Re: Interval arithmetic and Dependency problems
- To: mathgroup at smc.vnet.net
- Subject: [mg96896] Re: Interval arithmetic and Dependency problems
- From: Bill Rowe <readnews at sbcglobal.net>
- Date: Thu, 26 Feb 2009 08:03:14 -0500 (EST)
On 2/25/09 at 4:04 AM, aanderson at amherst.edu (Andy Anderson) wrote:
>Mathematica symbolizes a disjoint interval as Interval[], and this
>might result from, say
>IntervalIntersection[Interval[{0,1}], Interval[{2,4}]].
>While I would like to think of this as a "null" interval, one
>without any range (in particular Interval[{0,0}]),
But Interval[{0,0}] is not a "null" interval. Instead it is an
interval containing a single point on the real line. If
Interval[{0,0}] was defined to be a "null" interval, how would
you deal with
IntervalIntersection[ Interval[{-1,0}], Interval[{0,1}] ]
>Mathematica
>instead treats this object as the same thing as Interval[{Infinity,
>-Infinity}], i.e. completely off the real line (not explained in the
>documentation). Therefore,
>Interval[{0,1}] + Interval[] == Interval[] (rather than
>Interval[{0,1}])
Hmm... It looks like you are expecting Interval[{0,1}] +
Interval[] to represent the union of two intervals. If so, why
aren't you using the function IntervalUnion, i.e.,
In[12]:= IntervalUnion[Interval[{0, 1}], Interval[]]
Out[12]= Interval[{0,1}]
And if you aren't using Interval[{0,1}] + Interval[] to be the
union of the two intervals, what are you defining the sum of two
intervals to be?
>More importantly, I want to test if two intervals overlap or not. If
>they don't,
>IntervalIntersection[Interval[{0,1}], Interval[{2,4}]] == Interval[]
>returns True, which is fine, but if they *do* overlap,
>IntervalIntersection[Interval[{0,3}], Interval[{2,4}]] == Interval[]
>does *not* return False but rather leaves it undetermined.
Instead of Equal use SameQ, i.e.,
In[14]:= IntervalIntersection[Interval[{0, 3}], Interval[{2,
4}]] ===
Interval[]
Out[14]= False