Calculating the width of the interval defined by an inequality

6

$begingroup$

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

edited Jan 7 at 3:38

m_goldberg

85.2k872196

asked Jan 7 at 0:04

Monire JaliliMonire Jalili

311

$endgroup$

add a comment | 

6

$begingroup$

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

edited Jan 7 at 3:38

m_goldberg

85.2k872196

asked Jan 7 at 0:04

Monire JaliliMonire Jalili

311

$endgroup$

add a comment | 

6

6

6

$begingroup$

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

edited Jan 7 at 3:38

m_goldberg

85.2k872196

asked Jan 7 at 0:04

Monire JaliliMonire Jalili

311

$endgroup$

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

edited Jan 7 at 3:38

m_goldberg

85.2k872196

asked Jan 7 at 0:04

Monire JaliliMonire Jalili

311

share|improve this question

edited Jan 7 at 3:38

m_goldberg

85.2k872196

asked Jan 7 at 0:04

Monire JaliliMonire Jalili

311

share|improve this question

share|improve this question

edited Jan 7 at 3:38

m_goldberg

85.2k872196

edited Jan 7 at 3:38

m_goldberg

85.2k872196

edited Jan 7 at 3:38

m_goldberg

85.2k872196

85.2k872196

asked Jan 7 at 0:04

Monire JaliliMonire Jalili

311

asked Jan 7 at 0:04

Monire JaliliMonire Jalili

311

asked Jan 7 at 0:04

Monire JaliliMonire Jalili

311

311

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

10

$begingroup$

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

edited Jan 7 at 15:14

answered Jan 7 at 1:00

Henrik SchumacherHenrik Schumacher

51.3k469146

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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).
  $endgroup$
  – Roman
  Jan 7 at 10:22
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – Henrik Schumacher
  Jan 7 at 10:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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])?
  $endgroup$
  – Roman
  Jan 7 at 10:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes even better!
  $endgroup$
  – Roman
  Jan 7 at 10:32
  

  
  
  
  

add a comment | 

3

$begingroup$

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

edited Jan 7 at 9:04

answered Jan 7 at 0:55

Bill WattsBill Watts

3,2011519

$endgroup$

add a comment | 

2

$begingroup$

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

edited Jan 7 at 14:51

answered Jan 7 at 4:52

m_goldbergm_goldberg

85.2k872196

$endgroup$

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

10

$begingroup$

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

edited Jan 7 at 15:14

answered Jan 7 at 1:00

Henrik SchumacherHenrik Schumacher

51.3k469146

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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).
  $endgroup$
  – Roman
  Jan 7 at 10:22
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – Henrik Schumacher
  Jan 7 at 10:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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])?
  $endgroup$
  – Roman
  Jan 7 at 10:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes even better!
  $endgroup$
  – Roman
  Jan 7 at 10:32
  

  
  
  
  

add a comment | 

10

$begingroup$

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

edited Jan 7 at 15:14

answered Jan 7 at 1:00

Henrik SchumacherHenrik Schumacher

51.3k469146

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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).
  $endgroup$
  – Roman
  Jan 7 at 10:22
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – Henrik Schumacher
  Jan 7 at 10:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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])?
  $endgroup$
  – Roman
  Jan 7 at 10:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes even better!
  $endgroup$
  – Roman
  Jan 7 at 10:32
  

  
  
