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How to generate all commutative pairings of list elements?
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How to generate all commutative pairings of list elements?
Finding all elements within a certain range in a sorted listGenerating a non-convex polyhedron from a list of vertex coordinatesGenerating an ordered list of pairs of elements from ordered listsMining for matching elements of sets with equal divisorsWay to generate all multisetsHow to identify all the connected vertices from a list of coordinate pairs?Given list of associations, generate a list of all common associationsHow to generate all involutive permutations?
$begingroup$
Given a list with an even number of elements, e.g.
list = RandomSample[Array[e, 20]];
how can one generate a list of all different commutative pairings of the elements most efficiently in Mathematica?
A tiny example is:
list = e[1],e[4],e[3],e[2];
pairings[list]
e[1],e[2] , e[3],e[4] ,
e[1],e[3] , e[2],e[4] ,
e[1],e[4] , e[2],e[3] ,
Note how the commutativity of the pairings sets e.g. e[1],e[2] and e[2],e[1] to be the same pair, so that only one such term is generated.
EDIT:
Alternatively, one can ask this question in terms of graphs:
How to generate all distinct sets of disconnected un-directed edges from a list of vertices most efficiently?
list-manipulation function-construction
edited 1 hour ago
user64494
3,92111323
asked 9 hours ago
KagaratschKagaratsch
5,25141351
$endgroup$
|
show 6 more comments
$begingroup$
Given a list with an even number of elements, e.g.
list = RandomSample[Array[e, 20]];
how can one generate a list of all different commutative pairings of the elements most efficiently in Mathematica?
A tiny example is:
list = e[1],e[4],e[3],e[2];
pairings[list]
e[1],e[2] , e[3],e[4] ,
e[1],e[3] , e[2],e[4] ,
e[1],e[4] , e[2],e[3] ,
Note how the commutativity of the pairings sets e.g. e[1],e[2] and e[2],e[1] to be the same pair, so that only one such term is generated.
EDIT:
Alternatively, one can ask this question in terms of graphs:
How to generate all distinct sets of disconnected un-directed edges from a list of vertices most efficiently?
list-manipulation function-construction
edited 1 hour ago
user64494
3,92111323
asked 9 hours ago
KagaratschKagaratsch
5,25141351
$endgroup$
$begingroup$
Subsets[list, 2]?
$endgroup$
– Christopher Lamb
9 hours ago
$begingroup$
@ChristopherLamb This creates a list of all possible pairs. But starting with a list of2nelements, we are looking for a list of groups ofnpairs instead.
$endgroup$
– Kagaratsch
9 hours ago
2
$begingroup$
@Kagaratsch partition[l_, v_, comp_] := Flatten /@ Reap [ Scan [ Which[ comp[ v, #], Sow[#, -1], comp[v, #], Sow[#, 1], True, Sow[#,0]]&,l], -1,0,1][[2]] (* Three way partition function using and ordering function by sowing values with tags -1,0, or 1, depending on a relation. You could build up a list by specifying tags with Sow and patterns that match those tags in Reap. If you build the list piece by piece with recursion I would not recommend using Append instead an approach using Reap and Sow could be more effective to collect intermediate results*)
$endgroup$
– Schopenhauer
4 hours ago
1
$begingroup$
@Kagaratsch I’ve seen programs that use Fold as alternative to recursion. g[] = x; g[l_] = f[First[l], g[Rest[l]]; could be translated to g[l_]= Fold[f[#1,#2]&, x,l].
$endgroup$
– Schopenhauer
4 hours ago
2
$begingroup$
@Kagaratsch I would also check the ??Developer`* and ??Experimental`* contexts for hidden gems like PartitionMap.
$endgroup$
– Schopenhauer
4 hours ago
|
show 6 more comments
$begingroup$
Given a list with an even number of elements, e.g.
list = RandomSample[Array[e, 20]];
how can one generate a list of all different commutative pairings of the elements most efficiently in Mathematica?
A tiny example is:
list = e[1],e[4],e[3],e[2];
pairings[list]
e[1],e[2] , e[3],e[4] ,
e[1],e[3] , e[2],e[4] ,
e[1],e[4] , e[2],e[3] ,
Note how the commutativity of the pairings sets e.g. e[1],e[2] and e[2],e[1] to be the same pair, so that only one such term is generated.
EDIT:
Alternatively, one can ask this question in terms of graphs:
How to generate all distinct sets of disconnected un-directed edges from a list of vertices most efficiently?
list-manipulation function-construction
edited 1 hour ago
user64494
3,92111323
asked 9 hours ago
KagaratschKagaratsch
5,25141351
$endgroup$
Given a list with an even number of elements, e.g.
list = RandomSample[Array[e, 20]];
how can one generate a list of all different commutative pairings of the elements most efficiently in Mathematica?
A tiny example is:
list = e[1],e[4],e[3],e[2];
pairings[list]
e[1],e[2] , e[3],e[4] ,
e[1],e[3] , e[2],e[4] ,
e[1],e[4] , e[2],e[3] ,
Note how the commutativity of the pairings sets e.g. e[1],e[2] and e[2],e[1] to be the same pair, so that only one such term is generated.
EDIT:
Alternatively, one can ask this question in terms of graphs:
How to generate all distinct sets of disconnected un-directed edges from a list of vertices most efficiently?
list-manipulation function-construction
list-manipulation function-construction
edited 1 hour ago
user64494
3,92111323
asked 9 hours ago
KagaratschKagaratsch
5,25141351
edited 1 hour ago
user64494
3,92111323
asked 9 hours ago
KagaratschKagaratsch
5,25141351
edited 1 hour ago
user64494
3,92111323
edited 1 hour ago
user64494
3,92111323
edited 1 hour ago
user64494
3,92111323
3,92111323
asked 9 hours ago
KagaratschKagaratsch
5,25141351
asked 9 hours ago
KagaratschKagaratsch
5,25141351
asked 9 hours ago
KagaratschKagaratsch
5,25141351
5,25141351
$begingroup$
Subsets[list, 2]?
