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What is the identity for Tuples?

Duplicate TuplesSelecting from a list returned by TuplesSelect elements in a list of tuples of quaternions that satisfy a conditionTuples of elements from list excluding anything with repeated valuesDelete duplicates in tuples

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;

.everyonelovesstackoverflowposition:absolute;height:1px;width:1px;opacity:0;top:0;left:0;pointer-events:none;

5

1

$begingroup$

My question is, what list p in the following statement returns the list q?

Tuples[p,q]

If we think of Tuples as a binary operator then p would be the identity for Tuples.

I thought an empty list would work, but evaluating the following

Tuples[,1,2,3]

gives instead of 1,2,3 or 1,2,3 as I had hoped.

The following doesn't work either:

Tuples[[EmptySet],1,2,3]

Certainly, I could write a function like the following

altTuples[p_List,q_List]:= If[Length[p]==0,q,Tuples[p,q]

That does exactly what I want, but I want to know if I'm missing something. Is there in fact an identity for Tuples? Is there a way to do what I want with Outer? I've tried the obvious solutions with no luck.

tuples

share|improve this question

edited Jul 8 at 23:25

JAS

asked Jul 7 at 22:18

JASJAS

2633 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How do you want your altTuples[] to behave when the length of $p$ is not zero? For example altTuples[a, b, c, d, e, f] returns Tuples[a, b, c, d, e, f], probably not what you want.
  $endgroup$
  – mjw
  Jul 8 at 1:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I corrected altTuples. Actually I want it to return Tuples[a, b, c, d, e, f]. I think Nothing is what I was looking for.
  $endgroup$
  – JAS
  Jul 8 at 23:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What you have above is almost correct (up to a typo). Anyway seems to produce what you want. 'AltTuples[p,q]` gives the same output as Tuples[p,q] when the first argument is a list with non-zero length.
  $endgroup$
  – mjw
  Jul 8 at 23:36
  
  

  
  
  
  

add a comment
 | 

5

1

$begingroup$

My question is, what list p in the following statement returns the list q?

Tuples[p,q]

If we think of Tuples as a binary operator then p would be the identity for Tuples.

I thought an empty list would work, but evaluating the following

Tuples[,1,2,3]

gives instead of 1,2,3 or 1,2,3 as I had hoped.

The following doesn't work either:

Tuples[[EmptySet],1,2,3]

Certainly, I could write a function like the following

altTuples[p_List,q_List]:= If[Length[p]==0,q,Tuples[p,q]

That does exactly what I want, but I want to know if I'm missing something. Is there in fact an identity for Tuples? Is there a way to do what I want with Outer? I've tried the obvious solutions with no luck.

tuples

share|improve this question

edited Jul 8 at 23:25

JAS

asked Jul 7 at 22:18

JASJAS

2633 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How do you want your altTuples[] to behave when the length of $p$ is not zero? For example altTuples[a, b, c, d, e, f] returns Tuples[a, b, c, d, e, f], probably not what you want.
  $endgroup$
  – mjw
  Jul 8 at 1:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I corrected altTuples. Actually I want it to return Tuples[a, b, c, d, e, f]. I think Nothing is what I was looking for.
  $endgroup$
  – JAS
  Jul 8 at 23:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What you have above is almost correct (up to a typo). Anyway seems to produce what you want. 'AltTuples[p,q]` gives the same output as Tuples[p,q] when the first argument is a list with non-zero length.
  $endgroup$
  – mjw
  Jul 8 at 23:36
  
  

  
  
  
  

add a comment
 | 

$begingroup$

My question is, what list p in the following statement returns the list q?

Tuples[p,q]

If we think of Tuples as a binary operator then p would be the identity for Tuples.

I thought an empty list would work, but evaluating the following

Tuples[,1,2,3]

gives instead of 1,2,3 or 1,2,3 as I had hoped.

