Calculating the width of the interval defined by an inequality












5














I am looking for a Mathematica function that takes an inequality as the input and gives back the width defined by upper bound - lower bound:



Example: 



Fn[1 <= x <= 2.5]



1.5




If the inequality is evaluated to False (e.g., 2 <= x <= 1), then I need the function to return 0.



I truly appreciate your help.






function-construction inequalities







share|improve this question









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Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    5














    I am looking for a Mathematica function that takes an inequality as the input and gives back the width defined by upper bound - lower bound:



    Example: 



    Fn[1 <= x <= 2.5]



    1.5




    If the inequality is evaluated to False (e.g., 2 <= x <= 1), then I need the function to return 0.



    I truly appreciate your help.






    function-construction inequalities







    share|improve this question









    New contributor




    Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      5












      5








      5







      I am looking for a Mathematica function that takes an inequality as the input and gives back the width defined by upper bound - lower bound:



      Example: 



      Fn[1 <= x <= 2.5]



      1.5




      If the inequality is evaluated to False (e.g., 2 <= x <= 1), then I need the function to return 0.



      I truly appreciate your help.






      function-construction inequalities







      share|improve this question









      New contributor




      Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      I am looking for a Mathematica function that takes an inequality as the input and gives back the width defined by upper bound - lower bound:



      Example: 



      Fn[1 <= x <= 2.5]



      1.5




      If the inequality is evaluated to False (e.g., 2 <= x <= 1), then I need the function to return 0.



      I truly appreciate your help.






      function-construction inequalities




      function-construction inequalities






      share|improve this question









      New contributor




      Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









      New contributor




      Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question








      edited yesterday









      m_goldberg

      84.3k872195




      84.3k872195






      New contributor




      Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked yesterday









      Monire JaliliMonire Jalili

      261




      261




      New contributor




      Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          3 Answers
          3






          active

          oldest

          votes


















          9














          f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var], Length[Flatten[{var}]]]
          

          
f[1 <= x <= 2.5, x]



          1.5




          This works also for some systems of inequalities in several variables:



          f[{1 <= x <= 2.5, 0 <= y <= x}, {x, y}]



          2.625




          Edit:



          This one-argument version treats all symbols in the first argument as variables:



          f[ineq_] := f[ineq, DeleteDuplicates[Cases[ineq, _Symbol]]]





          share|improve this answer























          • When the dimension of the region is less than Length[var], for example a line embedded in the 2D plane, then RegionMeasure gives the measure in the reduced dimension: f[{1 <= x <= 2.5, y == 0}, {x, y}] gives 1.5 (the length of the line instead of its area), which is not what's usually expected. Fix this with f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var], Length[var]], so that now f[{1 <= x <= 2.5, y == 0}, {x, y}] gives 0 as expected (the line has zero area).
            – Roman
            20 hours ago












          • Good point! Thank your for remark; I fixed it. I should have been more cautious; these dimensional issues with regions is actually a frequent source of confusion.
            – Henrik Schumacher
            19 hours ago










          • I was too fast in commenting: the function now doesn't work for var=x since Length[x]=0. Maybe two separate definitions for var_Symbol (using dimension 1) and for var_List (using dimensions Length[var])?
            – Roman
            19 hours ago










          • Yes even better!
            – Roman
            19 hours ago



















          3














          fn[expr_] := Module[{},
          
  If[! expr, Return [0]];
          
  If[Head[expr] == Inequality, Return[Abs[expr[[5]] - expr[[1]]]]];
          
  Return[Abs[expr[[3]] - expr[[1]]]];
          
  ]
          

          
fn[2 <= x <= 1]
          
(*0*)
          

          
fn[1 <= x <= 2.5]
          
(*1.5*)
          

          
fn[2.5 > x > 1]
          
(*1.5*)


          Don't know if this works in all cases, but works in the simple cases you provide plus some.






          share|improve this answer































            2














            To get a function that would handle the all the kinds of arguments I want it to handle turned out to be more of a challenge than I anticipated, but here is what I came up with.



            Edit



            This version is handle expressions that evaluate to False more robustly.



