# Exercise 2.3.5

Referring back to the searching problem (see Exercise 2.1-3), observe that if the sequence A is sorted, we can check the midpoint of the sequence against $\nu$ and eliminate half of the sequence from further consideration. The

binary searchalgorithm repeats this procedure, halving the size of the remaining portion of the sequence each time. Write pseudocode, either iterative or recursive, for binary search. Argue that the worst-case running time of binary search is $\Theta(\lg{n})$.

Here's the presudocode:

```
BINARY-SEARCH(A, v):
low = 1
high = A.length
while low <= high
mid = (low + high) / 2
if A[mid] == v
return mid
if A[mid] < v
low = mid + 1
else
high = mid - 1
return NIL
```

The argument is fairly straightforward and I will make it brief:

$$ T(n+1) = T(n/2) + c $$

This is the recurrence shown in the chapter text, and we know, that this is logarithmic time.

### C code

// Indices in the C code are different int binary_search(int A[], int length, int v) { int low = 0; int high = length; int mid; while (low < high) { mid = (low + high) / 2; if (A[mid] == v) return mid; else if (A[mid] < v) low = mid + 1; else high = mid; } return -1; }