Section: Mathematical Operators
:
operator - the two argument form
y = a : c
and the three argument form
y = a : b : c
The two argument form is exactly equivalent to a:1:c
. The output y
is the vector
where a+nb <= c
. There is a third form of the colon operator, the
no-argument form used in indexing (see indexing
for more details).
y = a : b : c
The first question that one might need to answer is: how many points in this vector? If you answered
then you would be doing the straighforward, but not correct thing. because a, b, and c are all floating point values, there are errors associated with each of the quantities that can lead to n not being an integer. A better way (and the way FreeMat currently does the calculation) is to compute the bounding values (for b positive)
where
means we replace x by the floating point number that is closest to it in the direction of y. Once we have determined the number of points we have to compute the intermediate values
but one can readily verify for themselves that this may \emph{not} be the same as the vector
even for the case where
for some n. The reason is that the roundoff in the calculations may
be different depending on the nature of the sum. FreeMat uses the
following strategy to compute the double-colon vector:
\begin{enumerate}
\item The value n
is computed by taking the floor of the larger
value in the interval defined above.
\item If n
falls inside the interval defined above, then it is
assumed that the user intended c = a + n*b
, and the symmetric
algorithm is used. Otherwise, the nonsymmetric algorithm is used.
\item The symmetric algorithm computes the vector via
working symmetrically from both ends of the vector (hence the nomenclature), while the nonsymmetric algorithm computes
In practice, the entries are computed by repeated accumulation instead
of multiplying the step size by an integer.
\item The real interval calculation is modified so that we get the
exact same result with a:b:c
and c:-b:a
(which basically means
that instead of moving towards infinity, we move towards the signed
infinity where the sign is inherited from b
).
\end{enumerate}
If you think this is all very obscure, it is. But without it, you will
be confronted by mysterious vectors where the last entry is dropped,
or where the values show progressively larger amounts of accumulated
roundoff error.
--> y = 1:4 y = 1 2 3 4Now by half-steps:
--> y = 1:.5:4 y = 1.0000 1.5000 2.0000 2.5000 3.0000 3.5000 4.0000Now going backwards (negative steps)
--> y = 4:-.5:1 y = 4.0000 3.5000 3.0000 2.5000 2.0000 1.5000 1.0000If the endpoints are the same, one point is generated, regardless of the step size (middle argument)
--> y = 4:1:4 y = 4If the endpoints define an empty interval, the output is an empty matrix:
--> y = 5:4 y = Empty array 1x0