4,6,2,5,0,3,5};
int nonleapyearMonValue[] = {3,6,3,6,1,4,6,2,5,0,3,5};
void calculateDay(int date,int mon,int year)
{
int val,cent,monval,yr,i=3,temp,leapyear=0;
int splitcent[4],wrongInput=0;
if(year%4 == 0)
leapyear = 1;
if(mon>12 || mon <1)
wrongInput = 1;
if(date==30 && !(mon%2) || mon==7 || mon==1)
wrongInput = 1;
if(mon==2 && date>28 && !leapyear)
wrongInput = 1;
if(!wrongInput)
{
while(year)
{
splitcent[i--] = year%10;
year = year/10;
}
temp = splitcent[0]*10 + splitcent[1];
cent = 2*(3-(temp%4));
temp = splitcent[2]*10 + splitcent[3];
yr = temp + floor(temp/4);
if(leapyear)
monval = leapyearMonValue[mon-1];
else
monval = nonleapyearMonValue[mon-1];
val = cent + yr + monval + date;
val = val%7;
switch(val)
{
case 0:
{
printf("day is Sunday\n");
break;
}
case 1:
{
printf("day is Monday\n");
break;
}
case 2:
{
printf("day is Tuesday\n");
break;
}
case 3:
{
printf("day is Wednesday\n");
break;
}
case 4:
{
printf("day is Thursday\n");
break;
}
case 5:
{
printf("day is Friday\n");
break;
}
case 6:
{
printf("day is Saturday\n");
break;
}
default:
printf("wrong calculation\n");
}
}
else
printf("wrong Input... check the date, mon \n");
}
int main()
{
int date,mon,year;
printf("Enter the date in format DD/MM/YYYY\t eg: 14/11/2007\n");
scanf("%d/%d/%d",&date,&mon,&year);
calculateDay(date,mon,year);
getch();
return 0;
}
Courtesy:
Wikipedia
Q: Given a date, we have to find out the day of the week.
Solution:
1) C = 2(3-(cent%4))
where cent = first 2 digits of the century
2) y = year + floor(year/4)
where year = last two digits of the year given
3)m = choose from the table
Month | Value |
---|---|
January | 0(leap year-3) |
Febuary | 3(leap year-6) |
March | 3 |
April | 6 |
May | 1 |
June | 4 |
July | 6 |
August | 2 |
September | 5 |
October | 0 |
November | 3 |
December | 5 |
4)sum = c + y + m + date given
5)val = sum%7
6) sun - 0,mon-1,tue-2,... sat - 6, use value and find the day
Sample program:
#include
int leapyearMonValue[] = {0,3,3,6,1,4,6,2,5,0,3,5};
int nonleapyearMonValue[] = {3,6,3,6,1,4,6,2,5,0,3,5};
void calculateDay(int date,int mon,int year)
{
int val,cent,monval,yr,i=3,temp,leapyear=0;
int splitcent[4],wrongInput=0;
if(year%4 == 0)
leapyear = 1;
if(mon>12 || mon <1)
wrongInput = 1;
if(date==30 && !(mon%2) || mon==7 || mon==1)
wrongInput = 1;
if(mon==2 && date>28 && !leapyear)
wrongInput = 1;
if(!wrongInput)
{
while(year)
{
splitcent[i--] = year%10;
year = year/10;
}
temp = splitcent[0]*10 + splitcent[1];
cent = 2*(3-(temp%4));
temp = splitcent[2]*10 + splitcent[3];
yr = temp + floor(temp/4);
if(leapyear)
monval = leapyearMonValue[mon-1];
else
monval = nonleapyearMonValue[mon-1];
val = cent + yr + monval + date;
val = val%7;
switch(val)
{
case 0:
{
printf("day is Sunday\n");
break;
}
case 1:
{
printf("day is Monday\n");
break;
}
case 2:
{
printf("day is Tuesday\n");
break;
}
case 3:
{
printf("day is Wednesday\n");
break;
}
case 4:
{
printf("day is Thursday\n");
break;
}
case 5:
{
printf("day is Friday\n");
break;
}
case 6:
{
printf("day is Saturday\n");
break;
}
default:
printf("wrong calculation\n");
}
}
else
printf("wrong Input... check the date, mon \n");
}
int main()
{
int date,mon,year;
printf("Enter the date in format DD/MM/YYYY\t eg: 14/11/2007\n");
scanf("%d/%d/%d",&date,&mon,&year);
calculateDay(date,mon,year);
getch();
return 0;
}
Wikipedia
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