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Re: Interval arithmetic and Dependency problems

  • To: mathgroup at
  • Subject: [mg96896] Re: Interval arithmetic and Dependency problems
  • From: Bill Rowe <readnews at>
  • Date: Thu, 26 Feb 2009 08:03:14 -0500 (EST)

On 2/25/09 at 4:04 AM, aanderson at (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}] ]

>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

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}]] ===

Out[14]= False

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