The following doesn't work either:

Tuples[[EmptySet],1,2,3]

Certainly, I could write a function like the following

altTuples[p_List,q_List]:= If[Length[p]==0,q,Tuples[p,q]

That does exactly what I want, but I want to know if I'm missing something. Is there in fact an identity for Tuples? Is there a way to do what I want with Outer? I've tried the obvious solutions with no luck.

tuples

share|improve this question

edited Jul 8 at 23:25

JAS

asked Jul 7 at 22:18

JASJAS

2633 bronze badges

$endgroup$

Tuples[p,q]

Tuples[,1,2,3]

Tuples[[EmptySet],1,2,3]

altTuples[p_List,q_List]:= If[Length[p]==0,q,Tuples[p,q]

share|improve this question

edited Jul 8 at 23:25

JAS

asked Jul 7 at 22:18

JASJAS

2633 bronze badges

share|improve this question

edited Jul 8 at 23:25

JAS

asked Jul 7 at 22:18

JASJAS

2633 bronze badges

asked Jul 7 at 22:18

JASJAS

2633 bronze badges

asked Jul 7 at 22:18

JASJAS

2633 bronze badges

4 Answers
4

active

oldest

votes

7

$begingroup$

If 1,2,3 is fine, you can use Nothing:

Tuples[Nothing, 1, 2, 3]
(* 1, 2, 3 *)

If you want 1,2,3, you can Flatten the result, of course.

share|improve this answer

answered Jul 7 at 22:45

AccidentalFourierTransformAccidentalFourierTransform

7,07811 gold badge1414 silver badges4545 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, this does exactly what I need.
  $endgroup$
  – JAS
  Jul 8 at 23:20
  

  
  
  
  

add a comment
 | 

4

$begingroup$

You can use Inactive[Sequence][] as identity like this:

Tuples[Inactive[Sequence][],q]//Activate

q

share|improve this answer

answered Jul 7 at 22:36

Thies HeideckeThies Heidecke

7,9042626 silver badges4040 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

Use TagSetDelayed to define a function that behaves as desired:

ClearAll[iDentity]
iDentity /: iDentity[___], a : __ := iDentity[a]
iDentity /: Tuples[iDentity[a_]] := a

Tuples[iDentity[], q]

q

Tuples[iDentity[blah], 1, 2, 3]

1, 2, 3

Alternatively, define your function altTuples with two signatures:

ClearAll[altTuples]
altTuples[tuplesIdentity ] := a
altTuples[x_] := Tuples[x]

altTuples[x, y, 1, 2]

x, 1, x, 2, y, 1, y, 2

altTuples[tuplesIdentity, 1, 2]

1, 2

altTuples[, 1, 2]

1, 2

share|improve this answer

edited Jul 8 at 1:48

answered Jul 7 at 23:21

kglrkglr

224k1010 gold badges254254 silver badges511511 bronze badges

$endgroup$

add a comment
 | 

0

$begingroup$

I think this is what you want:

altTuples[p_List, q_List] := If[Length[p] == 0, q, Tuples[p, q]]

The statement

altTuples[, d, e, f]

returns

d, e, f

and

altTuples[a, b, c, d, e, f]

returns

a, d, a, e, a, f, b, d, b, e, b, f, c, d, c, e, c, f

share|improve this answer

answered Jul 8 at 23:34

mjwmjw

1,6051313 bronze badges

$endgroup$

add a comment
 | 

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7

$begingroup$

If 1,2,3 is fine, you can use Nothing:

Tuples[Nothing, 1, 2, 3]
(* 1, 2, 3 *)

If you want 1,2,3, you can Flatten the result, of course.

share|improve this answer

answered Jul 7 at 22:45

AccidentalFourierTransformAccidentalFourierTransform

7,07811 gold badge1414 silver badges4545 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, this does exactly what I need.
  $endgroup$
  – JAS
  Jul 8 at 23:20
  

  
  
  
  

add a comment
 | 

7

$begingroup$

If 1,2,3 is fine, you can use Nothing:

Tuples[Nothing, 1, 2, 3]
(* 1, 2, 3 *)

If you want 1,2,3, you can Flatten the result, of course.

