Time Complexity Analysis On For Loops

Question

I am confused on analyzing the time complexity on a 'for loop'. Do the contents in the loop not counted like example:

for(i=0;i<=n;i++){
   statement
       .
       .
}

Wouldnt i=0 excute once, i<=n excute (n+1), and i++ execute n times. So why is the time complexity only counted (n+1) times for the 'for loop'? shouln't the contents inside be summed like: 1+n+1+n? just confused since my prof summed them up but the book only counted (n+1).

Answer 1

What you are asking is what the asymptotic notation is all about.

Say that you have a program like the following

for (i = 0; i < n; ++i) {
    print("hello")
}

that compiles into a fictional pseudocode machine language in which each of the numbered items below are single machine instructions:

 1. let i -> 0 
 2. if (i >= n) goto 6. 
 3. print "hello" 
 4. increment i 
 5. goto 2. 
 6. ... (continue with the rest of the program)

then for a given n the total number instructions that will run when executing the loop is 4 n + 1, one instruction to initialize i and four instructions for the body of the loop which executes n times.

Big-O et. al. is about how a function like f( n ) = 4 n + 1 behaves as n grows. The definition of Big-O is basically that for large n 's some function f( n ) is O(g( n )) if there exists a constant k such that

f( n ) ≤ k g( n ).

In this case if g( n ) = n we can see that, for all values of n greater than zero, f( n ) ≤ 5 g( n ) so 5 is a valid k and we can say that f( n ) is O( n ).

This is the point of Big-O: it lets us talk about how functions grow ignoring additive and multiplicative constants.