            ClearAll[fn, helper1, helper2]
            

            
SetAttributes[fn, HoldFirst]
            
fn[expr_] := If[expr, helper1[expr], helper2[expr], helper1[expr]]
            

            
SetAttributes[helper1, HoldFirst]
            
helper1[expr : _Inequality | _Less | _LessEqual | _Greater | _GreaterEqual] :=
            
  Module[{args = List @@ Unevaluated[expr], a, b},
            
    {a, b} = MinMax[Select[args, NumericQ]];
            
    b - a]
            
helper1[___] = $Failed;
            

            
SetAttributes[helper2, HoldFirst]
            
helper2[expr : _Inequality | _Less | _LessEqual | _Greater | _GreaterEqual] := 0;
            
helper2[___] = $Failed;
            

            
###Tests
            

            
fn[1 < x <= 2.5]



            1.5




            fn[1 < x <= π]



            -1 + π




            fn[1 >= x > π]



            0




            fn[1 >= x > -1]



            2




            fn[-1 < 1 <= 2.5]



            3.5




            fn[1 < x < 3 < y < 5]



            4




            fn[1.5 < 2]



            0.5




            fn["garbage"]



            $Failed




            fn[1 == 1]



            $Failed




             fn[1 != 1]



            $Failed







            share|improve this answer























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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              9














              f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var], Length[Flatten[{var}]]]
              

              
f[1 <= x <= 2.5, x]



              1.5




              This works also for some systems of inequalities in several variables:



              f[{1 <= x <= 2.5, 0 <= y <= x}, {x, y}]



              2.625




              Edit:



              This one-argument version treats all symbols in the first argument as variables:



              f[ineq_] := f[ineq, DeleteDuplicates[Cases[ineq, _Symbol]]]





              share|improve this answer























              • When the dimension of the region is less than Length[var], for example a line embedded in the 2D plane, then RegionMeasure gives the measure in the reduced dimension: f[{1 <= x <= 2.5, y == 0}, {x, y}] gives 1.5 (the length of the line instead of its area), which is not what's usually expected. Fix this with f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var], Length[var]], so that now f[{1 <= x <= 2.5, y == 0}, {x, y}] gives 0 as expected (the line has zero area).
                – Roman
                20 hours ago












              • Good point! Thank your for remark; I fixed it. I should have been more cautious; these dimensional issues with regions is actually a frequent source of confusion.
                – Henrik Schumacher
                19 hours ago










              • I was too fast in commenting: the function now doesn't work for var=x since Length[x]=0. Maybe two separate definitions for var_Symbol (using dimension 1) and for var_List (using dimensions Length[var])?
                – Roman
                19 hours ago










              • Yes even better!
                – Roman
                19 hours ago
















              9














              f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var], Length[Flatten[{var}]]]
              

              
f[1 <= x <= 2.5, x]



              1.5




              This works also for some systems of inequalities in several variables:



              f[{1 <= x <= 2.5, 0 <= y <= x}, {x, y}]



              2.625




              Edit:



              This one-argument version treats all symbols in the first argument as variables:



              f[ineq_] := f[ineq, DeleteDuplicates[Cases[ineq, _Symbol]]]





              share|improve this answer























              • When the dimension of the region is less than Length[var], for example a line embedded in the 2D plane, then RegionMeasure gives the measure in the reduced dimension: f[{1 <= x <= 2.5, y == 0}, {x, y}] gives 1.5 (the length of the line instead of its area), which is not what's usually expected. Fix this with f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var], Length[var]], so that now f[{1 <= x <= 2.5, y == 0}, {x, y}] gives 0 as expected (the line has zero area).
                – Roman
                20 hours ago












              • Good point! Thank your for remark; I fixed it. I should have been more cautious; these dimensional issues with regions is actually a frequent source of confusion.
                – Henrik Schumacher
                19 hours ago










              • I was too fast in commenting: the function now doesn't work for var=x since Length[x]=0. Maybe two separate definitions for var_Symbol (using dimension 1) and for var_List (using dimensions Length[var])?
                – Roman
                19 hours ago










              • Yes even better!
                – Roman
                19 hours ago














              9












              9








              9






              f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var], Length[Flatten[{var}]]]
              

              
f[1 <= x <= 2.5, x]



              1.5




              This works also for some systems of inequalities in several variables:



              f[{1 <= x <= 2.5, 0 <= y <= x}, {x, y}]



              2.625




              Edit:



              This one-argument version treats all symbols in the first argument as variables:



              f[ineq_] := f[ineq, DeleteDuplicates[Cases[ineq, _Symbol]]]





              share|improve this answer














              f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var], Length[Flatten[{var}]]]
              

              
f[1 <= x <= 2.5, x]



