Number of addition and multiplication operators in this algorithm

Question

Consider the following algorithm:

i := 1
t := 0
    while i ≤ n
       t := t + i
       i := 2i

I'm interested in finding out how many addition and multiplication operations this algorithm executes; however, I'm running into trouble. I understand that the value of i doubles after each iteration, but I don't know how to generalize the algorithm to give a correct number of operations up until the value of n. If anyone can shed some light on the issue, I'd greatly appreciate it.

Thank you!

Each loop contains one addition and one multiplication, so you just have to figure out how many times the loop runs. — Some programmer dude, Dec 05 '12 at 06:13
I understand this. However, I'm having trouble at this point. — John Taylor, Dec 05 '12 at 06:14

score 1 · Accepted Answer · answered Dec 05 '12 at 06:12

1

Since the value of i doubles every loop and i <= n

i*2^x <= n

and maximizing x gives the number of loops. Since i = 1

2^x = n
x = floor log(n)

and we perform 1 addition and 1 multiplication per loop. I think you can take it from here :)

answered Dec 05 '12 at 06:12

daniel gratzer

52,833
11
94
134

Can you explain how you got the term 2^x? – John Taylor Dec 05 '12 at 06:15
The number of loops execute – daniel gratzer Dec 05 '12 at 06:17
OH (No pun intended)! I was misinterpreting the problem (as well as your response). Thanks for the help, really appreciate it :D – John Taylor Dec 05 '12 at 06:19
No problem, if my answer is good then please check the little green check beside it to mark this question as answered – daniel gratzer Dec 05 '12 at 06:20
In one minute, I will! You little handsome booger :P <3 – John Taylor Dec 05 '12 at 06:22

Number of addition and multiplication operators in this algorithm

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