$endgroup$
– Christopher Lamb
9 hours ago
$begingroup$
@ChristopherLamb This creates a list of all possible pairs. But starting with a list of2nelements, we are looking for a list of groups ofnpairs instead.
$endgroup$
– Kagaratsch
9 hours ago
2
$begingroup$
@Kagaratsch partition[l_, v_, comp_] := Flatten /@ Reap [ Scan [ Which[ comp[ v, #], Sow[#, -1], comp[v, #], Sow[#, 1], True, Sow[#,0]]&,l], -1,0,1][[2]] (* Three way partition function using and ordering function by sowing values with tags -1,0, or 1, depending on a relation. You could build up a list by specifying tags with Sow and patterns that match those tags in Reap. If you build the list piece by piece with recursion I would not recommend using Append instead an approach using Reap and Sow could be more effective to collect intermediate results*)
$endgroup$
– Schopenhauer
4 hours ago
1
$begingroup$
@Kagaratsch I’ve seen programs that use Fold as alternative to recursion. g[] = x; g[l_] = f[First[l], g[Rest[l]]; could be translated to g[l_]= Fold[f[#1,#2]&, x,l].
$endgroup$
– Schopenhauer
4 hours ago
2
$begingroup$
@Kagaratsch I would also check the ??Developer`* and ??Experimental`* contexts for hidden gems like PartitionMap.
$endgroup$
– Schopenhauer
4 hours ago
|
show 6 more comments
$begingroup$
Subsets[list, 2]?
$endgroup$
– Christopher Lamb
9 hours ago
$begingroup$
@ChristopherLamb This creates a list of all possible pairs. But starting with a list of2nelements, we are looking for a list of groups ofnpairs instead.
$endgroup$
– Kagaratsch
9 hours ago
2
$begingroup$
@Kagaratsch partition[l_, v_, comp_] := Flatten /@ Reap [ Scan [ Which[ comp[ v, #], Sow[#, -1], comp[v, #], Sow[#, 1], True, Sow[#,0]]&,l], -1,0,1][[2]] (* Three way partition function using and ordering function by sowing values with tags -1,0, or 1, depending on a relation. You could build up a list by specifying tags with Sow and patterns that match those tags in Reap. If you build the list piece by piece with recursion I would not recommend using Append instead an approach using Reap and Sow could be more effective to collect intermediate results*)
$endgroup$
– Schopenhauer
4 hours ago
1
$begingroup$
@Kagaratsch I’ve seen programs that use Fold as alternative to recursion. g[] = x; g[l_] = f[First[l], g[Rest[l]]; could be translated to g[l_]= Fold[f[#1,#2]&, x,l].
$endgroup$
– Schopenhauer
4 hours ago
2
$begingroup$
@Kagaratsch I would also check the ??Developer`* and ??Experimental`* contexts for hidden gems like PartitionMap.
$endgroup$
– Schopenhauer
4 hours ago
$begingroup$
Subsets[list, 2] ?$endgroup$
– Christopher Lamb
9 hours ago
$begingroup$
Subsets[list, 2] ?$endgroup$
– Christopher Lamb
9 hours ago
$begingroup$
@ChristopherLamb This creates a list of all possible pairs. But starting with a list of
2n elements, we are looking for a list of groups of n pairs instead.$endgroup$
– Kagaratsch
9 hours ago
$begingroup$
@ChristopherLamb This creates a list of all possible pairs. But starting with a list of
2n elements, we are looking for a list of groups of n pairs instead.$endgroup$
– Kagaratsch
9 hours ago
2
2
$begingroup$
@Kagaratsch partition[l_, v_, comp_] := Flatten /@ Reap [ Scan [ Which[ comp[ v, #], Sow[#, -1], comp[v, #], Sow[#, 1], True, Sow[#,0]]&,l], -1,0,1][[2]] (* Three way partition function using and ordering function by sowing values with tags -1,0, or 1, depending on a relation. You could build up a list by specifying tags with Sow and patterns that match those tags in Reap. If you build the list piece by piece with recursion I would not recommend using Append instead an approach using Reap and Sow could be more effective to collect intermediate results*)
$endgroup$
– Schopenhauer
4 hours ago
$begingroup$
@Kagaratsch partition[l_, v_, comp_] := Flatten /@ Reap [ Scan [ Which[ comp[ v, #], Sow[#, -1], comp[v, #], Sow[#, 1], True, Sow[#,0]]&,l], -1,0,1][[2]] (* Three way partition function using and ordering function by sowing values with tags -1,0, or 1, depending on a relation. You could build up a list by specifying tags with Sow and patterns that match those tags in Reap. If you build the list piece by piece with recursion I would not recommend using Append instead an approach using Reap and Sow could be more effective to collect intermediate results*)
$endgroup$
– Schopenhauer
4 hours ago
1
1
$begingroup$
@Kagaratsch I’ve seen programs that use Fold as alternative to recursion. g[] = x; g[l_] = f[First[l], g[Rest[l]]; could be translated to g[l_]= Fold[f[#1,#2]&, x,l].
$endgroup$
– Schopenhauer
4 hours ago
$begingroup$
@Kagaratsch I’ve seen programs that use Fold as alternative to recursion. g[] = x; g[l_] = f[First[l], g[Rest[l]]; could be translated to g[l_]= Fold[f[#1,#2]&, x,l].
$endgroup$
– Schopenhauer
4 hours ago
2
2
$begingroup$
@Kagaratsch I would also check the ??Developer`* and ??Experimental`* contexts for hidden gems like PartitionMap.
$endgroup$
– Schopenhauer
4 hours ago
$begingroup$
@Kagaratsch I would also check the ??Developer`* and ??Experimental`* contexts for hidden gems like PartitionMap.
$endgroup$
– Schopenhauer
4 hours ago
|
show 6 more comments
4 Answers
4
active
oldest
votes
$begingroup$
I think the number of such pairings is given by:
pairCounts[n_?EvenQ] := Multinomial @@ ConstantArray[2, n/2]/(n/2)!
