Wolfram Function Repository

Instant-use add-on functions for the Wolfram Language

Function Repository Resource:

 IntervalComplement

Source Notebook

Calculate the complement of intervals

Contributed by: Sander Huisman

ResourceFunction["IntervalComplement"][intervalall,interval1,interval2,…]

gives the interval representing all the points of intervalall that are not in any of the intervali.

Details and Options

ResourceFunction["IntervalComplement"] only works on Interval objects with bounds that are real and not symbolic.
ResourceFunction["IntervalComplement"] supports the intervals to have -∞ and +∞ bounds.

Examples

Basic Examples (2) 

Calculate the complement of two intervals:

In[1]:=
ResourceFunction["IntervalComplement"][Interval[{3, 10}], Interval[{0, 5}]]
Out[1]=

Intervals can be disjoint:

In[2]:=
ResourceFunction["IntervalComplement"][Interval[{0, 4}, {8, 12}], Interval[{2, 3}, {6, 11}]]
Out[2]=

Scope (6) 

"Subtract" two Interval objects from another interval object:

In[3]:=
ResourceFunction["IntervalComplement"][Interval[{0, 10}], Interval[{-2, 5}], Interval[{6, 11}]]
Out[3]=

Subtracting an interval might create a disjoint interval:

In[4]:=
ResourceFunction["IntervalComplement"][Interval[{0, 10}], Interval[{3, 7}]]
Out[4]=

Unbounded intervals can be used:

In[5]:=
ResourceFunction["IntervalComplement"][Interval[{0, 10}], Interval[{-\[Infinity], 5}]]
Out[5]=
In[6]:=
ResourceFunction["IntervalComplement"][Interval[{0, \[Infinity]}], Interval[{-5, 5}]]
Out[6]=
In[7]:=
ResourceFunction["IntervalComplement"][ Interval[{-\[Infinity], \[Infinity]}], Interval[{-5, 5}]]
Out[7]=

If there is full overlap an empty interval is returned:

In[8]:=
ResourceFunction["IntervalComplement"][Interval[{0, 10}], Interval[{-\[Infinity], \[Infinity]}]]
Out[8]=

An empty interval stays empty:

In[9]:=
ResourceFunction["IntervalComplement"][Interval[], Interval[{-5, 5}]]
Out[9]=

Complementing with an empty interval has no effect:

In[10]:=
ResourceFunction["IntervalComplement"][Interval[{-5, 5}], Interval[]]
Out[10]=

Applications (1) 

Calculate the absolute complement of the interval [-5,5] by taking the relative complement with the full interval:

In[11]:=
a = Interval[{5, -5}]; b = ResourceFunction["IntervalComplement"][ Interval[{-\[Infinity], \[Infinity]}], a]; Quiet[NumberLinePlot[{a, b}]]
Out[13]=

Properties and Relations (2) 

Compare the complement with the IntervalUnion and the IntervalIntersection of two intervals:

In[14]:=
a = Interval[{0, 10}]; b = Interval[{5, 15}]; NumberLinePlot[{a, b, IntervalUnion[a, b], IntervalIntersection[a, b], ResourceFunction["IntervalComplement"][a, b], ResourceFunction["IntervalComplement"][b, a]}, PlotLegends -> {"a", "b", "a\[MediumSpace]\[Union]\[MediumSpace]b", "a\[MediumSpace]\[Intersection]\[MediumSpace]b", "a\[MediumSpace]\\\[MediumSpace]b", "b\[MediumSpace]\\\[MediumSpace]a"}]
Out[16]=

Compare to regular sets:

In[17]:=
a = Range[0, 10]; b = Range[5, 15]; NumberLinePlot[{a, b, Union[a, b], Intersection[a, b], Complement[a, b], Complement[b, a]}, PlotLegends -> {"a", "b", "a\[MediumSpace]\[Union]\[MediumSpace]b", "a\[MediumSpace]\[Intersection]\[MediumSpace]b", "a\[MediumSpace]\\\[MediumSpace]b", "b\[MediumSpace]\\\[MediumSpace]a"}]
Out[19]=

Possible Issues (1) 

Intervals need to be numeric, if the input is symbolic it will stay unevaluated:

In[20]:=
ResourceFunction["IntervalComplement"][Interval[{0, 10}], Interval[{-2, c}]]
Out[20]=

Neat Examples (1) 

Visualize the "subtraction" of multiple intervals (red) from a base interval (green):

In[21]:=
all = Interval[{-2, 3}, {7, 8}, {11, 12}, {15, \[Infinity]}]; other = {Interval[{1.6, 3.1}, {3.5, 4}, {6, 7.5}], Interval[{7.5, 7.75}], Interval[{-\[Infinity], -1}], Interval[{0.1, 0.7}], Interval[{11.5, 17.5}]}; comp = ResourceFunction["IntervalComplement"][all, Sequence @@ other]; NumberLinePlot[ Join[{Style[all, Darker@Green]}, Style[#, Red] & /@ other, {Style[comp, Black]}], PlotLegends -> LineLegend[{Darker@Green, Red, Red, Red, Red, Black}, {"\!\(\*SubscriptBox[\(Interval\), \(all\)]\)", "\!\(\*SubscriptBox[\(Interval\), \(1\)]\)", "\!\(\*SubscriptBox[\(Interval\), \(2\)]\)", "\!\(\*SubscriptBox[\(Interval\), \(3\)]\)", "\!\(\*SubscriptBox[\(Interval\), \(4\)]\)", "Complement"}]]
Out[24]=

Publisher

SHuisman

Version History

  • 1.0.0 – 22 May 2019

License Information