Randomly distribute numbers to lists

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2
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I have 4 empty lists. I would like to randomly distribute numbers from 1 to 10 to these 4 empty lists. How can I add integers to lists randomly?



For example:



m = ;
n = ;
p = ;
r = ;
listz = Range[10];


1->m , 2->n , 3->p , 4->m , 5->r , 6->r, 7->n , 8->m, 9->p , 10->n (Randomly distributed)



Here is the code I tried:



m = ;
n = ;
p = ;
r = ;
listz = Range[10];
s = RandomChoice[m, n, p, r, 4];
Do[listm = AppendTo[s, listz[i]];, i, 1, 10]
listm


Wanted Solution:(Randomly distributed)



m = 1, 5, 8
n = 4, 6, 9
p = 7, 10
r = 2, 3





list-manipulation random education







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  • Why not just generate four random lists in the first place?
    – gothicVI
    2 hours ago














up vote
2
down vote

favorite












I have 4 empty lists. I would like to randomly distribute numbers from 1 to 10 to these 4 empty lists. How can I add integers to lists randomly?



For example:



m = ;
n = ;
p = ;
r = ;
listz = Range[10];


1->m , 2->n , 3->p , 4->m , 5->r , 6->r, 7->n , 8->m, 9->p , 10->n (Randomly distributed)



Here is the code I tried:



m = ;
n = ;
p = ;
r = ;
listz = Range[10];
s = RandomChoice[m, n, p, r, 4];
Do[listm = AppendTo[s, listz[i]];, i, 1, 10]
listm


Wanted Solution:(Randomly distributed)



m = 1, 5, 8
n = 4, 6, 9
p = 7, 10
r = 2, 3





list-manipulation random education







share|improve this question







New contributor




ithilquessirr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.



















  • Why not just generate four random lists in the first place?
    – gothicVI
    2 hours ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite











I have 4 empty lists. I would like to randomly distribute numbers from 1 to 10 to these 4 empty lists. How can I add integers to lists randomly?



For example:



m = ;
n = ;
p = ;
r = ;
listz = Range[10];


1->m , 2->n , 3->p , 4->m , 5->r , 6->r, 7->n , 8->m, 9->p , 10->n (Randomly distributed)



Here is the code I tried:



m = ;
n = ;
p = ;
r = ;
listz = Range[10];
s = RandomChoice[m, n, p, r, 4];
Do[listm = AppendTo[s, listz[i]];, i, 1, 10]
listm


Wanted Solution:(Randomly distributed)



m = 1, 5, 8
n = 4, 6, 9
p = 7, 10
r = 2, 3





list-manipulation random education







share|improve this question







New contributor




ithilquessirr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I have 4 empty lists. I would like to randomly distribute numbers from 1 to 10 to these 4 empty lists. How can I add integers to lists randomly?



For example:



m = ;
n = ;
p = ;
r = ;
listz = Range[10];


1->m , 2->n , 3->p , 4->m , 5->r , 6->r, 7->n , 8->m, 9->p , 10->n (Randomly distributed)



Here is the code I tried:



m = ;
n = ;
p = ;
r = ;
listz = Range[10];
s = RandomChoice[m, n, p, r, 4];
Do[listm = AppendTo[s, listz[i]];, i, 1, 10]
listm


Wanted Solution:(Randomly distributed)



m = 1, 5, 8
n = 4, 6, 9
p = 7, 10
r = 2, 3





list-manipulation random education




list-manipulation random education






share|improve this question







New contributor




ithilquessirr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question







New contributor




ithilquessirr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question






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Check out our Code of Conduct.









asked 2 hours ago









ithilquessirr

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New contributor





ithilquessirr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






ithilquessirr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • Why not just generate four random lists in the first place?
    – gothicVI
    2 hours ago
















  • Why not just generate four random lists in the first place?
    – gothicVI
    2 hours ago















Why not just generate four random lists in the first place?
– gothicVI
2 hours ago




Why not just generate four random lists in the first place?
– gothicVI
2 hours ago










