Calculating the width of the interval defined by an inequality
6
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 Jan 7 at 3:38
m_goldberg
84.5k872196
asked Jan 7 at 0:04
Monire JaliliMonire Jalili
311
New contributor
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add a comment |
6
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 Jan 7 at 3:38
m_goldberg
84.5k872196
asked Jan 7 at 0:04
Monire JaliliMonire Jalili
311
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 |
6
6
6
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 Jan 7 at 3:38
m_goldberg
84.5k872196
asked Jan 7 at 0:04
Monire JaliliMonire Jalili
311
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 Jan 7 at 3:38
m_goldberg
84.5k872196
asked Jan 7 at 0:04
Monire JaliliMonire Jalili
311
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 Jan 7 at 3:38
m_goldberg
84.5k872196
asked Jan 7 at 0:04
Monire JaliliMonire Jalili
311
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 Jan 7 at 3:38
m_goldberg
84.5k872196
edited Jan 7 at 3:38
m_goldberg
84.5k872196
edited Jan 7 at 3:38
m_goldberg
84.5k872196
84.5k872196
asked Jan 7 at 0:04
Monire JaliliMonire Jalili
311
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 Jan 7 at 0:04
Monire JaliliMonire Jalili
311
asked Jan 7 at 0:04
Monire JaliliMonire Jalili
311
311
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
10
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 Jan 7 at 1:00
Henrik SchumacherHenrik Schumacher
50.1k469144
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
Jan 7 at 10:22
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
Jan 7 at 10:28
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
Jan 7 at 10:31
Yes even better!
– Roman
Jan 7 at 10:32
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 Jan 7 at 0:55
Bill WattsBill Watts
3,0311517
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 Jan 7 at 4:52
m_goldbergm_goldberg
84.5k872196
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
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 Jan 7 at 1:00
Henrik SchumacherHenrik Schumacher
50.1k469144
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
Jan 7 at 10:22
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
Jan 7 at 10:28
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
Jan 7 at 10:31
Yes even better!
– Roman
Jan 7 at 10:32
add a comment |
10
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 Jan 7 at 1:00
Henrik SchumacherHenrik Schumacher
50.1k469144
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
Jan 7 at 10:22
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
Jan 7 at 10:28
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
Jan 7 at 10:31
Yes even better!
– Roman
Jan 7 at 10:32
add a comment |
10
10
10
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 Jan 7 at 1:00
Henrik SchumacherHenrik Schumacher
50.1k469144
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 Jan 7 at 1:00
Henrik SchumacherHenrik Schumacher
50.1k469144
edited Jan 7 at 15:14
answered Jan 7 at 1:00
Henrik SchumacherHenrik Schumacher
50.1k469144
answered Jan 7 at 1:00
Henrik SchumacherHenrik Schumacher
50.1k469144
answered Jan 7 at 1:00
Henrik SchumacherHenrik Schumacher
50.1k469144
50.1k469144
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
Jan 7 at 10:22
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
Jan 7 at 10:28
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
Jan 7 at 10:31
Yes even better!
– Roman
Jan 7 at 10:32
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
Jan 7 at 10:22
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
Jan 7 at 10:28
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
Jan 7 at 10:31
Yes even better!
– Roman
Jan 7 at 10:32
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
Jan 7 at 10:22
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
Jan 7 at 10:22
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
Jan 7 at 10:28
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
Jan 7 at 10:28
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
Jan 7 at 10:31
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
Jan 7 at 10:31
Yes even better!
– Roman
Jan 7 at 10:32
Yes even better!
– Roman
Jan 7 at 10:32
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 Jan 7 at 0:55
Bill WattsBill Watts
3,0311517
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 Jan 7 at 0:55
Bill WattsBill Watts
3,0311517
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 Jan 7 at 0:55
Bill WattsBill Watts
3,0311517
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 Jan 7 at 0:55
Bill WattsBill Watts
3,0311517
edited Jan 7 at 9:04
answered Jan 7 at 0:55
Bill WattsBill Watts
3,0311517
answered Jan 7 at 0:55
Bill WattsBill Watts
3,0311517
answered Jan 7 at 0:55
Bill WattsBill Watts
3,0311517
3,0311517
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 Jan 7 at 4:52
m_goldbergm_goldberg
84.5k872196
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 Jan 7 at 4:52
m_goldbergm_goldberg
84.5k872196
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 Jan 7 at 4:52
m_goldbergm_goldberg
84.5k872196
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 Jan 7 at 4:52
m_goldbergm_goldberg
84.5k872196
edited Jan 7 at 14:51
answered Jan 7 at 4:52
m_goldbergm_goldberg
84.5k872196
answered Jan 7 at 4:52
m_goldbergm_goldberg
84.5k872196
answered Jan 7 at 4:52
m_goldbergm_goldberg
84.5k872196
84.5k872196
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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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