So, you will get:
pairCounts[20]
654729075
which is a lot of pairings for a list of length 20. What do you plan to do with this list?
At any rate, here is a not very efficient method:
partitions[a_,b_] := a,b
partitions[a_,b__] := Catenate@Table[
Prepend[a, b[[i]]] /@ partitions[Delete[b, i]],
i, Length[b]
]
For example:
partitions[Range[4]]
partitions[Range[6]]
1, 2, 3, 4, 1, 3, 2, 4, 1, 4, 2, 3
1, 2, 3, 4, 5, 6, 1, 2, 3, 5, 4, 6, 1, 2, 3, 6, 4,
5, 1, 3, 2, 4, 5, 6, 1, 3, 2, 5, 4, 6, 1, 3, 2,
6, 4, 5, 1, 4, 2, 3, 5, 6, 1, 4, 2, 5, 3, 6, 1,
4, 2, 6, 3, 5, 1, 5, 2, 3, 4, 6, 1, 5, 2, 4, 3,
6, 1, 5, 2, 6, 3, 4, 1, 6, 2, 3, 4, 5, 1, 6, 2,
4, 3, 5, 1, 6, 2, 5, 3, 4
answered 9 hours ago
Carl WollCarl Woll
82.1k3105212
$endgroup$
$begingroup$
Thanks, this one is quicker than the one I could write! I plan to collapse some of the vertices to obtain connected graphs and determine their multiplicity.
$endgroup$
– Kagaratsch
8 hours ago
$begingroup$
By the waypairCounts[n_?EvenQ] := (n-1)!!
$endgroup$
– Kagaratsch
7 hours ago
add a comment |
$begingroup$
ClearAll[perfectMatchings]
perfectMatchings = Module[subs = Subsets[#, 2], l = Length @ #, matchings,
matchings = FindIndependentVertexSet[LineGraph[UndirectedEdge @@@ subs], l/2, All];
Extract[subs, List /@ matchings] ] &;
perfectMatchings[Range @ 4] // Grid // TeXForm
$smallbeginarraycc
1,4 & 2,3 \
1,3 & 2,4 \
1,2 & 3,4 \
endarray$
perfectMatchings[Range @ 6] // Grid // TeXForm
$smallbeginarrayccc
1,6 & 2,5 & 3,4 \
1,6 & 2,4 & 3,5 \
1,6 & 2,3 & 4,5 \
1,5 & 2,6 & 3,4 \
1,5 & 2,4 & 3,6 \
1,5 & 2,3 & 4,6 \
1,4 & 2,6 & 3,5 \
1,4 & 2,5 & 3,6 \
1,4 & 2,3 & 5,6 \
1,3 & 2,6 & 4,5 \
1,3 & 2,5 & 4,6 \
1,3 & 2,4 & 5,6 \
1,2 & 3,6 & 4,5 \
1,2 & 3,5 & 4,6 \
1,2 & 3,4 & 5,6 \
endarray$
Note: This is much slower than Carl's partition.
answered 9 hours ago
kglrkglr
196k10216440
$endgroup$
$begingroup$
Unfortunately, this seems to break down forlists that are longer than4elements. For example, withlist=e[1],e[2],e[3],e[4],e[5],e[6]the set of groupings of edges, among others, will have the two distinct elementse[5],e[6],e[1],e[2],e[3],e[4]ande[5],e[6],e[1],e[3],e[2],e[4], so that for example the edgee[5],e[6]shows up more than once in the groupings, butSubsetscan only generate the edge once, so that it might be a suboptimal starting point...
$endgroup$
– Kagaratsch
9 hours ago
$begingroup$
@Kagaratsch, right. I thinkGatheris not the right approach sinceDisjointQis not transitive. I will delete for now; and post an update if i find the right approach.
$endgroup$
– kglr
9 hours ago
add a comment |
$begingroup$
Here a recursive solution, which I suspect is similar to the one by Carl Woll:
pairings[list_, progress_] := Block[,
If[Length[list] > 1,
Flatten[
Table[
pairings[Drop[list[[2 ;;]], i - 1],
Append[progress, list[[1]], list[[i]]]]
, i, 2, Length[list]]
, 1]
,
p[progress]
]
]
With outputs
pairings[Range[4], ]
p[1, 2, 3, 4], p[1, 3, 2, 4], p[1, 4, 2, 3]
and
pairings[Range[6], ]
p[1, 2, 3, 4, 5, 6], p[1, 2, 3, 5, 4, 6],
p[1, 2, 3, 6, 4, 5], p[1, 3, 2, 4, 5, 6],
p[1, 3, 2, 5, 4, 6], p[1, 3, 2, 6, 4, 5],
p[1, 4, 2, 3, 5, 6], p[1, 4, 2, 5, 3, 6],
p[1, 4, 2, 6, 3, 5], p[1, 5, 2, 3, 4, 6],
p[1, 5, 2, 4, 3, 6], p[1, 5, 2, 6, 3, 4],
p[1, 6, 2, 3, 4, 5], p[1, 6, 2, 4, 3, 5],
p[1, 6, 2, 5, 3, 4]
Turns out, this one is a bit slower than partitions by Carl Woll:
pairings[Range[14], ] // Length // AbsoluteTiming
1.91637, 135135
partitions[Range[14]] // Length // AbsoluteTiming
1.1277, 135135
answered 8 hours ago
KagaratschKagaratsch
5,25141351
$endgroup$
add a comment |
$begingroup$
How is it that the comment by @ChristopherLamb is incorrect? Sets are order independent, so I would think this creates "commutative pairs."
set = Table[e[n], n, 4]
(* e[1],e[2],e[3],e[4] *)
Subsets[set, 2]
(* e[1],e[2],e[1],e[3],e[1],e[4],e[2],e[3],e[2],e[4],e[3],e[4] *)
answered 9 hours ago
David KeithDavid Keith
2,1801516
$endgroup$
$begingroup$
This output is missing a dimension in the list, which would partition this set of pairs into groups of pairs in which all elements show up exactly once. Also note that some edges will exist more than once within these groups when list dimension is larger than4.