4 Answers
4






active

oldest

votes

















up vote
2
down vote













on Mathematica 11.3 



(R=RandomSample)[Range@10,10]~TakeList~R@RandomChoice[Select[IntegerPartitions@10,Length@#==4&]]





Also I tried to fix your code 



a = ;
b = ;
c = ;
d = ;
Listz = a, b, c, d;
Do[t=RandomChoice@Range@4;Listz[[t]]=AppendTo[Listz[[t]],i],i,1,10]
Listz





share|improve this answer






















  • Have you tried that with a list of 100 elements?
    – Henrik Schumacher
    1 hour ago










  • how about 10^100?
    – J42161217
    1 hour ago










  • @J42161217 Because Select[IntegerPartitions@10, Length@# == 4 &] sorted, you cannot get Lenght=1or 2 set for m.
    – Okkes Dulgerci
    1 hour ago











  • @OkkesDulgerci fixed!
    – J42161217
    1 hour ago


















up vote
1
down vote













n = 100;
m = 4;
list = Range[n];
s = RandomChoice[Range[m], n];

SparseArray[Transpose[s, list] -> _, m, n]["AdjacencyLists"]


Or slower, but easier to understand:



Lookup[GroupBy[Transpose[s, list], First -> Last], Range[m]]





share|improve this answer






















  • Is this supposed to spew errors in 11.3?
    – user6014
    2 hours ago










  • Ah, sorry. Please try again.
    – Henrik Schumacher
    2 hours ago










  • Sorry, I couldn't understand what sparsearray function does here. But thanks this code is working.
    – ithilquessirr
    2 hours ago










  • Well, you might inspect MatrixForm@Normal[SparseArray[Transpose[s, listz] -> _, mm, nn]] to find that out.
    – Henrik Schumacher
    1 hour ago

















up vote
1
down vote













Edit: I realized that my previous solution doesn't produce the same length partitions. Here is the desired solution.



s = RandomSample[Range@10, 10];
t = RandomSample@RandomChoice@IntegerPartitions[10, 4]
m, n, p, r = TakeList[s, t]

(*or m, n, p, r = FoldPairList[TakeDrop, s, t]*)



1, 5, 10, 3, 7, 2, 9, 6, 8, 4




Original answer:



s = RandomSample[Range@10, 10];
t = RandomSample[Range@4, 4];
m, n, p, r = FoldPairList[TakeDrop, s, t]



6, 2, 9, 5, 1, 7, 10, 3, 8, 4







share|improve this answer





























    up vote
    1
    down vote













    listz = CharacterRange["A", "K"];
    k = 4;

    SeedRandom[1]
    rc = RandomChoice[Range @ k, Length @ listz];
    m, n, p, r = Pick[listz, rc, #]& /@ Range[k];



    "C", "F", "G", "H", "K", "B", "D", "E", "I", , "A", "J"




    Also



    Lookup[GroupBy[Transpose[rc, listz], First -> Last], #, ]& /@ Range[k]



    "C", "F", "G", "H", "K", "B", "D", "E", "I", , "A", "J"




    Update: To get k non-empty sublists:



    SeedRandom[1]
    rc = Module[r, While[Length[Counts[r = RandomChoice[Range@k, Length @ listz]]] < k]; r];
    Pick[listz , rc, # ] & /@ Range[k ]



    "A", "B", "C", "E", "H", "I", "F", "D", "G", "J", "K" 




    Lookup[GroupBy[Transpose[rc , listz], First -> Last], #, ] & /@ Range[k]



    "A", "B", "C", "E", "H", "I", "F", "D", "G", "J", "K"







    share|improve this answer






















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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      2
      down vote













      on Mathematica 11.3 



      (R=RandomSample)[Range@10,10]~TakeList~R@RandomChoice[Select[IntegerPartitions@10,Length@#==4&]]





      Also I tried to fix your code 



      a = ;
      b = ;
      c = ;
      d = ;
      Listz = a, b, c, d;
      Do[t=RandomChoice@Range@4;Listz[[t]]=AppendTo[Listz[[t]],i],i,1,10]
      Listz





      share|improve this answer






















      • Have you tried that with a list of 100 elements?
        – Henrik Schumacher
        1 hour ago