share|improve this answer

answered Jul 7 at 22:45

AccidentalFourierTransformAccidentalFourierTransform

7,07811 gold badge1414 silver badges4545 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, this does exactly what I need.
  $endgroup$
  – JAS
  Jul 8 at 23:20
  

  
  
  
  

add a comment
 | 

$begingroup$

If 1,2,3 is fine, you can use Nothing:

Tuples[Nothing, 1, 2, 3]
(* 1, 2, 3 *)

If you want 1,2,3, you can Flatten the result, of course.

share|improve this answer

answered Jul 7 at 22:45

AccidentalFourierTransformAccidentalFourierTransform

7,07811 gold badge1414 silver badges4545 bronze badges

$endgroup$

Tuples[Nothing, 1, 2, 3]
(* 1, 2, 3 *)

share|improve this answer

answered Jul 7 at 22:45

AccidentalFourierTransformAccidentalFourierTransform

7,07811 gold badge1414 silver badges4545 bronze badges

answered Jul 7 at 22:45

AccidentalFourierTransformAccidentalFourierTransform

7,07811 gold badge1414 silver badges4545 bronze badges

answered Jul 7 at 22:45

AccidentalFourierTransformAccidentalFourierTransform

7,07811 gold badge1414 silver badges4545 bronze badges

4

$begingroup$

You can use Inactive[Sequence][] as identity like this:

Tuples[Inactive[Sequence][],q]//Activate

q

share|improve this answer

answered Jul 7 at 22:36

Thies HeideckeThies Heidecke

7,9042626 silver badges4040 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

You can use Inactive[Sequence][] as identity like this:

Tuples[Inactive[Sequence][],q]//Activate

q

share|improve this answer

answered Jul 7 at 22:36

Thies HeideckeThies Heidecke

7,9042626 silver badges4040 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

You can use Inactive[Sequence][] as identity like this:

Tuples[Inactive[Sequence][],q]//Activate

q

share|improve this answer

answered Jul 7 at 22:36

Thies HeideckeThies Heidecke

7,9042626 silver badges4040 bronze badges

$endgroup$

Tuples[Inactive[Sequence][],q]//Activate

share|improve this answer

answered Jul 7 at 22:36

Thies HeideckeThies Heidecke

7,9042626 silver badges4040 bronze badges

answered Jul 7 at 22:36

Thies HeideckeThies Heidecke

7,9042626 silver badges4040 bronze badges

answered Jul 7 at 22:36

Thies HeideckeThies Heidecke

7,9042626 silver badges4040 bronze badges

4

$begingroup$

Use TagSetDelayed to define a function that behaves as desired:

ClearAll[iDentity]
iDentity /: iDentity[___], a : __ := iDentity[a]
iDentity /: Tuples[iDentity[a_]] := a

Tuples[iDentity[], q]

q

Tuples[iDentity[blah], 1, 2, 3]

1, 2, 3

Alternatively, define your function altTuples with two signatures:

ClearAll[altTuples]
altTuples[tuplesIdentity ] := a
altTuples[x_] := Tuples[x]

altTuples[x, y, 1, 2]

x, 1, x, 2, y, 1, y, 2

altTuples[tuplesIdentity, 1, 2]

1, 2

altTuples[, 1, 2]

1, 2

share|improve this answer

edited Jul 8 at 1:48

answered Jul 7 at 23:21

kglrkglr

224k1010 gold badges254254 silver badges511511 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

Use TagSetDelayed to define a function that behaves as desired:

ClearAll[iDentity]
iDentity /: iDentity[___], a : __ := iDentity[a]
iDentity /: Tuples[iDentity[a_]] := a

Tuples[iDentity[], q]

q

Tuples[iDentity[blah], 1, 2, 3]

1, 2, 3

Alternatively, define your function altTuples with two signatures:

ClearAll[altTuples]
altTuples[tuplesIdentity ] := a
altTuples[x_] := Tuples[x]

altTuples[x, y, 1, 2]

x, 1, x, 2, y, 1, y, 2

altTuples[tuplesIdentity, 1, 2]