$endgroup$
– Kagaratsch
9 hours ago
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think the number of such pairings is given by:
pairCounts[n_?EvenQ] := Multinomial @@ ConstantArray[2, n/2]/(n/2)!
So, you will get:
pairCounts[20]
654729075
which is a lot of pairings for a list of length 20. What do you plan to do with this list?
At any rate, here is a not very efficient method:
partitions[a_,b_] := a,b
partitions[a_,b__] := Catenate@Table[
Prepend[a, b[[i]]] /@ partitions[Delete[b, i]],
i, Length[b]
]
For example:
partitions[Range[4]]
partitions[Range[6]]
1, 2, 3, 4, 1, 3, 2, 4, 1, 4, 2, 3
1, 2, 3, 4, 5, 6, 1, 2, 3, 5, 4, 6, 1, 2, 3, 6, 4,
5, 1, 3, 2, 4, 5, 6, 1, 3, 2, 5, 4, 6, 1, 3, 2,
6, 4, 5, 1, 4, 2, 3, 5, 6, 1, 4, 2, 5, 3, 6, 1,
4, 2, 6, 3, 5, 1, 5, 2, 3, 4, 6, 1, 5, 2, 4, 3,
6, 1, 5, 2, 6, 3, 4, 1, 6, 2, 3, 4, 5, 1, 6, 2,
4, 3, 5, 1, 6, 2, 5, 3, 4
answered 9 hours ago
Carl WollCarl Woll
82.1k3105212
$endgroup$
$begingroup$
Thanks, this one is quicker than the one I could write! I plan to collapse some of the vertices to obtain connected graphs and determine their multiplicity.
$endgroup$
– Kagaratsch
8 hours ago
$begingroup$
By the waypairCounts[n_?EvenQ] := (n-1)!!
$endgroup$
– Kagaratsch
7 hours ago
add a comment |
$begingroup$
I think the number of such pairings is given by:
pairCounts[n_?EvenQ] := Multinomial @@ ConstantArray[2, n/2]/(n/2)!
So, you will get:
pairCounts[20]
654729075
which is a lot of pairings for a list of length 20. What do you plan to do with this list?
At any rate, here is a not very efficient method:
partitions[a_,b_] := a,b
partitions[a_,b__] := Catenate@Table[
Prepend[a, b[[i]]] /@ partitions[Delete[b, i]],
i, Length[b]
]
For example:
partitions[Range[4]]
partitions[Range[6]]
1, 2, 3, 4, 1, 3, 2, 4, 1, 4, 2, 3
1, 2, 3, 4, 5, 6, 1, 2, 3, 5, 4, 6, 1, 2, 3, 6, 4,
5, 1, 3, 2, 4, 5, 6, 1, 3, 2, 5, 4, 6, 1, 3, 2,
6, 4, 5, 1, 4, 2, 3, 5, 6, 1, 4, 2, 5, 3, 6, 1,
4, 2, 6, 3, 5, 1, 5, 2, 3, 4, 6, 1, 5, 2, 4, 3,
6, 1, 5, 2, 6, 3, 4, 1, 6, 2, 3, 4, 5, 1, 6, 2,
4, 3, 5, 1, 6, 2, 5, 3, 4
answered 9 hours ago
Carl WollCarl Woll
82.1k3105212
$endgroup$
$begingroup$
Thanks, this one is quicker than the one I could write! I plan to collapse some of the vertices to obtain connected graphs and determine their multiplicity.
$endgroup$
– Kagaratsch
8 hours ago
$begingroup$
By the waypairCounts[n_?EvenQ] := (n-1)!!
$endgroup$
– Kagaratsch
7 hours ago
add a comment |
$begingroup$
I think the number of such pairings is given by:
pairCounts[n_?EvenQ] := Multinomial @@ ConstantArray[2, n/2]/(n/2)!
So, you will get:
pairCounts[20]
654729075
which is a lot of pairings for a list of length 20. What do you plan to do with this list?
At any rate, here is a not very efficient method:
partitions[a_,b_] := a,b
partitions[a_,b__] := Catenate@Table[
Prepend[a, b[[i]]] /@ partitions[Delete[b, i]],
i, Length[b]
]
For example:
partitions[Range[4]]
partitions[Range[6]]
1, 2, 3, 4, 1, 3, 2, 4, 1, 4, 2, 3
1, 2, 3, 4, 5, 6, 1, 2, 3, 5, 4, 6, 1, 2, 3, 6, 4,
5, 1, 3, 2, 4, 5, 6, 1, 3, 2, 5, 4, 6, 1, 3, 2,
6, 4, 5, 1, 4, 2, 3, 5, 6, 1, 4, 2, 5, 3, 6, 1,
4, 2, 6, 3, 5, 1, 5, 2, 3, 4, 6, 1, 5, 2, 4, 3,
6, 1, 5, 2, 6, 3, 4, 1, 6, 2, 3, 4, 5, 1, 6, 2,
4, 3, 5, 1, 6, 2, 5, 3, 4
answered 9 hours ago
Carl WollCarl Woll
82.1k3105212
$endgroup$
I think the number of such pairings is given by:
pairCounts[n_?EvenQ] := Multinomial @@ ConstantArray[2, n/2]/(n/2)!
So, you will get:
pairCounts[20]
654729075
which is a lot of pairings for a list of length 20. What do you plan to do with this list?