      • how about 10^100?
        – J42161217
        1 hour ago










      • @J42161217 Because Select[IntegerPartitions@10, Length@# == 4 &] sorted, you cannot get Lenght=1or 2 set for m.
        – Okkes Dulgerci
        1 hour ago











      • @OkkesDulgerci fixed!
        – J42161217
        1 hour ago















      up vote
      2
      down vote













      on Mathematica 11.3 



      (R=RandomSample)[Range@10,10]~TakeList~R@RandomChoice[Select[IntegerPartitions@10,Length@#==4&]]





      Also I tried to fix your code 



      a = ;
      b = ;
      c = ;
      d = ;
      Listz = a, b, c, d;
      Do[t=RandomChoice@Range@4;Listz[[t]]=AppendTo[Listz[[t]],i],i,1,10]
      Listz





      share|improve this answer






















      • Have you tried that with a list of 100 elements?
        – Henrik Schumacher
        1 hour ago










      • how about 10^100?
        – J42161217
        1 hour ago










      • @J42161217 Because Select[IntegerPartitions@10, Length@# == 4 &] sorted, you cannot get Lenght=1or 2 set for m.
        – Okkes Dulgerci
        1 hour ago











      • @OkkesDulgerci fixed!
        – J42161217
        1 hour ago













      up vote
      2
      down vote










      up vote
      2
      down vote









      on Mathematica 11.3 



      (R=RandomSample)[Range@10,10]~TakeList~R@RandomChoice[Select[IntegerPartitions@10,Length@#==4&]]





      Also I tried to fix your code 



      a = ;
      b = ;
      c = ;
      d = ;
      Listz = a, b, c, d;
      Do[t=RandomChoice@Range@4;Listz[[t]]=AppendTo[Listz[[t]],i],i,1,10]
      Listz





      share|improve this answer














      on Mathematica 11.3 



      (R=RandomSample)[Range@10,10]~TakeList~R@RandomChoice[Select[IntegerPartitions@10,Length@#==4&]]





      Also I tried to fix your code 



      a = ;
      b = ;
      c = ;
      d = ;
      Listz = a, b, c, d;
      Do[t=RandomChoice@Range@4;Listz[[t]]=AppendTo[Listz[[t]],i],i,1,10]
      Listz






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 59 mins ago

























      answered 1 hour ago









      J42161217

      2,447218




      2,447218











      • Have you tried that with a list of 100 elements?
        – Henrik Schumacher
        1 hour ago










      • how about 10^100?
        – J42161217
        1 hour ago










      • @J42161217 Because Select[IntegerPartitions@10, Length@# == 4 &] sorted, you cannot get Lenght=1or 2 set for m.
        – Okkes Dulgerci
        1 hour ago











      • @OkkesDulgerci fixed!
        – J42161217
        1 hour ago

















      • Have you tried that with a list of 100 elements?
        – Henrik Schumacher
        1 hour ago










      • how about 10^100?
        – J42161217
        1 hour ago










      • @J42161217 Because Select[IntegerPartitions@10, Length@# == 4 &] sorted, you cannot get Lenght=1or 2 set for m.
        – Okkes Dulgerci
        1 hour ago











      • @OkkesDulgerci fixed!
        – J42161217
        1 hour ago
















      Have you tried that with a list of 100 elements?
      – Henrik Schumacher
      1 hour ago




      Have you tried that with a list of 100 elements?
      – Henrik Schumacher
      1 hour ago












      how about 10^100?
      – J42161217
      1 hour ago




      how about 10^100?
      – J42161217
      1 hour ago












      @J42161217 Because Select[IntegerPartitions@10, Length@# == 4 &] sorted, you cannot get Lenght=1or 2 set for m.
      – Okkes Dulgerci
      1 hour ago





      @J42161217 Because Select[IntegerPartitions@10, Length@# == 4 &] sorted, you cannot get Lenght=1or 2 set for m.
      – Okkes Dulgerci
      1 hour ago