              1.5




              This works also for some systems of inequalities in several variables:



              f[{1 <= x <= 2.5, 0 <= y <= x}, {x, y}]



              2.625




              Edit:



              This one-argument version treats all symbols in the first argument as variables:



              f[ineq_] := f[ineq, DeleteDuplicates[Cases[ineq, _Symbol]]]






              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited 15 hours ago

























              answered yesterday









              Henrik SchumacherHenrik Schumacher

              49.8k469142




              49.8k469142












              • When the dimension of the region is less than Length[var], for example a line embedded in the 2D plane, then RegionMeasure gives the measure in the reduced dimension: f[{1 <= x <= 2.5, y == 0}, {x, y}] gives 1.5 (the length of the line instead of its area), which is not what's usually expected. Fix this with f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var], Length[var]], so that now f[{1 <= x <= 2.5, y == 0}, {x, y}] gives 0 as expected (the line has zero area).
                – Roman
                20 hours ago












              • Good point! Thank your for remark; I fixed it. I should have been more cautious; these dimensional issues with regions is actually a frequent source of confusion.
                – Henrik Schumacher
                19 hours ago










              • I was too fast in commenting: the function now doesn't work for var=x since Length[x]=0. Maybe two separate definitions for var_Symbol (using dimension 1) and for var_List (using dimensions Length[var])?
                – Roman
                19 hours ago










              • Yes even better!
                – Roman
                19 hours ago


















              • When the dimension of the region is less than Length[var], for example a line embedded in the 2D plane, then RegionMeasure gives the measure in the reduced dimension: f[{1 <= x <= 2.5, y == 0}, {x, y}] gives 1.5 (the length of the line instead of its area), which is not what's usually expected. Fix this with f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var], Length[var]], so that now f[{1 <= x <= 2.5, y == 0}, {x, y}] gives 0 as expected (the line has zero area).
                – Roman
                20 hours ago












              • Good point! Thank your for remark; I fixed it. I should have been more cautious; these dimensional issues with regions is actually a frequent source of confusion.
                – Henrik Schumacher
                19 hours ago










              • I was too fast in commenting: the function now doesn't work for var=x since Length[x]=0. Maybe two separate definitions for var_Symbol (using dimension 1) and for var_List (using dimensions Length[var])?
                – Roman
                19 hours ago










              • Yes even better!
                – Roman
                19 hours ago
















              When the dimension of the region is less than Length[var], for example a line embedded in the 2D plane, then RegionMeasure gives the measure in the reduced dimension: f[{1 <= x <= 2.5, y == 0}, {x, y}] gives 1.5 (the length of the line instead of its area), which is not what's usually expected. Fix this with f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var], Length[var]], so that now f[{1 <= x <= 2.5, y == 0}, {x, y}] gives 0 as expected (the line has zero area).
              – Roman
              20 hours ago






              When the dimension of the region is less than Length[var], for example a line embedded in the 2D plane, then RegionMeasure gives the measure in the reduced dimension: f[{1 <= x <= 2.5, y == 0}, {x, y}] gives 1.5 (the length of the line instead of its area), which is not what's usually expected. Fix this with f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var], Length[var]], so that now f[{1 <= x <= 2.5, y == 0}, {x, y}] gives 0 as expected (the line has zero area).
              – Roman
              20 hours ago














              Good point! Thank your for remark; I fixed it. I should have been more cautious; these dimensional issues with regions is actually a frequent source of confusion.
              – Henrik Schumacher
              19 hours ago




              Good point! Thank your for remark; I fixed it. I should have been more cautious; these dimensional issues with regions is actually a frequent source of confusion.
              – Henrik Schumacher
              19 hours ago












              I was too fast in commenting: the function now doesn't work for var=x since Length[x]=0. Maybe two separate definitions for var_Symbol (using dimension 1) and for var_List (using dimensions Length[var])?
              – Roman
              19 hours ago




              I was too fast in commenting: the function now doesn't work for var=x since Length[x]=0. Maybe two separate definitions for var_Symbol (using dimension 1) and for var_List (using dimensions Length[var])?
              – Roman
              19 hours ago












              Yes even better!
              – Roman
              19 hours ago




              Yes even better!
              – Roman
              19 hours ago











              3














              fn[expr_] := Module[{},
              
  If[! expr, Return [0]];
              
  If[Head[expr] == Inequality, Return[Abs[expr[[5]] - expr[[1]]]]];
              
  Return[Abs[expr[[3]] - expr[[1]]]];
              
