Here's another binary search, this one without an explicit
test for a match in the loop, and without a variable
named found.

We're looking for number x in an array T, and then
we'll print the corresponding element of an array N.

The idea (trick?) of this algorithm is that once the
pointers have crossed, the left-hand one must point at
the desired entry, if it's there.

We assume here that if (b+e)/2 isn't an integer, the
extra .5 will be thrown away.  So for example
(16 + 17)/2 = 16


Get the arrays T and N and the size n
Get incoming phone number x

b <-- 1; e <-- n;

while (b <= e)
   { m <-- (b+e)/2;
     if (T[m] < x) b <-- m+1;
     else e <-- m-1; }

if (T[b] = x) Print N[b]; else Print "Sorry";