      @OkkesDulgerci fixed!
      – J42161217
      1 hour ago





      @OkkesDulgerci fixed!
      – J42161217
      1 hour ago











      up vote
      1
      down vote













      n = 100;
      m = 4;
      list = Range[n];
      s = RandomChoice[Range[m], n];

      SparseArray[Transpose[s, list] -> _, m, n]["AdjacencyLists"]


      Or slower, but easier to understand:



      Lookup[GroupBy[Transpose[s, list], First -> Last], Range[m]]





      share|improve this answer






















      • Is this supposed to spew errors in 11.3?
        – user6014
        2 hours ago










      • Ah, sorry. Please try again.
        – Henrik Schumacher
        2 hours ago










      • Sorry, I couldn't understand what sparsearray function does here. But thanks this code is working.
        – ithilquessirr
        2 hours ago










      • Well, you might inspect MatrixForm@Normal[SparseArray[Transpose[s, listz] -> _, mm, nn]] to find that out.
        – Henrik Schumacher
        1 hour ago














      up vote
      1
      down vote













      n = 100;
      m = 4;
      list = Range[n];
      s = RandomChoice[Range[m], n];

      SparseArray[Transpose[s, list] -> _, m, n]["AdjacencyLists"]


      Or slower, but easier to understand:



      Lookup[GroupBy[Transpose[s, list], First -> Last], Range[m]]





      share|improve this answer






















      • Is this supposed to spew errors in 11.3?
        – user6014
        2 hours ago










      • Ah, sorry. Please try again.
        – Henrik Schumacher
        2 hours ago










      • Sorry, I couldn't understand what sparsearray function does here. But thanks this code is working.
        – ithilquessirr
        2 hours ago










      • Well, you might inspect MatrixForm@Normal[SparseArray[Transpose[s, listz] -> _, mm, nn]] to find that out.
        – Henrik Schumacher
        1 hour ago












      up vote
      1
      down vote










      up vote
      1
      down vote









      n = 100;
      m = 4;
      list = Range[n];
      s = RandomChoice[Range[m], n];

      SparseArray[Transpose[s, list] -> _, m, n]["AdjacencyLists"]


      Or slower, but easier to understand:



      Lookup[GroupBy[Transpose[s, list], First -> Last], Range[m]]





      share|improve this answer














      n = 100;
      m = 4;
      list = Range[n];
      s = RandomChoice[Range[m], n];

      SparseArray[Transpose[s, list] -> _, m, n]["AdjacencyLists"]


      Or slower, but easier to understand:



      Lookup[GroupBy[Transpose[s, list], First -> Last], Range[m]]






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 1 hour ago

























      answered 2 hours ago









      Henrik Schumacher

      43.9k263129




      43.9k263129











      • Is this supposed to spew errors in 11.3?
        – user6014
        2 hours ago










      • Ah, sorry. Please try again.
        – Henrik Schumacher
        2 hours ago










      • Sorry, I couldn't understand what sparsearray function does here. But thanks this code is working.
        – ithilquessirr
        2 hours ago










      • Well, you might inspect MatrixForm@Normal[SparseArray[Transpose[s, listz] -> _, mm, nn]] to find that out.
        – Henrik Schumacher
        1 hour ago
















      • Is this supposed to spew errors in 11.3?
        – user6014
        2 hours ago










      • Ah, sorry. Please try again.
        – Henrik Schumacher
        2 hours ago










      • Sorry, I couldn't understand what sparsearray function does here. But thanks this code is working.
        – ithilquessirr
        2 hours ago










      • Well, you might inspect MatrixForm@Normal[SparseArray[Transpose[s, listz] -> _, mm, nn]] to find that out.
        – Henrik Schumacher
        1 hour ago















      Is this supposed to spew errors in 11.3?
      – user6014
      2 hours ago




      Is this supposed to spew errors in 11.3?
      – user6014
      2 hours ago












      Ah, sorry. Please try again.
      – Henrik Schumacher
      2 hours ago




      Ah, sorry. Please try again.
      – Henrik Schumacher
      2 hours ago












      Sorry, I couldn't understand what sparsearray function does here. But thanks this code is working.
      – ithilquessirr
      2 hours ago




