BinarySearch运行时间的渐近增长率

Question

In the following pseudo code for binarySearch

myfunction(A[1],A[2]..A[n]:array of ints,x:positive integer){
  i = 1
  j = n
 while i<=j do 
     k=floor( (i+j)/2)
    if (x==A[k])
      return k;
    else if (x<A[k])
      j=k-1
    else
      i=k+1
 return -1 
  }

Why is this inequality true j-i+1 <= n / 2^t where t is the number of iterations. All I get is that n /2^t with t increasing if n/2^t =1 then t will be the total number of iterations but the Left handed side : i-j+1 I don't get how we connect these statements to make inequality.What is the logic behind?

Answer 1

j-i+1 is at any point in time the number of values still to consider. It is clear that the inequality holds at the beginning: n-1+1 = n / 2^0 .

Now with each iteration you approximately half the interval to be searched. A careful analysis of the different cases shows the interval is even reduced a bit more than that. So each iteration gives you at least another factor of 1/2 for the interval size.

I will just do one case and leave the other three as exercise for you. Assume j-i+1 is equal to s and assume s is an odd number. Then i+j is even and k=(i+j)/2 points exactly to the middle element leaving (s-1)/2 elements to the left of it and to the right of it. Lets assume now x < A[k] so that we will set j to k-1 . The new value of j-i+1 is therefore k-1-i+1 = ki = (s-1)/2 < s/2 .