At any rate, here is a not very efficient method:
partitions[a_,b_] := a,b
partitions[a_,b__] := Catenate@Table[
Prepend[a, b[[i]]] /@ partitions[Delete[b, i]],
i, Length[b]
]
For example:
partitions[Range[4]]
partitions[Range[6]]
1, 2, 3, 4, 1, 3, 2, 4, 1, 4, 2, 3
1, 2, 3, 4, 5, 6, 1, 2, 3, 5, 4, 6, 1, 2, 3, 6, 4,
5, 1, 3, 2, 4, 5, 6, 1, 3, 2, 5, 4, 6, 1, 3, 2,
6, 4, 5, 1, 4, 2, 3, 5, 6, 1, 4, 2, 5, 3, 6, 1,
4, 2, 6, 3, 5, 1, 5, 2, 3, 4, 6, 1, 5, 2, 4, 3,
6, 1, 5, 2, 6, 3, 4, 1, 6, 2, 3, 4, 5, 1, 6, 2,
4, 3, 5, 1, 6, 2, 5, 3, 4
answered 9 hours ago
Carl WollCarl Woll
82.1k3105212
answered 9 hours ago
Carl WollCarl Woll
82.1k3105212
answered 9 hours ago
Carl WollCarl Woll
82.1k3105212
answered 9 hours ago
Carl WollCarl Woll
82.1k3105212
82.1k3105212
$begingroup$
Thanks, this one is quicker than the one I could write! I plan to collapse some of the vertices to obtain connected graphs and determine their multiplicity.
$endgroup$
– Kagaratsch
8 hours ago
$begingroup$
By the waypairCounts[n_?EvenQ] := (n-1)!!
$endgroup$
– Kagaratsch
7 hours ago
add a comment |
$begingroup$
Thanks, this one is quicker than the one I could write! I plan to collapse some of the vertices to obtain connected graphs and determine their multiplicity.
$endgroup$
– Kagaratsch
8 hours ago
$begingroup$
By the waypairCounts[n_?EvenQ] := (n-1)!!
$endgroup$
– Kagaratsch
7 hours ago
$begingroup$
Thanks, this one is quicker than the one I could write! I plan to collapse some of the vertices to obtain connected graphs and determine their multiplicity.
$endgroup$
– Kagaratsch
8 hours ago
$begingroup$
Thanks, this one is quicker than the one I could write! I plan to collapse some of the vertices to obtain connected graphs and determine their multiplicity.
$endgroup$
– Kagaratsch
8 hours ago
$begingroup$
By the way
pairCounts[n_?EvenQ] := (n-1)!!$endgroup$
– Kagaratsch
7 hours ago
$begingroup$
By the way
pairCounts[n_?EvenQ] := (n-1)!!$endgroup$
– Kagaratsch
7 hours ago
add a comment |
$begingroup$
ClearAll[perfectMatchings]
perfectMatchings = Module[subs = Subsets[#, 2], l = Length @ #, matchings,
matchings = FindIndependentVertexSet[LineGraph[UndirectedEdge @@@ subs], l/2, All];
Extract[subs, List /@ matchings] ] &;
perfectMatchings[Range @ 4] // Grid // TeXForm
$smallbeginarraycc
1,4 & 2,3 \
1,3 & 2,4 \
1,2 & 3,4 \
endarray$
perfectMatchings[Range @ 6] // Grid // TeXForm
$smallbeginarrayccc
1,6 & 2,5 & 3,4 \
1,6 & 2,4 & 3,5 \
1,6 & 2,3 & 4,5 \
1,5 & 2,6 & 3,4 \
1,5 & 2,4 & 3,6 \
1,5 & 2,3 & 4,6 \
1,4 & 2,6 & 3,5 \
1,4 & 2,5 & 3,6 \
1,4 & 2,3 & 5,6 \
1,3 & 2,6 & 4,5 \
1,3 & 2,5 & 4,6 \
1,3 & 2,4 & 5,6 \
1,2 & 3,6 & 4,5 \
1,2 & 3,5 & 4,6 \
1,2 & 3,4 & 5,6 \
endarray$
Note: This is much slower than Carl's partition.
answered 9 hours ago
kglrkglr
196k10216440
$endgroup$
$begingroup$
Unfortunately, this seems to break down forlists that are longer than4elements. For example, withlist=e[1],e[2],e[3],e[4],e[5],e[6]the set of groupings of edges, among others, will have the two distinct elementse[5],e[6],e[1],e[2],e[3],e[4]ande[5],e[6],e[1],e[3],e[2],e[4], so that for example the edgee[5],e[6]shows up more than once in the groupings, butSubsetscan only generate the edge once, so that it might be a suboptimal starting point...
$endgroup$
– Kagaratsch
9 hours ago
$begingroup$
@Kagaratsch, right. I thinkGatheris not the right approach sinceDisjointQis not transitive. I will delete for now; and post an update if i find the right approach.
$endgroup$
– kglr
9 hours ago
add a comment |
$begingroup$
ClearAll[perfectMatchings]
perfectMatchings = Module[subs = Subsets[#, 2], l = Length @ #, matchings,
matchings = FindIndependentVertexSet[LineGraph[UndirectedEdge @@@ subs], l/2, All];
Extract[subs, List /@ matchings] ] &;
perfectMatchings[Range @ 4] // Grid // TeXForm
$smallbeginarraycc
1,4 & 2,3 \
1,3 & 2,4 \
1,2 & 3,4 \
endarray$
perfectMatchings[Range @ 6] // Grid // TeXForm
$smallbeginarrayccc
1,6 & 2,5 & 3,4 \
1,6 & 2,4 & 3,5 \
1,6 & 2,3 & 4,5 \
1,5 & 2,6 & 3,4 \
1,5 & 2,4 & 3,6 \
1,5 & 2,3 & 4,6 \
1,4 & 2,6 & 3,5 \
1,4 & 2,5 & 3,6 \
1,4 & 2,3 & 5,6 \
1,3 & 2,6 & 4,5 \
1,3 & 2,5 & 4,6 \
1,3 & 2,4 & 5,6 \
1,2 & 3,6 & 4,5 \
1,2 & 3,5 & 4,6 \
1,2 & 3,4 & 5,6 \
endarray$
Note: This is much slower than Carl's partition.
answered 9 hours ago
kglrkglr
196k10216440
$endgroup$
$begingroup$
Unfortunately, this seems to break down forlists that are longer than4elements. For example, withlist=e[1],e[2],e[3],e[4],e[5],e[6]the set of groupings of edges, among others, will have the two distinct elementse[5],e[6],e[1],e[2],e[3],e[4]ande[5],e[6],e[1],e[3],e[2],e[4], so that for example the edgee[5],e[6]shows up more than once in the groupings, butSubsetscan only generate the edge once, so that it might be a suboptimal starting point...
