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
edited yesterday
m_goldberg
84.3k872195
asked yesterday
Monire JaliliMonire Jalili
261
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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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add a comment |
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
edited yesterday
m_goldberg
84.3k872195
asked yesterday
Monire JaliliMonire Jalili
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.
add a comment |
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
edited yesterday
m_goldberg
84.3k872195
asked yesterday
Monire JaliliMonire Jalili
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.
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
edited yesterday
m_goldberg
84.3k872195
asked yesterday
Monire JaliliMonire Jalili
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.
edited yesterday
m_goldberg
84.3k872195
asked yesterday
Monire JaliliMonire Jalili
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.
edited yesterday
m_goldberg
84.3k872195
edited yesterday
m_goldberg
84.3k872195
edited yesterday
m_goldberg
84.3k872195
84.3k872195
asked yesterday
Monire JaliliMonire Jalili
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.
asked yesterday
Monire JaliliMonire Jalili
261
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.
add a comment |
add a comment |
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]]]
answered yesterday
Henrik SchumacherHenrik Schumacher
49.8k469142
When the dimension of the region is less thanLength[var], for example a line embedded in the 2D plane, thenRegionMeasuregives the measure in the reduced dimension:f[{1 <= x <= 2.5, y == 0}, {x, y}]gives1.5(the length of the line instead of its area), which is not what's usually expected. Fix this withf[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var], Length[var]], so that nowf[{1 <= x <= 2.5, y == 0}, {x, y}]gives0as 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 forvar=xsinceLength[x]=0. Maybe two separate definitions forvar_Symbol(using dimension1) and forvar_List(using dimensionsLength[var])?
– Roman
19 hours ago
Yes even better!
– Roman
19 hours ago
add a comment |
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.
answered yesterday
Bill WattsBill Watts
2,9481516
add a comment |
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
answered yesterday
m_goldbergm_goldberg
84.3k872195
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
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]]]
answered yesterday
Henrik SchumacherHenrik Schumacher
49.8k469142
When the dimension of the region is less thanLength[var], for example a line embedded in the 2D plane, thenRegionMeasuregives the measure in the reduced dimension:f[{1 <= x <= 2.5, y == 0}, {x, y}]gives1.5(the length of the line instead of its area), which is not what's usually expected. Fix this withf[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var], Length[var]], so that nowf[{1 <= x <= 2.5, y == 0}, {x, y}]gives0as 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 forvar=xsinceLength[x]=0. Maybe two separate definitions forvar_Symbol(using dimension1) and forvar_List(using dimensionsLength[var])?
– Roman
19 hours ago
Yes even better!
– Roman
19 hours ago
add a comment |
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]]]
answered yesterday
Henrik SchumacherHenrik Schumacher
49.8k469142
When the dimension of the region is less thanLength[var], for example a line embedded in the 2D plane, thenRegionMeasuregives the measure in the reduced dimension:f[{1 <= x <= 2.5, y == 0}, {x, y}]gives1.5(the length of the line instead of its area), which is not what's usually expected. Fix this withf[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var], Length[var]], so that nowf[{1 <= x <= 2.5, y == 0}, {x, y}]gives0as 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 forvar=xsinceLength[x]=0. Maybe two separate definitions forvar_Symbol(using dimension1) and forvar_List(using dimensionsLength[var])?
– Roman
19 hours ago
Yes even better!
– Roman
19 hours ago
add a comment |
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]]]
answered yesterday
Henrik SchumacherHenrik Schumacher
49.8k469142
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]]]
answered yesterday
Henrik SchumacherHenrik Schumacher
49.8k469142
edited 15 hours ago
answered yesterday
Henrik SchumacherHenrik Schumacher
49.8k469142
answered yesterday
Henrik SchumacherHenrik Schumacher
49.8k469142
answered yesterday
Henrik SchumacherHenrik Schumacher
49.8k469142
49.8k469142
When the dimension of the region is less thanLength[var], for example a line embedded in the 2D plane, thenRegionMeasuregives the measure in the reduced dimension:f[{1 <= x <= 2.5, y == 0}, {x, y}]gives1.5(the length of the line instead of its area), which is not what's usually expected. Fix this withf[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var], Length[var]], so that nowf[{1 <= x <= 2.5, y == 0}, {x, y}]gives0as 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 forvar=xsinceLength[x]=0. Maybe two separate definitions forvar_Symbol(using dimension1) and forvar_List(using dimensionsLength[var])?
– Roman
19 hours ago
Yes even better!
– Roman
19 hours ago
add a comment |
When the dimension of the region is less thanLength[var], for example a line embedded in the 2D plane, thenRegionMeasuregives the measure in the reduced dimension:f[{1 <= x <= 2.5, y == 0}, {x, y}]gives1.5(the length of the line instead of its area), which is not what's usually expected. Fix this withf[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var], Length[var]], so that nowf[{1 <= x <= 2.5, y == 0}, {x, y}]gives0as 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 forvar=xsinceLength[x]=0. Maybe two separate definitions forvar_Symbol(using dimension1) and forvar_List(using dimensionsLength[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
add a comment |
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.
answered yesterday
Bill WattsBill Watts
2,9481516
add a comment |
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.
answered yesterday
Bill WattsBill Watts
2,9481516
add a comment |
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.
answered yesterday
Bill WattsBill Watts
2,9481516
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.
answered yesterday
Bill WattsBill Watts
2,9481516
edited 21 hours ago
answered yesterday
Bill WattsBill Watts
2,9481516
answered yesterday
Bill WattsBill Watts
2,9481516
answered yesterday
Bill WattsBill Watts
2,9481516
2,9481516
add a comment |
add a comment |
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
answered yesterday
m_goldbergm_goldberg
84.3k872195
add a comment |
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
answered yesterday
m_goldbergm_goldberg
84.3k872195
add a comment |
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
answered yesterday
m_goldbergm_goldberg
84.3k872195
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
answered yesterday
m_goldbergm_goldberg
84.3k872195
edited 15 hours ago
answered yesterday
m_goldbergm_goldberg
84.3k872195
answered yesterday
m_goldbergm_goldberg
84.3k872195
answered yesterday
m_goldbergm_goldberg
84.3k872195
84.3k872195
add a comment |
add a comment |
Monire Jalili is a new contributor. Be nice, and check out our Code of Conduct.
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Monire Jalili is a new contributor. Be nice, and check out our Code of Conduct.
Monire Jalili is a new contributor. Be nice, and check out our Code of Conduct.
Monire Jalili is a new contributor. Be nice, and check out our Code of Conduct.
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