      Sorry, I couldn't understand what sparsearray function does here. But thanks this code is working.
      – ithilquessirr
      2 hours ago












      Well, you might inspect MatrixForm@Normal[SparseArray[Transpose[s, listz] -> _, mm, nn]] to find that out.
      – Henrik Schumacher
      1 hour ago




      Well, you might inspect MatrixForm@Normal[SparseArray[Transpose[s, listz] -> _, mm, nn]] to find that out.
      – Henrik Schumacher
      1 hour ago










      up vote
      1
      down vote













      Edit: I realized that my previous solution doesn't produce the same length partitions. Here is the desired solution.



      s = RandomSample[Range@10, 10];
      t = RandomSample@RandomChoice@IntegerPartitions[10, 4]
      m, n, p, r = TakeList[s, t]

      (*or m, n, p, r = FoldPairList[TakeDrop, s, t]*)



      1, 5, 10, 3, 7, 2, 9, 6, 8, 4




      Original answer:



      s = RandomSample[Range@10, 10];
      t = RandomSample[Range@4, 4];
      m, n, p, r = FoldPairList[TakeDrop, s, t]



      6, 2, 9, 5, 1, 7, 10, 3, 8, 4







      share|improve this answer


























        up vote
        1
        down vote













        Edit: I realized that my previous solution doesn't produce the same length partitions. Here is the desired solution.



        s = RandomSample[Range@10, 10];
        t = RandomSample@RandomChoice@IntegerPartitions[10, 4]
        m, n, p, r = TakeList[s, t]

        (*or m, n, p, r = FoldPairList[TakeDrop, s, t]*)



        1, 5, 10, 3, 7, 2, 9, 6, 8, 4




        Original answer:



        s = RandomSample[Range@10, 10];
        t = RandomSample[Range@4, 4];
        m, n, p, r = FoldPairList[TakeDrop, s, t]



        6, 2, 9, 5, 1, 7, 10, 3, 8, 4







        share|improve this answer
























          up vote
          1
          down vote










          up vote
          1
          down vote









          Edit: I realized that my previous solution doesn't produce the same length partitions. Here is the desired solution.



          s = RandomSample[Range@10, 10];
          t = RandomSample@RandomChoice@IntegerPartitions[10, 4]
          m, n, p, r = TakeList[s, t]

          (*or m, n, p, r = FoldPairList[TakeDrop, s, t]*)



          1, 5, 10, 3, 7, 2, 9, 6, 8, 4




          Original answer:



          s = RandomSample[Range@10, 10];
          t = RandomSample[Range@4, 4];
          m, n, p, r = FoldPairList[TakeDrop, s, t]



          6, 2, 9, 5, 1, 7, 10, 3, 8, 4







          share|improve this answer














          Edit: I realized that my previous solution doesn't produce the same length partitions. Here is the desired solution.



          s = RandomSample[Range@10, 10];
          t = RandomSample@RandomChoice@IntegerPartitions[10, 4]
          m, n, p, r = TakeList[s, t]

          (*or m, n, p, r = FoldPairList[TakeDrop, s, t]*)



          1, 5, 10, 3, 7, 2, 9, 6, 8, 4




          Original answer:



          s = RandomSample[Range@10, 10];
          t = RandomSample[Range@4, 4];
          m, n, p, r = FoldPairList[TakeDrop, s, t]



          6, 2, 9, 5, 1, 7, 10, 3, 8, 4








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 1 hour ago

























          answered 2 hours ago









          Okkes Dulgerci

          3,4061716




          3,4061716




















              up vote
              1
              down vote













              listz = CharacterRange["A", "K"];
              k = 4;

              SeedRandom[1]
              rc = RandomChoice[Range @ k, Length @ listz];
              m, n, p, r = Pick[listz, rc, #]& /@ Range[k];



              "C", "F", "G", "H", "K", "B", "D", "E", "I", , "A", "J"




              Also



              Lookup[GroupBy[Transpose[rc, listz], First -> Last], #, ]& /@ Range[k]



              "C", "F", "G", "H", "K", "B", "D", "E", "I", , "A", "J"