$endgroup$
– Kagaratsch
9 hours ago
$begingroup$
@Kagaratsch, right. I thinkGatheris not the right approach sinceDisjointQis not transitive. I will delete for now; and post an update if i find the right approach.
$endgroup$
– kglr
9 hours ago
add a comment |
$begingroup$
ClearAll[perfectMatchings]
perfectMatchings = Module[subs = Subsets[#, 2], l = Length @ #, matchings,
matchings = FindIndependentVertexSet[LineGraph[UndirectedEdge @@@ subs], l/2, All];
Extract[subs, List /@ matchings] ] &;
perfectMatchings[Range @ 4] // Grid // TeXForm
$smallbeginarraycc
1,4 & 2,3 \
1,3 & 2,4 \
1,2 & 3,4 \
endarray$
perfectMatchings[Range @ 6] // Grid // TeXForm
$smallbeginarrayccc
1,6 & 2,5 & 3,4 \
1,6 & 2,4 & 3,5 \
1,6 & 2,3 & 4,5 \
1,5 & 2,6 & 3,4 \
1,5 & 2,4 & 3,6 \
1,5 & 2,3 & 4,6 \
1,4 & 2,6 & 3,5 \
1,4 & 2,5 & 3,6 \
1,4 & 2,3 & 5,6 \
1,3 & 2,6 & 4,5 \
1,3 & 2,5 & 4,6 \
1,3 & 2,4 & 5,6 \
1,2 & 3,6 & 4,5 \
1,2 & 3,5 & 4,6 \
1,2 & 3,4 & 5,6 \
endarray$
Note: This is much slower than Carl's partition.
answered 9 hours ago
kglrkglr
196k10216440
$endgroup$
ClearAll[perfectMatchings]
perfectMatchings = Module[subs = Subsets[#, 2], l = Length @ #, matchings,
matchings = FindIndependentVertexSet[LineGraph[UndirectedEdge @@@ subs], l/2, All];
Extract[subs, List /@ matchings] ] &;
perfectMatchings[Range @ 4] // Grid // TeXForm
$smallbeginarraycc
1,4 & 2,3 \
1,3 & 2,4 \
1,2 & 3,4 \
endarray$
perfectMatchings[Range @ 6] // Grid // TeXForm
$smallbeginarrayccc
1,6 & 2,5 & 3,4 \
1,6 & 2,4 & 3,5 \
1,6 & 2,3 & 4,5 \
1,5 & 2,6 & 3,4 \
1,5 & 2,4 & 3,6 \
1,5 & 2,3 & 4,6 \
1,4 & 2,6 & 3,5 \
1,4 & 2,5 & 3,6 \
1,4 & 2,3 & 5,6 \
1,3 & 2,6 & 4,5 \
1,3 & 2,5 & 4,6 \
1,3 & 2,4 & 5,6 \
1,2 & 3,6 & 4,5 \
1,2 & 3,5 & 4,6 \
1,2 & 3,4 & 5,6 \
endarray$
Note: This is much slower than Carl's partition.
answered 9 hours ago
kglrkglr
196k10216440
edited 1 min ago
answered 9 hours ago
kglrkglr
196k10216440
answered 9 hours ago
kglrkglr
196k10216440
answered 9 hours ago
kglrkglr
196k10216440
196k10216440
$begingroup$
Unfortunately, this seems to break down forlists that are longer than4elements. For example, withlist=e[1],e[2],e[3],e[4],e[5],e[6]the set of groupings of edges, among others, will have the two distinct elementse[5],e[6],e[1],e[2],e[3],e[4]ande[5],e[6],e[1],e[3],e[2],e[4], so that for example the edgee[5],e[6]shows up more than once in the groupings, butSubsetscan only generate the edge once, so that it might be a suboptimal starting point...
$endgroup$
– Kagaratsch
9 hours ago
$begingroup$
@Kagaratsch, right. I thinkGatheris not the right approach sinceDisjointQis not transitive. I will delete for now; and post an update if i find the right approach.
$endgroup$
– kglr
9 hours ago
add a comment |
$begingroup$
Unfortunately, this seems to break down forlists that are longer than4elements. For example, withlist=e[1],e[2],e[3],e[4],e[5],e[6]the set of groupings of edges, among others, will have the two distinct elementse[5],e[6],e[1],e[2],e[3],e[4]ande[5],e[6],e[1],e[3],e[2],e[4], so that for example the edgee[5],e[6]shows up more than once in the groupings, butSubsetscan only generate the edge once, so that it might be a suboptimal starting point...
$endgroup$
– Kagaratsch
9 hours ago
$begingroup$
@Kagaratsch, right. I thinkGatheris not the right approach sinceDisjointQis not transitive. I will delete for now; and post an update if i find the right approach.
