Find an upper and lower bound of following code

问题

I need to find a closest upper and lower bound of following code. I am beeginer in this so sorry for my mistake.

Upper bound for p() is O(log(n)), and lower bound is O(1)

Upper bound for notp() is O(log(n)), and lower bound is O(1)

I think that lower bound is O(1) because if I have n=4 then I get in the loop and since n%i==0 I call p() and it notice that is not a prime number so O(1) then since i=2 the other notp would not be executed. That is bast case scenario.

Worst case scenario that I go through loop so that log(n), and execute a p and a upper bound is O(log(n)) so it is O(log(n)^2) but I am not so sure that this is good, please tell me where I made a mistake?

int i;
for (i = 2; i*i <= n; i++) {
  if (n % i == 0) {
    p();
    break;
  }
}
if(i*i > n)
  notp();

回答1:

For order calculations, you would normally multiply them together when there is a loop.

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

That would be O(n), as the loop is a single case.

A nested loop would be

for( i = 0; i < n ; i++ ){ 
    for( j =0; j < n; j++ ){
       DoSomething(i,j);
    }
}

That becomes O( n^2 ) as they are additive.

If you have non-nested loops...

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


for( j = 0; j < sqrt(n) ; j++ ){
     DoSomething(j);
}

Then the O( x ), is the largest term. That is because for very large n, the smaller values of x don't matter.

So you have a loop in your case, which is O( sqrt( n ) ). That is because it is limited by i *i being less than n.

Then either one of p() or notp() will be called.

(I think p() and notp() are the wrong way round).

Re-factoring your code....

int i;
for (i = 2; i*i <= n; i++) {
  if (n % i == 0) {
    /* p(); */
    break;
  }
}
if(i*i > n)
  notp();
else
  p();

So we have the O( sqrt(n) ) time plus either of the p / notp functions which are O( log(n) )

O( sqrt(n ) + log(n) )

As the sqrt grows faster than n, it overpowers the log(n) wikipedia Big O, leaving it as the final value.

O( sqrt(n) )

回答2:

As the termination condition of the loop is i*i <= n, the most possible number of iteration of the loop is sqrt(n). As there is a break in the condition of n % i == 0, the worst case of this part is sqrt(n) + log(n) which is O(sqrt(n)). Also, eighter the second condition is true or not, as the worst case of nopt() is O(log(n)), the worst case of the algorithm in total is O(sqrt(n)).

来源：https://stackoverflow.com/questions/54177166/find-an-upper-and-lower-bound-of-following-code

标签

algorithm

complexity-theory