上限算法分析

Question

>Problem: n^4 + 100n^2 + 50. 

>Solution given: n^4 + 100n^2 + 50 <= 2n^4 for all n>=1
>                n^4 + 100n^2 + 50 = O(n^4) with c=2 and n0 = 100

But when n is 1, the above function will be "4+100+50 <= 2" which is not true. How can i derive the correct upper bound for this problem or let me know if the given solution is wrong.

The problem is from Data structures and algorithms made easy in java.

Answer 1

A correct way to state the solution would be

n^4 + 100n^2 + 50 <= c*n^4 for all n>=n0 with c=2 and n0 = 100

=> n^4 + 100n^2 + 50 = O(n^4)

Answer 2

We need to find, smallest rate of growth, g(n) such that

c g(n) >= f(n) for n>=k.

For some constant value of c and k, the above equation will hold true. We do not consider lower values of n. This means g(n) for low values of n is not important. For large values of n , g(n) will be the maximum rate of growth of f(n) .

Here, f(n)= n^4 + 100 n^2 + 50

When n is very large, g(n) = n^4

Find c and k , so that cn^4 >= n^4 + 100 n ^2 + 50

If we discard, lower terms 100 n^2 and 50 . we can say c should be 2 .

2 n^4 >= n^4 .

To find value of k , try substituting n^2 = t , n^4 = t^2 and c=2 ,

2t^2 >= t^2 + 100t + 50

t^2 >= 100t +50

If i start putting values of t from 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , and 10 and t^2 =100

At 10 , i still have

100,00 <= 100, 00 +50

At t=11 , and t^2 = 121 , i have below

14,641 >= 12150.

So my k will be 11 .

Similarly for the other equation, f(n) = 3n +8

g(n) will be n . Find c and k so that below is true .

c.g(n) >= f(n)

4n>=3n+ 8, discard constant 8 to find c , and insert constant 8 back in equation to find k .

At k=8 , we have 32>=32 .