Because a Bézier curve is a B-spline curve in a special form and because de Boor's algorithm works very similar to de Casteljau's algorithm, we expect that the latter may be a special case of the former. This indeed is the case.

Suppose we have a B-spline curve **C**(*u*) of degree *n*
defined by *n*+1 control points **P**_{0}, ...,
**P**_{n} and a knot vector of *p*+1 0's followed by
*p*+1 1's. We discussed on a
previous
page that **C**(*u*) is actually a Bézier curve of degree
*n*.

Consider the computation steps of de Boor's algorithm. We already
know **C**(0) = **P**_{0} and
**C**(1) = **P**_{n}. What we need to show is that
de Boor's algorithm and de Casteljau's algorithm act the same way and compute
the same result for every *u*. Let the knots be
*u*_{0} = *u*_{1} = ... =
*u*_{n} = 0 and
*u*_{n+1} = *u*_{n+2} = ... =
*u*_{2n+1} = 1. Given any *u* in (0,1),
it is in [*u*_{n}, *u*_{n+1}).
Consequently, the control points that involve in the knot insertion
computation are **P**_{n}, ...,
**P**_{n-n} = **P**_{0}, and the
intervals used for calculating the *a*_{i}'s are
[*u*_{n}, *u*_{n+n}) = [0,1),
[*u*_{n-1}, *u*_{n+n-1}) = [0,1),
..., and
[*u*_{n-n+1}, *u*_{n+1}) = [0,1).
Therefore, for each *i*, we have

Since each leg is divided with a ratio of
*a*_{i}:1-*a*_{i}, the above
expression shows that in this special case the dividing ratio is all the
same and is equal to *u*:1-*u*. Thus, the first knot insertion
in de Boor's algorithm computes exactly the first column in de Casteljau's
algorithm.

A similar argument shows that the dividing ratio in all subsequent insertion
steps is also equal to *u*:1-*u*. Consequently, in this
special case, the computation steps of de Boor's algorithm are exactly
identical to the corresponding steps of de Casteljau's algorithm.
Therefore, de Boor's algorithm reduces to de Casteljau's algorithm
when the knot vector has only two knots of multiplicity *n*+1.