Re: [algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-27 Thread saurabh modi
I think that this code can suffice. http://www.ideone.com/ESW8Z #includestdio.h int solve(int,int); int main() { printf(%d,solve(100,10)); return 0; } int solve(int a,int b) { int quotient=0,k=0; if(ab) return quotient; while((1k)*b=a) { k++; } k--;

[algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-27 Thread Dave
@Saurabh: According to the statement of the problem using multiplication is not allowed, but you can replace (1k)*b by bk to eliminate your multiplications. Dave On Aug 27, 2:49 am, saurabh modi saurabhmodi102...@gmail.com wrote: I think that this code can suffice. http://www.ideone.com/ESW8Z

Re: [algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-27 Thread saurabh modi
haha yeah okay..that can be done :-) i had forgotten abt * -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to

[algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-25 Thread WgpShashank
@Dave Yup, but Overall Complexity Will remain O(log(Quotient)) as y=logn^k=klogn=O(logn) where k is constant isn't it ? Also case of -Ive Numbers Can be handled easily :) *Thanks Shashank Mani Computer Science Birla Institute of Technology Mesra* -- You received this message because you

[algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-25 Thread Dave
@Shashank: Regarding the complexity, let's say that you are dividing 15 by 1. The on the original call, you will shift 4 times, on the first recursive call, 3 times, then 2 times, then 1 time. This is a total of ten shifts. This is log(quotient) * (log(quotient) - 1) / 2, which is

[algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-25 Thread WgpShashank
@Dave True :) *Thanks Shashank Mani Computer Science Birla Institute of Technology Mesra* -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/mSeEd-_wRcMJ.

[algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-23 Thread WgpShashank
@Dave , You are right , i mean to say we reduce the number of instruction or comparisons executed by the program. ,Never Mind here is recursive code for doing the same , Algorithm is already explained #includeiostream using namespace std; int dividend,divisor,remainder; int division(int

Re: [algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-23 Thread teja bala
//works only for ints #includestdio.h int div(int a,int b); main() { int q,w,x; scanf(%d %d,q,w); x=div(q,w); printf(%d,x); } int div(int x,int y) { int z=x,count=0; while(z=y) { z-=y; count++; } return count; } -- You received this message because you are subscribed to the

[algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-23 Thread Dave
@Teja: The complexity of your code is O(quotient). Code has been presented that is O(log(quotient)). Dave On Aug 23, 8:47 am, teja bala pawanjalsa.t...@gmail.com wrote: //works only for ints #includestdio.h  int div(int a,int b);  main()  {   int q,w,x;   scanf(%d %d,q,w);   x=div(q,w);

[algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-23 Thread Dave
@Shashank: Hmmm. In the recursive call, you use the variable divisor, but I don't see anything assigned to it. Furthermore, you don't handle negatives, which accounted for almost half of my code. In addition, you have to shift the divisor left multiple times in each recursive call to align it

Re: [algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-20 Thread WgpShashank
@Lucky sorry for delay, I am saying compelxity will be number if bits required to represent quioutent , i think you just try with exmple i have given @Dave I didn't See The Whole Code but i think Logic Will Remain The Same.i Think You forgot to give algorithm :P also we can reduce the line of

[algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-20 Thread Dave
@Shashank: I didn't know that reducing the number of lines of source code was the goal. Often-times, efficiency demands more code rather than less. For example, back in the days of the Cray-1 supercomputer, which had vector registers and could do an operation on up to 64 operand pairs in one

[algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-19 Thread WgpShashank
We can use bit logic to reduce the time complexity to O(logn) where n is Quotient Algorithm will be as follow as we know 1. Left shifting an unsigned number by 1 multiplies that number by 2. 2. Right shifting an unsigned number by 1 divides that number by 2. Therefore the procedure for the

Re: [algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-19 Thread Sanjay Rajpal
@Shashank : Nice solution :) *Regards Sanju Happy to Help :)* On Fri, Aug 19, 2011 at 2:36 AM, WgpShashank shashank7andr...@gmail.comwrote: We can use bit logic to reduce the time complexity to O(logn) where n is Quotient Algorithm will be as follow as we know 1. Left shifting an

