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Derivative of Sort

Retaining and reusing a one-to-one mapping from a sortSort+Union on a listIntegrate fails for generic parameters but works for any specific values (v. 10.0.x bug)How to sort colors properly?Sort a list of derivativesUnexpected result from Sort with symbolic valuesWhat is the Alphabetical Inequality Test?

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3

$begingroup$

I want to calculate gradient of Sort[x1,x2,x3,x4] but it seems like I'm getting a wrong answer. Specifically Plot[Sort[2, 1, x, 10][[4]], x, -20, 20] gives me

while D[Sort[2,1,x,10],x] returns 0,0,0,1, which is not true.

Is there a way to fix it?

list-manipulation calculus-and-analysis sorting

share|improve this question

asked Jul 20 at 22:43

stivstiv

16955 bronze badges

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Since Sort[2, 1, x, 10] yields 1, 2, 10, x and the derivative of 1, 2, 10, x is 0, 0, 0, 1, I'd say the derivative is correct. What do you think it should be?
  $endgroup$
  – Michael E2
  Jul 20 at 22:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As for the Plot[], since it holds its argument, it won't sort 2, 1, x, 10 and then plug in x; instead, it plugs in x and then sorts (then takes part 4, which will be the greatest of the four numbers. -- I'm not sure which order of operations (plug in then sort or the reverse) you want. Max[2,1,x,10] might be a better way to approach it.
  $endgroup$
  – Michael E2
  Jul 20 at 22:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Derivative is HevysideTheta(x-10) as you can see from the graph.
  $endgroup$
  – stiv
  Jul 20 at 22:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Try D[Max[2, 1, x, 10], x].
  $endgroup$
  – Michael E2
  Jul 20 at 22:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Max is a different function, which works, Sort doesn't. I need to differentiate Sort, not Max.
  $endgroup$
  – stiv
  Jul 20 at 22:56
  

  
  
  
  

add a comment
 | 

3

$begingroup$

I want to calculate gradient of Sort[x1,x2,x3,x4] but it seems like I'm getting a wrong answer. Specifically Plot[Sort[2, 1, x, 10][[4]], x, -20, 20] gives me

while D[Sort[2,1,x,10],x] returns 0,0,0,1, which is not true.

Is there a way to fix it?

list-manipulation calculus-and-analysis sorting

share|improve this question

asked Jul 20 at 22:43

stivstiv

16955 bronze badges

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Since Sort[2, 1, x, 10] yields 1, 2, 10, x and the derivative of 1, 2, 10, x is 0, 0, 0, 1, I'd say the derivative is correct. What do you think it should be?
  $endgroup$
  – Michael E2
  Jul 20 at 22:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As for the Plot[], since it holds its argument, it won't sort 2, 1, x, 10 and then plug in x; instead, it plugs in x and then sorts (then takes part 4, which will be the greatest of the four numbers. -- I'm not sure which order of operations (plug in then sort or the reverse) you want. Max[2,1,x,10] might be a better way to approach it.
  $endgroup$
  – Michael E2
  Jul 20 at 22:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Derivative is HevysideTheta(x-10) as you can see from the graph.
  $endgroup$
  – stiv
  Jul 20 at 22:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Try D[Max[2, 1, x, 10], x].
  $endgroup$
  – Michael E2
  Jul 20 at 22:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Max is a different function, which works, Sort doesn't. I need to differentiate Sort, not Max.
  $endgroup$
  – stiv
  Jul 20 at 22:56
  

  
  
  
  

add a comment
 | 

$begingroup$

I want to calculate gradient of Sort[x1,x2,x3,x4] but it seems like I'm getting a wrong answer. Specifically Plot[Sort[2, 1, x, 10][[4]], x, -20, 20] gives me

while D[Sort[2,1,x,10],x] returns 0,0,0,1, which is not true.

Is there a way to fix it?

list-manipulation calculus-and-analysis sorting

share|improve this question

asked Jul 20 at 22:43

stivstiv

16955 bronze badges

$endgroup$

share|improve this question

asked Jul 20 at 22:43

stivstiv

16955 bronze badges

share|improve this question

asked Jul 20 at 22:43

stivstiv

16955 bronze badges

asked Jul 20 at 22:43

stivstiv

16955 bronze badges

asked Jul 20 at 22:43

stivstiv

16955 bronze badges

2 Answers
2

active

oldest

votes

8

$begingroup$

Treating Sort as a function from the disjoint union of $bf R^n$, $n=0,1,2,dots$, to the same union, here is one way to define the derivative. I'll call the numerical sort NSort, just so I don't have to overwrite a built-in function.

