Is there a function to partition an integer set? Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Partition a set into subsets of size $k$Partition a set into $k$ non-empty subsetsHow to delete mirror symmetric point pair efficientlyPartition a range of integers into triplesSelecting last partition in integer partitionHow to partition a 2-D array properly?generating integer partitionsImproved a code to work once (no need to repeat)Partition a set of n objects into k subsets?Groupings of the Elements of a List with at Most $k$ Elements
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Is there a function to partition an integer set?
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Partition a set into subsets of size $k$Partition a set into $k$ non-empty subsetsHow to delete mirror symmetric point pair efficientlyPartition a range of integers into triplesSelecting last partition in integer partitionHow to partition a 2-D array properly?generating integer partitionsImproved a code to work once (no need to repeat)Partition a set of n objects into k subsets?Groupings of the Elements of a List with at Most $k$ Elements
$begingroup$
First I give an example. For an integer set $(0,1,2,3,4)$, there are eight kinds of subdivision or partition like this
$$(0,4);\~~(0,1)(1,4);~~(0,2)(2,4);~~(0,3)(3,4);\
(0,1)(1,2)(2,4);~~(0,1)(1,3)(3,4);~~(0,2)(2,3)(3,4);~~\(0,1)(1,2)(2,3)(3,4); $$
For a more general set $(0,1,2,...,n)$, there are $2^n-1$ kinds of partition.I believe that there must be a special name for this kind of partition mathematically. How can I realize it in MMA?
list-manipulation combinatorics partitions
asked Apr 1 at 7:26
Mark_PhysMark_Phys
1357
$endgroup$
add a comment |
$begingroup$
First I give an example. For an integer set $(0,1,2,3,4)$, there are eight kinds of subdivision or partition like this
$$(0,4);\~~(0,1)(1,4);~~(0,2)(2,4);~~(0,3)(3,4);\
(0,1)(1,2)(2,4);~~(0,1)(1,3)(3,4);~~(0,2)(2,3)(3,4);~~\(0,1)(1,2)(2,3)(3,4); $$
For a more general set $(0,1,2,...,n)$, there are $2^n-1$ kinds of partition.I believe that there must be a special name for this kind of partition mathematically. How can I realize it in MMA?
list-manipulation combinatorics partitions
asked Apr 1 at 7:26
Mark_PhysMark_Phys
1357
$endgroup$
add a comment |
$begingroup$
First I give an example. For an integer set $(0,1,2,3,4)$, there are eight kinds of subdivision or partition like this
$$(0,4);\~~(0,1)(1,4);~~(0,2)(2,4);~~(0,3)(3,4);\
(0,1)(1,2)(2,4);~~(0,1)(1,3)(3,4);~~(0,2)(2,3)(3,4);~~\(0,1)(1,2)(2,3)(3,4); $$
For a more general set $(0,1,2,...,n)$, there are $2^n-1$ kinds of partition.I believe that there must be a special name for this kind of partition mathematically. How can I realize it in MMA?
list-manipulation combinatorics partitions
asked Apr 1 at 7:26
Mark_PhysMark_Phys
1357
$endgroup$
First I give an example. For an integer set $(0,1,2,3,4)$, there are eight kinds of subdivision or partition like this
$$(0,4);\~~(0,1)(1,4);~~(0,2)(2,4);~~(0,3)(3,4);\
(0,1)(1,2)(2,4);~~(0,1)(1,3)(3,4);~~(0,2)(2,3)(3,4);~~\(0,1)(1,2)(2,3)(3,4); $$
For a more general set $(0,1,2,...,n)$, there are $2^n-1$ kinds of partition.I believe that there must be a special name for this kind of partition mathematically. How can I realize it in MMA?
