This is the simple pattern finding problem in which we have to find the
most frequent patterns of persons that listens/vote for the same band.
Here we can apply the frequent pattern algorithm like FP Tree or Apriori
algorithm. Link for the tutorial of FP tree is given below
i was thinking of character array. which is same as string.
Any thoughts on better alternatives?
On Friday, 6 July 2012 13:34:01 UTC-4, anshu wrote:
yes, We can have much better data structure for storing big integer
instead of string just for simplicity I have taken string.
On Fri, Jul 6,
Best will be storing BIG_INT in array of integers.
int [] val; //store actual number
int size; // required length of val array.
take the string as input for constructors and convert it into 2's
complement form, maintaining little endian order or big endian order.
Perform other operation keeping
You are given 2 dice. Both are fair. One of the dice has no numbers printed
on it. You have to label the unmarked dice such that when both the dice are
thrown, the sum on the faces is evenly distributed between 1 and 12 .
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plz explain ur ans .how it is calculated?
On Sat, Jul 7, 2012 at 12:54 PM, Amol Sharma amolsharm...@gmail.com wrote:
Three 0's and Three 6's.
36 possibilities and 12 possible values of sum, so each sum should come in
3 possibilities and hence three 0's and three 6's.
--
Amol Sharma
The next code, it's about the problem Eight Queens, i found it in internet,
but i'm new in LISP and i don´t know how to compile it...
Can you help me?
(defun find-queen (arr p d)
(destructuring-bind ((px py) (dx dy)) (list p d)
(do ((x px (+ x dx))
(y py (+ y dy)))
((not
Label 3 of the faces with 0 and other 3 faces with 6.
--
Prakhar Jain
IIIT Allahabad
B.Tech IT 3rd Year
Mob no: +91 9454992196
E-mail: rit2009...@iiita.ac.in
jprakha...@gmail.com
On Sat, Jul 7, 2012 at 12:52 AM, Hraday Sharma hradaysha...@gmail.comwrote:
You are given 2 dice.
@Raj: trace karke dekh na yaar when u have 3 0s and 3 6s.. the sum
distribution would look like this:
given below are the possibilities:
Combination of 1,2,3,4,5,6 with 0
1+0 = 1
2+0 = 2
3+0 = 3
OR
4+0 = 4
5+0 = 5
6+0 = 6
Combination of 1,2,3,4,5,6 with 6
1+6 = 7
2+6 = 8
3+6 = 9
OR
4+6
How do you find the intersection of two sets in a secured way?
Which means imagine a situation where there is a client who has got a set
S1={1,2,3}, and there is a server who has got a set S2[{2,3}. Client wants
to find the count of intersection which is |S1 intersection S2| = 2. But,
it
