Is there a function to partition an integer set?












5












$begingroup$


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







share|improve this question









$endgroup$

















    5












    $begingroup$


    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







    share|improve this question









    $endgroup$















      5












      5








      5





      $begingroup$


      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







      share|improve this question









      $endgroup$




      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






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 2 days ago









      user10709user10709

      1186




      1186






















          1 Answer
          1






          active

          oldest

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          8












          $begingroup$

          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}}}







          share|improve this answer











          $endgroup$













          • $begingroup$
            Thanks, it works well!!!
            $endgroup$
            – user10709
            2 days ago










          • $begingroup$
            @user10709 I'm glad I helped!
            $endgroup$
            – J42161217
            2 days ago










          • $begingroup$
            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 slightly pensive♦
            2 days ago










          • $begingroup$
            @J.M.isslightlypensive yes, you are right
            $endgroup$
            – J42161217
            2 days ago












          Your Answer





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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          8












          $begingroup$

          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}}}







          share|improve this answer











          $endgroup$













          • $begingroup$
            Thanks, it works well!!!
            $endgroup$
            – user10709
            2 days ago










          • $begingroup$
            @user10709 I'm glad I helped!
            $endgroup$
            – J42161217
            2 days ago










          • $begingroup$
            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 slightly pensive♦
            2 days ago










          • $begingroup$
            @J.M.isslightlypensive yes, you are right
            $endgroup$
            – J42161217
            2 days ago
















          8












          $begingroup$

          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}}}







          share|improve this answer











          $endgroup$













          • $begingroup$
            Thanks, it works well!!!
            $endgroup$
            – user10709
            2 days ago










          • $begingroup$
            @user10709 I'm glad I helped!
            $endgroup$
            – J42161217
            2 days ago










          • $begingroup$
            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 slightly pensive♦
            2 days ago










          • $begingroup$
            @J.M.isslightlypensive yes, you are right
            $endgroup$
            – J42161217
            2 days ago














          8












          8








          8





          $begingroup$

          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}}}







          share|improve this answer











          $endgroup$



          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}}}








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 2 days ago

























          answered 2 days ago









          J42161217J42161217

          4,118324




          4,118324












          • $begingroup$
            Thanks, it works well!!!
            $endgroup$
            – user10709
            2 days ago










          • $begingroup$
            @user10709 I'm glad I helped!
            $endgroup$
            – J42161217
            2 days ago










          • $begingroup$
            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 slightly pensive♦
            2 days ago










          • $begingroup$
            @J.M.isslightlypensive yes, you are right
            $endgroup$
            – J42161217
            2 days ago


















          • $begingroup$
            Thanks, it works well!!!
            $endgroup$
            – user10709
            2 days ago










          • $begingroup$
            @user10709 I'm glad I helped!
            $endgroup$
            – J42161217
            2 days ago










          • $begingroup$
            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 slightly pensive♦
            2 days ago










          • $begingroup$
            @J.M.isslightlypensive yes, you are right
            $endgroup$
            – J42161217
            2 days ago
















          $begingroup$
          Thanks, it works well!!!
          $endgroup$
          – user10709
          2 days ago




          $begingroup$
          Thanks, it works well!!!
          $endgroup$
          – user10709
          2 days ago












          $begingroup$
          @user10709 I'm glad I helped!
          $endgroup$
          – J42161217
          2 days ago




          $begingroup$
          @user10709 I'm glad I helped!
          $endgroup$
          – J42161217
          2 days ago












          $begingroup$
          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 slightly pensive♦
          2 days ago




          $begingroup$
          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 slightly pensive♦
          2 days ago












          $begingroup$
          @J.M.isslightlypensive yes, you are right
          $endgroup$
          – J42161217
          2 days ago




          $begingroup$
          @J.M.isslightlypensive yes, you are right
          $endgroup$
          – J42161217
          2 days ago


















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