hello____world's blog

Doubts on Ternary Search
By hello____world, history, 4 years ago, In English

I have two doubts regarding Ternary Search, I hope someone can help.

1. Complexity:

In every blog, I find complexty of ternary search as O(log3(N)). I know how we get this complexity but if we do something like this:

int i=0, j=MAX;
while(i<=j){
    int m=(i+j)/2;
    if(eval(m)<eval(m+1)){
        j=m-1;
    }else{
        i=m+1;
    }
}
cout<<j+1;

Wouldn’t complexity be O(log(2))? What are the limitations of this kind of implementation? I believe we can modify this implementation to work with decimal values as well.

2. Type of Function?

Can we use this algorithm work with non-strict decrease-increase function?
If I have a function with values: {5,5,4,3,3,3,2,3,4,4,5} i.e: f(0)=5, f(2)=4 and so on.
How should I implement ternary search on this function?
To be specific, what should I do if f(m1)==f(m2)?

If you have any blog which can help me with these doubts, please share.

  • Vote: I like it
  • +3
  • Vote: I do not like it

Comments Comments (4)
| Write comment?
»
4 years ago, # |
  Vote: I like it +8 Vote: I do not like it

In complexity all logs with constant base are equal: $$$\log_b a = \frac{\log a}{\log b}$$$

You can't use it with non-strictly decreasing/increasing functions.

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    In complexity all logs with constant base are equal

    Ah, yes! Thanks.
    Also is there any problem with using M, M+1 as two points? In almost every tutorial, they divide the array into 3 parts whereas dividing the array into 2 looks like a better solution.

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it +8 Vote: I do not like it

      Dividing the array into 2 is the integer ternary search, it doesn't work for decimal values. You need to split into 3 for decimal values.

      → Reply
»
4 years ago, # |
  Vote: I like it +13 Vote: I do not like it

Can we use this algorithm work with non-strict decrease-increase function?

No. Imagine you have an array that is $$$1$$$ everywhere except at one unknown point it is $$$0$$$. No matter how you query points, if it evaluates to $$$1$$$ every time you cannot guarantee to find the minimum with less than $$$n-1$$$ queries.

→ Reply