O（1）+ O（2）+ ... + O（n）的阶数之和

Question

What does the sum O(1)+O(2)+ .... +O(n) evaluate to?

I have seen its solution somewhere it was written:

O(n(n+1) / 2) = O(n^2)

but I am not satisfied with it because O(1) = O(2) = constant , so according to me it must evaluate to O(n) only. I have also seen this in Cormen:

Σ(i=1 to n) O(i)

In above expression there is only a single anonymous function. This function is not the same as O(1) + O(2) + ... + O(n) which doesn't really have a clean interpretation.

Answer 1

The question seems to be perfectly on-topic since there is a tag asymptotic_complexity ...

According to CLRS , p. 49,

"The number of anonymous functions in an expression is understood to be equal to the number of times the asymptotic notation appears. For example, in the expression

sum(O(i), i=1..n)

there is only a single anonymous function (a function of i). This expression is thus not the same as O(1) + O(2) + ... + O(n), which doesn't really have a clean interpretation"

Actually, in your formula, the constants behind the "O" notation may be all different, and their growth may change the asymptotic behaviour of the whole sum. Don't write this!

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To answer your question more completely, in sum(O(i), i=1..n), you can use the fact that (see GKP p. 450 for the following)

O(f(n)g(n)) = f(n) O(g(n))

Thus, O(i) = i O(1), this time with the same O(1) in your formula. Therefore,

sum(O(i), i=1..n) = sum(i, i=1, n) O(1)

=n(n+1)/2 O(1) = O(n^2)

This way you can eliminate your sum without trouble.

Answer 2

Go by the definition of the Big-Oh notation.

You have n functions f_i , each of which satisfies

a_i * i <= f_i(x) <= b_i * i; x > X_i

for some positive constants a_i, b_i and sufficiently large X_i . let X = max_i ( X_i ) and sum over the n inequalities to get

sum_i=1..n ( a_i * i ) <= sum_i=1..n ( f_i(x) ) <= sum_i=1..n ( b_i * i )

noting that

sum_i=1..n ( min(a_i) * i ) <= sum_i=1..n ( a_i * i )
sum_i=1..n ( b_i * i )      <= sum_i=1..n ( max(b_i) * i )

arriving at

    min(a_i) * 0.5*(n(n-1))  <= sum_i=1..n ( f_i(x) ) <= max(b_i) * 0.5*(n(n-1))
<=>       A       * (n(n-1)) <= sum_i=1..n ( f_i(x) ) <=        B      *(n(n-1))

thus the sum of the functions you started with is indeed O(n^2) .

Answer 3

sure, O(1) and O(2) is a constant, so we can forget about them. But is O(n/2) constant, I guess no. So try (for your homework) count just second half of the elements:

O(n/2) + O(n/2+1) + ... O(n) = ??

you will come up with O(n^2) :)

Answer 4

There is a trick to find the formula, really not that hard.

You have: 1 + 2 + 3 + 4 + .... + n

If you count the list twice (one time ordered from low to high one time from high to low you get twice the result)

((1 + 2 + 3 + 4 + ... + n) + (n + (n - 1) + (n - 2) ... + 2 + 1)) / 2

This is the same as

((1 + n) + (2 + n - 1) + (3 + n - 2) + ... + (n + 1)) / 2

And this is:

((n + 1) * n) / 2

or as we right in o notion O(n²).