Re: [algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-19 Thread Arun Vishwanathan
@dave: actually how did u get the approach to this? I mean why did u have to do the q|=1 in the if(a=b) condition and q=1 always in the loop? On Fri, Aug 19, 2011 at 1:46 PM, Sanjay Rajpal tosanjayraj...@gmail.comwrote: @Shashank : Would you throw some light on how you determined the complexity

[algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-19 Thread Dave
@Arun: Just think about doing long division with paper and pencil. The q=1 statement is the same as bringing down a new digit. Doing so also expands the quotient by one digit. The q|=1 statement is the same as writing down a nonzero digit in the quotient. Dave On Aug 19, 10:31 am, Arun

[algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-19 Thread Dave
@Sanjay: Shashank was just reiterating what I said in http://groups.google.com/group/algogeeks/msg/8590f8a2a8408d93. The algorithm Shashank described is what I had previously provided code for in http://groups.google.com/group/algogeeks/msg/735671f6a1e16eec, with a correction in

Re: [algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-19 Thread Sanjay Rajpal
@Dave : How did you approach this solution to this problem ? and how you saw that th complexity is O(log_2(Quotient)) ? *Regards Sanju Happy to Help :)* On Fri, Aug 19, 2011 at 8:44 AM, Dave dave_and_da...@juno.com wrote: @Sanjay: Shashank was just reiterating what I said in

[algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-18 Thread Don
exp(ln(a)-ln(b)) On Aug 18, 8:56 am, radha krishnan radhakrishnance...@gmail.com wrote: how to do using BIT manipulation ? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To

[algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-18 Thread Dave
@Don: Try it with a = b = -1. Dave On Aug 18, 9:45 am, Don dondod...@gmail.com wrote: exp(ln(a)-ln(b)) On Aug 18, 8:56 am, radha krishnan radhakrishnance...@gmail.com wrote: how to do using BIT manipulation ? -- You received this message because you are subscribed to the Google Groups

[algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-18 Thread Dave
@Radha: You could simulate long division. It would look something like this: int divide(int a, int b) { int i, k=0, q=0, s=1; // error checking if( b == 0 ) return 0 // return 0 for division by zero // handle signs if( a 0 ) { a = -a; s = -1; } if( b 0 )

Re: [algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-18 Thread aditya kumar
how abt subtracting . like a=a-b till a becomes zero . no of times subtraction is done is the answer . correct me if i am wrong ! On Thu, Aug 18, 2011 at 8:59 PM, Dave dave_and_da...@juno.com wrote: @Radha: You could simulate long division. It would look something like this: int divide(int

[algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-18 Thread DheerajSharma
wat about shifting 'a' right by floar(log2(b)) and adding 1 to it.. On Aug 18, 8:48 pm, aditya kumar aditya.kumar130...@gmail.com wrote: how abt subtracting . like a=a-b till a becomes zero . no of times subtraction is done is the answer . correct me if i am wrong ! On Thu, Aug 18,

[algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-18 Thread Dave
@Aditya: Not wrong, but inefficient and inelegant. Repeated subtraction has complexity O(quotient), while long division has complexity O(log quotient). Dave On Aug 18, 10:48 am, aditya kumar aditya.kumar130...@gmail.com wrote: how abt subtracting . like a=a-b till a becomes zero . no of times

[algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-18 Thread Dave
@Dheeraj: What about it? It doesn't give the quotient. What is it supposed to do? Dave On Aug 18, 11:06 am, DheerajSharma dheerajsharma1...@gmail.com wrote: wat about shifting 'a' right by floar(log2(b)) and adding 1 to it.. On Aug 18, 8:48 pm, aditya kumar aditya.kumar130...@gmail.com wrote:

Re: [algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-18 Thread Arun Vishwanathan
@dave: in your algorithm, I have a doubt in the second loop( for loop ). q=0 initially so the first q1 stays zero and then q|=1 makes q=1 now. 1 then becomes x 2 and then again with the OR 2 becomes 3. 3 becomes 6 and with the OR 6 becomes 7. for example if i need to do 24/3, according to the

[algogeeks] Re: calculate a/b without using ‘*’, ‘/’’ and ‘%’

2011-08-18 Thread Dave
@Arun: Oops. It looks like the while loop condition should be b = a instead of b a, and the if condition within the for loop should be a = b instead of a b. I'm glad I equivocated with the phrase would look something like. With a = 24 and b = 3, the while loop shifts b left and increments k