              Update: To get k non-empty sublists:



              SeedRandom[1]
              rc = Module[r, While[Length[Counts[r = RandomChoice[Range@k, Length @ listz]]] < k]; r];
              Pick[listz , rc, # ] & /@ Range[k ]



              "A", "B", "C", "E", "H", "I", "F", "D", "G", "J", "K" 




              Lookup[GroupBy[Transpose[rc , listz], First -> Last], #, ] & /@ Range[k]



              "A", "B", "C", "E", "H", "I", "F", "D", "G", "J", "K"







              share|improve this answer


























                up vote
                1
                down vote













                listz = CharacterRange["A", "K"];
                k = 4;

                SeedRandom[1]
                rc = RandomChoice[Range @ k, Length @ listz];
                m, n, p, r = Pick[listz, rc, #]& /@ Range[k];



                "C", "F", "G", "H", "K", "B", "D", "E", "I", , "A", "J"




                Also



                Lookup[GroupBy[Transpose[rc, listz], First -> Last], #, ]& /@ Range[k]



                "C", "F", "G", "H", "K", "B", "D", "E", "I", , "A", "J"




                Update: To get k non-empty sublists:



                SeedRandom[1]
                rc = Module[r, While[Length[Counts[r = RandomChoice[Range@k, Length @ listz]]] < k]; r];
                Pick[listz , rc, # ] & /@ Range[k ]



                "A", "B", "C", "E", "H", "I", "F", "D", "G", "J", "K" 




                Lookup[GroupBy[Transpose[rc , listz], First -> Last], #, ] & /@ Range[k]



                "A", "B", "C", "E", "H", "I", "F", "D", "G", "J", "K"







                share|improve this answer
























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  listz = CharacterRange["A", "K"];
                  k = 4;

                  SeedRandom[1]
                  rc = RandomChoice[Range @ k, Length @ listz];
                  m, n, p, r = Pick[listz, rc, #]& /@ Range[k];



                  "C", "F", "G", "H", "K", "B", "D", "E", "I", , "A", "J"




                  Also



                  Lookup[GroupBy[Transpose[rc, listz], First -> Last], #, ]& /@ Range[k]



                  "C", "F", "G", "H", "K", "B", "D", "E", "I", , "A", "J"




                  Update: To get k non-empty sublists:



                  SeedRandom[1]
                  rc = Module[r, While[Length[Counts[r = RandomChoice[Range@k, Length @ listz]]] < k]; r];
                  Pick[listz , rc, # ] & /@ Range[k ]



                  "A", "B", "C", "E", "H", "I", "F", "D", "G", "J", "K" 




                  Lookup[GroupBy[Transpose[rc , listz], First -> Last], #, ] & /@ Range[k]



                  "A", "B", "C", "E", "H", "I", "F", "D", "G", "J", "K"







                  share|improve this answer














                  listz = CharacterRange["A", "K"];
                  k = 4;

                  SeedRandom[1]
                  rc = RandomChoice[Range @ k, Length @ listz];
                  m, n, p, r = Pick[listz, rc, #]& /@ Range[k];



                  "C", "F", "G", "H", "K", "B", "D", "E", "I", , "A", "J"




                  Also



                  Lookup[GroupBy[Transpose[rc, listz], First -> Last], #, ]& /@ Range[k]



                  "C", "F", "G", "H", "K", "B", "D", "E", "I", , "A", "J"




                  Update: To get k non-empty sublists:



                  SeedRandom[1]
                  rc = Module[r, While[Length[Counts[r = RandomChoice[Range@k, Length @ listz]]] < k]; r];
                  Pick[listz , rc, # ] & /@ Range[k ]



                  "A", "B", "C", "E", "H", "I", "F", "D", "G", "J", "K" 




                  Lookup[GroupBy[Transpose[rc , listz], First -> Last], #, ] & /@ Range[k]



                  "A", "B", "C", "E", "H", "I", "F", "D", "G", "J", "K"








                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 8 mins ago

























                  answered 57 mins ago









                  kglr

                  169k8192396




                  169k8192396




















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