This is the third article about a seemingly naïve question: how do we write a binary search program? The first article appeared here.
The first program attempt was wrong. The second article introduced a new one:
— Program attempt #2.
from
i := 1 ; j := n
until i = j or Result > 0 loop
m := (i + j) // 2 — Integer division
if t [m] ≤ x then
i := m + 1
elseif t [m] = x then
Result := m
else — In this case t [m] > x
j := m – 1
end
end
The question was: it right?
Again no. A trivial example disproves it: n = 1, the array t contains a single element t [1] = 0, x = 0. Then the initialization sets both i and j to 1, i = j holds on entry to the loop which stops immediately, but Result is zero whereas it should be 1 (the place where x appears).
Here now is attempt #3, let us see it if fares better:
— Program attempt #3.
from
i := 1 ; j := n
until i = j loop
m := (i + j + 1) // 2
if t [m] ≤ x then
i := m + 1
else
j := m
end
end
if 1 ≤ i and i ≤ n then Result := i end
— If not, Result remains 0.
What about this one? Answer on Wednesday.
Post-publication note: the announced fourth ("Wednesday") article was published here.
Bertrand Meyer is chief technology officer of Eiffel Software (Goleta, CA), professor and provost at the Schaffhausen Institute of Technology (Switzerland), and head of the software engineering lab at Innopolis University (Russia).
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