list-manipulation combinatorics partitions
list-manipulation combinatorics partitions
asked Apr 1 at 7:26
Mark_PhysMark_Phys
1357
asked Apr 1 at 7:26
Mark_PhysMark_Phys
1357
asked Apr 1 at 7:26
Mark_PhysMark_Phys
1357
asked Apr 1 at 7:26
Mark_PhysMark_Phys
1357
asked Apr 1 at 7:26
Mark_PhysMark_Phys
1357
1357
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
P[n] will return the set you are asking
P[n_] := Partition[Join[0, #, n], 2, 1] & /@ Subsets[Range[n - 1]]
P[4]
0, 4, 0, 1, 1, 4, 0, 2, 2, 4, 0, 3, 3, 4, 0,
1, 1, 2, 2, 4, 0, 1, 1, 3, 3, 4, 0, 2, 2, 3, 3,
4, 0, 1, 1, 2, 2, 3, 3, 4
answered Apr 1 at 7:47
J42161217J42161217
4,513324
$endgroup$
$begingroup$
Thanks, it works well!!!
$endgroup$
– Mark_Phys
Apr 1 at 9:08
$begingroup$
@user10709 I'm glad I helped!
$endgroup$
– J42161217
Apr 1 at 9:10
$begingroup$
Of course, you can combine the functions so that only one application ofMap[]is needed:Partition[Join[0, #, n], 2, 1] & /@ Subsets[Range[n - 1]].
$endgroup$
– J. M. is away♦
Apr 1 at 14:35
$begingroup$
@J.M.isslightlypensive yes, you are right
$endgroup$
– J42161217
Apr 1 at 14:52
add a comment |
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1 Answer
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1 Answer
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$begingroup$
P[n] will return the set you are asking
P[n_] := Partition[Join[0, #, n], 2, 1] & /@ Subsets[Range[n - 1]]
P[4]
0, 4, 0, 1, 1, 4, 0, 2, 2, 4, 0, 3, 3, 4, 0,
1, 1, 2, 2, 4, 0, 1, 1, 3, 3, 4, 0, 2, 2, 3, 3,
4, 0, 1, 1, 2, 2, 3, 3, 4
answered Apr 1 at 7:47
J42161217J42161217
4,513324
$endgroup$
$begingroup$
Thanks, it works well!!!
$endgroup$
– Mark_Phys
Apr 1 at 9:08
$begingroup$
@user10709 I'm glad I helped!
$endgroup$
– J42161217
Apr 1 at 9:10
$begingroup$
Of course, you can combine the functions so that only one application ofMap[]is needed:Partition[Join[0, #, n], 2, 1] & /@ Subsets[Range[n - 1]].
$endgroup$
– J. M. is away♦
Apr 1 at 14:35
$begingroup$
@J.M.isslightlypensive yes, you are right
$endgroup$
– J42161217
Apr 1 at 14:52
add a comment |
$begingroup$
P[n] will return the set you are asking
P[n_] := Partition[Join[0, #, n], 2, 1] & /@ Subsets[Range[n - 1]]
P[4]
0, 4, 0, 1, 1, 4, 0, 2, 2, 4, 0, 3, 3, 4, 0,
1, 1, 2, 2, 4, 0, 1, 1, 3, 3, 4, 0, 2, 2, 3, 3,
4, 0, 1, 1, 2, 2, 3, 3, 4
answered Apr 1 at 7:47
J42161217J42161217
4,513324
$endgroup$
$begingroup$
Thanks, it works well!!!
$endgroup$
– Mark_Phys
Apr 1 at 9:08
$begingroup$
@user10709 I'm glad I helped!
$endgroup$
– J42161217
Apr 1 at 9:10
$begingroup$
Of course, you can combine the functions so that only one application ofMap[]is needed:Partition[Join[0, #, n], 2, 1] & /@ Subsets[Range[n - 1]].
