algorithm - Trouble figuring out these tough Big-O examples -

i'm trying study upcoming quiz big-o notation. i've got few examples here they're giving me trouble. seem little advanced lot of basic examples find online help. here problems i'm stuck on.

1.     `for (i = 1; <= n/2; = * 2) {             sum = sum + product;             (j= 1; j < i*i*i; j = j + 2) {                 sum++;                 product += sum;             }        }` 

for one, i = * 2 in outer loop implies o(log(n)), , don't think i <= n/2 condition changes because of how ignore constants. outer loop stays o(log(n)). inner loops condition j < i*i*i confuses me because in terms of 'i' , not 'n'. big-o of inner loop o(i^3)? , big-o entire problem o( (i^3) * log(n) )?

2.      `for (i = n; >= 1; = /2) {              sum = sum + product              (j = 1; j < i*i; j = j + 2) {                  sum ++;                      (k = 1 ; k < i*i*j; k++)                      product *= * j;              }         }` 

for one, outermost loop implies o(log(n)). middle loop implies, again unsure, o(i^2)? , innermost loop implies o(i^2*j)? i've never seen examples before i'm guessing @ point. big-o notation problem o(i^4 * n * j)?

3.     `for (i = 1; < n*n; = i*2) {             (j = 0; j < i*i; j++) {                  sum ++;                 (k = i*j; k > 0; k = k - 2)                     product *= * j;             }        }` 

the outermost loop 1 has n^2 condition, logarithmic increment, think cancels out regular o(n). middle loop o(i^2), , innermost loop think o(n) , trying trick you. problem big-o notation o(n^2 * i^2)?

4.     `int = 1, j = 2;             while (i <= n) {                 sum += 1;                 = * j;                 j = j * 2;        }`  1 did few iterations better see happening: = 1,    j = 2 = 2,    j = 4 = 8,    j = 8 = 64,   j = 16 = 1024, j = 32  

so clearly, 'i' grows quickly, , condition met quickly. i'm not sure kind of big-o notation is.

any pointers or hints can give appreciated, guys.

you can't add or j o-notation, must converted n.

for first one:

let k log 2 i.

then inner loop done 2^(k*3)/2=2^(3k-1) times each iteration of outer loop.

k goes 1 log2n. total number of iterations sum of 2^(3k-1) k 1 log 2 n 4/7(n^3-1) according wolfram alpha, o(n^3).

for last one, i=j1*j2*j3*...jk, , jm=2^m

i=2^1*2^2*...2^k=2^(1+2+...k)

so 1+2+3+...+k=log 2 n

(k+1)k/2 = log 2 n

which o(sqrt(log n))

btw, log n^2 not n. question better ask @ computer science here.