I'm using Schaum's Outline of Programming With Fortran 77 book, and there's an example about binary search using bracketing group of values method. First of all here's the code :
    INTEGER X(100)
    INTEGER RANGE
    INTEGER START , FINISH
    PRINT *, 'Number of values ?'
    READ *, N
    DO 10 I = 1, N
        READ *, X(I)
    END DO 
    PRINT *, 'Enter Value'
    READ *, VAL
    START =  1 
    FINISH = N
    RANGE = FINISH - START 
    MID = (START + FINISH) /2
    DO WHILE( X(MID) .NE. VAL .AND. RANGE .NE. 0)
      IF (VAL .GT. X(MID))THEN
        START = MID 
      ELSE 
        FINISH = MID
      END IF
      RANGE = FINISH - START
      MID = (START + FINISH)/2
    END DO 
    IF( X(MID) .NE. VAL) THEN
      PRINT *, VAL, 'NOT FOUND'
    ELSE
      PRINT *, 'VALUE AT' , MID
    END IF
    END
The problem is, when i enter 7 values array like
2 | 9 | 11 | 23 | 49 | 55 | 66
And search for 66 for example, when
MID = 5
, the new MID for the next loop becomes 6 . But when it's 6, it can't get incremented for the next loop because
MID = (START + FINISH)/2 = (6+7)/2 = 6
Where it should be 7 of course.
It still on 6. And my program never give me an output of course. What shall I do here ?
So, the answer is NO, it is not possible to use or implement Binary Search on unsorted arrays or lists, because, the repeated targeting of the mid element of one half depends on the sorted order of data structure.
Binary search works on sorted arrays.
Binary search can be implemented only on a sorted list of items. If the elements are not sorted already, we need to sort them first.
This is just a typo, or maybe someone got confused when they translated it from a C version and had to to change indexing.
The key invariant in the loop is that the value, if it's in the list, must fall in the array somewhere from indices start to finish, inclusive. But once you've excluded mid, it should be taken out the list. But it's not here, so the list is always too long and you run into the problem you see.
A correct version sets start to mid+1, or finish to mid-1, to exclude mid. The corrected code, written in a Fortran 90 style:
program binarysearch
    implicit none
    integer, allocatable, dimension(:) ::  x
    integer :: range, start, finish, mid
    integer :: i, n, val
    print *, 'Number of values ?'
    read *, N
    allocate( x(N) )
    do i = 1, n
        READ *, x(i)
    end do
    print *, 'Enter Value'
    read *, VAL
    start =  1
    finish = N
    range = finish - start
    mid = (start + finish) /2
    do while( x(mid) /= val .and. range >  0)
      if (val > x(mid)) then
        start = mid + 1
      else
        finish = mid - 1
      end if
      range = finish - start
      mid = (start + finish)/2
    end do
    if( x(mid) /= val) then
      print *, val, 'NOT FOUND'
    else
      print *, 'VALUE AT' , mid
    end if
    deallocate( x )
end program binarysearch
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