NSort[list_?(VectorQ[#, NumericQ] &)] := NumericalSort[list];

Derivative[orders_List][NSort][list_List] /; Length[orders] == Length[list] :=
 With[args = Array[Slot, Length@orders],
 Evaluate@Piecewise[
 Function[
 p, D[p, Sequence @@ Transpose@args, orders], 
 Less @@ p] /@ Permutations[args],
 Indeterminate
 ] & @@ list]

OP's example:

df = D[NSort[2, 1, x, 10], x]

Plot[Indexed[df, 4], x, -20, 20] 

Another example:

D[NSort[2, 1, 3 x, x^2], x, 2] // PiecewiseExpand

share|improve this answer

edited Jul 21 at 1:31

answered Jul 21 at 1:26

Michael E2Michael E2

161k1313 gold badges219219 silver badges521521 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  COOL! Thank you so much!
  $endgroup$
  – stiv
  Jul 21 at 1:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @stiv You're welcome.
  $endgroup$
  – Michael E2
  Jul 21 at 3:10
  

  
  
  
  

add a comment
 | 

4

$begingroup$

Although this is definitely an XY problem, here is a possible solution:

list = 2, 1, x, 10;
listDer = D[list, x];

Plot[
 listDer[[Ordering[list, -1]]]
 , x, -20, 20
]

share|improve this answer

answered Jul 20 at 23:38

AccidentalFourierTransformAccidentalFourierTransform

7,07811 gold badge1414 silver badges4545 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This seems likely but the insistence on avoid Max makes me think the OP is interested in other components of the sorted vector as well.
  $endgroup$
  – Michael E2
  Jul 21 at 0:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelE2 Yeah, probably. That's easy though: just drop the -1 in Ordering. Hard to tell what exactly OP wants anyway.
  $endgroup$
  – AccidentalFourierTransform
  Jul 21 at 0:45
  

  
  
  
  

add a comment
 | 

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8

$begingroup$

Treating Sort as a function from the disjoint union of $bf R^n$, $n=0,1,2,dots$, to the same union, here is one way to define the derivative. I'll call the numerical sort NSort, just so I don't have to overwrite a built-in function.

NSort[list_?(VectorQ[#, NumericQ] &)] := NumericalSort[list];

Derivative[orders_List][NSort][list_List] /; Length[orders] == Length[list] :=
 With[args = Array[Slot, Length@orders],
 Evaluate@Piecewise[
 Function[
 p, D[p, Sequence @@ Transpose@args, orders], 
 Less @@ p] /@ Permutations[args],
 Indeterminate
 ] & @@ list]

OP's example:

df = D[NSort[2, 1, x, 10], x]

Plot[Indexed[df, 4], x, -20, 20] 

Another example:

D[NSort[2, 1, 3 x, x^2], x, 2] // PiecewiseExpand

share|improve this answer

edited Jul 21 at 1:31

answered Jul 21 at 1:26

Michael E2Michael E2

161k1313 gold badges219219 silver badges521521 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  COOL! Thank you so much!
  $endgroup$
  – stiv
  Jul 21 at 1:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @stiv You're welcome.
  $endgroup$
  – Michael E2
  Jul 21 at 3:10
  

  
  
  
  

add a comment
 | 

8

$begingroup$

Treating Sort as a function from the disjoint union of $bf R^n$, $n=0,1,2,dots$, to the same union, here is one way to define the derivative. I'll call the numerical sort NSort, just so I don't have to overwrite a built-in function.

NSort[list_?(VectorQ[#, NumericQ] &)] := NumericalSort[list];

Derivative[orders_List][NSort][list_List] /; Length[orders] == Length[list] :=
 With[args = Array[Slot, Length@orders],
 Evaluate@Piecewise[
 Function[
 p, D[p, Sequence @@ Transpose@args, orders], 
 Less @@ p] /@ Permutations[args],
 Indeterminate
 ] & @@ list]

OP's example:

df = D[NSort[2, 1, x, 10], x]

Plot[Indexed[df, 4], x, -20, 20] 

Another example:

D[NSort[2, 1, 3 x, x^2], x, 2] // PiecewiseExpand

share|improve this answer

edited Jul 21 at 1:31

answered Jul 21 at 1:26

Michael E2Michael E2

161k1313 gold badges219219 silver badges521521 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  COOL! Thank you so much!
  $endgroup$
  – stiv
  Jul 21 at 1:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @stiv You're welcome.
  $endgroup$
  – Michael E2
  Jul 21 at 3:10
  

  
  
  
  

add a comment
 | 

$begingroup$

Treating Sort as a function from the disjoint union of $bf R^n$, $n=0,1,2,dots$, to the same union, here is one way to define the derivative. I'll call the numerical sort NSort, just so I don't have to overwrite a built-in function.

