This is the fourth article about a seemingly naïve question: how do we write a binary search program? The first article appeared here.
The first two program attempts were wrong. The third article introduced a new variant reading as follows:
— 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.
The question was: is it right?
I know, you have every right to be upset at me, but the answer here too is no.
Consider a two-element array t = [0 0] (so n = 2, remember that our arrays are indexed from 1 by convention) and a search value x = 1. The successive values of the variables and expressions are:
m i j i + j + 1
After initialization: 1 2 4
i ≠ j, so enter loop: 2 3 2 6 — First branch of "if" since t [2] < x
i ≠ j, enter loop again: 3 ⚠ — Out-of-bounds memory access!
— (trying to access non-existent t [3])
Oops!
Note that we could hope to get rid of the array overflow by initializing i to 0 rather than 1. This variant (version #3′) is left as a bonus question to the patient reader. (Hint: it is also not correct. Find a counter-example.)
OK, this has to end at some point. What about the following version (#4): is it right?
— Program attempt #4.
from
i := 0 ; j := n + 1
until i = j loop
m := (i + j) // 2
if t [m] ≤ x then
i := m + 1
else
j := m
end
end
if 1 ≤ i and i ≤ n then Result := i end
Answer on Friday.
Post-publication note: the announced fifth ("Friday") 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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