  ]
              

              
fn[2 <= x <= 1]
              
(*0*)
              

              
fn[1 <= x <= 2.5]
              
(*1.5*)
              

              
fn[2.5 > x > 1]
              
(*1.5*)


              Don't know if this works in all cases, but works in the simple cases you provide plus some.






              share|improve this answer




























                3














                fn[expr_] := Module[{},
                
  If[! expr, Return [0]];
                
  If[Head[expr] == Inequality, Return[Abs[expr[[5]] - expr[[1]]]]];
                
  Return[Abs[expr[[3]] - expr[[1]]]];
                
  ]
                

                
fn[2 <= x <= 1]
                
(*0*)
                

                
fn[1 <= x <= 2.5]
                
(*1.5*)
                

                
fn[2.5 > x > 1]
                
(*1.5*)


                Don't know if this works in all cases, but works in the simple cases you provide plus some.






                share|improve this answer


























                  3












                  3








                  3






                  fn[expr_] := Module[{},
                  
  If[! expr, Return [0]];
                  
  If[Head[expr] == Inequality, Return[Abs[expr[[5]] - expr[[1]]]]];
                  
  Return[Abs[expr[[3]] - expr[[1]]]];
                  
  ]
                  

                  
fn[2 <= x <= 1]
                  
(*0*)
                  

                  
fn[1 <= x <= 2.5]
                  
(*1.5*)
                  

                  
fn[2.5 > x > 1]
                  
(*1.5*)


                  Don't know if this works in all cases, but works in the simple cases you provide plus some.






                  share|improve this answer














                  fn[expr_] := Module[{},
                  
  If[! expr, Return [0]];
                  
  If[Head[expr] == Inequality, Return[Abs[expr[[5]] - expr[[1]]]]];
                  
  Return[Abs[expr[[3]] - expr[[1]]]];
                  
  ]
                  

                  
fn[2 <= x <= 1]
                  
(*0*)
                  

                  
fn[1 <= x <= 2.5]
                  
(*1.5*)
                  

                  
fn[2.5 > x > 1]
                  
(*1.5*)


                  Don't know if this works in all cases, but works in the simple cases you provide plus some.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 21 hours ago

























                  answered yesterday









                  Bill WattsBill Watts

                  2,9481516




                  2,9481516























                      2














                      To get a function that would handle the all the kinds of arguments I want it to handle turned out to be more of a challenge than I anticipated, but here is what I came up with.



                      Edit



                      This version is handle expressions that evaluate to False more robustly.



                      ClearAll[fn, helper1, helper2]
                      

                      
SetAttributes[fn, HoldFirst]
                      
fn[expr_] := If[expr, helper1[expr], helper2[expr], helper1[expr]]
                      

                      
SetAttributes[helper1, HoldFirst]
                      
helper1[expr : _Inequality | _Less | _LessEqual | _Greater | _GreaterEqual] :=
                      
  Module[{args = List @@ Unevaluated[expr], a, b},
                      
    {a, b} = MinMax[Select[args, NumericQ]];
                      
    b - a]
                      
helper1[___] = $Failed;
                      

                      
SetAttributes[helper2, HoldFirst]
                      
helper2[expr : _Inequality | _Less | _LessEqual | _Greater | _GreaterEqual] := 0;
                      
helper2[___] = $Failed;
                      

                      
###Tests
                      

                      
fn[1 < x <= 2.5]



                      1.5




                      fn[1 < x <= π]



                      -1 + π




                      fn[1 >= x > π]



                      0




                      fn[1 >= x > -1]



                      2




                      fn[-1 < 1 <= 2.5]



                      3.5




                      fn[1 < x < 3 < y < 5]



                      4




                      fn[1.5 < 2]



                      0.5




                      fn["garbage"]



                      $Failed




                      fn[1 == 1]



                      $Failed




                       fn[1 != 1]



                      $Failed







                      share|improve this answer




























                        2














                        To get a function that would handle the all the kinds of arguments I want it to handle turned out to be more of a challenge than I anticipated, but here is what I came up with.



                        Edit



                        This version is handle expressions that evaluate to False more robustly.



