在算法分析过程中找到上限有什么不同的行为？

Question

I'm learning Algorithm Analysis and Big-O notation. There are these exercise examples for reference:

Example-1: Find upper bound for f(n) = 3n + 8

Solution: 3n + 8 ≤ 4n, for all n ≥ 8

       ∴ 3n + 8 = O(n) with c = 4 and n0 = 8

Another One,

Example-2: Find upper bound for f(n) = n^2 + 1

Solution: n^2 + 1 ≤ 2(n^2), for all n ≥ 1

       ∴ n^2 + 1 = O(n^2) with c = 2 and n0 = 1

Now here comes the next example and the one that's bugging me,

Example-3: Find upper bound for f(n) = 2n^3 – 2n^2

Solution: 2n^3 – 2n^2 ≤ 2n^3, for all n ≥ 1

       ∴ 2n^3 – 2n^2 = O(n^3 ) with c = 2 and n0 = 1

Why did we use 2n^3 for comparison in the last example?

Means In every example we used greater values ie In first example we used 4n because the equation has 3n as maximum limit,

In second example we used 2(n^2), because n^2 was the maximum in that equation.

Now that means in the third equation we should use 3(n^3) instead of 2(n^2).

Maybe I'm not getting something here, Can you elaborate the missing pieces?

And what is the need for c and n0 here. n0 is the point from which we consider the rate of growth for given algorithm but why c? I'm new to algorithmic analysis.

Answer 1

The terms are replaced so that all the exponents are the same. For an upper-bound, you want to replace them with larger values. Increasing the exponent will produce larger values for positive terms, but for negative terms then you can replace them with 0 instead, removing the term.

For 3n + 8 , 3n + n is an upper-bound because 8 <= n for n > n0 .

For n^2 + 1 , n^2 + n^2 is an upper-bound because 1 <= n^2 for n > n0 .

For 3n^3 - 2n^2 , 3n^3 + 0 is an upper-bound because -2n^2 <= 0 for n > n0 .

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c and n0 are needed because they're part of the definition of Big-O:

`f(n) = O(g(n))` means that `f(n) <= c.g(n)` for some c and large enough n

By finding values of c and n0 , you can show that some functions fit this definition where c is what you need to multiply g by to make it larger than f .