       Re: Re: Interval arithmetic and Dependency problems

• To: mathgroup at smc.vnet.net
• Subject: [mg96804] Re: [mg95483] Re: Interval arithmetic and Dependency problems
• From: Andy Anderson <aanderson at amherst.edu>
• Date: Wed, 25 Feb 2009 04:04:40 -0500 (EST)
• References: <gl1bng\$998\$1@smc.vnet.net> <200901201043.FAA16621@smc.vnet.net>

```Nice to see this discussion here. I was just trying out the Interval
functions for a project where I need to compare two time intervals,
and noticed what is, to me, unexpected behavior.

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}]), 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}])

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. This seems
incorrect to me under either interpretation of Interval[]. Can anyone
justify this?

In any case, my workaround is to instead use:

Min[IntervalIntersection[Interval[{0,1}], Interval[{2,4}]]] == Infinity

which behaves consistently. This turns out to be more than twice as
fast as actually comparing endpoints for overlap.

P.S. You know what would be nice? The option to define end points as
exclusive, e.g. [2,4) which might be defined as
Interval[{2,4},Exclusive->Max].

-- Andy

On Jan 20, 2009, at 5:43 AM, Jean-Marc Gulliet wrote:

> In article <gl1bng\$998\$1 at smc.vnet.net>, magma <maderri2 at gmail.com>
> wrote:
> [...]
>> For example evaluate
>>
>> f[x_] := x^2 + x
>> f[Interval[{-1, 1}]]
>>
>> and you get
>>
>> Interval[{-1, 2}]
>>
>> which is not really very good.
> [...]
>
> Intervals are assumed to be independent, say, like two independent
> variables. Therefore, what Mathematica computes in your example is
>
> In:= Interval[{-1, 1}]^2 + Interval[{-1, 1}]
>
> Out= Interval[{-1, 2}]
>
> You could think of the endpoints of the resulting interval as being
> the
> min and max values of a function of two variables over the region
> -1<=x|y<=1.
>
> In:= Minimize[{x^2 + y, -1 <= x <= 1, -1 <= y <= 1}, {x, y}]
>
> Out= {-1, {x -> 0, y -> -1}}
>
> In:= Maximize[{x^2 + y, -1 <= x <= 1, -1 <= y <= 1}, {x, y}]
>
> Out= {2, {x -> -1, y -> 1}}
>
> This also explains the difference between x^2 (one variable) and x*x
> which is rally computed as x*y (two independent variables).
>
> In:= Minimize[{x*y, -1 <= x <= 1, -1 <= y <= 1}, {x, y}]
>
> Out= {-1, {x -> -1, y -> 1}}
>
> In:= Maximize[{x*y, -1 <= x <= 1, -1 <= y <= 1}, {x, y}]
>
> Out= {1, {x -> -1, y -> -1}}
>
> Regards,
> --Jean-Marc
>

```

• Prev by Date: Re: newbie: how to define, typeset a multi-rule function?
• Next by Date: Re: newbie: programmatic sequence of plots?
• Previous by thread: Re: voices for version 7 text-to-speech
• Next by thread: Re: Interval arithmetic and Dependency problems