算法分析 - 渐近分析

Question

Hi i have started learning algorithm analysis. Here i have a doubt in asymptotic analysis. Let's say i have a function f(n) = 5n^3 + 2n^2 + 23. Now i need to find the Big-Oh, Big-Omega and Theta Notations for the above function,

Big-Oh: 
    f(n) <= (5 + 2 + 23) n^3  // raising all the n's to the power of 3 will give us a value which will be always greater than f(n)
    f(n) <= 30n^3
    f(n) belongs to Big-Oh(n^3)

Big-Omega:
    n^3 <= f(n)
    f(n) belongs to Big-Omega(n^3)

Theta: 
    n^3 <= f(n) <= 30 n^3 
    f(n) belongs to Theta ( n^3)
So here, 
    f(n) belongs to Big-Oh(n^3)
    f(n) belongs to Big-Omega(n^3)
    f(n) belongs to Theta(n^3)

Like this for any polynomial,the order of growth for Oh,Omega and Theta Notations are same(in our case it is order of n^3). When order of growth will be same for all the notations, then what is the use of showing them with different notations and where exactly it can be used? Kindly give me a practical example if possible.

Answer 1

Big theta (Θ) is when our upper bound (O) and lower bound (Ω) are the same, in other words it's a tight bound then. That's one reason to show both O and Ω (or well all three).

Why is this useful? Because Θ is a tight bound - it's much stronger than O . With O you can say that the above is O(n^1000) and you are still technically correct. A lot of times O != Ω and you don't have that tight bound.

Why are we usually talking about O usually, then? Well because in most cases we are interested in the upper bound (of the worst case scenario) of our algorithm. Sometimes we simply don't know the Θ and we go with O instead. Also it's important to notice that many people simply misuse those symbols, don't fully understand them and/or are simply not precise enough and use O in places where Θ could be used.

For instance quicksort does not have a tight bound (unless we are talking specifically about either best/average or worst case analysis) as it's Ω(nlogn) in the best and average cases but O(n^2) in the worst case. On the other hand mergesort is both Ω(nlogn) and O(nlogn) therefore it's also Θ(nlogn).

All in all it's all theoretical as quicksort in practics is in most cases faster as you usually don't hit the worst case and the operations done by quicksort are easier.

Answer 2

In the example you gave, the execution time is known and seems to be fixed. So it is in O(n^3), Ω(n^3) and thus in Θ(n^3) for all cases.

However, for an algorithm, the execution time may, and not so rarely, depend on the input the algorithm is running on.

Eg.: to search a key in a linked list takes going through all the list members in the worst case and that's linear time. In the best case, the key you're looking for at that time is in the very beginning of the list and that's constant time. So the algorithm is in O(n) and in Ω(1)=O(1). There's no valid f(n) to specify Θ(f(n)) for this algorithm.

Answer 3

Above Function Running Time can be computed in the following manner

• Is it omega and big o of n ? 5n^3+2n^2+23 >=cn 5n^2+2n+23/n >= c as n grows and it tends to infinite such constant c can exist which is smaller than or equals to left hand side of the inequality so the function running time is omega of n. 5n^2+2n+23/n <= c when n tends to infinite this inequality doesn't hold true because such constant can not exist which is equal to or greater than Left hand side of the inequality so the function running time is not big o of n.
• is it omega and Big o of n ^3 ? 5n^3+2n^2+23 >=cn^3 5 + 2/n + 23/n^3 >=c this inequality holds true so it's omega of n^3. 5 + 2/n + 23/n^3 <=c this inequality also holds true so it's big o of n^3. Since it's omega and big o of n^3 hence it's theta of n^3 as well. Similarly its omega of n^ 2 but not big o of n^2.