NSort[list_?(VectorQ[#, NumericQ] &)] := NumericalSort[list];

Derivative[orders_List][NSort][list_List] /; Length[orders] == Length[list] :=
 With[args = Array[Slot, Length@orders],
 Evaluate@Piecewise[
 Function[
 p, D[p, Sequence @@ Transpose@args, orders], 
 Less @@ p] /@ Permutations[args],
 Indeterminate
 ] & @@ list]

OP's example:

df = D[NSort[2, 1, x, 10], x]

Plot[Indexed[df, 4], x, -20, 20] 

Another example:

D[NSort[2, 1, 3 x, x^2], x, 2] // PiecewiseExpand

share|improve this answer

edited Jul 21 at 1:31

answered Jul 21 at 1:26

Michael E2Michael E2

161k1313 gold badges219219 silver badges521521 bronze badges

$endgroup$

NSort[list_?(VectorQ[#, NumericQ] &)] := NumericalSort[list];

Derivative[orders_List][NSort][list_List] /; Length[orders] == Length[list] :=
 With[args = Array[Slot, Length@orders],
 Evaluate@Piecewise[
 Function[
 p, D[p, Sequence @@ Transpose@args, orders], 
 Less @@ p] /@ Permutations[args],
 Indeterminate
 ] & @@ list]

df = D[NSort[2, 1, x, 10], x]

Plot[Indexed[df, 4], x, -20, 20] 

D[NSort[2, 1, 3 x, x^2], x, 2] // PiecewiseExpand

share|improve this answer

edited Jul 21 at 1:31

answered Jul 21 at 1:26

Michael E2Michael E2

161k1313 gold badges219219 silver badges521521 bronze badges

answered Jul 21 at 1:26

Michael E2Michael E2

161k1313 gold badges219219 silver badges521521 bronze badges

answered Jul 21 at 1:26

Michael E2Michael E2

161k1313 gold badges219219 silver badges521521 bronze badges

4

$begingroup$

Although this is definitely an XY problem, here is a possible solution:

list = 2, 1, x, 10;
listDer = D[list, x];

Plot[
 listDer[[Ordering[list, -1]]]
 , x, -20, 20
]

share|improve this answer

answered Jul 20 at 23:38

AccidentalFourierTransformAccidentalFourierTransform

7,07811 gold badge1414 silver badges4545 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This seems likely but the insistence on avoid Max makes me think the OP is interested in other components of the sorted vector as well.
  $endgroup$
  – Michael E2
  Jul 21 at 0:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelE2 Yeah, probably. That's easy though: just drop the -1 in Ordering. Hard to tell what exactly OP wants anyway.
  $endgroup$
  – AccidentalFourierTransform
  Jul 21 at 0:45
  

  
  
  
  

add a comment
 | 

4

$begingroup$

Although this is definitely an XY problem, here is a possible solution:

list = 2, 1, x, 10;
listDer = D[list, x];

Plot[
 listDer[[Ordering[list, -1]]]
 , x, -20, 20
]

share|improve this answer

answered Jul 20 at 23:38

AccidentalFourierTransformAccidentalFourierTransform

7,07811 gold badge1414 silver badges4545 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This seems likely but the insistence on avoid Max makes me think the OP is interested in other components of the sorted vector as well.
  $endgroup$
  – Michael E2
  Jul 21 at 0:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelE2 Yeah, probably. That's easy though: just drop the -1 in Ordering. Hard to tell what exactly OP wants anyway.
  $endgroup$
  – AccidentalFourierTransform
  Jul 21 at 0:45
  

  
  
  
  

add a comment
 | 

$begingroup$

Although this is definitely an XY problem, here is a possible solution:

list = 2, 1, x, 10;
listDer = D[list, x];

Plot[
 listDer[[Ordering[list, -1]]]
 , x, -20, 20
]

share|improve this answer

answered Jul 20 at 23:38

AccidentalFourierTransformAccidentalFourierTransform

7,07811 gold badge1414 silver badges4545 bronze badges

$endgroup$

list = 2, 1, x, 10;
listDer = D[list, x];

Plot[
 listDer[[Ordering[list, -1]]]
 , x, -20, 20
]

share|improve this answer

answered Jul 20 at 23:38

AccidentalFourierTransformAccidentalFourierTransform

7,07811 gold badge1414 silver badges4545 bronze badges

answered Jul 20 at 23:38

AccidentalFourierTransformAccidentalFourierTransform

7,07811 gold badge1414 silver badges4545 bronze badges

answered Jul 20 at 23:38

AccidentalFourierTransformAccidentalFourierTransform

7,07811 gold badge1414 silver badges4545 bronze badges