1, 2

altTuples[, 1, 2]

1, 2

share|improve this answer

edited Jul 8 at 1:48

answered Jul 7 at 23:21

kglrkglr

224k1010 gold badges254254 silver badges511511 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Use TagSetDelayed to define a function that behaves as desired:

ClearAll[iDentity]
iDentity /: iDentity[___], a : __ := iDentity[a]
iDentity /: Tuples[iDentity[a_]] := a

Tuples[iDentity[], q]

q

Tuples[iDentity[blah], 1, 2, 3]

1, 2, 3

Alternatively, define your function altTuples with two signatures:

ClearAll[altTuples]
altTuples[tuplesIdentity ] := a
altTuples[x_] := Tuples[x]

altTuples[x, y, 1, 2]

x, 1, x, 2, y, 1, y, 2

altTuples[tuplesIdentity, 1, 2]

1, 2

altTuples[, 1, 2]

1, 2

share|improve this answer

edited Jul 8 at 1:48

answered Jul 7 at 23:21

kglrkglr

224k1010 gold badges254254 silver badges511511 bronze badges

$endgroup$

ClearAll[iDentity]
iDentity /: iDentity[___], a : __ := iDentity[a]
iDentity /: Tuples[iDentity[a_]] := a

Tuples[iDentity[], q]

Tuples[iDentity[blah], 1, 2, 3]

ClearAll[altTuples]
altTuples[tuplesIdentity ] := a
altTuples[x_] := Tuples[x]

altTuples[x, y, 1, 2]

altTuples[tuplesIdentity, 1, 2]

altTuples[, 1, 2]

share|improve this answer

edited Jul 8 at 1:48

answered Jul 7 at 23:21

kglrkglr

224k1010 gold badges254254 silver badges511511 bronze badges

answered Jul 7 at 23:21

kglrkglr

224k1010 gold badges254254 silver badges511511 bronze badges

answered Jul 7 at 23:21

kglrkglr

224k1010 gold badges254254 silver badges511511 bronze badges

0

$begingroup$

I think this is what you want:

altTuples[p_List, q_List] := If[Length[p] == 0, q, Tuples[p, q]]

The statement

altTuples[, d, e, f]

returns

d, e, f

and

altTuples[a, b, c, d, e, f]

returns

a, d, a, e, a, f, b, d, b, e, b, f, c, d, c, e, c, f

share|improve this answer

answered Jul 8 at 23:34

mjwmjw

1,6051313 bronze badges

$endgroup$

add a comment
 | 

0

$begingroup$

I think this is what you want:

altTuples[p_List, q_List] := If[Length[p] == 0, q, Tuples[p, q]]

The statement

altTuples[, d, e, f]

returns

d, e, f

and

altTuples[a, b, c, d, e, f]

returns

a, d, a, e, a, f, b, d, b, e, b, f, c, d, c, e, c, f

share|improve this answer

answered Jul 8 at 23:34

mjwmjw

1,6051313 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

I think this is what you want:

altTuples[p_List, q_List] := If[Length[p] == 0, q, Tuples[p, q]]

The statement

altTuples[, d, e, f]

returns

d, e, f

and

altTuples[a, b, c, d, e, f]

returns

a, d, a, e, a, f, b, d, b, e, b, f, c, d, c, e, c, f

share|improve this answer

answered Jul 8 at 23:34

mjwmjw

1,6051313 bronze badges

$endgroup$

altTuples[p_List, q_List] := If[Length[p] == 0, q, Tuples[p, q]]

altTuples[, d, e, f]

d, e, f

altTuples[a, b, c, d, e, f]

a, d, a, e, a, f, b, d, b, e, b, f, c, d, c, e, c, f

share|improve this answer

answered Jul 8 at 23:34

mjwmjw

1,6051313 bronze badges

answered Jul 8 at 23:34

mjwmjw

1,6051313 bronze badges

answered Jul 8 at 23:34

mjwmjw

1,6051313 bronze badges