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Open Intervals

Following up on this question, I was wondering whether Mma can handle open intervals. For example, the union of the intervals, $$1<x<5$$ and $$5<x<8$$

should not include the number 5. This is easy enough to do in one's head, but how can it be done, if at all, computationally?

Interval Complement

Also, is there a way to find the complement of two intervals? IntervalComplement[int1,int2,int3] should contain all the points in int1 that are not in the other intervals.


Edit:

Let's take Mark McClure's data as an example.

int1 = x < -2 || -1 <= x < 1 || x == 3 || 4 < x <= Pi^2;
int2 = -3 <= x < 0 || x > 1;

The intervals are shown below:

intervals

The Interval Complement (drawn above in blue on the x-axis) would seem to be:

x < -3 || 0 <= x < 1
share|improve this question
    
    
@belisarius Yes, Andrzej Kozlowski's IntervalComplement does work for closed intervals. Very nice! It seems like his code does not handle open intervals, relying, as it does, on Interval for input. –  David Carraher Oct 1 '12 at 2:57
    
Read thru the end :) forums.wolfram.com/mathgroup/archive/2006/Oct/msg00347.html –  belisarius Oct 1 '12 at 3:02
1  
I have now read through to the end. I did not feel that anything was resolved. I did not find compelling the attempted explanation of IntervalComplement's inherent contradiction. Furthermore, the discussions about inclusion or not of endpoints in intervals was biased by the fact that mms does not (apparently) recognize open intervals. Would you like to present your take on the discussion? –  David Carraher Oct 1 '12 at 3:18
    
I had have a little domestic accident and I'm typing single handed and painfully. Perhaps tomorrow,sorry. –  belisarius Oct 1 '12 at 5:41

3 Answers 3

up vote 18 down vote accepted

I'd represent the sets using inequalities and/or equalities and then apply Reduce. Here's an example:

set1 = x < -2 || -1 <= x < 1 || x == 3 || 4 < x <= Pi^2;
set2 = -3 <= x < 0 || x > 1;
Reduce[set1 && set2]

enter image description here

Here's the complement of the union of the two intervals.

Reduce[!(set1 || set2)]

(* Out: x==1 *)

We might define an interval complement function as follows:

intervalComplement[bigInt_, moreInts__] := 
  Reduce[bigInt && (! (Or @@ {moreInts}))];

For example:

intervalComplement[-10 < x <= 10, -8 < x <= -6, 
  0 <= x <= 2, x == 3]
(* Out: -10 < x <= -8 || -6 < x < 0 || 2 < x < 3 || 3 < x <= 10 *)
share|improve this answer
    
You seem to have generated the intersection of the intervals instead of the complement of the intervals. Consider Graphics[{Arrow[{{-2, 1}, {-8, 1}}], Line[{{-1, 1}, {1, 1}}], Point[{3, 1}], Line[{{4, 1}, {Pi^2, 1}}], Line[{{-3, -1}, {0, -1}}], Arrow[{{1, -1}, {12, -1}}]}, Axes -> {True, True}, GridLinesStyle -> Dashed, BaseStyle -> 14, GridLines -> {{-2, -1, 1, 3, 4}, None}, Ticks -> {Range[-8, 10], None}] You gave the points common to set1, set2. –  David Carraher Oct 1 '12 at 3:44
    
Let's see. In any case I probably won't be awarding the accept till tomorrow. It wouldn't surprise me if another came in. There are actually two interrelated questions: one related to open/closed intervals, the other to interval complements. –  David Carraher Oct 1 '12 at 4:13
    
You might want to take a look at the explanation and picture I placed in the question. –  David Carraher Oct 1 '12 at 4:31
    
@DavidCarraher I did. I also checked that my intervalComplement gave your desired answer. –  Mark McClure Oct 1 '12 at 4:34
    
Very nice solution. It handles open intervals! –  David Carraher Oct 1 '12 at 16:04

Here's an implementation of interval complement that is meant to be used with Interval expressions. Interval represents closed intervals according to the documentation, and this is consistent with the things the built-in functions do with intervals. However, in this implementation of the interval complement I simply ignore whether an interval is open or closed. I realize that this is not exactly what you asked for. I wrote this because I needed it, and I thought it'd be useful to post it.


intervalInverse[Interval[]] := Interval[{-Infinity, Infinity}]
intervalInverse[Interval[int__]] :=
 Interval @@ Partition[
   Replace[Flatten[{int}],
    {{-Infinity, mid___, Infinity} :> {mid},
     {-Infinity, mid__} :> {mid, Infinity},
     {mid__, Infinity} :> {-Infinity, mid},
     {mid___} :> {-Infinity, mid, Infinity}
     }
    ], 2]

intervalComplement[a_Interval, b__Interval] := 
 IntervalIntersection[a, intervalInverse@IntervalUnion[b]]

intervalInverse[a] will compute the complement $(-\infty, \infty) \setminus a$.

intervalComplement[a,b,c,...] will compute $a \setminus (b \cup c \cup \ldots )$.


Example usage:

In[]:= intervalInverse[Interval[{1, 2}]]
Out[]= Interval[{-Infinity, 1}, {2, Infinity}]

In[]:= intervalComplement[Interval[{0, 10}], Interval[{2, 3}]]
Out[]= Interval[{0, 2}, {3, 10}]
share|improve this answer
    
Nice. It surprises me that something like this is not native to Mathematica, yet. –  David Carraher Feb 21 at 13:36
    
@DavidCarraher My guess is that it's not implemented because the complement of two closed intervals is going to be an open interval, which can't be represented using Interval (if we keep the assumption that it always represents a closed ones). I was sloppy and ignored all this because it didn't matter for my application. That sloppyness results in things like this intervalInverse@intervalInverse[Interval[{1, 1}]] --> Interval[]. By taking the "inverse" twice I lost the endpoints of this zero-length non-empty closed interval and ended up with an empty interval. –  Szabolcs Feb 21 at 18:09
    
Yes, but is an empty interval any stranger than an empty set? Granted, it is an interval without location. On the other hand, an empty set is a set with no elements. –  David Carraher Feb 21 at 19:48

This is my code

{a = x > 1 && x < 5, b = x > 5 && x < 8}
{Reduce[a && b]}  

and

{a = x > 1 && x < 5, b = x >= 5 && x < 8}
{Reduce[a || b]}

Edit

Some examples

{a = x > 0 && x < 3, b = x > -1 && x < 2, c = x > -2 && x < 1} {Reduce[a && (b || c)],
Reduce[(a && b) || (a && c)], Reduce[a || (b && c)], Reduce[(a || b) && (a || c)]} 
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