$endgroup$
– kglr
9 hours ago
$begingroup$
Unfortunately, this seems to break down for
lists that are longer than 4 elements. For example, with list=e[1],e[2],e[3],e[4],e[5],e[6] the set of groupings of edges, among others, will have the two distinct elements e[5],e[6],e[1],e[2],e[3],e[4] and e[5],e[6],e[1],e[3],e[2],e[4], so that for example the edge e[5],e[6] shows up more than once in the groupings, but Subsets can only generate the edge once, so that it might be a suboptimal starting point...$endgroup$
– Kagaratsch
9 hours ago
$begingroup$
Unfortunately, this seems to break down for
lists that are longer than 4 elements. For example, with list=e[1],e[2],e[3],e[4],e[5],e[6] the set of groupings of edges, among others, will have the two distinct elements e[5],e[6],e[1],e[2],e[3],e[4] and e[5],e[6],e[1],e[3],e[2],e[4], so that for example the edge e[5],e[6] shows up more than once in the groupings, but Subsets can only generate the edge once, so that it might be a suboptimal starting point...$endgroup$
– Kagaratsch
9 hours ago
$begingroup$
@Kagaratsch, right. I think
Gather is not the right approach since DisjointQ is not transitive. I will delete for now; and post an update if i find the right approach.$endgroup$
– kglr
9 hours ago
$begingroup$
@Kagaratsch, right. I think
Gather is not the right approach since DisjointQ is not transitive. I will delete for now; and post an update if i find the right approach.$endgroup$
– kglr
9 hours ago
add a comment |
$begingroup$
Here a recursive solution, which I suspect is similar to the one by Carl Woll:
pairings[list_, progress_] := Block[,
If[Length[list] > 1,
Flatten[
Table[
pairings[Drop[list[[2 ;;]], i - 1],
Append[progress, list[[1]], list[[i]]]]
, i, 2, Length[list]]
, 1]
,
p[progress]
]
]
With outputs
pairings[Range[4], ]
p[1, 2, 3, 4], p[1, 3, 2, 4], p[1, 4, 2, 3]
and
pairings[Range[6], ]
p[1, 2, 3, 4, 5, 6], p[1, 2, 3, 5, 4, 6],
p[1, 2, 3, 6, 4, 5], p[1, 3, 2, 4, 5, 6],
p[1, 3, 2, 5, 4, 6], p[1, 3, 2, 6, 4, 5],
p[1, 4, 2, 3, 5, 6], p[1, 4, 2, 5, 3, 6],
p[1, 4, 2, 6, 3, 5], p[1, 5, 2, 3, 4, 6],
p[1, 5, 2, 4, 3, 6], p[1, 5, 2, 6, 3, 4],
p[1, 6, 2, 3, 4, 5], p[1, 6, 2, 4, 3, 5],
p[1, 6, 2, 5, 3, 4]
Turns out, this one is a bit slower than partitions by Carl Woll:
pairings[Range[14], ] // Length // AbsoluteTiming
1.91637, 135135
partitions[Range[14]] // Length // AbsoluteTiming
1.1277, 135135
answered 8 hours ago
KagaratschKagaratsch
5,25141351
$endgroup$
add a comment |
$begingroup$
Here a recursive solution, which I suspect is similar to the one by Carl Woll:
pairings[list_, progress_] := Block[,
If[Length[list] > 1,
Flatten[
Table[
pairings[Drop[list[[2 ;;]], i - 1],
Append[progress, list[[1]], list[[i]]]]
, i, 2, Length[list]]
, 1]
,
p[progress]
]
]
With outputs
pairings[Range[4], ]
p[1, 2, 3, 4], p[1, 3, 2, 4], p[1, 4, 2, 3]
and
pairings[Range[6], ]
p[1, 2, 3, 4, 5, 6], p[1, 2, 3, 5, 4, 6],
p[1, 2, 3, 6, 4, 5], p[1, 3, 2, 4, 5, 6],
p[1, 3, 2, 5, 4, 6], p[1, 3, 2, 6, 4, 5],
p[1, 4, 2, 3, 5, 6], p[1, 4, 2, 5, 3, 6],
p[1, 4, 2, 6, 3, 5], p[1, 5, 2, 3, 4, 6],
p[1, 5, 2, 4, 3, 6], p[1, 5, 2, 6, 3, 4],
p[1, 6, 2, 3, 4, 5], p[1, 6, 2, 4, 3, 5],
p[1, 6, 2, 5, 3, 4]
Turns out, this one is a bit slower than partitions by Carl Woll:
pairings[Range[14], ] // Length // AbsoluteTiming
1.91637, 135135
partitions[Range[14]] // Length // AbsoluteTiming
1.1277, 135135
answered 8 hours ago
KagaratschKagaratsch
5,25141351
$endgroup$
add a comment |
$begingroup$
Here a recursive solution, which I suspect is similar to the one by Carl Woll:
pairings[list_, progress_] := Block[,
If[Length[list] > 1,
Flatten[
Table[
pairings[Drop[list[[2 ;;]], i - 1],
Append[progress, list[[1]], list[[i]]]]
, i, 2, Length[list]]
, 1]
,
p[progress]
]
]
With outputs
pairings[Range[4], ]
p[1, 2, 3, 4], p[1, 3, 2, 4], p[1, 4, 2, 3]
and
pairings[Range[6], ]
p[1, 2, 3, 4, 5, 6], p[1, 2, 3, 5, 4, 6],
p[1, 2, 3, 6, 4, 5], p[1, 3, 2, 4, 5, 6],
p[1, 3, 2, 5, 4, 6], p[1, 3, 2, 6, 4, 5],
p[1, 4, 2, 3, 5, 6], p[1, 4, 2, 5, 3, 6],
p[1, 4, 2, 6, 3, 5], p[1, 5, 2, 3, 4, 6],
p[1, 5, 2, 4, 3, 6], p[1, 5, 2, 6, 3, 4],
p[1, 6, 2, 3, 4, 5], p[1, 6, 2, 4, 3, 5],
p[1, 6, 2, 5, 3, 4]
Turns out, this one is a bit slower than partitions by Carl Woll:
pairings[Range[14], ] // Length // AbsoluteTiming
1.91637, 135135
partitions[Range[14]] // Length // AbsoluteTiming
1.1277, 135135
answered 8 hours ago
KagaratschKagaratsch
5,25141351
$endgroup$
Here a recursive solution, which I suspect is similar to the one by Carl Woll:
pairings[list_, progress_] := Block[,
If[Length[list] > 1,
Flatten[
Table[
pairings[Drop[list[[2 ;;]], i - 1],
Append[progress, list[[1]], list[[i]]]]
, i, 2, Length[list]]
, 1]
,
p[progress]
]
]
With outputs
pairings[Range[4], ]
p[1, 2, 3, 4], p[1, 3, 2, 4], p[1, 4, 2, 3]
and
pairings[Range[6], ]
p[1, 2, 3, 4, 5, 6], p[1, 2, 3, 5, 4, 6],
p[1, 2, 3, 6, 4, 5], p[1, 3, 2, 4, 5, 6],
p[1, 3, 2, 5, 4, 6], p[1, 3, 2, 6, 4, 5],
p[1, 4, 2, 3, 5, 6], p[1, 4, 2, 5, 3, 6],
p[1, 4, 2, 6, 3, 5], p[1, 5, 2, 3, 4, 6],
p[1, 5, 2, 4, 3, 6], p[1, 5, 2, 6, 3, 4],
p[1, 6, 2, 3, 4, 5], p[1, 6, 2, 4, 3, 5],
p[1, 6, 2, 5, 3, 4]
Turns out, this one is a bit slower than partitions by Carl Woll:
pairings[Range[14], ] // Length // AbsoluteTiming
1.91637, 135135
partitions[Range[14]] // Length // AbsoluteTiming
1.1277, 135135
answered 8 hours ago
KagaratschKagaratsch
5,25141351
edited 8 hours ago
answered 8 hours ago
KagaratschKagaratsch
5,25141351
answered 8 hours ago
KagaratschKagaratsch
5,25141351
answered 8 hours ago
KagaratschKagaratsch
5,25141351
5,25141351
add a comment |
add a comment |
$begingroup$
How is it that the comment by @ChristopherLamb is incorrect? Sets are order independent, so I would think this creates "commutative pairs."
