arftdipto's blog

Binary Search on Double
By arftdipto, history, 5 years ago, In English

Can anyone suggest me a good blog for binary search on double ? I found these type of implementation on different blog but failed to manage explanation. Could anyone explain it please ?

Type 1:

#include<bits/stdc++.h>
using namespace std;

//We want to find an element which a<0.0000001

const float target = 0.0000001;

int main()
{
    float l=0.00000000,r=100000.00000000;
    cout << l << " " << r;
    while((r-l)>0.0000000001){
        float mid = (float)((l+r)/2);
        cout << mid << endl;
        if(mid>target) r=mid;
        else l=mid;
    }cout << l;
}

Type 2:

int iterations = 0;
while(iterations < 300)
{
  // your binary search logic
  iterations++;
}
  • Vote: I like it
  • 0
  • Vote: I do not like it

Comments Comments (7)
| Write comment?
»
5 years ago, # |
  Vote: I like it 0 Vote: I do not like it

This one is good

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Thanks :D

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Pashka goes into a lot of detail regarding the implementation issue you raised in your post. Look at his posts on EDU Binary Search. I suffered from the same EPS nightmare until I went through his tutorials.

First register for the EDU. Then: Link for the post

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Do you know why 100 iterations will always be sufficient?

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Because the maximal value for $$$long \space long$$$ is $$$2^{63}-1$$$. Even though binary search works in $$$O(\log n)$$$, there is not estimated value.

      Simple binary search on array for finding specific value may do more operations than exact $$$\log n$$$ value. For example, $$$a = [1, 3, 4, 7, 9, 14], \space x = 7$$$ does 3 operations, while $$$\log 6 = 2$$$. Since, this is very small testcase, the difference is small too.

      → Reply
      • »
        »
        »
        »
        4 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Can you explain with refernce to real numbers?

        → Reply
        • »
          »
          »
          »
          »
          4 years ago, # ^ |
            Vote: I like it +1 Vote: I do not like it

          let us say I want to search for a number in a range of width W.
          x lies somewhere in the range [0,W]. If I were to choose any random number in this range then max error possible is W.
          Everytime I choose the midpoint of the range and will discard a range with width exactly half of the current search range.

          Now width of search range after 1 iteration is W/2, so max error possible is W/2.

          After K-iterations max error possible is W/(2^(K-1))

          → Reply