  
  

add a comment | 

10

10

10

$begingroup$

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

edited Jan 7 at 15:14

answered Jan 7 at 1:00

Henrik SchumacherHenrik Schumacher

51.3k469146

$endgroup$

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

edited Jan 7 at 15:14

answered Jan 7 at 1:00

Henrik SchumacherHenrik Schumacher

51.3k469146

share|improve this answer

share|improve this answer

edited Jan 7 at 15:14

edited Jan 7 at 15:14

edited Jan 7 at 15:14

answered Jan 7 at 1:00

Henrik SchumacherHenrik Schumacher

51.3k469146

answered Jan 7 at 1:00

Henrik SchumacherHenrik Schumacher

51.3k469146

answered Jan 7 at 1:00

Henrik SchumacherHenrik Schumacher

51.3k469146

51.3k469146

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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).
  $endgroup$
  – Roman
  Jan 7 at 10:22
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – Henrik Schumacher
  Jan 7 at 10:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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])?
  $endgroup$
  – Roman
  Jan 7 at 10:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes even better!
  $endgroup$
  – Roman
  Jan 7 at 10:32
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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).
  $endgroup$
  – Roman
  Jan 7 at 10:22
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – Henrik Schumacher
  Jan 7 at 10:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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])?
  $endgroup$
  – Roman
  Jan 7 at 10:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes even better!
  $endgroup$
  – Roman
  Jan 7 at 10:32
  

  
  
  
  

$begingroup$
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).
$endgroup$
– Roman
Jan 7 at 10:22

$begingroup$
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).
$endgroup$
– Roman
Jan 7 at 10:22

$begingroup$
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.
$endgroup$
– Henrik Schumacher
Jan 7 at 10:28

$begingroup$
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.
$endgroup$
– Henrik Schumacher
Jan 7 at 10:28

$begingroup$
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])?
$endgroup$
– Roman
Jan 7 at 10:31

$begingroup$
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])?
$endgroup$
– Roman
Jan 7 at 10:31

$begingroup$
Yes even better!
$endgroup$
– Roman
Jan 7 at 10:32

$begingroup$
Yes even better!
$endgroup$
– Roman
Jan 7 at 10:32

add a comment | 

3

$begingroup$

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

edited Jan 7 at 9:04

answered Jan 7 at 0:55

Bill WattsBill Watts

3,2011519

$endgroup$

add a comment | 

3

$begingroup$

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

edited Jan 7 at 9:04

answered Jan 7 at 0:55

Bill WattsBill Watts

3,2011519

$endgroup$

add a comment | 

3

3

3

$begingroup$

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

edited Jan 7 at 9:04

answered Jan 7 at 0:55

Bill WattsBill Watts

3,2011519

$endgroup$

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

edited Jan 7 at 9:04

answered Jan 7 at 0:55

Bill WattsBill Watts

3,2011519

share|improve this answer

share|improve this answer

edited Jan 7 at 9:04

edited Jan 7 at 9:04

edited Jan 7 at 9:04

answered Jan 7 at 0:55

Bill WattsBill Watts

3,2011519

answered Jan 7 at 0:55

Bill WattsBill Watts

3,2011519

answered Jan 7 at 0:55

Bill WattsBill Watts

3,2011519

3,2011519

add a comment | 

add a comment | 

2

$begingroup$

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

edited Jan 7 at 14:51

answered Jan 7 at 4:52

m_goldbergm_goldberg

85.2k872196

$endgroup$

add a comment | 

2

$begingroup$

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

edited Jan 7 at 14:51

answered Jan 7 at 4:52

m_goldbergm_goldberg

85.2k872196

$endgroup$

add a comment | 

2

2

2

$begingroup$

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

edited Jan 7 at 14:51

answered Jan 7 at 4:52

m_goldbergm_goldberg

85.2k872196

$endgroup$

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

edited Jan 7 at 14:51

answered Jan 7 at 4:52

m_goldbergm_goldberg

85.2k872196

share|improve this answer

share|improve this answer

edited Jan 7 at 14:51

edited Jan 7 at 14:51

edited Jan 7 at 14:51

answered Jan 7 at 4:52

m_goldbergm_goldberg

85.2k872196

answered Jan 7 at 4:52

m_goldbergm_goldberg

85.2k872196

answered Jan 7 at 4:52

m_goldbergm_goldberg

85.2k872196

85.2k872196

add a comment | 

add a comment | 

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