$endgroup$
– J. M. is away♦
Apr 1 at 14:35
$begingroup$
@J.M.isslightlypensive yes, you are right
$endgroup$
– J42161217
Apr 1 at 14:52
add a comment |
$begingroup$
P[n] will return the set you are asking
P[n_] := Partition[Join[0, #, n], 2, 1] & /@ Subsets[Range[n - 1]]
P[4]
0, 4, 0, 1, 1, 4, 0, 2, 2, 4, 0, 3, 3, 4, 0,
1, 1, 2, 2, 4, 0, 1, 1, 3, 3, 4, 0, 2, 2, 3, 3,
4, 0, 1, 1, 2, 2, 3, 3, 4
answered Apr 1 at 7:47
J42161217J42161217
4,513324
$endgroup$
P[n] will return the set you are asking
P[n_] := Partition[Join[0, #, n], 2, 1] & /@ Subsets[Range[n - 1]]
P[4]
0, 4, 0, 1, 1, 4, 0, 2, 2, 4, 0, 3, 3, 4, 0,
1, 1, 2, 2, 4, 0, 1, 1, 3, 3, 4, 0, 2, 2, 3, 3,
4, 0, 1, 1, 2, 2, 3, 3, 4
answered Apr 1 at 7:47
J42161217J42161217
4,513324
edited Apr 1 at 14:52
answered Apr 1 at 7:47
J42161217J42161217
4,513324
answered Apr 1 at 7:47
J42161217J42161217
4,513324
answered Apr 1 at 7:47
J42161217J42161217
4,513324
4,513324
$begingroup$
Thanks, it works well!!!
$endgroup$
– Mark_Phys
Apr 1 at 9:08
$begingroup$
@user10709 I'm glad I helped!
$endgroup$
– J42161217
Apr 1 at 9:10
$begingroup$
Of course, you can combine the functions so that only one application ofMap[]is needed:Partition[Join[0, #, n], 2, 1] & /@ Subsets[Range[n - 1]].
$endgroup$
– J. M. is away♦
Apr 1 at 14:35
$begingroup$
@J.M.isslightlypensive yes, you are right
$endgroup$
– J42161217
Apr 1 at 14:52
add a comment |
$begingroup$
Thanks, it works well!!!
$endgroup$
– Mark_Phys
Apr 1 at 9:08
$begingroup$
@user10709 I'm glad I helped!
$endgroup$
– J42161217
Apr 1 at 9:10
$begingroup$
Of course, you can combine the functions so that only one application ofMap[]is needed:Partition[Join[0, #, n], 2, 1] & /@ Subsets[Range[n - 1]].
$endgroup$
– J. M. is away♦
Apr 1 at 14:35
$begingroup$
@J.M.isslightlypensive yes, you are right
$endgroup$
– J42161217
Apr 1 at 14:52
$begingroup$
Thanks, it works well!!!
$endgroup$
– Mark_Phys
Apr 1 at 9:08
$begingroup$
Thanks, it works well!!!
$endgroup$
– Mark_Phys
Apr 1 at 9:08
$begingroup$
@user10709 I'm glad I helped!
$endgroup$
– J42161217
Apr 1 at 9:10
$begingroup$
@user10709 I'm glad I helped!
$endgroup$
– J42161217
Apr 1 at 9:10
$begingroup$
Of course, you can combine the functions so that only one application of
Map[] is needed: Partition[Join[0, #, n], 2, 1] & /@ Subsets[Range[n - 1]].$endgroup$
– J. M. is away♦
Apr 1 at 14:35
$begingroup$
Of course, you can combine the functions so that only one application of
Map[] is needed: Partition[Join[0, #, n], 2, 1] & /@ Subsets[Range[n - 1]].$endgroup$
– J. M. is away♦
Apr 1 at 14:35
$begingroup$
@J.M.isslightlypensive yes, you are right
$endgroup$
– J42161217
Apr 1 at 14:52
$begingroup$
@J.M.isslightlypensive yes, you are right
$endgroup$
– J42161217
Apr 1 at 14:52
add a comment |
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