                        ClearAll[fn, helper1, helper2]
                        

                        
SetAttributes[fn, HoldFirst]
                        
fn[expr_] := If[expr, helper1[expr], helper2[expr], helper1[expr]]
                        

                        
SetAttributes[helper1, HoldFirst]
                        
helper1[expr : _Inequality | _Less | _LessEqual | _Greater | _GreaterEqual] :=
                        
  Module[{args = List @@ Unevaluated[expr], a, b},
                        
    {a, b} = MinMax[Select[args, NumericQ]];
                        
    b - a]
                        
helper1[___] = $Failed;
                        

                        
SetAttributes[helper2, HoldFirst]
                        
helper2[expr : _Inequality | _Less | _LessEqual | _Greater | _GreaterEqual] := 0;
                        
helper2[___] = $Failed;
                        

                        
###Tests
                        

                        
fn[1 < x <= 2.5]



                        1.5




                        fn[1 < x <= π]



                        -1 + π




                        fn[1 >= x > π]



                        0




                        fn[1 >= x > -1]



                        2




                        fn[-1 < 1 <= 2.5]



                        3.5




                        fn[1 < x < 3 < y < 5]



                        4




                        fn[1.5 < 2]



                        0.5




                        fn["garbage"]



                        $Failed




                        fn[1 == 1]



                        $Failed




                         fn[1 != 1]



                        $Failed







                        share|improve this answer


























                          2












                          2








                          2






                          To get a function that would handle the all the kinds of arguments I want it to handle turned out to be more of a challenge than I anticipated, but here is what I came up with.



                          Edit



                          This version is handle expressions that evaluate to False more robustly.



                          ClearAll[fn, helper1, helper2]
                          

                          
SetAttributes[fn, HoldFirst]
                          
fn[expr_] := If[expr, helper1[expr], helper2[expr], helper1[expr]]
                          

                          
SetAttributes[helper1, HoldFirst]
                          
helper1[expr : _Inequality | _Less | _LessEqual | _Greater | _GreaterEqual] :=
                          
  Module[{args = List @@ Unevaluated[expr], a, b},
                          
    {a, b} = MinMax[Select[args, NumericQ]];
                          
    b - a]
                          
helper1[___] = $Failed;
                          

                          
SetAttributes[helper2, HoldFirst]
                          
helper2[expr : _Inequality | _Less | _LessEqual | _Greater | _GreaterEqual] := 0;
                          
helper2[___] = $Failed;
                          

                          
###Tests
                          

                          
fn[1 < x <= 2.5]



                          1.5




                          fn[1 < x <= π]



                          -1 + π




                          fn[1 >= x > π]



                          0




                          fn[1 >= x > -1]



                          2




                          fn[-1 < 1 <= 2.5]



                          3.5




                          fn[1 < x < 3 < y < 5]



                          4




                          fn[1.5 < 2]



                          0.5




                          fn["garbage"]



                          $Failed




                          fn[1 == 1]



                          $Failed




                           fn[1 != 1]



                          $Failed







                          share|improve this answer














                          To get a function that would handle the all the kinds of arguments I want it to handle turned out to be more of a challenge than I anticipated, but here is what I came up with.



                          Edit



                          This version is handle expressions that evaluate to False more robustly.



                          ClearAll[fn, helper1, helper2]
                          

                          
SetAttributes[fn, HoldFirst]
                          
fn[expr_] := If[expr, helper1[expr], helper2[expr], helper1[expr]]
                          

                          
SetAttributes[helper1, HoldFirst]
                          
helper1[expr : _Inequality | _Less | _LessEqual | _Greater | _GreaterEqual] :=
                          
  Module[{args = List @@ Unevaluated[expr], a, b},
                          
    {a, b} = MinMax[Select[args, NumericQ]];
                          
    b - a]
                          
helper1[___] = $Failed;
                          

                          
SetAttributes[helper2, HoldFirst]
                          
helper2[expr : _Inequality | _Less | _LessEqual | _Greater | _GreaterEqual] := 0;
                          
helper2[___] = $Failed;
                          

                          
###Tests
                          

                          
fn[1 < x <= 2.5]



                          1.5




                          fn[1 < x <= π]



                          -1 + π




                          fn[1 >= x > π]



                          0




                          fn[1 >= x > -1]



                          2




                          fn[-1 < 1 <= 2.5]



                          3.5




                          fn[1 < x < 3 < y < 5]



                          4




                          fn[1.5 < 2]



                          0.5




                          fn["garbage"]



                          $Failed




                          fn[1 == 1]



                          $Failed




                           fn[1 != 1]



                          $Failed








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                          edited 15 hours ago

























                          answered yesterday









                          m_goldbergm_goldberg

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