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August 1, 2014

August 1, 2014

Posted by **Elizabeth** on Monday, October 12, 2009 at 1:32pm.

n= 1 running time= 1

3 1+1= 2

7 1+2= 3

15 1+3= 4

31 1+4=5

63 1+5= 6

127 7

255 8

511 9

1023 10

How in the world does this happen?

It has something to do with log n steps. I don't understand this concept.

What does it mean log n= <<n for large n?

Also, what is O(n)?

- Computer Science -
**Elizabeth**, Monday, October 12, 2009 at 1:33pmRunning time is 1,2,3,4,5,6,7,8,9,10. It sll got put together, but really those numbers are separate from the n numbers

- Computer Science -
**MathMate**, Monday, October 12, 2009 at 2:15pmSorry to be nagging, but next time, when the teacher writes something on the board that you do not understand, raise your hand and ask for explanation. There may be 6 other students who don't understand and they would thank you for asking.

Back to the board.

If you have to calculate X=2^{n}-1, where n=10, you can start with k=1, n=2¹-1=1

for k=2, you double the previous x and add 1 to get 3. For k=3, again double the previous x(=3) and add 1 to get 7.

If you do this 10 times, you will get 2^{10}-1 in 10 steps, i.e. up to k=n.

Since 1024 is equal to 2^{10}, therefore log_{2}(1024)=10.

The 'cost' of doing this calculation is therefore proportional to n, or approximate equal to n, or order n, or O(n).

For an explanation of the Big-O notation, you can read up:

http://en.wikipedia.org/wiki/Big_O_notation

I do not know if algebra is a prerequisite for this algorithm course, but definitely you will need to know quite well logarithms (to the base 2 and base e). Textbooks usually start off with a thorough revision of math required.

The idea behind all this is to be able to compare the efficiency of different algorithms, or way of calculating or evaluating formulas. An algorithm of O(log n) performs better that of O(n), which in turn performs better than that of O(n²), etc. This is the object of the very interesting course.

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