algorithm - Order Of Growth complicated for loops -


for following code fragment, order of growth in terms of n?

int sum = 0; (int = 1; <= n; = i*2)   (int j = 1; j <= n; j = j*2)     (int k = 1; k <= i; k++)         sum++;

i have figured there lgn term, stuck on evaluating part : lgn(1 + 4 + 8 + 16 + ....). last term of sequence be? need last term calculate sum.

you have geometric progression in outer loops, there closed form sum of want take log:

1 + 2 + 4 + ... + 2^n = 2^(n+1) - 1

to precise, sum is

1 + ... + 2^(floor(ld(n))

with ld denoting logarithm base 2.

the outer 2 loops independent each other, while innermost loop depends on i. there single operation (increment) in innermost loop, means number of visits innermost loop equals summation result.

  \sum_i=1..( floor(ld(n)) ) {       \sum_j=1..( floor(ld(n)) ) {           \sum_k=1..2^i { 1 }       }   }      // adjust innermost summation bounds    = \sum_i=1..( floor(ld(n)) ) {       \sum_j=1..( floor(ld(n)) ) {           -1 + \sum_k=0..2^i { 1 }       }   }      // swap outer summations , resolve innermost summation = \sum_j=1..( floor(ld(n)) ) {       \sum_i=1..( floor(ld(n)) ) {           2^i       }   }     // resolve inner summation = \sum_j=1..( floor(ld(n)) ) {       2^(floor(ld(n)) + 1) - 2   }     // resolve outer summation = ld(n) * n - 2 * floor(ld(n))

this amounts o(n log n) ( second term in expression vanishes asymptotically wrt first ) in big-oh notation.


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