set = Table[e[n], n, 4]
(* e[1],e[2],e[3],e[4] *)
Subsets[set, 2]
(* e[1],e[2],e[1],e[3],e[1],e[4],e[2],e[3],e[2],e[4],e[3],e[4] *)
answered 9 hours ago
David KeithDavid Keith
2,1801516
$endgroup$
$begingroup$
This output is missing a dimension in the list, which would partition this set of pairs into groups of pairs in which all elements show up exactly once. Also note that some edges will exist more than once within these groups when list dimension is larger than4.
$endgroup$
– Kagaratsch
9 hours ago
add a comment |
$begingroup$
How is it that the comment by @ChristopherLamb is incorrect? Sets are order independent, so I would think this creates "commutative pairs."
set = Table[e[n], n, 4]
(* e[1],e[2],e[3],e[4] *)
Subsets[set, 2]
(* e[1],e[2],e[1],e[3],e[1],e[4],e[2],e[3],e[2],e[4],e[3],e[4] *)
answered 9 hours ago
David KeithDavid Keith
2,1801516
$endgroup$
$begingroup$
This output is missing a dimension in the list, which would partition this set of pairs into groups of pairs in which all elements show up exactly once. Also note that some edges will exist more than once within these groups when list dimension is larger than4.
$endgroup$
– Kagaratsch
9 hours ago
add a comment |
$begingroup$
How is it that the comment by @ChristopherLamb is incorrect? Sets are order independent, so I would think this creates "commutative pairs."
set = Table[e[n], n, 4]
(* e[1],e[2],e[3],e[4] *)
Subsets[set, 2]
(* e[1],e[2],e[1],e[3],e[1],e[4],e[2],e[3],e[2],e[4],e[3],e[4] *)
answered 9 hours ago
David KeithDavid Keith
2,1801516
$endgroup$
How is it that the comment by @ChristopherLamb is incorrect? Sets are order independent, so I would think this creates "commutative pairs."
set = Table[e[n], n, 4]
(* e[1],e[2],e[3],e[4] *)
Subsets[set, 2]
(* e[1],e[2],e[1],e[3],e[1],e[4],e[2],e[3],e[2],e[4],e[3],e[4] *)
answered 9 hours ago
David KeithDavid Keith
2,1801516
answered 9 hours ago
David KeithDavid Keith
2,1801516
answered 9 hours ago
David KeithDavid Keith
2,1801516
answered 9 hours ago
David KeithDavid Keith
2,1801516
2,1801516
$begingroup$
This output is missing a dimension in the list, which would partition this set of pairs into groups of pairs in which all elements show up exactly once. Also note that some edges will exist more than once within these groups when list dimension is larger than4.
$endgroup$
– Kagaratsch
9 hours ago
add a comment |
$begingroup$
This output is missing a dimension in the list, which would partition this set of pairs into groups of pairs in which all elements show up exactly once. Also note that some edges will exist more than once within these groups when list dimension is larger than4.
$endgroup$
– Kagaratsch
9 hours ago
$begingroup$
This output is missing a dimension in the list, which would partition this set of pairs into groups of pairs in which all elements show up exactly once. Also note that some edges will exist more than once within these groups when list dimension is larger than
4.$endgroup$
– Kagaratsch
9 hours ago
$begingroup$
This output is missing a dimension in the list, which would partition this set of pairs into groups of pairs in which all elements show up exactly once. Also note that some edges will exist more than once within these groups when list dimension is larger than
4.$endgroup$
– Kagaratsch
9 hours ago
add a comment |
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$begingroup$
Subsets[list, 2]?$endgroup$
– Christopher Lamb
9 hours ago
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@ChristopherLamb This creates a list of all possible pairs. But starting with a list of
2nelements, we are looking for a list of groups ofnpairs instead.$endgroup$
– Kagaratsch
9 hours ago
2
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@Kagaratsch partition[l_, v_, comp_] := Flatten /@ Reap [ Scan [ Which[ comp[ v, #], Sow[#, -1], comp[v, #], Sow[#, 1], True, Sow[#,0]]&,l], -1,0,1][[2]] (* Three way partition function using and ordering function by sowing values with tags -1,0, or 1, depending on a relation. You could build up a list by specifying tags with Sow and patterns that match those tags in Reap. If you build the list piece by piece with recursion I would not recommend using Append instead an approach using Reap and Sow could be more effective to collect intermediate results*)
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– Schopenhauer
4 hours ago
1
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@Kagaratsch I’ve seen programs that use Fold as alternative to recursion. g[] = x; g[l_] = f[First[l], g[Rest[l]]; could be translated to g[l_]= Fold[f[#1,#2]&, x,l].
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– Schopenhauer
4 hours ago
2
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@Kagaratsch I would also check the ??Developer`* and ??Experimental`* contexts for hidden gems like PartitionMap.
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– Schopenhauer
4 hours ago